Math Jams

DPatrick 2014-03-28 19:00:09
Welcome to the 2014 AIME II Math Jam!

DPatrick 2014-03-28 19:00:15
I'm Dave Patrick, and I'll be leading our discussion tonight.

DPatrick 2014-03-28 19:00:23
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.

DPatrick 2014-03-28 19:00:38
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.

DPatrick 2014-03-28 19:00:46
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.

DPatrick 2014-03-28 19:01:01
There are a lot of students here! As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your responses do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!

DPatrick 2014-03-28 19:01:10
We do have two teaching assistants with us tonight to help answer your questions: Luis (Duelist) and Jessie (numberdance).

DPatrick 2014-03-28 19:01:23
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

DPatrick 2014-03-28 19:01:37
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics.

DPatrick 2014-03-28 19:01:42
Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest.

DPatrick 2014-03-28 19:01:50
So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be acknowledged.

DPatrick 2014-03-28 19:02:01
Let's get started! We're going to work through all 15 problems from the 2014 AIME II, in order.

DPatrick 2014-03-28 19:02:07
1. Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.

DPatrick 2014-03-28 19:02:24
You'll note that I'll always put the current problem under discussion in a window at the top of the classroom.

DPatrick 2014-03-28 19:02:42
You can resize that top area by dragging on the horizontal bar between it and the main classroom panel.

DPatrick 2014-03-28 19:02:46
What's the key to solving long wordy problems like this?

ChilledLemonade 2014-03-28 19:03:08
Reading!

antler 2014-03-28 19:03:08
translate the words into math

IsabeltheCat 2014-03-28 19:03:08
turn words into math!

AndrewK 2014-03-28 19:03:08
Read the question

DPatrick 2014-03-28 19:03:16
First and most important: READ THE PROBLEM CAREFULLY!

DPatrick 2014-03-28 19:03:25
Then we want to translate the words into mathematical expressions.

DPatrick 2014-03-28 19:03:38
What is a general technique for "work" problems like this?

NormanWho 2014-03-28 19:04:01
a rate problem

hotstuffFTW 2014-03-28 19:04:01
Finding rates

Mathgeek727 2014-03-28 19:04:01
rate?

AndrewK 2014-03-28 19:04:01
Find the rate of working for each worker

poweroftwo 2014-03-28 19:04:01
express each person as work/hr ratio

XxAndreixX 2014-03-28 19:04:01
Find expressions for the painting speed of each person.

DPatrick 2014-03-28 19:04:10
Yes, it's usually best if we express the work rates in fractions of a completed job per unit time.

DPatrick 2014-03-28 19:04:24
For example, rather than say "Abe can paint the room in 15 hours," how would we describe Abe's work rate?

hnkevin42 2014-03-28 19:04:47
1/15 room/hr

Modest_Ked 2014-03-28 19:04:47
1/15 done in 1 hour

Mathgeek727 2014-03-28 19:04:47
1/15 rooms per hour

joey8189681 2014-03-28 19:04:47
1/15 per hour

tluo5458 2014-03-28 19:04:47
r=1/15

speck 2014-03-28 19:04:47
He paints $\frac{1}{15}$ of the room in an hour

chezbgone2 2014-03-28 19:04:47
Abe paints 1 per 15 hours, so 1/15

Bomist0 2014-03-28 19:04:47
1/15 job per hour

AndrewK 2014-03-28 19:04:47
$\frac{1}{15}$ room/hour

DPatrick 2014-03-28 19:04:52
He paints the entire room in 15 hours, so he paints $\dfrac{1}{15}$ rooms per hour.

DPatrick 2014-03-28 19:04:58
How fast do the other two paint?

tluo5458 2014-03-28 19:05:30
Bea: 1/10 Coe: 2/15

hotstuffFTW 2014-03-28 19:05:30
1/10 for Bea and 2/15 for Coe

mjlove 2014-03-28 19:05:30
1/10 and 2/15

danusv 2014-03-28 19:05:30
1/10 room per hour and 2/15 room per hour

UrInvalid 2014-03-28 19:05:30
1/10 for bea, 2/15 for coe

ChilledLemonade 2014-03-28 19:05:30
Bea paints at 1/10 per hour and Coe at 2/15 per hour

DPatrick 2014-03-28 19:05:35
Bea paints $50\%$ faster than Abe, so she paints $(1.5)\dfrac{1}{15} = \dfrac{1}{10}$ rooms per hour.

DPatrick 2014-03-28 19:05:44
Coe paints twice as fast as Abe, so he paints $\dfrac{2}{15}$ rooms per hour.

DPatrick 2014-03-28 19:05:51
Now we can translate the rest of the problem.

DPatrick 2014-03-28 19:05:58
"Abe begins to paint the room and works alone for the first hour and a half."
What does this tell us?

poweroftwo 2014-03-28 19:06:30
he gets 1/10 done in 90 minutes

mathwizard888 2014-03-28 19:06:30
he painted 1/10 of the room

hotstuffFTW 2014-03-28 19:06:30
He completes 1.5/15 = 1/10 of the job

simon1221 2014-03-28 19:06:30
he does 1/10 of the total work

CornSaltButter 2014-03-28 19:06:30
He paints 1/10 of the room in that time

dmagician34 2014-03-28 19:06:30
He painted 1.5 x 1/15 = 1/10 of the house

peterdragonking 2014-03-28 19:06:33
He finished1/10 of it

DPatrick 2014-03-28 19:06:37
Right. After 1.5 hours, Abe has painted $(1.5)\dfrac{1}{15} = \dfrac{1}{10}$ of the room.

DPatrick 2014-03-28 19:06:44
Next: "Then Bea joins Abe, and they work together until half the room is painted."

DPatrick 2014-03-28 19:06:52
How fast do they work together?

tluo5458 2014-03-28 19:07:25
1/10+1/15=1/6

summitwei 2014-03-28 19:07:25
1/15+1/10

Bomist0 2014-03-28 19:07:25
1/10 + 1/15 = 1/6

danusv 2014-03-28 19:07:25
Combined they work at the rate of 1/15+ 1/10= 1/6 room per hour

speck 2014-03-28 19:07:25
They paint at $\dfrac{1}{6}$ of the total per hour

jap23 2014-03-28 19:07:25
1/6 per hour

hwl0304 2014-03-28 19:07:25
1/6 rph

CornSaltButter 2014-03-28 19:07:25
1/15+1/10=5/30= 1/6 of the room in 1 hour

wu2481632 2014-03-28 19:07:25
they work 1/15+1/10 per hour which is 1/6 per hour

simon1221 2014-03-28 19:07:25
1/6 room/hour (1/15 + 1/10)

DPatrick 2014-03-28 19:07:31
Yes: Together they paint $\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{1}{6}$ of the room per hour.

DPatrick 2014-03-28 19:07:42
And how much of the room do they have to paint together?

wu2481632 2014-03-28 19:08:00
4/10 of the room

AndrewK 2014-03-28 19:08:00
4/10 of the room

mathwizard888 2014-03-28 19:08:00
1/2-1/10=2/5

joey8189681 2014-03-28 19:08:00
2/5 of the room

danusv 2014-03-28 19:08:00
1/2-1/10=2/5 of the room

ChilledLemonade 2014-03-28 19:08:00
1/2-1/10=2/5

DPatrick 2014-03-28 19:08:04
They continue until $\dfrac12$ the room is done, and Abe already did $\dfrac{1}{10}$, so they have $\dfrac12 - \dfrac{1}{10} = \dfrac25$ of the room that they have to paint together.

DPatrick 2014-03-28 19:08:09
So how long does it take them?

speck 2014-03-28 19:08:45
So $\dfrac{x}{6} = \dfrac{2}{5}$, $x=2.4$

hotstuffFTW 2014-03-28 19:08:45
12/5 or 2.4

jap23 2014-03-28 19:08:45
2.4 hours

wu2481632 2014-03-28 19:08:45
2/5 / 1/6 which is 2/5*6 which is 12/5

poweroftwo 2014-03-28 19:08:45
2/5 divided by 1/6 is 12/5

DPatrick 2014-03-28 19:08:52
Right: To paint $\dfrac25$ of the room at a rate of $\dfrac16$ room per hour, it takes them
$$ \dfrac{\frac25}{\frac16} = \dfrac{12}{5} $$ hours.

DPatrick 2014-03-28 19:09:03
Next: "Then Coe joins Abe and Bea, and they work together until the entire room is painted."

DPatrick 2014-03-28 19:09:07
How fast do they all work together?

mathwizard888 2014-03-28 19:09:40
1/6+2/15=3/10

abishek99 2014-03-28 19:09:40
3/10 room per hour

UrInvalid 2014-03-28 19:09:40
1/6 + 2/15 = 3/10

wu2481632 2014-03-28 19:09:40
they work at 1/15+1/10+2/15 which is 9/30 which is 3/10

SaladQ 2014-03-28 19:09:40
1/10 + 1/15 + 2/15 = 3/10

CornSaltButter 2014-03-28 19:09:40
1/15+1/10+2/15= 3/10 of the room per hour

hjl00 2014-03-28 19:09:40
3/10 of the room per hour

DPatrick 2014-03-28 19:09:44
Together they work $\dfrac{1}{15} + \dfrac{1}{10} + \dfrac{2}{15} = \dfrac{9}{30} = \dfrac{3}{10}$ rooms per hour.

DPatrick 2014-03-28 19:09:49
So how long does it take them to finish?

hnkevin42 2014-03-28 19:10:12
5/3 hours?

wu2481632 2014-03-28 19:10:12
1/2 / 3/10 which is 10/6 = 5/3

UrInvalid 2014-03-28 19:10:12
1/2 / 3/10 = 5/3

CornSaltButter 2014-03-28 19:10:12
1/2 divided by 3/10, which is 5/3 hours

Bomist0 2014-03-28 19:10:12
½ ÷ 3/10 = 5/3 hours

IsabeltheCat 2014-03-28 19:10:12
5/3 hr

DPatrick 2014-03-28 19:10:16
To paint the remaining $\dfrac12$ room, it takes them
$$ \dfrac{\frac12}{\frac{3}{10}} = \dfrac{5}{3} $$ hours.

DPatrick 2014-03-28 19:10:22
And to finish: "Find the number of minutes after Abe begins for the three of them to finish painting the room."

DPatrick 2014-03-28 19:10:41
Abe spends 1.5 hours, or 90 minutes.
Abe and Bea spend $\dfrac{12}{5}$ hours, or 144 minutes.
All three spend $\dfrac53$ hours, or 100 minutes.

hjl00 2014-03-28 19:11:00
334 minutes

hotstuffFTW 2014-03-28 19:11:00
This means 5/3+12/5+3/2= 5 17/30 or 334 min

mathwizard888 2014-03-28 19:11:00
total 334 ans

ChilledLemonade 2014-03-28 19:11:00
90+100+144 = 334

dantx5 2014-03-28 19:11:00
334 minutes total

poweroftwo 2014-03-28 19:11:00
add them you get 334!

tluo5458 2014-03-28 19:11:00
so it's 334!

jmann123456789 2014-03-28 19:11:00
334 minutes

DPatrick 2014-03-28 19:11:06
Right: all told, they take $90 + 144 + 100 = \boxed{334}$ minutes.

DPatrick 2014-03-28 19:11:29
The AIME almost always has a work or rate problem like this among the first few problems.

DPatrick 2014-03-28 19:11:47
It's worth practicing how to do them quickly for future contests.

DPatrick 2014-03-28 19:11:52
Onwards!

DPatrick 2014-03-28 19:11:56
2. Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac13.$ The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

DPatrick 2014-03-28 19:12:10
Yikes, two wordy problems in a row to start us off!

poweroftwo 2014-03-28 19:12:32
venn diagram!!

abishek99 2014-03-28 19:12:32
venn diagram

dantx5 2014-03-28 19:12:32
venn diagram

hnkevin42 2014-03-28 19:12:37
"Given problems..." A Venn diagram would really help.

AndrewK 2014-03-28 19:12:37
Assume there are 100 men and then draw a Venn Diagram

DPatrick 2014-03-28 19:12:49
Good idea. Since we're trying to essentially count men with some of A, B, or C, a 3-circle Venn Diagram might help a lot.

DPatrick 2014-03-28 19:12:55

DPatrick 2014-03-28 19:13:05
Let's fill these in with percentages. How can we start?

poweroftwo 2014-03-28 19:13:36
just A, just B, or just C is .1

abishek99 2014-03-28 19:13:36
outer regions have 10 people each

wu2481632 2014-03-28 19:13:36
only one risk factor is 0.1

AndrewK 2014-03-28 19:13:36
Fill in only A, only B, and only C

UrInvalid 2014-03-28 19:13:36
each of the parts in only one circle is 0.10

hnkevin42 2014-03-28 19:13:36
Fill the outsides with 0.1 since those people have no other risk factors

Bomist0 2014-03-28 19:13:36
Ones only in A,B,C are 0.1 or 10%

want2learn 2014-03-28 19:13:36
"For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1."

DPatrick 2014-03-28 19:13:43
Having exactly one risk factor is 10% (for each), so we'll put 10 inside of each of the 3 circles (but outside the overlap):

DPatrick 2014-03-28 19:13:48

DPatrick 2014-03-28 19:14:16
I'm just going to put the numbers (without percent signs), but you should think of them as percentages. (Or you can think of there being exactly 100 men.) The numbers ought to total to 100 at the end.

DPatrick 2014-03-28 19:14:18
Now what?

abishek99 2014-03-28 19:14:42
14% in intersections of only two circles

Vinni 2014-03-28 19:14:42
14 overlaps

Mathgeek727 2014-03-28 19:14:42
in overlapping for 2 circles not 3, put 14??

joey8189681 2014-03-28 19:14:42
Put 14% on the two overlapping areas

hwl0304 2014-03-28 19:14:42
do the same for 14

wu2481632 2014-03-28 19:14:42
.14 for two of the risks

CornSaltButter 2014-03-28 19:14:42
Now look at the number of people with exactly two risk factors= 14 people for each region.

danusv 2014-03-28 19:14:42
Put 14 in the union of A and B, A and C, and B and C

DPatrick 2014-03-28 19:14:49
Good -- having exactly two risk factors is 14% (for each pair), so we'll put 14 inside of each of the overlap regions of two circles:

DPatrick 2014-03-28 19:14:54

DPatrick 2014-03-28 19:15:05
Next -- How do we use this data:
"The probability that a randomly selected man has all three risk factors, given that he has A and B, is $\frac13.$"

DPatrick 2014-03-28 19:15:13
What region of the Venn Diagram corresponds to "given that he has A and B"?

AndrewK 2014-03-28 19:15:31
The intersection of A and B

dmagician34 2014-03-28 19:15:31
The intersection between A and B

summitwei 2014-03-28 19:15:31
the intersection of A and B...

ChilledLemonade 2014-03-28 19:15:31
The intersection of all three and the intersection of A and B

bengals 2014-03-28 19:15:31
the intersection of A and B

DPatrick 2014-03-28 19:15:35
It's the area inside both the A and B circles, shaded below:

DPatrick 2014-03-28 19:15:40

DPatrick 2014-03-28 19:15:52
So what does the quoted sentence from the problem tell us?

