| Transcript
for the Math
Jam "2009 AMC 10/12 B Math Jam"
on Feb 26. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick19:00:55
Welcome to the 2009 AMC 10B/12B Math Jam! I'm Dave Patrick, your moderator for tonight's session.
DPatrick19:01:05
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.
DPatrick19:01:18
The classroom is moderated, meaning that students can type into the classroom, but only the moderator (that's me!) can choose which comments are displayed the classroom. This helps keep the class organized and on track. This also means that only well-written comments will be displayed into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick19:01:37
There are a lot of students here! As I said, only well-written comments will be passed to the entire group. Please do not take it personally if your comments 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!!
DPatrick19:02:11
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go (and certainly not to this large a group of students -- this Math Jam won't be as orderly or as thorough as our typical online classes.)! If you do not know a particular concept necessary for a problem, I probably will not be able to fully explain it to you tonight.
DPatrick19:02:32
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics.
DPatrick19:02:46
So please, when a problem is posted, do not simply respond with the answer. That's not why we're here. We're going to work through the problems step-by-step, and people who post comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, are going to be ignored.
DPatrick19:03:02
The Math Jam will proceed as follows:
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B. After that, time permitting, I will take requests for some other problems for discussion, and answer general questions.
DPatrick19:03:27
Let's start with Problem 21 from the AMC 10B:
DPatrick19:03:33
DPatrick19:04:10
As you'll see, I'll post the questions at the top of the screen. You can make that top-screen area larger or smaller by clicking-and-dragging on the horizontal gray bar above.
dafjbomb19:04:30
work in mod 8
karatemagic719:04:30
take it mod 8
ritwik_anand19:04:30
divisibility rule for 8!
Newanda19:04:30
3 to an even power is 1 mod 8, 3 to an odd power is 3 mod 8
dafjbomb19:04:30
3^0= 1 mod 8, 3^1= 3 mod 8, 3^2= 1 mod 8, 3^3= 3 mod 8... the pattern repeats every 2
DPatrick19:04:53
RIght -- since we care about divisibility by 8, we can work modulo 8, meaning we only consider the remainders when dividing by 8.
remy114019:05:04
try modular arithmetic, so 1mod8+3mod8+1mod8+3mod8+...+1mod8.
alligator11219:05:04
the remainder of the powers of 3 when divided by 8 go in this pattern: 1,3,1,3,1,3,1,3... where 3^0/8 is 1 mod 8, 3^1/8 is 3mod 8...
DPatrick19:05:18
3^0 = 1 (mod 8)
3^1 = 3 (mod 8)
3^2 = 9 = 1 (mod 8)
3^3 = 3 (mod 8)
etc.
DPatrick19:05:35
In other words, the terms alternate between 1 and 3 (mod 8).
DPatrick19:05:42
karatemagic719:06:01
You get 1005*(1+3).
Complex_Ninja19:06:01
( 1 + 3)1005 = 4020
erictao20050019:06:01
so there are 2010 numbers, so the sum is 4* 1005 = 4020
DPatrick19:06:26
Indeed, our sum is 1005 pairs of the form (1+3), so the sum is (1005*4) (mod 8).
DPatrick19:06:40
We don't actually have to multiply out 1005*4, though.
DPatrick19:07:07
There are an odd number of 4's. So the total is 4 (mod 8). (An even number of 4's would sum to 0 (mod 8).)
DPatrick19:07:23
Hence the answer is (D) 4.
DPatrick19:07:58
If the problem asks "what is the remainder when (blah) is divisible by n", it usually pays to work modulo n. That way you can eliminate a lot of unnecessary information.
DPatrick19:08:10
Next is #22 from the AMC 10B:
DPatrick19:08:19
DPatrick19:08:40
DPatrick19:09:14
(You might not fit both the problem and the picture up at the top, but if you resize it by dragging the horizontal gray bar, a scroll bar should appear)
DPatrick19:09:22
What is the key fact about this picture that we notice?
gh62519:09:52
A and B are similar!
Complex_Ninja19:09:52
similar triangles
tenniskidperson319:09:52
that B~A
DPatrick19:09:59
Correct: A and B are similar right triangles.
DPatrick19:10:22
(Note that the angles in the upper-right corner are complementary, and the similarity follows from that.)
darksigma19:10:31
So we can set up a proportion
DPatrick19:10:35
Exactly.
DPatrick19:10:44
We know the dimensions of A, and we know one side of B.
DPatrick19:11:10
What is the ratio of the side lengths in A to those in B?
gh62519:11:46
sqrt(5) : 2
Meta_Knight19:11:47
sqrt5/2
tenniskidperson319:11:47
DPatrick19:12:00
Right. The legs of A are 2 and 1, so the hypotenuse of A is sqrt(5).
DPatrick19:12:09
But we also know that the hypotenuse of B is 2.
DPatrick19:12:19
So the ratio of side lengths in A to those in B is sqrt(5)/2.
DPatrick19:12:31
What does that tell us about their areas?
alligator11219:12:58
the ratio of the squares is 5:4
echen19:12:58
in the ratio 5:4
Complex_Ninja19:12:58
are in 5 : 4
erictao20050019:12:58
they are 5:4
DPatrick19:13:10
Right. The ratio of the areas is the square of the ratio of the lengths.
DPatrick19:13:18
So (Area of A)/(Area of B) = 5/4.
DPatrick19:13:21
Thus, what is the area of B?
darksigma19:13:50
The area of B is 4/5
erictao20050019:13:50
4/5?
rubik19:13:50
4/5
Meta_Knight19:13:50
4/5
Jadesymdragon19:13:50
4/5
skaterpigsusa19:13:57
4/5 * area of A = 4/5 * (2*1/2)
DPatrick19:14:05
The area of A is 1, so the area of B is 1/(5/4) = 4/5.
DPatrick19:14:25
This is all the information we need to get the volume of cake and the area of the icing.