DPatrick 2014-03-28 19:15:58
"The probability that a randomly selected man has all three risk factors, given that he has A and B, is $\frac13.$"

DPatrick 2014-03-28 19:15:59
?

poweroftwo 2014-03-28 19:16:39
1/3 of it is in the very center

AKAL3 2014-03-28 19:16:39
7 is the intersection of all 3

AndrewK 2014-03-28 19:16:39
It tells us the other 2/3 is 14 so the 1/3 with all 3 is 7

jap23 2014-03-28 19:16:39
that there are 7 in the center!

summitwei 2014-03-28 19:16:39
the intersection of everything is 7

Readingrocks88 2014-03-28 19:16:39
middle section probability is .7

AnneHahnlheem 2014-03-28 19:16:39
if x is the percentage in the middle area, x/x+14 should = 1/3

UrInvalid 2014-03-28 19:16:39
center/blue = 1/3

DPatrick 2014-03-28 19:16:51
Right. The part of that region with all three risk factors -- that is, the part in the center -- is $\frac13$ of the blue shaded region.

DPatrick 2014-03-28 19:17:19
So if we let the number in the center be $x$, then we need to solve $\dfrac{x}{x+14} = \dfrac13.$

Allan Z 2014-03-28 19:17:27
so 14=2/3 of blue region

DPatrick 2014-03-28 19:17:42
Or that's another good way to think about it: if the center is 1/3 of the blue region, then 14 is the other 2/3 of the blue regio.

DPatrick 2014-03-28 19:17:49
...region,

DPatrick 2014-03-28 19:17:52
...region!

DPatrick 2014-03-28 19:18:05
So the blue region is 21 total, and what need to be in the center is 7.

DPatrick 2014-03-28 19:18:14

DPatrick 2014-03-28 19:18:21
And that leaves what?

mathwizard888 2014-03-28 19:18:41
21 have none

tluo5458 2014-03-28 19:18:41
21 outside

poweroftwo 2014-03-28 19:18:41
21 on the outside

Bomist0 2014-03-28 19:18:41
outside, as 21%

Lemon123 2014-03-28 19:18:41
21 with none

AnneHahnlheem 2014-03-28 19:18:41
21 with no risk factors

Mathgeek727 2014-03-28 19:18:41
21 people with nothing

ChilledLemonade 2014-03-28 19:18:41
21 that dont have any risk factors

summitwei 2014-03-28 19:18:41
21 left

DPatrick 2014-03-28 19:18:47
We have 3(10) + 3(14) + 7 = 30 + 42 + 7 = 79 inside the circles, so that leaves 21 percent outside the circles:

DPatrick 2014-03-28 19:18:55

DPatrick 2014-03-28 19:19:00
And how do we finish?

DPatrick 2014-03-28 19:19:08
We want "the probability that a man has none of the three risk factors given that he does not have risk factor A".

DPatrick 2014-03-28 19:19:11
What percentage of the men don't have risk factor A?

poweroftwo 2014-03-28 19:19:31
55 don't have

AnneHahnlheem 2014-03-28 19:19:31
55

mathwizard888 2014-03-28 19:19:31
10+14+10+21=55

HYP135peppers 2014-03-28 19:19:31
55

poweroftwo 2014-03-28 19:19:31
55 don't have A

IsabeltheCat 2014-03-28 19:19:31
55%

Mathgeek727 2014-03-28 19:19:31
55?

danusv 2014-03-28 19:19:31
55 percent

DPatrick 2014-03-28 19:19:35
We add all the numbers outside of circle A, and get 21+10+14+10 = 55.

DPatrick 2014-03-28 19:19:44
So 55 don't have risk factor A.

DPatrick 2014-03-28 19:19:53
Hence, what is our probability?

AndrewK 2014-03-28 19:20:12
$\frac{21}{55}$!

AnneHahnlheem 2014-03-28 19:20:12
thus the probability is 21/55

hnkevin42 2014-03-28 19:20:12
21/55

coldsummer 2014-03-28 19:20:12
21/55

hotstuffFTW 2014-03-28 19:20:12
21/55 or answer is 076

Lemon123 2014-03-28 19:20:12
21/55

atmath2011 2014-03-28 19:20:12
21/55

jmann123456789 2014-03-28 19:20:12
21/55

DPatrick 2014-03-28 19:20:25
Right. Of these 55, only 21 don't have any of the three risk factors.

DPatrick 2014-03-28 19:20:31
Thus, our probability is $\dfrac{21}{55}.$

DPatrick 2014-03-28 19:20:38
This is already in lowest terms, so our final answer is $21 + 55 = \boxed{076}.$

DPatrick 2014-03-28 19:21:06
This problem is an example of conditional probability. We'll come back to this topic in problem #6.

DPatrick 2014-03-28 19:21:12
But in the meantime...

DPatrick 2014-03-28 19:21:15
3. A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2.$

DPatrick 2014-03-28 19:21:26

DPatrick 2014-03-28 19:21:43
We'll start easy: what's the area of the rectangle?

abishek99 2014-03-28 19:21:57
36a

antler 2014-03-28 19:21:57
36a

wu2481632 2014-03-28 19:21:57
36a

dantx5 2014-03-28 19:21:57
36a

byunstephanie 2014-03-28 19:21:57
36a

CornSaltButter 2014-03-28 19:21:57
36*a= 36a

dmagician34 2014-03-28 19:21:57
36a

DPatrick 2014-03-28 19:22:13
...and about 30 others of you also correctly said $36a.$

DPatrick 2014-03-28 19:22:20
So we need to find $a$ so that $36a$ is also the area of the hexagon.

DPatrick 2014-03-28 19:22:30
Let's zoom in and label the known lengths of the hexagon:

DPatrick 2014-03-28 19:22:35

DPatrick 2014-03-28 19:22:45
Now what?

byunstephanie 2014-03-28 19:23:14
drop altitudes for the triangles

ChilledLemonade 2014-03-28 19:23:14
Find area !

ScottBusche 2014-03-28 19:23:14
Find the area of the triangles.

want2learn 2014-03-28 19:23:14
Find the area of one of the side triangles.

hnkevin42 2014-03-28 19:23:14
finding the heights of the two triangles would help.

Lemon123 2014-03-28 19:23:14
find height of the triangles

DPatrick 2014-03-28 19:23:23
We can draw heights of the two triangles on the left and right sides:

DPatrick 2014-03-28 19:23:27

DPatrick 2014-03-28 19:23:47
Now we have 4 little right triangles with one side 12 and hypotenuse 18.

DPatrick 2014-03-28 19:23:56
What's the third side of those little right triangles?

AKAL3 2014-03-28 19:24:16
6sqrt5 is height

HYP135peppers 2014-03-28 19:24:16
Therefore, the altitude is $6\sqrt{5}$

tluo5458 2014-03-28 19:24:16
so the altitude is $6\sqrt{5}$!

AndrewK 2014-03-28 19:24:16
The height is $6 \sqrt{5}$

Awesome_1 2014-03-28 19:24:16
6sqrt5

Readingrocks88 2014-03-28 19:24:16
6sqrt5

simon1221 2014-03-28 19:24:16
6sqrt5

UrInvalid 2014-03-28 19:24:16
use Pythagorean theorem: sqrt(18^2-12^2)=6sqrt5

rubberballs 2014-03-28 19:24:16
6sqrt5

tfmtoto 2014-03-28 19:24:16
$\sqrt{18^2 - 12^2}$

DPatrick 2014-03-28 19:24:23
By the Pythagorean Theorem, those sides have length $$\sqrt{18^2 - 12^2} = \sqrt{324-144} = \sqrt{180} = 6\sqrt5.$$

DPatrick 2014-03-28 19:24:39
So what is the area of the hexagon?

hwl0304 2014-03-28 19:25:12
24a+144sqrt5

AndrewK 2014-03-28 19:25:12
$144 \sqrt{5} + 24a$

Bomist0 2014-03-28 19:25:12
6√5 * 24 + 24a

danusv 2014-03-28 19:25:12
Area is 24 a + 144 root 5

AnneHahnlheem 2014-03-28 19:25:12
24a+144sqrt5

tluo5458 2014-03-28 19:25:12
24a+144sqrt5

HYP135peppers 2014-03-28 19:25:12
$144\sqrt{5}+24a$

DPatrick 2014-03-28 19:25:18
The rectangle in the center has area $24a$, and the triangular areas sum to $144\sqrt5.$ (Each little triangle is a right triangle with legs 12 and $6\sqrt5$, so each has area $36\sqrt5.$)

DPatrick 2014-03-28 19:25:27
So the total area is $24a + 144\sqrt5.$
And this must equal $36a.$

IsabeltheCat 2014-03-28 19:25:54
a=12root5

abishek99 2014-03-28 19:25:54
a=12sqrt{5}

Readingrocks88 2014-03-28 19:25:54
a=12sqrt5

AndrewK 2014-03-28 19:25:54
a = $12 \sqrt{5}$

poweroftwo 2014-03-28 19:25:54
12a=144sqrt5

jap23 2014-03-28 19:25:54
So a = 12sqrt5 because 12a = 144sqrt5

DPatrick 2014-03-28 19:26:06
Right. We get $12a = 144\sqrt5,$ hence $a = 12\sqrt5,$ and thus our final answer is $a^2 = 144 \cdot 5 = \boxed{720}.$

DPatrick 2014-03-28 19:26:29
This might have been the easiest problem on the contest...or it might be the next one in fact:

DPatrick 2014-03-28 19:26:34
4. The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $$0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37},$$where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. Find the three-digit number $abc.$

DPatrick 2014-03-28 19:26:57
What can we do first?

rubberballs 2014-03-28 19:27:14
ab/99+abc/999

summitwei 2014-03-28 19:27:14
ab/99+abc/999=33/37

math-rules 2014-03-28 19:27:14
write as fractions

hwl0304 2014-03-28 19:27:14
convert the repeating decimals to fractions

tluo5458 2014-03-28 19:27:14
ab/99+abc/999= 33/37

joey8189681 2014-03-28 19:27:14
turn these to fractions

IIIIIII 2014-03-28 19:27:14
Expand the fraction

dmagician34 2014-03-28 19:27:14
Turn each into a fraction

tfmtoto 2014-03-28 19:27:14
Express as fractions.

DPatrick 2014-03-28 19:27:25
That's certainly one option, and that leads to an algebra problem.

DPatrick 2014-03-28 19:27:30
But there's an even easier solution.

hnkevin42 2014-03-28 19:27:38
Find 33/37 in decimal form, which is 0.891, with the 891 repeating

nsun48 2014-03-28 19:27:38
Find the decimal representation of 33/37

danusv 2014-03-28 19:27:41
line them up so that ABABAB+ABCABC = 0.891891

DPatrick 2014-03-28 19:27:45
Yeah. Since the left side is a repeating decimal, it might help if the right side is too.

DPatrick 2014-03-28 19:27:59
Since $37 \cdot 27 = 999,$ we have $\dfrac{33}{37} = \dfrac{33 \cdot 27}{37 \cdot 27} = \dfrac{891}{999}.$

DPatrick 2014-03-28 19:28:07
Therefore, $\dfrac{33}{37} = 0.891891\overline{891}.$

DPatrick 2014-03-28 19:28:34
As an aside, the fact that 999 = 27 * 37 seems to come up a lot in these problems. It's a useful random fact to know.

math-rules 2014-03-28 19:28:43
ababab + abcabc = 891891

ChilledLemonade 2014-03-28 19:28:43
so ababab+abcabc=891

DPatrick 2014-03-28 19:28:49
Right, we have:
\begin{align*}
& 0.ababab\ldots \\
+ & 0.abcabc\ldots \\ \hline
& 0.891891\ldots
\end{align*}

DPatrick 2014-03-28 19:28:55
What does this tell us?

math-rules 2014-03-28 19:29:15
a=4

danusv 2014-03-28 19:29:15
A = 4

lucylai 2014-03-28 19:29:15

Bomist0 2014-03-28 19:29:15
a=4

TheStrangeCharm 2014-03-28 19:29:15
a = 4

tluo5458 2014-03-28 19:29:19
a=4 and b = 4?

DPatrick 2014-03-28 19:29:23
Right! Just look at the first two digits to the right of the decimal point.

DPatrick 2014-03-28 19:29:31
We can see that $0.ab$ is a little less than half of $0.89.$

DPatrick 2014-03-28 19:30:01
In fact we must have $0.ab = 0.44.$ (If $0.ab = 0.43$ then the sum is too small, and if $0.ab = 0.45$ then the sum is too big.) So $a=b=4.$

rubberballs 2014-03-28 19:30:25
c=7?

stani95 2014-03-28 19:30:25
c=7

firemike 2014-03-28 19:30:25
so c = 7

64138luc 2014-03-28 19:30:25
a=b=4, c=7

math-rules 2014-03-28 19:30:25
c=7 then

Mathgeek727 2014-03-28 19:30:25
then c is 7?

jmann123456789 2014-03-28 19:30:25
c=7

DPatrick 2014-03-28 19:30:29
This must make $c=7$, so that (looking at the first three digits) $$aba+abc=444+447=891,$$and (looking at the next three digits) $$bab+abc=444+447=891.$$

DPatrick 2014-03-28 19:30:44
Thus our answer is just $\boxed{447}.$

DPatrick 2014-03-28 19:31:02
That problem should have been #1 in my opinion.

DPatrick 2014-03-28 19:31:18
Well, we've put it off long enough...

DPatrick 2014-03-28 19:31:22
5. Real numbers $r$ and $s$ are roots of $p(x) = x^3 + ax + b,$ and $r+4$ and $s-3$ are roots of $q(x) = x^3 + ax + b + 240.$ Find the sum of all possible values of $|b|.$

DPatrick 2014-03-28 19:31:40
I'm not going to lie: this problem is messy, and it seems (to me) too computationally-icky to be a #5.

DPatrick 2014-03-28 19:31:51
What's "hard" about it is getting from start to finish without making any algebraic mistakes.

DPatrick 2014-03-28 19:31:56
What's going to be our main tool in this problem?

coldsummer 2014-03-28 19:32:10
Vieta's Formulas

Nitzuga 2014-03-28 19:32:10
Vieta's

mathman500 2014-03-28 19:32:10
vieta's

bestwillcui1 2014-03-28 19:32:10
Vieta's Formulae

HYP135peppers 2014-03-28 19:32:10
Vieta's!

UrInvalid 2014-03-28 19:32:10
vieta's formulas

hjl00 2014-03-28 19:32:10
viettas

tapir1729 2014-03-28 19:32:10
vieta

DPatrick 2014-03-28 19:32:14
Vieta's Formulas!

DPatrick 2014-03-28 19:32:33
In particular, one of the formulas gives us some strong information right off the bat. What do we notice about these cubics?

math-rules 2014-03-28 19:32:55
sum of roots is 0

simon1221 2014-03-28 19:32:55
no x^2

lucylai 2014-03-28 19:32:55
the sum of roots is 0

UrInvalid 2014-03-28 19:32:55
no x^2 term

IIIIIII 2014-03-28 19:32:55
they have no x^2

XxAndreixX 2014-03-28 19:32:55
Sum of the roots is 0.

ChilledLemonade 2014-03-28 19:32:55
they have no x^2 term

danusv 2014-03-28 19:32:55
The sum of roots is 0

hugzy2013 2014-03-28 19:32:55
they don't have a term that is to the second power

TheStrangeCharm 2014-03-28 19:32:55
the roots sum to 0 because there is no $x^2$ term.

DPatrick 2014-03-28 19:33:02
There's no $x^2$ term.

DPatrick 2014-03-28 19:33:07
This means that the sum of the roots is 0.

DPatrick 2014-03-28 19:33:16
So if we call the third root of $p(x)$ to be $t,$ then we know that $r+s+t = 0.$

DPatrick 2014-03-28 19:33:30
And what is the third root of $q(x)$?

UrInvalid 2014-03-28 19:34:00
t-1

sojourner1 2014-03-28 19:34:00
t-1

dmagician34 2014-03-28 19:34:00
t-1

tluo5458 2014-03-28 19:34:00
t-1

lucylai 2014-03-28 19:34:00

AnneHahnlheem 2014-03-28 19:34:00
t-1

TheStrangeCharm 2014-03-28 19:34:00
$-r-s-1$.

hjl00 2014-03-28 19:34:00
-r-s-1

atmath2011 2014-03-28 19:34:00
t-1

DPatrick 2014-03-28 19:34:06
Since they also have to sum to zero, the third root of $q$ must be $t-1,$ since $$(r+4)+(s-3)+(t-1) = r+s+t = 0.$$

DPatrick 2014-03-28 19:34:18
So $p(x)$ has roots $r,s,t$, and $q(x)$ has roots $r+4,s-3,t-1.$

DPatrick 2014-03-28 19:34:27
Now what?

math-rules 2014-03-28 19:34:44
use other 2 vieta formulas

bestwillcui1 2014-03-28 19:34:49
random vieta bash gg

lucylai 2014-03-28 19:34:53
use a and b

DPatrick 2014-03-28 19:35:02
Right, we've got two more Vieta's Formulas to use.

DPatrick 2014-03-28 19:35:16
Let's just go left-to-right down the cubic, and look at the Vieta's Formula for the $ax$ term in both $p$ and $q.$

DPatrick 2014-03-28 19:35:25
From $p$, we get $rs + rt + st = a.$

DPatrick 2014-03-28 19:35:31
From $q$, we get $(r+4)(s-3) + (r+4)(t-1) + (s-3)(t-1) = a.$

DPatrick 2014-03-28 19:35:48
(These are both the sum of products of pairs of roots.)

DPatrick 2014-03-28 19:36:03
They're both $a$, so...?

kevin38017 2014-03-28 19:36:15
set the two equal?

hnkevin42 2014-03-28 19:36:15
set them equal

Readingrocks88 2014-03-28 19:36:15
they equal each other!

sojourner1 2014-03-28 19:36:15
set them equal to each other

mathwizard888 2014-03-28 19:36:15
set them equal

antler 2014-03-28 19:36:15
put them equal to each other

DPatrick 2014-03-28 19:36:22
Sure, why not? Equating them, we get $$rs + rt + st = (r+4)(s-3) + (r+4)(t-1) + (s-3)(t-1).$$

DPatrick 2014-03-28 19:36:31
Now comes the first of the messy parts...we need to simplify this.

DPatrick 2014-03-28 19:36:38
There's no point in us slogging through all the work now -- I'll just tell you that it simplifies to $0 = -4r + 3s + t - 13.$

Bomist0 2014-03-28 19:36:50
t=-r-s

DPatrick 2014-03-28 19:37:00
Right, we also know that $r+s+t = 0,$ so we can substitute $t = -r-s.$

DPatrick 2014-03-28 19:37:08
This gives $0 = -5r + 2s - 13.$

DPatrick 2014-03-28 19:37:19
That's about as simple as we can get it for now.

DPatrick 2014-03-28 19:37:40
Let's just leave it like that, and go to the third Vieta Formula, the one involving the product of all three roots.

math-rules 2014-03-28 19:37:54
now do the b terms

danusv 2014-03-28 19:37:54
-b= rst and -b-240 = (r+4)(s-3)(-r-s-1)

DPatrick 2014-03-28 19:37:57
From $p$, we get $rst = -b.$

DPatrick 2014-03-28 19:38:01
From $q$, we get $(r+4)(s-3)(t-1) = -b-240.$

DPatrick 2014-03-28 19:38:18
Putting these together gives $$rst - 240 = (r+4)(s-3)(t-1).$$

CornSaltButter 2014-03-28 19:38:32
Expand again?!

DPatrick 2014-03-28 19:38:37
Yes: again, we have to slog through expanding the right side. I'll save you the trouble.

DPatrick 2014-03-28 19:38:42
The $rst$ term will cancel, and we'll end up with
$$0 = -rs -3rt +4st + 3r - 4s -12t + 252.$$

DPatrick 2014-03-28 19:38:54
Now what?

kevin38017 2014-03-28 19:39:04
t=-r-s

lucylai 2014-03-28 19:39:04

Bomist0 2014-03-28 19:39:04
sub t in again?!!!

hshiems 2014-03-28 19:39:04
t=-r-s

pedronr 2014-03-28 19:39:04
substitute for t

UrInvalid 2014-03-28 19:39:04
t=-r-s

DPatrick 2014-03-28 19:39:12
Sure, now we can again use $t = -r-s.$ So we have
$$0=-rs-3r(-r-s)+4s(-r-s)+3r-4s-12(-r-s)+252.$$

DPatrick 2014-03-28 19:39:21
This (after more simplifying!) becomes
$$0 = 3r^2 - 4s^2 - 2rs + 15r + 8s + 252.$$

DPatrick 2014-03-28 19:39:35
Now what?

lucylai 2014-03-28 19:39:43

mathman500 2014-03-28 19:39:43
Substitute 2s=5r+13

DPatrick 2014-03-28 19:39:53
Good choice. From earlier we know that $2s = 5r + 13,$ so we can substitute that in too:
$$0 = 3r^2 - (5r+13)^2 - r(5r+13) + 15r + 4(5r+13) + 252.$$(Note that we're substituting for $2s,$ not just $s.$)

DPatrick 2014-03-28 19:40:20
After yet more algebra(!), this simplifies down to $$0 = -27r^2 - 108r + 135.$$

DPatrick 2014-03-28 19:40:31
Ooh, we can divide by -27 and have just $$0 = r^2 + 4r - 5.$$

hugzy2013 2014-03-28 19:40:43
factor?

joey8189681 2014-03-28 19:40:43
factor

summitwei 2014-03-28 19:40:43
yay a quadratic!

patl02 2014-03-28 19:40:43
Quadratics yay

summitwei 2014-03-28 19:40:48
(r-1)(r+5)

math-rules 2014-03-28 19:40:48
(r+5)(r-1)=0

DPatrick 2014-03-28 19:40:55
This factors as $0 = (r+5)(r-1),$ so the solutions are $r=-5$ and $r=1.$

QuantumandMath 2014-03-28 19:41:15
substitute in

HYP135peppers 2014-03-28 19:41:24
now we use $2s=5r-13$

pedronr 2014-03-28 19:41:24
find s and t now

danusv 2014-03-28 19:41:24
Plug in the answers into s and t

DPatrick 2014-03-28 19:41:31
Right, going back and using $2s = 5r+13$ and $r+s+t = 0,$ we get the solutions
$$(r,s,t) = (-5,-6,11) \quad\text{and}\quad (r,s,t) = (1,9,-10).$$

DPatrick 2014-03-28 19:41:48
We can double-check if these work if we want, but we've already spent enough time on this problem.

LOTRFan123 2014-03-28 19:42:04
now find b

mathwizard888 2014-03-28 19:42:04
b is -330 or 90, so 330+90=420 ans

ChilledLemonade 2014-03-28 19:42:04
Find product of the roots

DPatrick 2014-03-28 19:42:17
Right. $b=-rst$, so the first triple gives $b = -330$ and the second triple gives $b = 90.$

DPatrick 2014-03-28 19:42:23
So our final answer is $330 + 90 = \boxed{420}.$

DPatrick 2014-03-28 19:42:53
I have to say that I HATED that problem. It was fairly uninteresting and just a lot of algebra-bashing. I had to do it 4 times until I got it right.

DPatrick 2014-03-28 19:43:06
So on to better pastures...

DPatrick 2014-03-28 19:43:11
6. Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\frac23,$ and each of the other five sides has probability $\frac{1}{15}.$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Bomist0 2014-03-28 19:43:31
conditional probability

hwl0304 2014-03-28 19:43:31
Conditional probability??

UrInvalid 2014-03-28 19:43:31
conditional probability

DPatrick 2014-03-28 19:43:41
Indeed, this is another conditional probability problem.

DPatrick 2014-03-28 19:44:04
There's a fancy formula called Bayes Theorem that you can use, but it's just as easy to reason it out.

bestwillcui1 2014-03-28 19:44:13
find probability that it used fair die and prob that biased die sa was used

DPatrick 2014-03-28 19:44:39
Right: we first have to determine the probabilities that, after 2 rolls, we had the fair die or the biased die.

DPatrick 2014-03-28 19:44:44
What's the probability that the fair die rolls two sixes?

byunstephanie 2014-03-28 19:44:56
1/36

dmagician34 2014-03-28 19:44:56
1/36

bluebandit21 2014-03-28 19:44:56
1/36

tluo5458 2014-03-28 19:44:56
1/36

forthegreatergood 2014-03-28 19:44:56
(1/6)^2

firemike 2014-03-28 19:44:56
1/36

hjl00 2014-03-28 19:44:56
1/36

hnkevin42 2014-03-28 19:44:56
1/36

speck 2014-03-28 19:44:56
$\dfrac{1}{36}$

eswa2000 2014-03-28 19:44:56
1/36

DPatrick 2014-03-28 19:45:01
The fair die rolls 6-6 with probability $\dfrac16 \cdot \dfrac16 = \dfrac{1}{36}.$

DPatrick 2014-03-28 19:45:06
What's the probability that the biased die rolls two sixes?

LOTRFan123 2014-03-28 19:45:24
4/9 = 16/36

poweroftwo 2014-03-28 19:45:24
16/36=4/9

forthegreatergood 2014-03-28 19:45:24
(2/3)^2= 4/9 = 16/36

CornSaltButter 2014-03-28 19:45:24
(2/3)^2=4/9

hnkevin42 2014-03-28 19:45:24
4/9 = 16/36

DivideBy0 2014-03-28 19:45:24
(2/3)^2

DPatrick 2014-03-28 19:45:27
The biased die rolls 6-6 with probability $\dfrac46 \cdot \dfrac46 = \dfrac{16}{36}.$ (Let's leave it over 36 because it's easier to compare the two dice that way.)

DPatrick 2014-03-28 19:45:35
So if we roll two sixes, what's the likelihood that we have the fair die, and what's the likelihood that we have the biased die?

math-rules 2014-03-28 19:46:03
1/17 and 16/17

UrInvalid 2014-03-28 19:46:03
1/17 and 16/17

j2002 2014-03-28 19:46:03
1/17 and 16/17

poweroftwo 2014-03-28 19:46:03
16/17 biased, 1/17 fair

alex31415 2014-03-28 19:46:03
1/17,16/17

hwl0304 2014-03-28 19:46:03
1/17, and 16/17

Readingrocks88 2014-03-28 19:46:03
1/17 and 16/17

hnkevin42 2014-03-28 19:46:03
1/17 for the fair, and 16/17 for the biased?

AndrewK 2014-03-28 19:46:03
16/17 is the biased die and 1/17 the regular die

wu2481632 2014-03-28 19:46:03
1/17,m 16/17

speck 2014-03-28 19:46:03
Fair: 1/17, biased: 16/17

IIIIIII 2014-03-28 19:46:06
1/17 and 16/17

DPatrick 2014-03-28 19:46:12
Right. We have the fair die with probability $\dfrac{\frac{1}{36}}{\frac{1}{36}+\frac{16}{36}} = \dfrac{1}{17},$ and we have the biased die with probability $\dfrac{\frac{16}{36}}{\frac{1}{36}+\frac{16}{36}} = \dfrac{16}{17}.$

DPatrick 2014-03-28 19:46:27
(If you don't quite see this sort of conditional probability calculation, imagine that we list all 72 possibilities. In 36 of those, we'll pick the fair die, of which 1 time we'll get 6-6. In the other 36, we'll pick the biased die, of which 16 times we'll get 6-6. These are all equally likely. So of the 17 times we end up with 6-6, 1 of those comes from the fair die and 16 come from the biased die.)

DPatrick 2014-03-28 19:46:47
So how do we use this data to finish the problem?

math-rules 2014-03-28 19:47:15
1/17 * 1/6 + 16/17 * 4/6

danusv 2014-03-28 19:47:15
So it is 16/17 * 2/3 + 1/17 * 1/6

urmilla 2014-03-28 19:47:15
(1/17)(1/6) + (16/17)(2/3) = 65/102

byunstephanie 2014-03-28 19:47:15
1/17 * 1/6 + 16/17 * 2/3

Bomist0 2014-03-28 19:47:15
1/17*1/6 + 16/17 * 2/3

bluebandit21 2014-03-28 19:47:15
16/17*2/3 + 1/17*1/6

hnkevin42 2014-03-28 19:47:15
(1/17)(1/6) + (16/17)(2/3) = 65/102

simon1221 2014-03-28 19:47:15
16/17*2/3 + 1/17*1/6

DPatrick 2014-03-28 19:47:23
In semi-English, its P(we have the fair die) * P(we roll a 6 on the fair die) + P(we have the biased die) * P(we roll a 6 on the biased die).

DPatrick 2014-03-28 19:47:43
(This is basically casework in which the two cases are not equally likely.)

DPatrick 2014-03-28 19:47:50
So, we roll a third six with probability $$\frac{1}{17} \cdot \frac{1}{6} + \frac{16}{17} \cdot \frac{4}{6}.$$

DPatrick 2014-03-28 19:47:57
This simplifies to $\dfrac{65}{102}$, which is in lowest terms, so our answer is $65 + 102 = \boxed{167}.$

DPatrick 2014-03-28 19:48:18
Rolling on...

DPatrick 2014-03-28 19:48:23
7. Let $f(x) = (x^2 + 3x + 2)^{\cos(\pi x)}.$ Find the sum of all positive integers $n$ for which
$$\left| \sum_{k=1}^n \log_{10}f(k)\right| = 1.$$

DPatrick 2014-03-28 19:48:43
It looks super-hard, but there's a lot of smoke and mirrors in this problem.

DPatrick 2014-03-28 19:49:24
There are a lot of things that we can do to remove the smoke and the mirrors.