DPatrick19:14:30
First the volume. What is it?
alligator11219:15:07
4/5 * 2
skaterpigsusa19:15:10
4/5 * 2 = 8/5
DPatrick19:15:13
Right.
DPatrick19:15:34
Piece "B" has a cross-section area of 4/5 (that we just computed) and a height of 2. So its volume is 2*(4/5) = 8/5.
DPatrick19:15:52
So c = 8/5.
DPatrick19:16:15
For s (the area of the icing), we already know that there's 4/5 area of icing on top. What other icing is there, and how much?
darksigma19:16:46
The area of the icing, s, is 4/5+2^2=24/5 because we have the top face of B and the side of the cube
gh62519:16:46
2*2=4 on the side
Yongyi78119:16:46
There is a side icing with an area of 4
zserf19:16:46
s=4/5+4=24/5
webmath19:16:46
2*2 for the side
skaterpigsusa19:16:46
2*2 = 4 on the side
DPatrick19:17:03
Right: there's another 2*2 = 4 square inches of icing on the side of the cake.
DPatrick19:17:20
So the total icing on the "B" piece is 4/5 + 4 = 24/5.
DPatrick19:17:28
Thus s = 24/5.
darksigma19:17:43
There fore, c+s=32/5
alligator11219:17:44
24/5 + 8/5 = 32/5, which is B
rubik19:17:44
and c+s = 32/5
DPatrick19:17:58
To finish, we just compute c+s = 8/5 + 24/5 = 32/5. Answer (B).
DPatrick19:18:36
The key was observing the similar triangles. Often, when a construction produces two right triangles, it pays to check if they are similar.
DPatrick19:19:44
Quick announcement before we continue: if you're here for a regular AoPS class (Algebra), you're in the wrong room (this is the Math Jam). Please log out and wait 2 minutes before returning.
DPatrick19:19:54
For the rest of us, let's go on to AMC 10B #23:
DPatrick19:20:00
DPatrick19:20:25
This was also AMC 12B #18.
Ghostmaster19:20:37
Diagram might help maybe?
DPatrick19:20:57
A diagram will definitely help (it helped me), but we need to do a little figuring-out first before we can draw the right diagram.
Complex_Ninja19:21:15
find where they are at 10 min and 11 min
mitofaith19:21:17
covert the 10 minutes into seconds
Meta Knight19:21:25
Figuring out the positions of Rachel and Robert at the beginning of the 10th minute would help.
DPatrick19:21:40
I agree. I started with 10 minutes = 600 seconds and 11 minutes = 660 seconds.
DPatrick19:21:56
So we can figure out where on the track the two people start and stop between 600 and 660 seconds.
DPatrick19:22:02
Let's start with Rachel.
Just4Math19:22:41
why not conver the seconds to fractional minutes?
alligator11219:22:42
at 630 seconds, Rachel is on the start line
webmath19:22:42
Rachel is at the starting line after 630 seconds.
DPatrick19:23:01
We could definitely do this using fractional minutes...we'll probably make that conversion before the end of the problem.
DPatrick19:23:15
But I found it easiest to compute 600 = 6*90 + 60 and 660 = 7*90 + 30.
DPatrick19:24:12
(I should mention, in general, that there are usually more than one good method for these problems. We don't have time to discuss all the different varieties, and I'm presenting the method that is most natural to me, but that doesn't mean that there's anything wrong with your method!)
yankeesrule00719:25:02
at 660 seconds, she'll be 1/3 of the way across the track
ritwik_anand19:25:02
so she is 1/3 finished at 660 seconds
Wraithbasher19:25:02
2/3 of the way at 10 min ; 1/3 of the way at 11 min
DPatrick19:25:05
So at 10 minutes, Rachel starts 60/90 = 2/3 of the way around the track, and at 11 minutes she's at 30/90 = 1/3 of the way around.
DPatrick19:25:28
Let's do the same computation for Robert, then I'll put them on a diagram.
Complex_Ninja19:26:16
600 = 7*80 + 40 for Robert and he is 40/80 1/2 way around
Meta Knight19:26:16
Robert is halfway around the track at the beginning of the 10th minute and ends 1/4 of the way around the track.
zserf19:26:16
at 600 seconds he's 1/2 way around the track
yankeesrule00719:26:16
at 10:00 he'll be halfway done with the track, at 11:00 he'll be a quarter way done with the track
DPatrick19:26:26
Right, for Robert the relevant calculations are 600 = 7*80 + 40 and 660 = 8*80 + 20.
DPatrick19:26:45
So at 10 min, he's at 40/80 = 1/2 way around the track, and at 11 min he's at 20/80 = 1/4 of the way around the track.
DPatrick19:26:53
At this point, a picture might clarify things:
DPatrick19:27:00
DPatrick19:27:33
The photographer's range is shaded: he has the quarter of the track centered at the start line.
DPatrick19:27:49
The red line is Rachel's path from 10-11 minutes, and the blue is Robert's path.
Meta Knight19:28:10
So now we need to figure out the fraction of the 10th minute that they're both in that quarter track.
zserf19:28:10
So we need to figure out when they are in the photographer's range
DPatrick19:28:30
Right, we need to now determine what portion of the minute they are in the shaded area.
DPatrick19:29:00
What fraction of the minute is Rachel (the red path) in the shaded area?
Meta Knight19:29:45
3/8
Wraithbasher19:29:45
3/8
Meta_Knight19:29:45
3/8
ThinkFlow19:29:45
3/8
rubik19:29:45
(1/4)/(2/3) = 3/8
DPatrick19:30:03
Right. Rachel runs 2/3 of the track, and the photograph captures 1/4 of the track.
DPatrick19:30:13
So that's (1/4)/(2/3) = 3/8 of the minute.
DPatrick19:30:39
In terms of fractions of a minute, what interval (between 10 and 11) does this cover?
Meta Knight19:31:05
Since her path is centered around the photographer, we now that the times she enters and leaves his viewing area are (1/2 +/- 3/16)
DPatrick19:31:18
Exactly. Rachel is on camera in the middle 3/8 of the minute.