Bomist0 2014-03-28 19:49:40
cos(πx) = 1 when x is even, -1 when x is odd

UrInvalid 2014-03-28 19:49:40
cos(n*pi)=(-1)^n for int n

CornSaltButter 2014-03-28 19:49:40
Wait ${cos(\pi{x})}$ is -1 or 1...

joey8189681 2014-03-28 19:49:40
cos(pi*x) is either equal to 1 or -1

atmath2011 2014-03-28 19:49:40
cos (pi x) = 1 when even

atmath2011 2014-03-28 19:49:40
-1 odd

QuantumandMath 2014-03-28 19:49:40
cos(pi x)is just 1 for even, -1 for odd

DPatrick 2014-03-28 19:49:57
Right: one thing to notice is that we only care about what this function looks like when $x$ is an integer.

DPatrick 2014-03-28 19:50:22
And $\cos(\pi k)$ is simple when $k$ is an integer: It's 1 if $k$ is even and -1 if $k$ is odd.

DPatrick 2014-03-28 19:50:31
Thus our function is $f(k) = (k^2 + 3k + 2)^{(-1)^k}$ for positive integers $k.$

TheStrangeCharm 2014-03-28 19:50:43
Factor as $(x + 1)(x + 2)$

alex31415 2014-03-28 19:50:43
x^2+3x+2=(x+1)(x+2)

hugzy2013 2014-03-28 19:50:43
factor x^2+3x+2

QuantumandMath 2014-03-28 19:50:43
x^2+3x+2=(x+1)(x+2)

DPatrick 2014-03-28 19:50:54
We also see that $k^2 + 3k + 2 = (k+1)(k+2).$

DPatrick 2014-03-28 19:51:03
So the function is just
$$f(k) = \left\{
\begin{array}{ll}
(k+1)(k+2) & \text{if $k$ is even} \\[2ex]
\dfrac{1}{(k+1)(k+2)} & \text{if $k$ is odd}
\end{array}\right.$$

DPatrick 2014-03-28 19:51:14
That is,
\begin{align*}
f(1) &= \dfrac{1}{2 \cdot 3}, \\[1.5ex]
f(2) &= 3 \cdot 4, \\[1.5ex]
f(3) &= \dfrac{1}{4 \cdot 5}, \\[1.5ex]
f(4) &= 5 \cdot 6,
\end{align*}etc.

DPatrick 2014-03-28 19:51:28
Ok, let's look at our equation now:
$$\left| \sum_{k=1}^n \log_{10}f(k)\right| = 1.$$How can we simplify this?

DivideBy0 2014-03-28 19:51:58
take product sum log rule

UrInvalid 2014-03-28 19:51:58
sum of logs is log of product

colinhy 2014-03-28 19:51:58
log(a) + lob(b) = log(ab)

XxAndreixX 2014-03-28 19:51:58
Use log sum to product identity.

byunstephanie 2014-03-28 19:51:58
well sum for logs is multiplying the f(k)s

DPatrick 2014-03-28 19:52:07
Right: a sum of logs is just the log of the product!

DPatrick 2014-03-28 19:52:12
So
$$\left| \sum_{k=1}^n \log_{10}f(k)\right| = \left| \log_{10} \prod_{k=1}^n f(k) \right| = 1.$$

DPatrick 2014-03-28 19:52:23
And what does that tell us about the product?

ChilledLemonade 2014-03-28 19:52:47
f(k) must equal 0.1 or 10

IIIIIII 2014-03-28 19:52:47
log f1f2f3 fn equals 1 or 1/10

fortenforge 2014-03-28 19:52:47
10 or 1/10

ChilledLemonade 2014-03-28 19:52:47
It is eqaul to 0.1 or 10

simon1221 2014-03-28 19:52:47
10 or 1/10?

hnkevin42 2014-03-28 19:52:47
has to be 10 or 1/10?

abishek99 2014-03-28 19:52:47
the product is 10 or 1/10

eswa2000 2014-03-28 19:52:47
it's 10 or 1/10

DPatrick 2014-03-28 19:52:51
Right!

DPatrick 2014-03-28 19:52:55
If we have $\left| \log_{10} y \right| = 1,$ then either $\log_{10} y = 1$ or $\log_{10} y = -1.$

DPatrick 2014-03-28 19:53:00
These respectively mean that $y = 10$ or $y = \frac{1}{10}.$

DPatrick 2014-03-28 19:53:22
(NOT -10! You can never take the log of a negative number.)

DPatrick 2014-03-28 19:53:30
So we have
$$\prod_{k=1}^n f(k) = 10 \text{ or } \frac{1}{10}.$$

DPatrick 2014-03-28 19:53:37
But what does $\displaystyle\prod_{k=1}^n f(k)$ equal?

DPatrick 2014-03-28 19:53:59
Listing a few might help.

bestwillcui1 2014-03-28 19:54:13
1/(2*3) * 3*4 etc

eswa2000 2014-03-28 19:54:13
it telescopes

DPatrick 2014-03-28 19:54:17
If $n=1$, then the product is $$f(1) = \frac{1}{2 \cdot 3}.$$

DPatrick 2014-03-28 19:54:25
If $n=2$, then the product is $$\frac{1}{2 \cdot 3} \cdot f(2) = \frac{1}{2 \cdot 3} \cdot (3 \cdot 4) = \frac{4}{2}.$$

DPatrick 2014-03-28 19:54:34
If $n=3$, then the product is $$\frac{4}{2} \cdot f(3) = \frac{4}{2} \cdot \frac{1}{4 \cdot 5} = \frac{1}{2 \cdot 5}.$$

DPatrick 2014-03-28 19:54:42
If $n=4$, then the product is $$\frac{1}{2 \cdot 5} \cdot f(4) = \frac{1}{2 \cdot 5} \cdot (5 \cdot 6) = \frac{6}{2}.$$

bestwillcui1 2014-03-28 19:54:52
n=3 works!!

Tan 2014-03-28 19:54:52
son=3 is a solution

DPatrick 2014-03-28 19:55:13
Indeed, while doing this we saw that when $n=3$ we get a product of 1/10. So $n=3$ is one of our answers.

DPatrick 2014-03-28 19:55:21
And do we see the general pattern?

IIIIIII 2014-03-28 19:55:40
y=n/2+1 for y being even.

LOTRFan123 2014-03-28 19:55:40
It alternates between bigger than 1 and smaller than 1

tluo5458 2014-03-28 19:55:40
for evens it's just (n+2)/2

UrInvalid 2014-03-28 19:55:52
for even numbers product=(N+2)/2

DPatrick 2014-03-28 19:56:03
Right: if $n$ is even, then the product is $\dfrac{n+2}{2}.$

IIIIIII 2014-03-28 19:56:20
the odd ones only go down.

DivideBy0 2014-03-28 19:56:29
(n+2)/2 or 1/(2n+4)

DPatrick 2014-03-28 19:56:43
Indeed, if $n$ is odd, then the product is $\dfrac{1}{2(n+2)}.$ They decrease and we already found that $n=3$ gives us 1/10.

DPatrick 2014-03-28 19:56:48
What $n$ gives us 10?

joey8189681 2014-03-28 19:57:02
18 is the other answer

summitwei 2014-03-28 19:57:02
so n=18!

speck 2014-03-28 19:57:02
So $n=18$ is a solution as well

alex31415 2014-03-28 19:57:02
18

mathwizard888 2014-03-28 19:57:02
18

speck 2014-03-28 19:57:02
$18$

DPatrick 2014-03-28 19:57:20
We're looking for an even $n$ such that $\dfrac{n+2}{2} = 10.$ Clearly we want $n=18.$

Bomist0 2014-03-28 19:57:33
only solutions, so 18+3

eswa2000 2014-03-28 19:57:33
so 3+18=21

DPatrick 2014-03-28 19:57:34
So $n \in \{3,18\}$ are our solutions, and the final answer is $3 + 18 = \boxed{021}.$

DPatrick 2014-03-28 19:57:54
Note that we could formally prove our formula for the product by induction if we wanted to, but I think the pattern is clear. It's a weird sort of telescoping product.

DPatrick 2014-03-28 19:58:14
OK, on to the first of several harder geometry problems:

DPatrick 2014-03-28 19:58:20
8. Circle $C$ with radius 2 has diameter $\overline{AB}.$ Circle $D$ is internally tangent to circle $C$ at $A.$ Circle $E$ is internally tangent to circle $C,$ externally tangent to circle $D,$ and tangent to $\overline{AB}.$ The radius of circle $D$ is three times the radius of circle $E$ and can be written in the form $\sqrt{m} - n,$ where $m$ and $n$ are positive integers. Find $m+n.$

hugzy2013 2014-03-28 19:58:33
diagram

ChilledLemonade 2014-03-28 19:58:33
Diagram please!

tluo5458 2014-03-28 19:58:33
draw a diagram!

bestwillcui1 2014-03-28 19:58:33
Diagram!!! xD

kevin38017 2014-03-28 19:58:33
Draw a diagram!

DPatrick 2014-03-28 19:58:39
Sure, let's sketch a picture.

DPatrick 2014-03-28 19:58:44

DPatrick 2014-03-28 19:58:55
The big outer circle is $C$, the larger circle inside is $D$, and the smaller circle is $E$.

DPatrick 2014-03-28 19:59:02
What else do we probably want to draw?

forthegreatergood 2014-03-28 19:59:12
connect some centers/points of tangency/etc

eswa2000 2014-03-28 19:59:12
connect the centers

kevin38017 2014-03-28 19:59:12
centers of the circles

math-rules 2014-03-28 19:59:12
radii!

hnkevin42 2014-03-28 19:59:12
the radii?

hjl00 2014-03-28 19:59:12
the radii

alex31415 2014-03-28 19:59:12
Radii, points of tangency

DPatrick 2014-03-28 19:59:16
We probably want radii to all the points of tangency:

DPatrick 2014-03-28 19:59:23

DPatrick 2014-03-28 19:59:36
Note that it is not a straight line from the center of $D$ to the center of $E$ to the point of tangency with $C$! In fact, where does the segment from the center of $E$ to its point of tangency with $C$ go if we extend it in the other direction?

sillyd 2014-03-28 20:00:10
to the center of C

kevin38017 2014-03-28 20:00:10
cetner of C

math-rules 2014-03-28 20:00:10
center of c

sojourner1 2014-03-28 20:00:10
the center of C

lucylai 2014-03-28 20:00:10
the center of C

eswa2000 2014-03-28 20:00:10
center of C

Readingrocks88 2014-03-28 20:00:10
center of the circle

DPatrick 2014-03-28 20:00:16
It goes to the center of $C$. Let me add that too, and label some points. I'll label the centers of the circles to be $C,$ $D,$ and $E,$ and a couple more points too:

DPatrick 2014-03-28 20:00:25

DPatrick 2014-03-28 20:00:36
Again, we note that $C$, $E$, and $F$ are collinear.

DPatrick 2014-03-28 20:00:53
Now we can start chasing lengths.

math-rules 2014-03-28 20:00:58
let radius of E be r, so radius of D is 3r.

alex31415 2014-03-28 20:00:58
Let the radius of E be r and the radius of D be 3r

DPatrick 2014-03-28 20:01:04
Good call. Let's set the radius of circle $E$ to be $r,$ so that the radius of $D$ is $3r.$ IMPORTANT TO NOT FORGET: our final answer will be $3r,$ NOT $r.$ (But setting up the variables this way will avoid fractions.)

DPatrick 2014-03-28 20:01:28
Can we chase lengths in a way that will let us set up an equation to solve for $r$?

DPatrick 2014-03-28 20:01:31
How can we start?

math-rules 2014-03-28 20:02:03
use pythag on CEG and DEG

HYP135peppers 2014-03-28 20:02:03
pythag on DEG

Bomist0 2014-03-28 20:02:03
pythagoras on DEG

DivideBy0 2014-03-28 20:02:03
Pythag!

bestwillcui1 2014-03-28 20:02:03
Pythagorean Theorem

byunstephanie 2014-03-28 20:02:03
Pythagorean on the triangles?

joey8189681 2014-03-28 20:02:03
right triangle with radii EG and DE

DPatrick 2014-03-28 20:02:22
Right: certainly we've got a couple of right triangles that we'll be able to use the Pythagorean Theorem on.

DPatrick 2014-03-28 20:02:54
For example, what sides of $DEG$ do we know?

danzhi 2014-03-28 20:03:19
EG=r, DE = 4r

Bomist0 2014-03-28 20:03:19
DE = 4r, EG = r, so DG = √15 r

AndrewK 2014-03-28 20:03:19
DE = 4r and EG = r

dmagician34 2014-03-28 20:03:19
DE = 4r and EG = r

kevin38017 2014-03-28 20:03:19
we know DE=r+3r=4r and EG=r

vinayak-kumar 2014-03-28 20:03:19
All except DG

twin77 2014-03-28 20:03:19
DE = 4r, EG = r

DPatrick 2014-03-28 20:03:29
Good. We know that $DE = 4r$ (the sum of the two radii) and $EG = r.$

DPatrick 2014-03-28 20:03:41
That gives us
$$ DG = \sqrt{(DE)^2 - (EG)^2} = \sqrt{(4r)^2 - r^2} = r\sqrt{15}.$$

DPatrick 2014-03-28 20:03:52
How about $CEG$? What sides do we know?

byunstephanie 2014-03-28 20:04:19
We also know that CE is 2-r

mathwizard888 2014-03-28 20:04:19
CE=2-r, EG=r

tluo5458 2014-03-28 20:04:19
eg=r and ce=2-r

lucylai 2014-03-28 20:04:29

DPatrick 2014-03-28 20:04:41
Right. $CF = 2$ and $EF = r,$ so $CE = 2-r.$ And we already had $EG = r.$

DPatrick 2014-03-28 20:04:46
That makes
$$ CG = \sqrt{(CE)^2 - (EG)^2} = \sqrt{(2-r)^2 - r^2} = \sqrt{4-4r}.$$

DPatrick 2014-03-28 20:05:03
So now we have $DG$ and $CG$ -- what does that give us?

twin77 2014-03-28 20:05:18
CD

Bomist0 2014-03-28 20:05:18
DC?

TheStrangeCharm 2014-03-28 20:05:18
DC

DaChickenInc 2014-03-28 20:05:24
CD

ChilledLemonade 2014-03-28 20:05:24
DC

DPatrick 2014-03-28 20:05:34
Right, we can subtract them to get
$$ DC = DG - CG = r\sqrt{15} - \sqrt{4-4r}.$$

DPatrick 2014-03-28 20:05:39
Why is that helpful?

eswa2000 2014-03-28 20:05:58
CD=2-3r

dmagician34 2014-03-28 20:05:58
and 2 - DC = radius of D

Readingrocks88 2014-03-28 20:05:58
DC+3r=2

kevin38017 2014-03-28 20:05:58
because DC also equals 2-3r

TheStrangeCharm 2014-03-28 20:05:58
But that is also equal to 2 - 3r.

lucylai 2014-03-28 20:05:58

alex31415 2014-03-28 20:05:58
DC is also 2-3r; then we can solve for r.

DPatrick 2014-03-28 20:06:16
Aha! We know $AC = 2$ and $AD = 3r$, so $DC = 2-3r.$

DPatrick 2014-03-28 20:06:36
So now we have an equation equating our two calculations for $DC$:
$$
r\sqrt{15} - \sqrt{4-4r} = 2 - 3r.
$$

DPatrick 2014-03-28 20:06:46
How do we solve this for $r$?

Bomist0 2014-03-28 20:07:04
isolate the radical

alex31415 2014-03-28 20:07:04
Move the r*sqrt15 to the other side, then square

summitwei 2014-03-28 20:07:10
subtract rsqrt15 from both sides then square

DPatrick 2014-03-28 20:07:15
Right, we want to isolate the term with the $r$ inside the radical.

DPatrick 2014-03-28 20:07:20
So we rewrite it as $$\sqrt{4-4r} = r\sqrt{15} + 3r - 2 = (3 + \sqrt{15})r - 2.$$

DPatrick 2014-03-28 20:07:33
Now we can square both sides:
$$ 4 - 4r = \left(3+\sqrt{15}\right)^2r^2 - 4(3+\sqrt{15})r + 4.$$

DPatrick 2014-03-28 20:07:46
This simplifies to
$$-4r = (24 + 6\sqrt{15})r^2 - (12 + 4\sqrt{15})r.$$

kevin38017 2014-03-28 20:08:03
divide both sides by r since r doesn't equal 0

bryanxqchen 2014-03-28 20:08:03
divide by r

DPatrick 2014-03-28 20:08:18
It looks like a quadratic, but it's not. We can divide by $r$ since we know $r >0:$

DPatrick 2014-03-28 20:08:36
Doing this and then solving for $r$ eventually gives:
$$
r = \frac{4 + 2\sqrt{15}}{12 + 3\sqrt{15}}.
$$

DPatrick 2014-03-28 20:08:51
We could simplify this now, or recall at this point that the quantity we really want is
$$ 3r = \frac{4 + 2\sqrt{15}}{4 + \sqrt{15}}.$$

summitwei 2014-03-28 20:09:01
rationalize the denominator...

Bomist0 2014-03-28 20:09:01
use conjugates

joey8189681 2014-03-28 20:09:01
rationalize the denominator

bestwillcui1 2014-03-28 20:09:01
RATIONALISE THE DENOMInator yeah!!1

DPatrick 2014-03-28 20:09:16
Right. We multiply numerator and denominator by $4 - \sqrt{15}$ and notice that the denominator becomes $4^2 - 15 = 1$! Therefore,
$$
3r = (4+2\sqrt{15})(4-\sqrt{15}) = 16 + 4\sqrt{15} - 30.
$$

alex31415 2014-03-28 20:09:34
this is sqrt240-14 so the answer is 254

Bomist0 2014-03-28 20:09:34
4√15 = √240

DPatrick 2014-03-28 20:09:36
This becomes $3r = \sqrt{240} - 14$ when written in the proper format, so our final answer is $240 + 14 = \boxed{254}.$

DPatrick 2014-03-28 20:09:56
On to more counting:

DPatrick 2014-03-28 20:10:01
9. Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

DPatrick 2014-03-28 20:10:30
As with most counting problems, there are likely lots of ways to approach this, and not necessarily a single "best" approach.

math-rules 2014-03-28 20:10:44
complementary casework

bestwillcui1 2014-03-28 20:10:44
Complementary counting

hugzy2013 2014-03-28 20:10:44
complementary counting

DPatrick 2014-03-28 20:10:53
I thought it would be easier to count the complement: count subsets that do not contain three adjacent seats. (Although it is definitely also possible to solve by counting possible arrangements directly.)

DPatrick 2014-03-28 20:11:02
The circularness of the table makes it a bit tricky. How do we account for the circularness?

hugzy2013 2014-03-28 20:11:23
fix one chair

CornSaltButter 2014-03-28 20:11:23
Set one chair at the top of the circle

Bomist0 2014-03-28 20:11:23
fix a chair

DPatrick 2014-03-28 20:11:30
One way is to focus on a particular chair -- say the chair at the top of the table in the picture below:

DPatrick 2014-03-28 20:11:37

DPatrick 2014-03-28 20:11:48
What are the possibilities involving that chair?

DPatrick 2014-03-28 20:11:53
(Remember that we're counting the complement -- subsets without 3 chairs in a row.)

tluo5458 2014-03-28 20:12:30
on or off

hnkevin42 2014-03-28 20:12:30
itself or one next to it

tluo5458 2014-03-28 20:12:30
that alone

Bomist0 2014-03-28 20:12:30
two in a row or 1 forever alone

tluo5458 2014-03-28 20:12:30
that with left or right

kevin38017 2014-03-28 20:12:30
surrounded by one chair on left or right or no chairs at all

lucylai 2014-03-28 20:12:30
there is a person on the left or the right or neither

DPatrick 2014-03-28 20:12:44
That chair can be occupied (by "occupied" I mean "in the subset" -- I'm thinking of the subset as the set of chairs that people sit in)...
...and can have a single neighbor on either side, but not both. If there's a neighbor, then the chair beyond the neighbor must be empty.

DPatrick 2014-03-28 20:12:51
Or the chair can be unoccupied.

DPatrick 2014-03-28 20:12:59
So that makes 4 cases, as shown in the 4 pictures below. (An "X" indicates an occupied chair, the known empty chairs are shown, and a "?" indicates an unknown chair):

DPatrick 2014-03-28 20:13:05

DPatrick 2014-03-28 20:13:11

DPatrick 2014-03-28 20:13:17

DPatrick 2014-03-28 20:13:24

DPatrick 2014-03-28 20:13:38
What's left to count in each of these cases?

hugzy2013 2014-03-28 20:13:59
the question marks

msinghal 2014-03-28 20:13:59
a line of chairs

hjl00 2014-03-28 20:13:59
the ?'s

kevin38017 2014-03-28 20:13:59
the other chairs

Bomist0 2014-03-28 20:13:59
the ?

Awesome_1 2014-03-28 20:13:59
location of other occupants

ChilledLemonade 2014-03-28 20:13:59
The possible arrangements of the other chairs

DPatrick 2014-03-28 20:14:05
Right.

DPatrick 2014-03-28 20:14:06
In the first two cases, we need to count ways to select from 6 chairs in a row with no 3 adjacent.

DPatrick 2014-03-28 20:14:13
In the third case, we need to count ways to select from 7 chairs in a row with no 3 adjacent.

DPatrick 2014-03-28 20:14:19
In the fourth case, we need to count ways to select from 9 chairs in a row with no 3 adjacent.

DPatrick 2014-03-28 20:14:29
This sounds like the same problem, but with different numbers of chairs.

msinghal 2014-03-28 20:14:41
recursion?

lucylai 2014-03-28 20:14:41
recursion

vinayak-kumar 2014-03-28 20:14:41
recursion?

sojourner1 2014-03-28 20:14:41
recursion type problem?

DPatrick 2014-03-28 20:14:49
Indeed, it does suggest trying to set up a recursion.

DPatrick 2014-03-28 20:14:58
Let's call $f(n)$ to be the number of ways to select from $n$ chairs in a row with no 3 adjacent.

DPatrick 2014-03-28 20:15:13
So what we want to count for our complement is $2f(6) + f(7) + f(9).$

QuantumandMath 2014-03-28 20:15:29
small cases

DPatrick 2014-03-28 20:15:35
$f(1) = 2$ since either the chair is included or not.

DPatrick 2014-03-28 20:15:41
$f(2) = 4$ since any subset of the 2 chairs is legal.

DPatrick 2014-03-28 20:15:46
What's $f(3)$?

tluo5458 2014-03-28 20:15:59
7

math-rules 2014-03-28 20:15:59
7

kevin38017 2014-03-28 20:15:59
8-1

bestwillcui1 2014-03-28 20:15:59
8-1

eswa2000 2014-03-28 20:15:59
7

DPatrick 2014-03-28 20:16:12
$f(3) = 7$ since any subset except all 3 chairs is legal. So $2^3 - 1 = 7.$

DPatrick 2014-03-28 20:16:19
How about $f(n)$ for $n \ge 4$?

Bomist0 2014-03-28 20:16:40
f(x) = f(x-1) + f(x-2) + f(x-3)

Allan Z 2014-03-28 20:16:40
fn-1+fn-2+fn-3

DPatrick 2014-03-28 20:16:49
Exactly! Think of $n$ chairs in a row.

DPatrick 2014-03-28 20:16:54
If the first chair is not included, then there are $f(n-1)$ ways to pick a subset from the remaining $n-1$ chairs.

DPatrick 2014-03-28 20:17:00
If the first chair is included but the second isn't, then there are $f(n-2)$ ways to pick a subset from the remaining $n-2$ chairs.

DPatrick 2014-03-28 20:17:07
If the first two chairs are included, then the third chair must be excluded. Then there are $f(n-3)$ ways to pick a subset from the remaining $n-3$ chairs.

DPatrick 2014-03-28 20:17:16
So $f(n) = f(n-1) + f(n-2) + f(n-3)$ for all $n \ge 4.$ (It's a tribonacci sequence!)

DPatrick 2014-03-28 20:17:34
So now we just turn the crank...

DPatrick 2014-03-28 20:17:42
$f(4) = 2+4+7 = 13.$
$f(5) = 4+7+13 = 24.$
$f(6) = 7+13+24 = 44.$
$f(7) = 13+24+44 = 81.$
$f(8) = 24+44+81 = 149.$
$f(9) = 44+81+149 = 274.$

DPatrick 2014-03-28 20:17:59
And I helpfully stickied what we wanted to count up top...

bestwillcui1 2014-03-28 20:18:05
44*2+81+274

DPatrick 2014-03-28 20:18:12
$2f(6) + f(7) + f(9) = 88 + 81 + 274 = 443.$

bestwillcui1 2014-03-28 20:18:19
that's our complement however

msinghal 2014-03-28 20:18:23
complement!

kevin38017 2014-03-28 20:18:23
but thats the complement

msinghal 2014-03-28 20:18:34
subrtact from 1024

mathwizard888 2014-03-28 20:18:34
1024-443=581 ans

alex31415 2014-03-28 20:18:34
The answer is 2^10-443=581

atmath2011 2014-03-28 20:18:34
1024- 443

DPatrick 2014-03-28 20:18:36
Right! There are $2^{10} = 1024$ possible subsets, and 443 of them don't work, so $1024 - 443 = \boxed{581}$ do work.

DPatrick 2014-03-28 20:18:57
As I said, there are certainly other ways to set up the counting in this problem. I don't even claim what I did is best.

DPatrick 2014-03-28 20:19:11
But it's what came most naturally to me (for whatever reason) and it seemed pretty straightforward.

DPatrick 2014-03-28 20:19:38
Let's do #10 (which is fairly quick), then I'm going to take a 5-min rest break before we go on to 11-15.

DPatrick 2014-03-28 20:19:43
10. Let $z$ be a complex number with $|z| = 2014.$ Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}.$ Then the area enclosed by $P$ can be written in the form $n\sqrt3,$ where $n$ is an integer. Find the remainder when $n$ is divided by 1000.

DPatrick 2014-03-28 20:20:00
As I said on the first AIME Math Jam a couple of weeks ago: The AIME always seems to have one of this sort of problem, that's fairly easy if you know how to use the geometry of the complex plane, and nearly impossible if you don't.

DPatrick 2014-03-28 20:20:21
How can we better write that equation?

msinghal 2014-03-28 20:20:40
multiply out

math-rules 2014-03-28 20:20:45
(z^2 + zw + w^2) = 0

DPatrick 2014-03-28 20:20:55
Sure, let's clear all the denominators by multiplying through by $zw(z+w)$.

DPatrick 2014-03-28 20:21:01
Now our equation is $zw = w(z+w) + z(z+w).$

DPatrick 2014-03-28 20:21:07
This simplifies further to $z^2 +zw + w^2 = 0.$

DPatrick 2014-03-28 20:21:13
Any further simplification?

math-rules 2014-03-28 20:21:29
multiply both sides by (z-w) to get (z-w)^3 = 0

lucylai 2014-03-28 20:21:29
multiply by z-w

bryanxqchen 2014-03-28 20:21:29
multiply by z-w

sojourner1 2014-03-28 20:21:32
we could multiply through by z-w for fun!

DPatrick 2014-03-28 20:21:36
If we multiply both sides by $z-w$, then it's just $z^3 - w^3 = 0.$

DPatrick 2014-03-28 20:21:47
Wow! What does that tell us about $w?$

LOTRFan123 2014-03-28 20:22:00
z^3=w^3

lucylai 2014-03-28 20:22:10
w is z rotated about origin by 120 or 240 degs

math-rules 2014-03-28 20:22:15
the polygon P is a triangle

mathwizard888 2014-03-28 20:22:22
w=z*(a cube root of unity)

DPatrick 2014-03-28 20:22:34
Right. We see that $w^3 = z^3.$ So $w$ must be a cube root of $z^3$.

DPatrick 2014-03-28 20:22:40
In other words, $w = \zeta z,$ where $\zeta$ is a cube root of 1.

DPatrick 2014-03-28 20:23:04
So the possible $w$'s are just $z$ rotated 120 degrees (in either direction) about the origin.

DPatrick 2014-03-28 20:23:40
(Like I said: if you know complex number geometry, this is probably easy. If not, it's probably complete nonsense.)

alex31415 2014-03-28 20:23:50
P is an equilateral triangle with side length 2014sqrt3

bestwillcui1 2014-03-28 20:23:50
we know that the magnitude of z is 2014

DPatrick 2014-03-28 20:24:03
Exactly. We see that $P$ is an equilateral triangle that's inscribed in a circle of radius 2014, with sides of length $2014\sqrt{3}.$

DPatrick 2014-03-28 20:24:27
So what's its area?

kevin38017 2014-03-28 20:24:40
area of an equilateral triangle= s^2 * rt3 / 4

DPatrick 2014-03-28 20:24:53
The area is $\dfrac{(2014\sqrt3)^2\sqrt3}{4} = \dfrac{3(2014)^2\sqrt3}{4}.$

DPatrick 2014-03-28 20:25:03
So we have $n = \dfrac{3(2014)^2}{4} = 3(1007)^2.$

Bomist0 2014-03-28 20:25:21
-->3*7^2

alex31415 2014-03-28 20:25:21
1007=7mod 1000, so answer is 3*7^2=147

kevin38017 2014-03-28 20:25:21
we only need mod 1000

DPatrick 2014-03-28 20:25:29
Right. The last three digits of $n$ are just $3 \cdot 7^2 = 3 \cdot 49 = \boxed{147}.$

DPatrick 2014-03-28 20:29:56
OK, ready to resume! 10 down, 5 to go.

DPatrick 2014-03-28 20:30:01
On to more geometry:

DPatrick 2014-03-28 20:30:05
11. In $\triangle RED,$ $RD=1,$ $\angle DRE = 75^\circ,$ and $\angle RED = 45^\circ.$ Let $M$ be the midpoint of segment $\overline{RD}.$ Point $C$ lies on side $\overline{ED}$ such that $\overline{RC} \perp \overline{EM}.$ Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA = AR.$ Then $AE = \dfrac{a - \sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c.$

DPatrick 2014-03-28 20:30:19
Here's a picture:

DPatrick 2014-03-28 20:30:26

DPatrick 2014-03-28 20:30:36
There's one more line begging to be drawn...