DPatrick19:31:35
So it's the time period between 5/16 past the minute and 11/16 past the minute.
DPatrick19:31:44
DPatrick19:31:51
How about Robert?
DPatrick19:31:57
What fraction of the minute is he on camera?
Meta_Knight19:32:13
robert is in for 1/3 of the time
Aeroalon19:32:13
(1/4)/(3/4) = 1/3
DPatrick19:32:30
Right, he runs 3/4 of a lap, and for 1/4 of a lap he's on camera. So he's on camera for (1/4)/(3/4) = 1/3 of a minute.
DPatrick19:32:48
What part of the minute exactly is he on camera?
ThinkFlow19:33:17
1/2 to 5/6
alligator11219:33:17
10 and 1/2 minutes - 10 and 5/6 minutes
DPatrick19:33:40
Right. We see that he starts on camera exactly half-way through the minute (this is immediate from the diagram).
DPatrick19:33:57
So he's on camera between 1/2 past and 5/6 past the 10-minute mark.
ThinkFlow19:34:03
So now we find the area in common.
DPatrick19:34:06
Right.
DPatrick19:34:15
Rachel is on camera from 5/16 to 11/16.
DPatrick19:34:24
Robert is on camera from 1/2 to 5/6.
alligator11219:34:35
it goes from 1/2 of a minute up to 11/16 of a minute, 11/16 - 1/2 = 3/16
ThinkFlow19:34:36
We are looking at the interval 1/2 to 11/16 for both.
DPatrick19:34:50
So they are both on camera between 1/2 and 11/16.
Meta_Knight19:35:08
11/16-1/2=3/16
Meta_Knight19:35:08
11/16-1/2=3/16
Just4Math19:35:08
the time that they are both in the same area at 8/16 to 11/16 so the probability is 3/16
ritwik_anand19:35:08
3/16
Meta_Knight19:35:08
the answer is C
DPatrick19:35:27
This is 11/16 - 1/2 = 3/16 of the minute, so they are both on camera for 3/16 of the time, and the answer is (C).
DPatrick19:35:51
I thought this was the hardest problem on the AMC 10B (24 and 25 are easier in my opinion).
DPatrick19:36:19
Although the math isn't that hard (it's just arithmetic!), keeping track of all the data and organizing it in a way that makes you comfortable is what makes this problem tricky.
DPatrick19:36:45
Let's continue on with AMC 10B #24:
DPatrick19:36:57
DPatrick19:37:05
DPatrick19:37:29
This problem is actually a lot easier than it looks.
MrBob19:37:48
it's half of an 18 gon
rubik19:37:48
the inside is basically half of a 18-gon
dafjbomb19:37:48
the whole figure would have 9*2= 18 sides
DPatrick19:38:13
One approach is to extend the whole figure around (basically take a mirror-image across the horizontal line), so that the interior region becomes a regular 18-sided polygon.
DPatrick19:38:38
That's easy, but I did something slightly different.
futbill202519:38:45
the inside is a decagon
DPatrick19:39:11
Yes, I just used the fact that the picture (as given) has a 10-sided polygon in it.
DPatrick19:39:24
And we know all its angles in terms of x.
DPatrick19:40:23
What are the 8 angles formed by the intersection of two trapezoids?
Yongyi78119:40:43
360 - 2x
erictao20050019:40:43
360-2x
darksigma19:40:43
360-2x
DPatrick19:41:06
Right, they are just 360-2x (since they are 360 minus two of the larger trapezoid angles).
DPatrick19:41:18
And the two angles at the base of arch?
darksigma19:41:32
180-x
MathPdx19:41:32
180 - x
ThinkFlow19:41:32
The two smaller angles are 180-x
DPatrick19:41:36
Right, those are just 180-x.
Meta Knight19:41:48
So we can set up an equation adding to the sum of the interior angles of a decagon and solve for x.
DPatrick19:41:54
Right. The total is 2(180-x) + 8(360-2x) .
DPatrick19:42:18
But we also know that the total angles of a decagon sum up to 8*180.
DPatrick19:42:28
(The general formula for an n-gon is 180*(n-2).)
DPatrick19:42:51
So we solve 2(180-x)+8(360-2x) = 8*180.
ThinkFlow19:43:10
Setting 2(180-x) + 8(360-2x) = 8*180, we find x = 100, so our answer is C.
Yongyi78119:43:10
So 18(180 - x) = 8*180, so 180 - x = 80, and x = 100
DPatrick19:43:17
We get x=100, so the answer is (A).
ThinkFlow19:43:43
oops...
alligator11219:43:43
it is A, not C
DPatrick19:43:54
There are a couple of equally simple ways to solve this problem.
alligator11219:44:00
one way is to extend all of the sides of the trapezoids, and they will all meet at one place
DPatrick19:44:04
That works nicely too.
DPatrick19:44:26
That is the AMC's "official" solution I believe.
DPatrick19:44:40
Let's now do AMC 10B #25:
DPatrick19:44:46
DPatrick19:44:57
This is also AMC 12B #17.
ritwik_anand19:45:48
find all possible combinations first
alligator11219:45:48
there are two stripes for each face which is 2^6 = 64
DPatrick19:46:05
Right, we can first easily could the number of different ways to stripe each side of the cube.
DPatrick19:46:27
There are 2 possibilities for each side, and 6 sides, so there are 2^6 = 64 different (equally likely) ways to paint the stripes.
DPatrick19:46:52
Now we have to count how many of these produce a continuous stripe encircling the cube.
yankeesrule00719:47:02
There are 3 different continous stripes that can go around a cube.
whylime19:47:02
there are three dimensions in which to encircle the cube
DPatrick19:47:30
Right, we first notice that there are 3 different possible continuous stripes.
ThinkFlow19:48:05
A ring must go around the entire cube, leaving 2 uncircled faces.
yankeesrule00719:48:05
For each of those stripes, there are two extra faces which can have any stripe orientation.
alligator11219:48:05
but for each continuous stripe, there are 4 because the 2 faces to the side do not matter which gives us 2*2 = 4
illusiat19:48:05
for each of them there are 4 ways to put the other stripes
DPatrick19:48:39
Right. Once we have the big stripe going all the way around, taking up 4 faces, we still have 2 faces left, which can have their stripes in either direction.
DPatrick19:48:49
So there are 2*2 = 4 ways to stripe the remaining two faces.
yankeesrule00719:48:57
So there are 3x4=12 possible combinations with a continous stripe.
ThinkFlow19:48:58
We choose one configuration, then choose how to stripe the other two faces, for 3 * 2 * 2 = 12 choices.
DPatrick19:49:35
Exactly: we have 3 choices for the big stripe all the way around the cube, then 4 ways to stripe the remaining two faces, for 3*4 = 12 successful stripings with a continuous stripe around the cube.
yankeesrule00719:49:53
Thus the probability is 12/64 or 3/16, B.
everestshi19:49:53
so(3*4)/64=3/16
Meta_Knight19:49:53
12/64=3/16 (B)
whylime19:49:53
so the answer is 12/64 = 3/16, which is (B)
darksigma19:49:53
Therefore the answer is 12/64=3/16, B
Yongyi78119:49:53
And finally, 12/64 = 3/16 (B).
DPatrick19:50:14
We have 12 successful stripings out of 64 total stripings, for a probability of 12/64 = 3/16. Answer (B).
DPatrick19:50:53
Like I said, I think that #24 and #25 were more straightforward than #23.
DPatrick19:51:14
Let's now move on to the AMC 12B!
DPatrick19:51:23
We'll start with AMC 12B #21.
DPatrick19:51:30
ThinkFlow19:52:20
Recursion.
Yongyi78119:52:21
Recursion!
DPatrick19:52:30
There is in fact a recursion here, but it may not be obvious right away.
DPatrick19:52:57
So let's approach it more naively.
DPatrick19:53:27
When we're trying to count something, it's usually best to look at the most restricive element of our problem first.
webmath19:53:33
Each woman can switch places with the woman to her left or right, or stay at their original site.
DPatrick19:53:40
True, but for two of women, they are more restricted.
ThinkFlow19:53:53
Far left or far right.
Just4Math19:53:57
the two on the end can only go to 2 chairs
DPatrick19:54:14
Right. The two women at the ends of the rows have fewer possibilities, so they are more restricted.
DPatrick19:54:24
So we can try to focus on one of them first.
DPatrick19:54:37
Let's focus on the woman at the left end. What are her options?
Just4Math19:54:54
stay, or right
ThinkFlow19:54:54
Switch or not.
schnook19:54:54
she could stay or go to the right
skaterpigsusa19:54:54
same seat or switch
DPatrick19:55:06
Right: she can stay put, or she can move one seat to the right.
DPatrick19:55:22
...but if she moves to the right, what else must happen?
MathPdx19:55:43
someone must move back to her seat
stupidityismygam19:55:43
if she moves the person to her right must switch
ThinkFlow19:55:43
The woman to her left switches as well.
Just4Math19:55:43
the 2nd woman has to sit on her chair
DPatrick19:56:09
Right: if the left-most woman moves her seat, then the woman sitting next to her has to take the empty seat on the end.
DPatrick19:56:29
So the only possibilities are (i) the woman at the left end stays in her seat, or (ii) the two women at the left end switch seats.
ThinkFlow19:56:50
And each one reduces to a smaller version of the original problem.
Newanda19:57:00
so this reduces the problem to the same conditions but with 8 or 9 women
DPatrick19:57:07
Exactly: that's the key observation (and that's what I mean by recursion).
DPatrick19:57:27
If the left-end woman stays put, then we have the same problem, but with only 9 women.
DPatrick19:57:37
If the two left-end women switch, then we have the same problem, but with only 8 women.
Mewto5555519:57:48
so f(n)=f(n-1)+f(n-2) where f(n) is the number of ways with n seats
DPatrick19:58:31
DPatrick19:58:44
This is what is known as a recurrence relation.
stupidityismygam19:58:55
f(1)=1 f(2)=2, go upwards from there
webmath19:58:58
where f(1) = 1 and f(2) = 2
DPatrick19:59:13
Right: to finish, we solve the problem for n=1 and n=2, and use our formula to crank it up to n=10.
DPatrick19:59:38
Clearly f(1) = 1 (one woman by herself can do nothing but stay put) and f(2) = 2 (two women can either stay put or switch seats).
DPatrick19:59:47
We know crank the handle on the equation.
DPatrick19:59:52
f(3) = f(1) + f(2)= 1+2 = 3
DPatrick19:59:58
f(4) = f(3) + f(2) = 3+2 = 5
DPatrick20:00:02
f(5) = f(4) + f(3) = 5+3 = 8
webmath20:00:09
So 89?
ThinkFlow20:00:09
After calculations, we find 89, or A.
ThinkFlow20:00:09
Calculating, we find that 10 women can sit in 89 ways. A
Smartguy20:00:17
f(10)=89 A!
DPatrick20:00:30
We get 1,2,3,5,8,13,21,34,55,89. So f(10) = 89 and the answer is (A).
DPatrick20:00:36
Of course, many of you recognize these numbers!
ThinkFlow20:00:54
This is the FIbbonacci relationship
darksigma20:00:54
I see fibonacci!
Mewto5555520:00:54
Or recognize that this is fibonacci and 89 is a fibonacci number.
UberYoda20:00:54
Fibonacci!
skaterpigsusa20:00:54
fibonacci sequence
carberry20:00:54
fibonnoci
pieater20:00:54
Fibonnacci Sequence
DPatrick20:01:07
This is the famous Fibonacci Sequence.
DPatrick20:01:28
By the way, this problem is also problem 9.12 in my Intermediate Counting & Probability textbook. :)
DPatrick20:01:39
It is one of the classic Fibonacci numbers problems.
rubik20:01:46
isn't there a formula for the sum of n terms of a fibonacci sequence...?