Bomist0 2014-03-28 20:30:57
the altitude

kevin38017 2014-03-28 20:30:57
perp from R to DA

math-rules 2014-03-28 20:30:57
altitude

DPatrick 2014-03-28 20:31:02
If we drop a perpendicular from $R$ down to $\overline{DE},$ it splits triangle $RED$ into a 30-60-90 triangle on the left (in blue) and a 45-45-90 triangle on the right (in red):

DPatrick 2014-03-28 20:31:12

DPatrick 2014-03-28 20:31:36
So let's chase lengths. What lengths do we quickly get, starting from $RD = 1$?

math-rules 2014-03-28 20:31:59
OD=1/2

antler 2014-03-28 20:31:59
do=1/2, ro=sqrt3/2

tluo5458 2014-03-28 20:31:59
do=1/2

bestwillcui1 2014-03-28 20:31:59
DO, RO, OE, RE

tluo5458 2014-03-28 20:31:59
ro=sqrt3/2

math-rules 2014-03-28 20:31:59
RO = rt3/2

tluo5458 2014-03-28 20:32:05
oe=sqrt3/2

hwl0304 2014-03-28 20:32:05
RO=sqrt3/2

twin77 2014-03-28 20:32:05
RO = sqrt3/2, DO = 1/2

tluo5458 2014-03-28 20:32:11
re=sqrt6/2

math-rules 2014-03-28 20:32:11
OE=RO=rt3/2

Bomist0 2014-03-28 20:32:11
DO = ½, RO = √3/2

DPatrick 2014-03-28 20:32:15
$RD = 1,$ so that makes $OD = \dfrac12$ and $OR = \dfrac{\sqrt3}{2}.$

DPatrick 2014-03-28 20:32:26
Then $OE = \dfrac{\sqrt3}{2}$ as well, and hence $RE = \dfrac{\sqrt6}{2}.$

DPatrick 2014-03-28 20:32:52
So much for the easy lengths. And we still don't seem to know much about point $A$.

DPatrick 2014-03-28 20:33:08
How do we get any info about point $A$?

Bomist0 2014-03-28 20:33:23
CA = AR

bestwillcui1 2014-03-28 20:33:23
Well CA=AR

tluo5458 2014-03-28 20:33:23
ca=ar

Allan Z 2014-03-28 20:33:28
draw the perpendicular bisector of RC

DPatrick 2014-03-28 20:33:42
Indeed, $CAR$ is an isosceles triangle, so maybe drawing its median/altitude will help. Let's label the foot of the altitude from $A$ to $\overline{CR}$ as $N$, and note that $N$ is also the midpoint of $\overline{CR}.$

DPatrick 2014-03-28 20:33:48

DPatrick 2014-03-28 20:33:58
Notice anything interesting?

kevin38017 2014-03-28 20:34:22
NA and ME are parallel

tanishq1 2014-03-28 20:34:22
AN parallel to ME...

tluo5458 2014-03-28 20:34:22
na and me

math-rules 2014-03-28 20:34:22
parallel lines!

ashgabat 2014-03-28 20:34:26
AN || EM ??

DPatrick 2014-03-28 20:34:31
$\overline{AN} \parallel \overline{ME}$, because they're both perpendicular to $\overline{CR}$.

vinayak-kumar 2014-03-28 20:34:40
ME||AN

math-rules 2014-03-28 20:34:40
ME || NA ?!

DaChickenInc 2014-03-28 20:34:40
$MN\parallel AE$, $AN\parallel EM$, AEMN is a parallelogram

DPatrick 2014-03-28 20:34:50
Why are $MN$ and $AE$ parallel?

eswa2000 2014-03-28 20:35:22
M and N are midpoints

DPatrick 2014-03-28 20:35:42
Right! $\overline{MN} \parallel \overline{CD}$, because $\overline{MN}$ is the segment connecting the midpoints of $\overline{RD}$ and $\overline{RC}$. (Another way to say this is that $RMN$ and $RDC$ are similar triangles, in ratio $1:2.$)

twin77 2014-03-28 20:35:59
so therefore MN = EA

DPatrick 2014-03-28 20:36:10
Exactly. So that means $MNAE$ is a parallelogram!

DPatrick 2014-03-28 20:36:17
In particular $MN = AE$, and as we just discussed, $2MN = CD$.

Allan Z 2014-03-28 20:36:26
or AE=1/2DC!

msinghal 2014-03-28 20:36:26
so AE = CD/2

DPatrick 2014-03-28 20:36:36
Yes. The answer that we're looking for is just $AE = MN = \frac12CD.$

DPatrick 2014-03-28 20:36:43
That's great, because now we don't need point $A$ anymore! Let me remove it, as it just clutters the diagram:

DPatrick 2014-03-28 20:36:53

DPatrick 2014-03-28 20:37:11
Our goal is to find $\frac12 CD.$ Now what?

Allan Z 2014-03-28 20:37:44
find CO

antler 2014-03-28 20:37:44
use similar triangles?

DPatrick 2014-03-28 20:37:53
Certainly if we could find $OC$ we'd be done, since $CD = OD - OC$ and we know $OD = \frac12.$

DPatrick 2014-03-28 20:38:15
So we might try to get at triangle $ROC.$ Is there a triangle similar to $ROC$ in the picture? Or can we make one?

Lord.of.AMC 2014-03-28 20:38:54
RC perpendicular to EM...

Bomist0 2014-03-28 20:38:54
EM perp to RC

Allan Z 2014-03-28 20:39:01
drop the altitude from M to DE

DPatrick 2014-03-28 20:39:46
Since $EM$ is perpendicular to $RC$, it makes sense to look for a triangle similar to $ROC$ that uses side $\overline{EM}$ somehow.

DPatrick 2014-03-28 20:39:58
And it has to be a right triangle, of course.

DPatrick 2014-03-28 20:40:18
So let's draw the perpendicular from $M$ down to $\overline{DE}$, and call the foot of that perpendicular $P$:

DPatrick 2014-03-28 20:40:27

Lord.of.AMC 2014-03-28 20:40:42
RCO similar to EMP

DPatrick 2014-03-28 20:40:50
Now triangles $MPE$ and $COR$ are similar right triangles! I'll label them in green so that they're clearer:

DPatrick 2014-03-28 20:40:57

DPatrick 2014-03-28 20:41:08
To see why they're similar, note that $\overline{ME}$, $\overline{MP}$, and $\overline{PE}$ are each perpendicular, respectively, to $\overline{CR}$, $\overline{CO}$, and $\overline{OR}.$ So in particular if we were to rotate $MEP$ 90 degrees, its sides would all be parallel to the corresponding sides in $CRO.$

chessderek 2014-03-28 20:41:30
we know MP

DPatrick 2014-03-28 20:41:47
Indeed we know a couple of the lengths in $MEP$!

Lord.of.AMC 2014-03-28 20:42:03
and DMP similar to DRO with factor 1/2

DPatrick 2014-03-28 20:42:12
Right, so in particular, $MP = \dfrac12 RO = \dfrac{\sqrt3}{4}.$ (This is because triangle $MPD$ is half of triangle $ROD,$ since $M$ and $P$ are midpoints.)

DPatrick 2014-03-28 20:42:33
And $EP = EO + \dfrac12 OD = \dfrac{\sqrt3}{2} + \dfrac14 = \dfrac{2\sqrt3 + 1}{4}.$

chessderek 2014-03-28 20:43:30
use ratios to solve for OC?

eswa2000 2014-03-28 20:43:30
side ratios to find CO

kevin38017 2014-03-28 20:43:30
we have the stuff we need to find CO with similarity

DPatrick 2014-03-28 20:43:47
Right, the side ratios of $MPE \sim COR$ give us $$\dfrac{OC}{OR} = \dfrac{PM}{PE} = \dfrac{\sqrt3}{2\sqrt3 + 1}.$$

DPatrick 2014-03-28 20:44:07
And we already know (from near the beginning -- it's still stickied up top) that $OR = \dfrac{\sqrt3}{2}.$

DPatrick 2014-03-28 20:44:23
So $$ OC = OR \cdot \dfrac{\sqrt3}{2\sqrt3 + 1} = \dfrac{\sqrt3}{2} \cdot \dfrac{\sqrt3}{2\sqrt3 + 1} = \dfrac{3}{4\sqrt3 + 2}.$$

DPatrick 2014-03-28 20:44:53
And this makes
$$CD = OD - OC = \frac12 - \frac{3}{4\sqrt3 + 2}.$$

DPatrick 2014-03-28 20:45:02
(It's just algebra from here...)

DPatrick 2014-03-28 20:45:15
This simplifies to $$CD = \frac{2\sqrt3 + 1 - 3}{4\sqrt3 + 2} = \frac{2\sqrt3-2}{4\sqrt3+2} = \frac{\sqrt3 - 1}{2\sqrt3 + 1}.$$

DPatrick 2014-03-28 20:45:31
And finally what we want is $$AE = \frac12CD = \frac{\sqrt3 - 1}{4\sqrt3 + 2}.$$

QuantumandMath 2014-03-28 20:45:42
rationalize the denominator

Bomist0 2014-03-28 20:45:42
rationalize denom

ChilledLemonade 2014-03-28 20:45:42
Rationalize

DPatrick 2014-03-28 20:45:49
Right, we still need to rationalize the denominator to get it into the final form.

DPatrick 2014-03-28 20:45:57
$$AE = \frac{\sqrt3-1}{4\sqrt3+2} \cdot \frac{4\sqrt3-2}{4\sqrt3-2}.$$

DPatrick 2014-03-28 20:46:07
This simplifies to
$$ AE = \frac{14-6\sqrt3}{48 - 4} = \frac{7 - 3\sqrt3}{22} = \frac{7-\sqrt{27}}{22}.$$

bestwillcui1 2014-03-28 20:46:25
7+27+22=56

twin77 2014-03-28 20:46:25
so 7 + 27 + 22 = 056

DPatrick 2014-03-28 20:46:29
Therefore, our final answer is $7 + 27 + 22 = \boxed{056}.$

DPatrick 2014-03-28 20:46:59
There are many other ways to do this problem. I just set up coordinates the first time and bashed through the calculations.

DPatrick 2014-03-28 20:47:27
Anyway, on to a different sort of geometry problem, one that's not really geometry at all:

DPatrick 2014-03-28 20:47:32
12. Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A) + \cos(3B) + \cos(3C) = 1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

DPatrick 2014-03-28 20:47:47
Let's set $a = 3A$ and $b = 3B$ and $c = 3C$ so I don't have to keep writing "3" all the time.

DPatrick 2014-03-28 20:47:55
So we have $a+b+c = 3\pi$ and $\cos a + \cos b + \cos c = 1.$

DPatrick 2014-03-28 20:48:08
What do we know?

Bomist0 2014-03-28 20:48:36
c = 3π -a-b

DPatrick 2014-03-28 20:49:03
Let's combine our equations by getting rid of $c$ by substituting:
$$\cos a + \cos b + \cos(3\pi - a - b) = 1.$$

DPatrick 2014-03-28 20:49:14
What does that third term simplify to?

Allan Z 2014-03-28 20:49:39
cos(3pi-a-b)=-cos(a+b)

kevin38017 2014-03-28 20:49:39
-cos(a+b)

zhuangzhuang 2014-03-28 20:49:39
-cos(a+b)

DPatrick 2014-03-28 20:50:02
$3\pi$ inside a cosine is the same as $\pi$, and we use the identity $\cos(\pi - \theta) = -\cos\theta.$ So now we have
$$\cos a + \cos b - \cos(a+b) = 1.$$

DPatrick 2014-03-28 20:50:16
Let's write it without the minus sign: $$\cos a + \cos b = \cos(a+b) + 1.$$

DPatrick 2014-03-28 20:50:27
Any ideas what we can do with this now?

DPatrick 2014-03-28 20:50:59
(Note: there are many ways to proceed, depending on how exotic a trig identity you want to use.)

summitwei 2014-03-28 20:51:13
cos summation property

tluo5458 2014-03-28 20:51:13
sum of cosines

zhuangzhuang 2014-03-28 20:51:34
sum to product

DPatrick 2014-03-28 20:51:35
Well, I see a sum of cosines on both sides. (using $\cos 0 = 1$) so the cosine sum-to-product formula might be helpful:
$$\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).$$

DPatrick 2014-03-28 20:51:57
When we apply that to our equation, we get
$$2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) = 2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a+b}{2}\right).$$

DPatrick 2014-03-28 20:52:12
Aha! What does this tell us?

twin77 2014-03-28 20:52:33
so cos((a-b)/2) = cos((a+b)/2)

Bomist0 2014-03-28 20:52:33
cos ((a-b)/2) = cos ((a+b)/2)

DPatrick 2014-03-28 20:52:38
Well...not so fast.

zhuangzhuang 2014-03-28 20:52:44
cos(a-b/2)=cos(a+b/2) or cos(a+b/2)=0

DPatrick 2014-03-28 20:52:53
Yeah, that's a possibility too.

DPatrick 2014-03-28 20:52:58
Either $\cos\left(\dfrac{a+b}{2}\right) = 0,$ or we can divide by it (and 2) to get $$\cos\left(\frac{a-b}{2}\right) = \cos\left(\frac{a+b}{2}\right).$$

DPatrick 2014-03-28 20:53:14
What happens in the first case, if $\cos\left(\dfrac{a+b}{2}\right) = 0$?

XxAndreixX 2014-03-28 20:53:47
a + b = pi

DPatrick 2014-03-28 20:54:25
That's true, but it skips an important step. Initially all we know is that $\dfrac{a+b}{2}$ must be a half-multiple of $\pi$.

DPatrick 2014-03-28 20:54:37
But we know that $0 < a+b < 3\pi$, so $0 < \dfrac{a+b}{2} < \dfrac32\pi$, and the only possibility is $\dfrac{a+b}{2} = \dfrac\pi2.$

DPatrick 2014-03-28 20:54:51
And indeed this means that $a+b = \pi.$

zhuangzhuang 2014-03-28 20:55:02
so C=120 degrees

kevin38017 2014-03-28 20:55:10
this means the thrid angle of the triangle is 120

DPatrick 2014-03-28 20:55:28
Right. This makes $c = 2\pi$ and $C = \dfrac23\pi.$ (Or if you prefer, $C = 120^\circ.$)

DPatrick 2014-03-28 20:55:39
What about the other case: $\cos\left(\dfrac{a-b}{2}\right) = \cos\left(\dfrac{a+b}{2}\right)$?