DPatrick20:02:02
There is, but it's ugly (it involves the square root of 5) and is not too useful for this sort of problem.
DPatrick20:02:13
Let's move on to #22:
DPatrick20:02:23
DPatrick20:02:39
This is an interesting problem because it's a good mix of geometry, number theory, and combinatorics.
Newanda20:03:19
coordinate bash!
stupidityismygam20:03:19
find the area of the parallelogram
DPatrick20:03:40
Yes, we'll probably want to start by writing coordinates for B and D, and then finding the area of the parallelogram in terms of those coordinates.
ritwik_anand20:03:49
we already have the area
awesomemathdude820:03:49
the area is 1000000
DPatrick20:04:08
Right: we can set out coordinate-based formula for the area equal to 1000000, and get an equation that we can solve.
DPatrick20:04:23
What are good choices for the coordinates of B and D?
math15420:04:48
B(b,b) and D(d,kd)
DPatrick20:05:00
That's what I did. :)We can say B = (b,b) and D = (d,kd) for some positive integers b and d.
DPatrick20:05:27
So if those two points and (0,0) are 3 vertices of a parallelogram, how do we find the area?
rubik20:05:52
height times base
Ispxye20:05:52
distance formula??
DPatrick20:06:07
Unfortunately, we don't have the height or the base, and to compute them would be ugly.
tinytim20:06:29
shoelace?
Paboga20:06:35
Take the determinant
ThinkFlow20:06:36
Shoelace.
DPatrick20:06:37
Aha.
DPatrick20:06:45
There is a clever gadget called the Shoelace Theorem.
DPatrick20:06:55
DPatrick20:07:03
We take the determinant of that 2x2 matrix.
rubik20:07:25
bkd-bd = bd(k-1)
DPatrick20:07:38
Right, we get that the area is bkd - bd = bd(k-1).
DPatrick20:07:56
But perhaps you didn't know the Shoelace Theorem. All is not lost!
DPatrick20:08:03
DPatrick20:08:55
Chopping up the area into nice-and-easy pieces, we see that the area of ABD (which is half the parallelogram) is:
(area of AED) + (area of BDEF) - (area of ABF)
DPatrick20:09:49
DPatrick20:10:17
(the middle piece is a trapezoid: b-d is the height, and 1/2(b+kd) is the average of the base lengths.)
DPatrick20:10:36
DPatrick20:10:57
Double it (since we only computed the area of ABD), and we get the area of ABCD is bd(k-1), just as we got with the Shoelace Theorem.
tinytim20:11:18
which means we have to find the number of solutions to bd(k-1)=1000000 where b and d are integers and k>1
awesomemathdude820:11:18
so we set that equal to 1000000?
DPatrick20:11:44
Indeed, now we have to count the number of solutions to 1000000=bd(k-1), where b,d are positive integers, and k>1.
DPatrick20:12:02
We're done with the geometry, now on to counting and number theory. :)
tinytim20:12:16
factoring 1000000 makes it easier
Meta Knight20:12:19
Note that 1,000,000 = 2^6 * 5^6
DPatrick20:12:39
Yes, so we must count solutions to 2^6 * 5^6 = bd(k-1).
rubik20:12:51
so we want three factors?
DPatrick20:13:15
Exactly: all we need to do is count the number of ways to factor 1000000 into three (ordered) positive factors.
DPatrick20:13:27
(noting b, d, and k-1 are all positive integers)
Yongyi78120:13:39
And we have balls and urns
james4l20:13:41
balls and urns to divide up the "2" and "5" factors
tinytim20:13:41
balls and urns
Smartguy20:13:41
use balls and urns
DPatrick20:13:50
Aha: many of you know the magic words: "balls and urns" :)
DPatrick20:14:07
Allow me to explain...
DPatrick20:14:23
Each of the three factors is going to be of the form 2^p * 5^q.
DPatrick20:15:01
So what we need to do is "distribute" the 6 factors of 2 to the three factors b, d, and k-1.
DPatrick20:15:17
The mental picture is that we have 6 balls labeled "2", and 3 urns labeled "b", "d", and "k-1".
DPatrick20:15:34
We put the balls in the urns, and that tells us how the 2's get distributed.
Meta Knight20:15:38
And another 6 balls labeled "5".
DPatrick20:15:54
Right, we'll separately do the same computation for six balls labeled "5".
DPatrick20:16:11
So we have to distribute 6 balls into 3 urns. How do we do that?
DPatrick20:16:23
Specifically, how do we count how many ways in which we can do that?
funtwo20:16:44
would that be 8 objects choose 2 dividers? or is it different
rubik20:16:44
8C2 ways (dividers)
ThinkFlow20:16:44
Put 2 dividers among the 6 balls, for 8C2 ways.
Twin Prime Conjecture20:16:44
8 choose 2
Yongyi78120:16:44
Let there be 2 dividers among the 6 balls
DPatrick20:16:57
Right. This is another common combinatorics gadget.
DPatrick20:17:07
Imagine the 6 balls sitting in a line.
DPatrick20:17:18
We need to insert two "dividers" to divide the balls into 3 groups.
DPatrick20:17:30
For example, in our context:
* * | * | * * *
means put a factor of 2^2 in the first number, a factor of 2^1 in the second number, and a factor of 2^3 in the third number.
DPatrick20:17:59
The six *'s are the "balls" representing factors of 2, and the vertical lines are the "dividers" splitting them into three groups.
DPatrick20:18:14
As another example, * | | * * * * *
means distribute as 2^1, 2^0, 2^5
DPatrick20:18:47
So to count this: we need count the number of ways to arrange 8 items in a line: 6 *'s and 2 |'s.
Meta_Knight20:19:03
8C2=28
tinytim20:19:04
which is simply 8C2
DPatrick20:19:09
This is 8!/6!2!, or more commonly C(8,2), which equals 28.
DPatrick20:19:22
So there are 28 ways to distribute the factors of 2.
tinytim20:19:38
Since it's the same for the factors of 5, we square 8C2=28 to get the answer
Meta_Knight20:19:38
and another 28 for the 5s
funtwo20:19:38
square that for the powers of 5 to get 784, which is c
DPatrick20:19:41
Right.
DPatrick20:20:05
We also distribute the factors of 5 the same way, via an identical process, getting 28 ways to distribute the factors of 5.
DPatrick20:20:26
So that's a total of 28^2 = 784 ways to factor 1000000 into 3 (ordered) pieces, and hence 784 (C) is our answer!
DPatrick20:21:18
I liked this problem: it's a fun mix of geometry, number theory, and combinatorics. :)
DPatrick20:21:39
Let's move on to #23 on the AMC 12B:
DPatrick20:21:47
DPatrick20:22:14
First of all, what is the region S?
ThinkFlow20:22:30
Square
math15420:22:30
A square.
funtwo20:22:30
a square?
gf484820:22:30
2*2 square
DPatrick20:22:35
It's a square of side length 2, centered at the origin:
DPatrick20:22:40
DPatrick20:23:03
OK, that was the easy step. :)
DPatrick20:23:16
When multiplying complex numbers, what's a better way to write them?
joeislittle20:23:46
polar form
james4l20:23:48
polar form
worthawholebean20:23:48
Polar form.
gf484820:23:48
cis?
Yongyi78120:23:48
Polar form
rubik20:23:48
trig form?
DPatrick20:24:14
Right. We can rewrite it using exponential (or polar) coordinates. The advantage of this is that we get a nice geometric interpretation of multiplication.
DPatrick20:24:21
DPatrick20:24:31
(I've labeled this number c so we can refer to it later.)
darksigma20:25:05
3/4+3/4i=3/2sqrt(2)*cis(pi/4)
joeschmidt20:25:05
3sqrt2/4cis45
joeislittle20:25:05
(3sqrt2/4,45)
worthawholebean20:25:05
DPatrick20:25:15
Right. These are all different ways of writing the same thing.
DPatrick20:25:38
DPatrick20:25:59
(Don't let the pi/4 throw you. pi/4 radians is 45 degrees, so replace pi/4 by 45 degrees if that makes you more comfortable.)
DPatrick20:26:36
DPatrick20:26:58
Therefore, geometrically, what is the result of multiplying by c?
joeislittle20:27:56
a dilation and rotation of the square
Yongyi78120:27:56
Rotating by pi/4 and scaling by 3/(2sqrt(2))
Twin Prime Conjecture20:27:56
enlarge it by 3/(2sqrt2)
ThinkFlow20:27:56
Magnitude increases by 3\sqrt{2}/4 and angle increases by 45?
DPatrick20:28:00
Right!
DPatrick20:28:32
The 3/(2*sqrt(2)) part means we increase the magnitude of z by 3/(2*sqrt2): that is, we move it away from the origin by a factor of 3/(2*sqrt(2)).
DPatrick20:28:43
The cis(pi/4) part means we rotate it (clockwise) by pi/4, or 45 degrees.
ThinkFlow20:29:00
This shows us how to draw our successful region.
DPatrick20:29:42
Exactly. If cz must end up in S, then we reverse this operation: that is, we take our region, rotate it 45 degrees, and multiply it by a factor of 2sqrt(2)/3, to see where z has to start to end up in S:
DPatrick20:29:48
DPatrick20:30:03
In other words, z has to start in the brown area in order to end up in S after we multiply by c.
DPatrick20:30:25
(If all this sort of stuff is unfamiliar with you, you should learn it when you take a trigonometry and/or precalculus class.)
DPatrick20:30:39
What are the vertices of the brown square region?
tinytim20:31:26
(4/3,4/3i)
Yongyi78120:31:26
(+-4/3, +-4/3)
ThinkFlow20:31:26
4/3,0 for everything, etc.?
DPatrick20:31:44
We can just track what happens to one of the corners of the original gray region.
DPatrick20:32:07
We take 1+i, rotate it 45 degrees to sqrt(2)i, then multiply by 2sqrt(2)/3 to get 4/3.
DPatrick20:32:16
So the top vertex of the brown square is (4/3)i.
DPatrick20:32:25
ThinkFlow20:32:44
Find the area of the orange square that is also in the grey square.
DPatrick20:33:04
Right. The "success" region is the region in the overlap of the two squares. So we need the area of the brown that is also in the gray.
tinytim20:33:14
this can be done by subtracting the areas of the issoceles right traingles
DPatrick20:33:31
Right, the easiest thing to do is to compute the areas of the four little gray triangles that are outside of the brown region.
tinytim20:34:00
each of them has a side length of 2/3
DPatrick20:34:31
Right. (For example, you can easily check that the intersection points in the first quadrant are 1+2/3i and 1/3 + i).
DPatrick20:34:41
So the little gray triangles each have area 2/9.
DPatrick20:35:06
We can use this to compute the overlap area, or...
tinytim20:35:12
subtracting that from 1 gives 7/9
DPatrick20:35:50
Right. We can just look in the first quadrant (since the other three quadrants are the same). The total gray area is 1, and the excluded gray area (in the little triangle) is 2/9. So the allowed area is 7/9.
DPatrick20:36:02
Hence the overlap with the brown shaded area makes up 7/9 of the original square S.
DPatrick20:36:14
So the answer is 7/9, answer (D).
DPatrick20:36:44
This is a hard problem, and if you don't have experience using trigonometry to multiply complex numbers, it's a really hard problem!
joeislittle20:37:00
Was this too technical for your liking?
DPatrick20:37:20
No, it's still interesting enough, and there should always be one hard trig/complex number problem on the AMC 12.
DPatrick20:37:34
On to #24, another trig problem :)
DPatrick20:37:42
DPatrick20:38:15
We need to figure out what is really going on here.
worthawholebean20:38:34
Graph it?
DPatrick20:38:44
That's a plan, but we have to figure out what the heck these function are first.
math15420:38:51
For the given range, cos^-1(cos x)=x
funtwo20:38:51
and the rhs is always x.