DPatrick 2014-03-28 20:55:50
When do two angles have the same cosine?

kevin38017 2014-03-28 20:56:37
b=0

Allan Z 2014-03-28 20:56:37
same or add up to 2pi

nalaxone44 2014-03-28 20:56:37
when they add to 360

tluo5458 2014-03-28 20:56:37
when the sum to 360

nalaxone44 2014-03-28 20:56:37
when they add to 2pi

bestwillcui1 2014-03-28 20:56:37
x and 360-x

kevin38017 2014-03-28 20:56:46
when they differ by a multiple of 2pi, are the same angle, add up to 2pi

DPatrick 2014-03-28 20:57:00
I think between you, you may have all the cases. More generally, their sum or difference has to be a multiple of $2\pi.$

DPatrick 2014-03-28 20:57:44
But in our equation, the sum of the angles is $a$ and their difference is $b$. The only way that one of these can be a multiple of $2\pi$ is if one of them itself is $2\pi$. (We know they're not 0, and $4\pi$ is too big.)

tluo5458 2014-03-28 20:58:15
so one is 120 degrees

DPatrick 2014-03-28 20:58:16
So either $a = 2\pi$ or $b = 2\pi$, and that means either $A = \frac23\pi = 120^\circ$ or $B = \frac23\pi = 120^\circ.$

kevin38017 2014-03-28 20:58:41
so either way our biggest angle is 120 degrees

DPatrick 2014-03-28 20:58:49
Right. In summary, we've proved that our identity holds if and only if one of the angles in the triangle is $\frac23\pi$ (or $120^\circ$).

DPatrick 2014-03-28 20:59:08
So what does that tell us about maximizing the length of the remaining side?

math-rules 2014-03-28 20:59:23
law of cosines

TheStrangeCharm 2014-03-28 20:59:23
it should be opposite the 120 degree angle

kevin38017 2014-03-28 20:59:26
its opposite angle is 120 degrees

abishek99 2014-03-28 20:59:32
it has to be opposite the 120 degree angle

Bomist0 2014-03-28 20:59:32
use cos law, but the side is opposite 120 degrees

zhuangzhuang 2014-03-28 20:59:32
let the 120 be between the 10, 13

DPatrick 2014-03-28 20:59:58
Right. We want to have a triangle with sides 10 and 13 and an angle of $\frac{2\pi}{3}$ between them. The side opposite the obtuse angle will be the longest side.

tanishq1 2014-03-28 21:00:10
now we jsut do law of cosines on 10 and 13

simon1221 2014-03-28 21:00:10
law of cosines, get remaining side

Bomist0 2014-03-28 21:00:13
x^2 = 10^2 + 13^2 - 2*10*13*cos120

DPatrick 2014-03-28 21:00:32
By the Law of Cosines:
$$ m = 10^2 + 13^2 - 2(10)(13)\left(\cos\frac{2\pi}{3}\right).$$

DPatrick 2014-03-28 21:00:58
This simplifies to $$m = 100 + 169 - 260(-0.5) = \boxed{399}.$$

DPatrick 2014-03-28 21:01:31
Not too bad. On to #13:

DPatrick 2014-03-28 21:01:36
13. Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5,$ no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

DPatrick 2014-03-28 21:01:53
Let's number the adults 1 through 10, and call their shoes L1, L2, ..., L10 and R1, R2, ..., R10 so that we can more easily discuss them.

DPatrick 2014-03-28 21:02:09
We can also denote the child's pairs as $(Ln, Rf(n))$ for some function $f$. (For example, if the child pairs L1 with R3, then $f(1) = 3$ and (L1,R3) is one of our pairs.)

DPatrick 2014-03-28 21:02:25
First, we can count the denominator of our eventual probability. How many possible pairings can the child make?

Allan Z 2014-03-28 21:02:55
10!

kevin38017 2014-03-28 21:02:55
10!

mathwizard888 2014-03-28 21:02:55
10!

math-rules 2014-03-28 21:02:55
10!

DPatrick 2014-03-28 21:02:59
She has 10 choices for which right shoe to pair with L1.

DPatrick 2014-03-28 21:03:05
Then she has 9 choices for which remaining right shoe to pair with L2.

DPatrick 2014-03-28 21:03:10
Then she has 8 choices for which remaining right shoe to pair with L3.

DPatrick 2014-03-28 21:03:16
And so on.

DPatrick 2014-03-28 21:03:22
So there are $10!$ possible pairings.

DPatrick 2014-03-28 21:04:00
(A fancy way to say this is that the function $f$ that I described above gives us a permutation of the 10 right shoes, and there are $10!$ permutations of 10 objects.)

DPatrick 2014-03-28 21:04:12
OK, on to the numerator.

DPatrick 2014-03-28 21:04:26
Let's call a subset of the adults "good" if their shoes are all matched among each other. For example, if (L1,R8), (L8,R3), and (L3,R1) were all pairs, then the subset {1,3,8} would be good.

DPatrick 2014-03-28 21:04:46
...because those 3 adults could grab those 3 pairs, and each retrieve their own shoes.

DPatrick 2014-03-28 21:04:53
So the problem is asking for the probability that there are no good subsets of size 1, 2, 3, or 4.

DPatrick 2014-03-28 21:04:59
But if this is the case, what else do we know?

kevin38017 2014-03-28 21:05:27
we can have either 2 good subsets of 5 or 1 good subset of 10

msinghal 2014-03-28 21:05:31
it is 5-5 or 10

DPatrick 2014-03-28 21:05:36
Why are those the only possibilities?

kevin38017 2014-03-28 21:05:41
because if we have a good subset of size 6,7,8,9 then we must have one of 1,2,3,4

DPatrick 2014-03-28 21:05:50
Right. There are no good subsets of size 6, 7, 8, or 9 either! Because if some of the adults form a good subset, then all the rest of the adults form a good subset too. (Continuing our example above, if {1,3,8} is good, then we know that {2,4,5,6,7,9,10} must have all their shoes paired among themselves as well, so it's a good subset too.)

DPatrick 2014-03-28 21:06:32
So either there's a 5-person good subset, in which case the 10 people break up into two 5-person good subsets, or there isn't, in which case there are no proper good subsets other than the entire group of 10 people.

DPatrick 2014-03-28 21:06:45
So let's count the possible pairings in these two cases.

DPatrick 2014-03-28 21:06:56
How many pairings have a 5-person good subset? How can we construct one?

tluo5458 2014-03-28 21:07:13
10c5

bestwillcui1 2014-03-28 21:07:13
Choose five people from 10

DPatrick 2014-03-28 21:07:20
First, we have to choose 5 people to be our good subset. There are $\binom{10}{5}$ ways to do this.

DPatrick 2014-03-28 21:07:28
Then how many ways can the child arrange the shoes?

DPatrick 2014-03-28 21:07:49
Let's think about how the child can construct a 5-person good pairing that doesn't have any 1,2,3,4-good subset.

DPatrick 2014-03-28 21:07:58
She has 4 choices for which right to pair with the first left -- she can't pick the matching shoe.

DPatrick 2014-03-28 21:08:08
Then what should she do next?

bestwillcui1 2014-03-28 21:08:29
Pick one of the other 4 shoes

kevin38017 2014-03-28 21:08:36
figure out how many choices the person whose right shoe she has's left shoe can be paired

DPatrick 2014-03-28 21:08:48
Right: she tries to pair the left shoe that corresponds to the right shoe that she just picked. (For example, if she started by pairing L1 with R3, then she next tries to pair L3.)

DPatrick 2014-03-28 21:09:01
She then has 3 choices for which right to pair with the next left -- she can't pick the matching right shoe from the first pair, or otherwise she'd create a 2-person good subset. (For example, if the 5-person set was {1,2,3,4,5} and she paired L1 with R3, then she can't pair L3 with R1, because then {1,3} would be a 2-person good subset.)

DPatrick 2014-03-28 21:09:26
Similarly, at each step she can choose any shoe except for the right shoe corresponding to the first left shoe -- because otherwise she'd make a smaller good set.

DPatrick 2014-03-28 21:09:40
So she has $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1 = 4!$ choices for pairing the shoes within the 5-person good subset.

DPatrick 2014-03-28 21:10:05
Another way to think about it: imagine the 5 people sitting around a table, and each hands their right shoe to the person on their right. This creates a 5-good pairing of shoes.

DPatrick 2014-03-28 21:10:21
...and there are $4!$ ways to seat 5 people around a round table.

bestwillcui1 2014-03-28 21:10:30
the other five also have to have a 5-person good subset though

DPatrick 2014-03-28 21:10:41
Right: the other 5 people also make a good subset, so there are $4!$ choices for pairing the shoes within that subset.

Doink 2014-03-28 21:10:57
Do we need to divide C(10, 5) by 2 because the order we chose the subsets does not matter?

Allan Z 2014-03-28 21:10:57
but wait, picking ABCDE is same as picking FGHIJ in 10C5, we overcounted by a multiple of 2

DPatrick 2014-03-28 21:11:16
You're right! We've double-counted!

DPatrick 2014-03-28 21:11:25
For example, we counted {1,2,3,4,5} as one of our $\binom{10}{5}$ good subsets. But that leaves {6,7,8,9,10} as a good subset, which we also counted as one of our $\binom{10}{5}$ good subsets.

DPatrick 2014-03-28 21:11:37
So we have to divide by 2.

DPatrick 2014-03-28 21:11:49
Therefore, collecting all our counts, there are $\dfrac12\binom{10}{5}(4!)^2$ arrangements with two 5-person good subsets.

DPatrick 2014-03-28 21:12:13
And the other case: how many pairings with no proper good subsets at all?

math-rules 2014-03-28 21:12:32
9!?

TheStrangeCharm 2014-03-28 21:12:32
9!

brzhang97 2014-03-28 21:12:32
9!

bestwillcui1 2014-03-28 21:12:32
Same logic gives 9!

eswa2000 2014-03-28 21:12:32
9!

XxAndreixX 2014-03-28 21:12:38
9!

kevin38017 2014-03-28 21:12:38
9!

morninglight 2014-03-28 21:12:38
9!

QuantumandMath 2014-03-28 21:12:38
9!

twin77 2014-03-28 21:12:38
9!

DPatrick 2014-03-28 21:12:46
Right. We can count them just like we counted the arrangements of the 5-person good subsets.

DPatrick 2014-03-28 21:12:52
The child can pair L1 with any of R2,...,R10, for 9 choices.

DPatrick 2014-03-28 21:12:58
The child can pair the next left with any of 8 remaining right shoes (she can't pick L1 until the end, otherwise she creates a smaller good subset), for 8 choices.

DPatrick 2014-03-28 21:13:06
The child can pair the next left with any of 7 remaining right shoes.

DPatrick 2014-03-28 21:13:14
And so on. At the end, she'll pair the remaining left shoe with R1. So there are $9!$ pairings in this case.

math-rules 2014-03-28 21:13:28
or just pass your right shoe around the table

eswa2000 2014-03-28 21:13:28
or do the table sitting argument

DPatrick 2014-03-28 21:13:41
Right: the same analogy shows that there are $9!$ pairings.

DPatrick 2014-03-28 21:13:48
Combining our two cases and dividing by the total number of pairings, we get that the probability is
$$ \frac{\frac12\binom{10}{5}(4!)^2 + 9!}{10!}.$$

DPatrick 2014-03-28 21:14:05
How do we efficiently simplify this?

Bomist0 2014-03-28 21:14:21
divide into two parts

kevin38017 2014-03-28 21:14:21
split the fraction

hjl00 2014-03-28 21:14:29
separate the 9! term

TheStrangeCharm 2014-03-28 21:14:29
break into two parts

DPatrick 2014-03-28 21:14:45
We could do that, since $\dfrac{9!}{10!} = \dfrac{1}{10}$ is easy.

DPatrick 2014-03-28 21:15:19
That leaves $ \dfrac{\frac12\binom{10}{5}(4!)^2}{10!}.$ How do we simplify this?

Bomist0 2014-03-28 21:15:33
use definition of 10C5 = 10!/5!/5!

DPatrick 2014-03-28 21:15:48
The first step is to write out the binomial coefficient in terms of factorials:
$$\dfrac{\frac{10!4!4!}{2 \cdot 5!5!}}{10!}.$$

antler 2014-03-28 21:16:08
cut off the five using the (4!)^2

DaChickenInc 2014-03-28 21:16:08
cancel the 10!

Bomist0 2014-03-28 21:16:08
10! cancels

DPatrick 2014-03-28 21:16:26
...and now most everything cancels. $10!$ cancels and $4!/5! = 1/5$ (twice).

mssmath 2014-03-28 21:16:39
1/50

twin77 2014-03-28 21:16:39
(4!)^2/(2*(5!)^2) gives 1/50

DivideBy0 2014-03-28 21:16:39
1/50

DPatrick 2014-03-28 21:16:46
So it's just $\dfrac{1}{50}.$

DPatrick 2014-03-28 21:17:05
So our final answer is $\dfrac{1}{50} + \dfrac{1}{10} = \dfrac{6}{50} = \dfrac{3}{25},$ so our final answer is $3 + 25 = \boxed{028}.$

DPatrick 2014-03-28 21:17:25
One more geometry:

DPatrick 2014-03-28 21:17:31
14. In $\triangle ABC,$ $AB=10,$ $\angle A = 30^\circ,$ and $\angle C = 45^\circ.$ Let $H,$ $D,$ and $M$ be points on the line $\overline{BC}$ such that $\overline{AH} \perp \overline{BC},$ $\angle BAD = \angle CAD,$ and $BM = CM.$ Point $N$ is the midpoint of the segment $\overline{HM},$ and point $P$ is on ray $AD$ such that $\overline{PN} \perp \overline{BC}.$ Then $AP^2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

DPatrick 2014-03-28 21:17:57
This should be a little reminiscent of #11. Again we've got a triangle that splits in 30-60-90 and 45-45-90.

DPatrick 2014-03-28 21:18:05

DPatrick 2014-03-28 21:18:26
Note that $M$ is the midpoint of $\overline{BC}$ and $N$ is the midpoint of $\overline{HM}.$ (This is hard to label in the diagram.)

DPatrick 2014-03-28 21:18:34
What else do you notice?

Allan Z 2014-03-28 21:19:01
HAD is 30-60-90

XxAndreixX 2014-03-28 21:19:06
<HAP = 30 degrees

Bomist0 2014-03-28 21:19:10
parallel lines

ashgabat 2014-03-28 21:19:15
AHC is a 45-45-90 Triangle

Mathgeek727 2014-03-28 21:19:15
30 60 90 triangle

antler 2014-03-28 21:19:15
triangle ahc is 45 45 90

DPatrick 2014-03-28 21:19:19
Indeed, all these are good!

DPatrick 2014-03-28 21:19:29
$AHC$ is a 45-45-90 triangle, so $\angle HAB = 15^\circ.$

DPatrick 2014-03-28 21:19:42
That means that $AHD$ is 30-60-90.

DPatrick 2014-03-28 21:19:53
But we've also got parallel lines $\overline{AH}$ and $\overline{PN}$.

DPatrick 2014-03-28 21:20:08
So $PND$ is 30-60-90 too!

DPatrick 2014-03-28 21:20:17
How does that help?

DPatrick 2014-03-28 21:20:49
Remember, keep your eye on the prize -- we're trying to find $AP.$ Can we relate it to any lengths that are easier to find?

DaChickenInc 2014-03-28 21:21:05
$AP:HN=2$

DPatrick 2014-03-28 21:21:18
Right: we know that $AD = 2HD$ and $PD = 2ND$.