DPatrick20:39:01
Aha, that helps!
DPatrick20:39:40
DPatrick20:39:52
That's easy to graph. :) I won't bother drawing it yet.
darksigma20:40:15
But the lhs isn't, depending on the value of x
DPatrick20:40:20
Right.
DPatrick20:40:32
tinytim20:40:44
then take the sin of both sides giving sin6x=sinx
gh62520:40:44
Take the sine of both sides to get sin6x = sinx
DPatrick20:41:09
This is one possible solution...but I find it a little messy. I prefer the visual solution that I'm about to present.
DPatrick20:41:26
I'd rather try to graph f(x).
DPatrick20:41:57
darksigma20:42:07
maybe find when arcsin(sin(6x)) does equal x
DPatrick20:42:47
What does f(x) equal for very small values of x?
ThinkFlow20:43:20
6x
funtwo20:43:20
6x
DPatrick20:43:36
Right. When 6x is less than pi/2, sin^-1 and sin are inverses of each other, so they cancel out.
DPatrick20:43:58
DPatrick20:44:11
DPatrick20:44:54
When I solved this problem, I did not write a formula for this step. Just think a broad sense what the function does.
math15420:44:58
It goes back down to 0 with slope -6
DPatrick20:45:02
Precisely.
DPatrick20:45:22
DPatrick20:45:36
DPatrick20:45:54
DPatrick20:46:41
As we go from pi/6 to pi/4, the value of the function continues to decrease from 0 to -pi/2. Then it starts going back up...
ThinkFlow20:46:45
Take our graph, see where it intersects the line g(x) = x.
DPatrick20:46:53
Right, now we have enough info to sketch a graph:
DPatrick20:47:06
The graph of f(x) is a "triangle wave" of line segments with slopes 6 and -6:
DPatrick20:47:11
DPatrick20:47:20
The graph of f is shown in blue, and the graph of y=x is shown in black.
DPatrick20:47:43
(Again, I should point out that to solve the problem, you don't need to prove that the segments are straight lines, except for the first segment from 0 to pi/12. You just need to know the "corners" and that it's strictly increasing or decreasing on the correct intervals.)
Just4Math20:48:06
why are we in a im vs real graph setting?
DPatrick20:48:16
oops, I forgot to change the axis labels from the last problem :)
ThinkFlow20:48:28
Our answer is clearly 4, which is choice B.
Yongyi78120:48:28
4 points of intersection
tinytim20:48:28
the lines intersect 4 times
darksigma20:48:28
so the number of intersections is 4, B
DPatrick20:48:34
We can clearly see 4 intersection points between the graph of f and the graph of y=x.
DPatrick20:48:40
So there are 4 solutions. Answer (B).
DPatrick20:49:28
I like this "visual" solution because we don't get bogged down in messy trig computations. We just need to understand the general behavior of the function f without getting too mired in the details.
DPatrick20:50:04
Finally, the monster, #25:
DPatrick20:50:12
DPatrick20:50:20
DPatrick20:50:46
Yikes. Where do we start?
Yongyi78120:51:10
Each vertex must be in a different quadrant
DPatrick20:51:37
It should be reasonably clear that each vertex has to be in one of the four quadrants. We could prove this rigorously but I don't really want to spend the time doing so.
tinytim20:52:00
First count the number of squares with sides parallel to the axes?
gf484820:52:00
horizontal squares?
tinytim20:52:00
split into 2 cases: (i) squares with sides parallel to the axes (ii) other squares
DPatrick20:52:10
We can start by counting the easy ones -- the squares with sides parallel to the axes.
DPatrick20:52:20
The smallest such square is 6x6. How many are there?
Meta Knight20:52:32
1
schnook20:52:32
1
karatemagic720:52:32
1
ThinkFlow20:52:32
1
pieater20:52:32
1
skaterpigsusa20:52:32
1
DPatrick20:52:39
There's only room for 1:
DPatrick20:52:43
DPatrick20:52:59
It's pretty clear from this picture that we can't fit it anywhere else.
DPatrick20:53:03
How many 7x7 are there?
gf484820:53:12
4
Yongyi78120:53:12
4
karatemagic720:53:12
4
Ketvoman20:53:12
4
rubik20:53:12
4
ArtofMath20:53:12
4
DPatrick20:53:24
There are 4. Here's one:
DPatrick20:53:30
DPatrick20:53:45
We can see that we can slide this square 1 unit right, 1 unit down, or both.
DPatrick20:54:09
In other words, each of the 4 points in the 2x2 square in the lower right of the upper left grid (shown in red below) can be the upper left corner of a 7x7 square:
DPatrick20:54:14
DPatrick20:54:43
We can't move the upper-left corner any further up or left, since then we'll go off the lower-right grid.
schnook20:54:51
There are 9 of the 8x8
pieater20:54:51
9 ways for an 8x8
DPatrick20:54:57
DPatrick20:55:04
By essentially the same reasoning, we can see that any of the red squares shown in the 3x3 grid of red dots can be the upper left vertex of a valid 8x8 square, so there are 3^2 = 9 squares that are 8x8.
DPatrick20:55:24
We now have a pattern... :)
skaterpigsusa20:55:29
1^2 + 2^2 + 3^2...?
pieater20:55:29
9 ways for an 8x8, 16 ways for a 9x9, 25 ways for a 10x10, 36 ways for the 11x11, and so on
DPatrick20:55:44
Sort of...
tinytim20:55:50
1^2+2^2+3^2+4^2+5^2+4^2+3^2+2^2+1^2
math15420:55:50
But after 25 it starts going back down.
DPatrick20:55:52
Right.
DPatrick20:56:17
The pattern continues up to 10x10: a 10x10 square can have its upper-left corner on any dot in the upper-left grid.