DPatrick 2014-03-28 21:21:52
That means that the same ratio holds for their difference too $AP = AD - PD = 2HD - 2ND = 2HN.$

Allan Z 2014-03-28 21:22:10
AP=2HN=HM

DPatrick 2014-03-28 21:22:20
Right, even better: since $N$ is the midpoint of $\overline{HM}$: it means that $AP = 2HN = HM.$

DPatrick 2014-03-28 21:22:32
So we can try to find $HM$ instead of $AP.$ That seems a lot easier.

DPatrick 2014-03-28 21:23:00
Any ideas?

eswa2000 2014-03-28 21:23:08
HM=HB+1/2BC

DPatrick 2014-03-28 21:23:21
Right, we can use point $B$ to split it up as $HM = HB + BM.$ And both of those are easy to find!

DPatrick 2014-03-28 21:24:00
We chase lengths around the original triangle (using the 30-60-90 and 45-45-90 split) to find $BM,$ basically just like we did in problem #11: $AB = 10$, so $OB = 5$, so $BC = 5\sqrt2$, so $BM = \frac12BC = \dfrac{5\sqrt2}{2}.$

DPatrick 2014-03-28 21:24:16
(I did it quick because we did the same thing three problems ago!)

DPatrick 2014-03-28 21:24:24
And how do we find $HB$?

Allan Z 2014-03-28 21:24:39
10sin15

kevin38017 2014-03-28 21:24:39
10sin 15 degrees

Bomist0 2014-03-28 21:24:39
sin15?

DPatrick 2014-03-28 21:24:47
Sure. Using right triangle $AHB$, we have $HB = 10 \sin 15^\circ.$

DPatrick 2014-03-28 21:25:01
You might have $\sin 15^\circ$ memorized. (It seems to come up a lot.) I don't, so I just compute $\sin(45^\circ - 30^\circ)$ using the sine subtraction formula. (I prefer this to the half-angle formula because I don't have to take a square root.)

DPatrick 2014-03-28 21:25:14
$$\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt6}{4} - \frac{\sqrt2}{4} = \frac{\sqrt6-\sqrt2}{4}.$$

DPatrick 2014-03-28 21:25:29
So $HB = 10 \sin 15^\circ = \dfrac{5(\sqrt6 - \sqrt2)}{2}.$

DPatrick 2014-03-28 21:25:51
How convenient when we add!
$$AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}.$$

DPatrick 2014-03-28 21:26:18
And to finish, we compute $(AP)^2 = \dfrac{150}{4} = \dfrac{75}{2}.$ So our final answer is $75 + 2 = \boxed{077}.$

DPatrick 2014-03-28 21:26:44
And last but not least:

DPatrick 2014-03-28 21:26:50
15. For any integer $k \ge 1,$ let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k) > 2,$ and $X(k) = 1$ if $p(k) = 2.$ Let $\{x_n\}$ be the sequence defined by $x_0 = 1,$ and $x_{n+1}X(x_n) = x_np(x_n)$ for $n \ge 0.$ Find the smallest positive integer $t$ such that $x_t = 2090.$

DPatrick 2014-03-28 21:27:22
How can we get started?

Bomist0 2014-03-28 21:27:29
patterns, patterns, patterns

Bomist0 2014-03-28 21:27:34
List a few values?

Allan Z 2014-03-28 21:27:34
start with small cases

ashgabat 2014-03-28 21:27:34
Find le pattern

DPatrick 2014-03-28 21:27:42
Absent any better ideas, we can try to compute the first few terms of the sequence, and try to find a pattern.

DPatrick 2014-03-28 21:27:52
We are given that $x_0 = 1.$

DPatrick 2014-03-28 21:27:59
Let's rearrange the equation a little so that it looks more like a "formula" for the next term:

DPatrick 2014-03-28 21:28:04
$$ x_{n+1} = \frac{x_n p(x_n)}{X(x_n)} \qquad \text{for all } n \ge 0.$$

DPatrick 2014-03-28 21:28:17
What is $x_1?$

lucylai 2014-03-28 21:28:44
2

sillyd 2014-03-28 21:28:44
2

math-rules 2014-03-28 21:28:44
2

mathwizard888 2014-03-28 21:28:44
2

dantx5 2014-03-28 21:28:44
2

DPatrick 2014-03-28 21:28:54
From the given formula, $\displaystyle x_1 = \frac{x_0 p(x_0)}{X(x_0)}= \frac{1\cdot p(1)}{X(1)} = \frac{1 \cdot 2}{1} = 2.$

DPatrick 2014-03-28 21:29:31
Note: $p(1)$ is the smallest prime not dividing 1, which is 2, and $X(1) = 1$ by the "exception" rule when $p = 2.$

DPatrick 2014-03-28 21:29:42
What is $x_2$?

mapletree14 2014-03-28 21:29:56
3

lucylai 2014-03-28 21:29:56
3

dantx5 2014-03-28 21:29:56
3

bengals 2014-03-28 21:29:56
3

DPatrick 2014-03-28 21:30:06
By the formula we have that $\displaystyle x_2 = \frac{x_1 p(x_1)}{X(x_1)}= \frac{2\cdot p(2)}{X(2)}.$

DPatrick 2014-03-28 21:30:21
We find that $p(2) = 3$ and $X(2) = 2$ is the product of all primes less than 3.

DPatrick 2014-03-28 21:30:29
Thus we have $\displaystyle x_2 = \frac{2 p(2)}{X(2)} = \frac{2 \cdot 3}{2} = 3.$

DPatrick 2014-03-28 21:30:39
What is $x_3$?

lucylai 2014-03-28 21:30:53
6=2*3

math-rules 2014-03-28 21:30:53
6

dantx5 2014-03-28 21:30:53
6

eswa2000 2014-03-28 21:30:53
6

bengals 2014-03-28 21:30:53
6

tluo5458 2014-03-28 21:30:53
6

kevin38017 2014-03-28 21:30:53
6

DPatrick 2014-03-28 21:30:59
We have that $\displaystyle x_3 = \frac{x_2 p(x_2)}{X(x_2)} = \frac{3 p(3)}{X(3)}.$

DPatrick 2014-03-28 21:31:08
Since $p(3) = 2$ we get $X(3) = 1$. Therefore $\displaystyle x_3 = \frac{3 p(3)}{X(3)} = \frac{3 \cdot 2}{1} = 6.$

DPatrick 2014-03-28 21:31:16
Let's do a couple more, but I'll speed it up a bit...

DPatrick 2014-03-28 21:31:22
We have $\displaystyle x_4 = \frac{x_3 p(x_3)}{X(x_3)} = \frac{6 p(6)}{X(6)} = \frac{6 \cdot 5}{6} = 5.$

DPatrick 2014-03-28 21:31:28
And we have that $\displaystyle x_5 = \frac{x_4 p(x_4)}{X(x_4)} = \frac{5 p(5)}{X(5)} = \frac{5 \cdot 2}{1} = 10.$

DPatrick 2014-03-28 21:31:43
In fact, here are the first few terms of the sequence:\[
\begin{array}{c|c}
n & x_n \\ \hline
0 & 1 \\
1 & 2 \\
2 & 3 \\
3 & 6 \\
4 & 5 \\
5 & 10 \\
6 & 15 \\
7 & 30 \\
8 & 7 \\
9 & 14 \\
10 & 21
\end{array}
\]

DPatrick 2014-03-28 21:31:49
Do you notice any patterns?

lucylai 2014-03-28 21:32:10
factorize them

dantx5 2014-03-28 21:32:10
prime factorization

lucylai 2014-03-28 21:32:10
factorize the x_n

ABCDE 2014-03-28 21:32:10
prime factorize

DPatrick 2014-03-28 21:32:17
Since we are multiplying and dividing by primes to get to the next term, we can try finding the prime factorization of each term.

DPatrick 2014-03-28 21:32:25
\[
\begin{array}{c|c}
n & x_n \\ \hline
0 & 1 \\
1 & 2\\
2 & 3 \\
3 & 2 \cdot 3 \\
4 & 5 \\
5 & 2 \cdot 5 \\
6 & 3 \cdot 5\\
7 & 2 \cdot 3 \cdot 5 \\
8 & 7 \\
9 & 2 \cdot 7 \\
10 & 3 \cdot 7
\end{array}
\]

DPatrick 2014-03-28 21:32:35
Now we really see some interesting things.

ABCDE 2014-03-28 21:32:56
binary

Allan Z 2014-03-28 21:32:56
binary by index of prime

bengals 2014-03-28 21:32:56
sort of like binary

DPatrick 2014-03-28 21:33:14
Yes! Redrawing my chart might make this a little clearer: \[
\begin{array}{c|c|c|c|r||c|c}
7 & 5 & 3 & 2 & \text{binary} & n & x_n \\ \hline
&&&& 0 & 0 & 1 \\
&&&\bullet& 1 & 1 & 2 \\
&&\bullet&& 10 & 2 & 3 \\
&&\bullet&\bullet& 11 & 3 & 6 \\
&\bullet&&& 100 & 4 & 5 \\
&\bullet&&\bullet& 101 & 5 & 10 \\
&\bullet&\bullet&& 110 & 6 & 15 \\
&\bullet&\bullet&\bullet& 111 & 7 & 30 \\
\bullet&&&& 1000 & 8 & 7 \\
\bullet&&&\bullet& 1001 & 9 & 14 \\
\bullet&&\bullet&& 1010 & 10 & 21 \\
\end{array}\]

DPatrick 2014-03-28 21:33:28
All I did was put dots in the columns corresponding to the binary representation of $n,$ and then multiplied the primes with dots in their column to get $x_n.$

DPatrick 2014-03-28 21:33:43
It looks like what's happening is that we're taking the binary representation of $n$, and then we're multiplying the primes that correspond to digits that are 1.

DPatrick 2014-03-28 21:33:55
For example, $6$ is $110$ in binary, so we multiply the 2nd the 3rd primes to get $x_6 = 3 \cdot 5.$

DPatrick 2014-03-28 21:34:15
This is just a conjecture, but the evidence is pretty strong, so on a contest like the AIME I'd be tempted to assume that it's true and use it to finish the problem. Let's do that now, and then we can go back and try to prove it.

ashgabat 2014-03-28 21:34:24
2090=2*5*11*19

math-rules 2014-03-28 21:34:24
2090=2*5*11*19

DPatrick 2014-03-28 21:34:37
We want $t$ such that $x_t = 2090 = 2 \cdot 5 \cdot 11 \cdot 19.$

DPatrick 2014-03-28 21:34:44
Which primes are these?

math-rules 2014-03-28 21:35:09
19,17,13,11,7,5,3,2...............10010101

ashgabat 2014-03-28 21:35:09
So we have 10010101

kevin38017 2014-03-28 21:35:17
2 is first 5 is 3rd 11 is 5th 19 is 8th

tluo5458 2014-03-28 21:35:23
first, third, fifth, eighth

Allan Z 2014-03-28 21:35:23
1,3,5,8

tluo5458 2014-03-28 21:35:23
10010101

DPatrick 2014-03-28 21:35:31
The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, ...
So 2090 is the product of the 1st, 3rd, 5th, and 8th primes.

DPatrick 2014-03-28 21:35:49
So assume that our conjecture is true, we want $t$ to be the number whose binary representation is 10010101 -- that is, it has 1's in the first, third, fifth, and eighth digits (counting from the right).

mathwizard888 2014-03-28 21:36:00
2090=2*5*11*19, which goes to 10010101 in binary, which is 149 ans

alex31415 2014-03-28 21:36:00
128+16+4+1=149 is our answer

DPatrick 2014-03-28 21:36:07
So $t = 128 + 16 + 4 + 1 = \boxed{149}$ is the answer.

DPatrick 2014-03-28 21:36:21
...we think. Is the conjecture actually true?

DPatrick 2014-03-28 21:36:33
(Note: I feel like this is somewhat flawed as an AIME problem, because you can observe the conjecture based on a few computations, and you don't need to prove it to answer the problem.)

Allan Z 2014-03-28 21:37:03
prove it by induction!

math-rules 2014-03-28 21:37:03
induction???

DPatrick 2014-03-28 21:37:05
We can try prove this using induction, since the sequence $\{x_n\}$ is defined recursively.

DPatrick 2014-03-28 21:37:15
The result holds for the base case $n = 0,$ so let's assume that it holds for some $x_n.$ That is, $x_n$ is the product of the primes that correspond to digits of 1 in the binary representation of $n.$

DPatrick 2014-03-28 21:37:34
Then what is $p(x_n),$ the smallest prime that does not divide $x_n$?

XxAndreixX 2014-03-28 21:37:59
The last 0 in the binary representation.

lucylai 2014-03-28 21:38:06
the rightmost 0

DPatrick 2014-03-28 21:38:39
Yes. The smallest prime that does not divide $x_n$ is the prime corresponding to the rightmost 0 in the binary representation of $n.$ Let's call this prime $P$. (Note that if $n$ has all 1's, then we might have to go to a leading zero: for example, if $n = 7,$ then the rightmost zero is the 0 in 0111, and $P=7,$ the fourth prime.)

DPatrick 2014-03-28 21:38:54
What is $X(x_n)$, the product of all the primes less than $p(x_n)$?

math-rules 2014-03-28 21:39:34
all the 1s to the right of the 0

mathwizard888 2014-03-28 21:39:42
the 1s at the end

DPatrick 2014-03-28 21:40:08
It's the product of all the primes before $P.$ Notice that these primes all correspond to all the 1's in the binary representation of $n$ that are to the right of the rightmost zero.

DPatrick 2014-03-28 21:40:32
So $n$ ends with something like 011....1. The 0 is corresponds to $p(x_n)$, and the 1's all correspond to $X(x_n).$

DPatrick 2014-03-28 21:40:38
And what happens when we go from $n$ to $n+1$?

math-rules 2014-03-28 21:41:07
so you are basically making the rightmost 0 a 1, and all of the 1s to the right of it 0, which is how you count up in binary

mathwizard888 2014-03-28 21:41:07
011....1 goes to 100....0

lucylai 2014-03-28 21:41:07
the rightmost 0 changes to a 1 and the 1's to the right of it change to 0

DPatrick 2014-03-28 21:41:25
Right. Our 011....1 becomes 100....0

DPatrick 2014-03-28 21:42:18
So all the primes that make up $X(x_n)$ get taken out when we go from $x_n$ to $x_{n+1}$ (since all those 1s changed to 0s), and the new prime that corresponds to $p(x_n)$ gets multiplied in when we go from $x_n$ to $x_{n+1}$ (since the rightmost 0 changed to a 1).

DPatrick 2014-03-28 21:42:54
But that's exactly what our recursive formula tells us to do! We divide by $X(x_n)$ (getting rid of those primes) and we multiply by $p(x_n)$ (adding in that new prime).

DPatrick 2014-03-28 21:43:08
Hence, the prime factorization of $x_{n + 1}$ corresponds to the binary representation of $n + 1.$

DPatrick 2014-03-28 21:43:19
So the result holds for $n + 1,$ and by induction, it holds for all positive integers $n.$

DPatrick 2014-03-28 21:43:40
That's all for the AIME II Math Jam! I hope you enjoyed it. Have a nice weekend!

2014 AIME II Math Jam

Facilitator: Dave Patrick