DPatrick20:56:33
But when we go up to 11x11, we must have the upper-left corner in the upper 4x4 grid of dots:
DPatrick20:56:38
DPatrick20:56:58
If we try to place an 11x11 on one of the black dots in the upper-left grid in the above diagram, it's too big to fit.
tinytim20:57:16
you have to go back down
ThinkFlow20:57:16
But there's only 1 14x14
DPatrick20:57:27
Right. By the same reasoning, there are 3^2 12x12 squares, 2^2 13x13 squares, and 1^2 14x14 square.
ArtofMath20:57:34
85 total
skaterpigsusa20:57:34
2* (1^2 + 2^2 + 3^2 + 4^2) + 5^2 ?
UberYoda20:57:38
so its 1+4+9+16+25+16+9+4+1
pieater20:57:38
85 squares parallel to axis
DPatrick20:57:48
So we add all these up, and we have 5^2 + 2(1^2+2^2+3^2+4^2) squares whose sides are parallel to the axes.
DPatrick20:57:54
This sums to 85.
DPatrick20:58:12
Certainly we're not sure, since 85 is smaller than all of the answer choices.
ArtofMath20:58:21
now for the messy work
tinytim20:58:21
now for the hard part
rubik20:58:29
there are diagonal squares
schnook20:58:29
We need to do the slanted ones
kthxbai20:58:29
tilted squares
DPatrick20:58:35
We have to deal with the squares with diagonal sides, such as the one below:
DPatrick20:58:40
DPatrick20:58:49
These are harder to deal with.
DPatrick20:59:07
We might think to use casework, but we'll need a very careful way to go through the cases.
DPatrick20:59:18
How can we organize our cases in a helpful way?
Aeroalon20:59:50
Side lengths?
karatemagic720:59:50
side length?
rubik20:59:50
side lengths
pieater20:59:50
by side length?
gf484820:59:50
by side length?
alligator11220:59:50
side lengths
rubik20:59:50
sort by side length
DPatrick21:00:05
Using side length seems messy, since we'll have all sorts of square root side lengths to deal with.
smokewisps21:00:14
different slopes
bluebear21:00:15
slope of sides?
darksigma21:00:15
slopes?
DPatrick21:00:30
Slopes is a better idea, since the slopes are rational.
DPatrick21:01:10
The way we thought of doing it (and you might have done something different) is to count cases using the "vector" formed by going from the upper left to the upper right vertex:
DPatrick21:01:15
DPatrick21:01:43
We can classify each square by the amount "over" (a) and "up" (b) we go when we go from the upper left to upper right vertex. For example, above, we have (a,b) = (9,3).
DPatrick21:02:04
We've already tackled all the (a,0) squares. We found the following:
DPatrick21:02:08
ArtofMath21:02:35
b=1
tinytim21:02:40
now to count b=1,2,3,4
DPatrick21:02:51
Right. We can note that we don't have to worry about b < 0, since we can just double the number of squares we find for b > 0. The squares for nonzero b come in pairs:
DPatrick21:02:57
DPatrick21:03:15
So we'll next look at b=1.
DPatrick21:03:22
What the smallest possible value of a?
gf484821:03:39
7
Aeroalon21:03:39
7?
DPatrick21:03:41
Right.
DPatrick21:03:50
Why didn't (a,b) = (6,1) work?
DPatrick21:03:59
Here's the picture again:
DPatrick21:04:05
rubik21:04:22
it won't touch the grid in the 3rd quadrant
math15421:04:22
The vertices won't all be in the quadrants
darksigma21:04:22
oh, the vertices would fall off the lattice points...
DPatrick21:04:30
Yeah, (a,b) = (6,1) is too small.
DPatrick21:04:53
More generally, is there a simple algebraic condition to tell when (a,b) will be too small?
kthxbai21:05:21
a-b>5
karatemagic721:05:22
a-b>=6
gf484821:05:22
a-b>=6
Mewto5555521:05:22
a-b>=6
DPatrick21:05:36
Right. We must have a-b >= 6. We can see this by adding more green lines to our picture:
DPatrick21:05:41
DPatrick21:05:59
The vertical distance between vertices A and C (upper left and lower right) is a-b. This must be at least 6.
DPatrick21:06:13
So we must have a-b >= 6. This lets us zero out a lot of entries in our table:
DPatrick21:06:18
DPatrick21:06:44
As some of you mentioned earlier, there's only one square for (a,b) = (7,1), and here it is:
DPatrick21:06:49
DPatrick21:07:06
You can see that this square can't "slide" in any direction.
DPatrick21:07:20
How about (a,b) = (8,1)?
DPatrick21:07:32
Here's an example:
DPatrick21:07:37
Yongyi78121:07:48
4?
gf484821:07:48
4
ThinkFlow21:07:48
4
karatemagic721:07:48
4
DPatrick21:07:50
Why 4?
skaterpigsusa21:08:19
slide to right by 1 and up by 1
Yongyi78121:08:19
*up, right, or both
Mewto5555521:08:19
you can shift 1 to the right, 1 up, both, or neither
DPatrick21:08:33
Right. We can see that this can slide 1 unit up, or 1 unit to the right, or both, and still be legal.
DPatrick21:08:38
In other words, its upper left corner can be one of the four green dots shown:
DPatrick21:08:43
joeislittle21:08:58
hmm, maybe they count themselves in the same fashion as the others
ArtofMath21:08:58
b=1 | 0 1 4 9 16 9 4 1 0
DPatrick21:09:06
Aha, we seem to see the same pattern!
DPatrick21:09:16
We can look at (a,b) = (9,1):
DPatrick21:09:24
DPatrick21:09:44
The upper-left corner can be in any of the green dots, so we get 3^2 of them.
DPatrick21:10:08
Yeah, we can see where this is headed...
schnook21:10:22
The pattern continues...
DPatrick21:10:24
The next one, (a,b)=(10,1) will give us 4^2 possibilities:
DPatrick21:10:28
DPatrick21:10:41
Then it will start to shrink back down...
rubik21:10:44
then 3^2 again?
DPatrick21:10:52
DPatrick21:11:02
This is (a,b) = (11,1). Here, we only have 3^2 possibilities. Notice how our grid of green dots is now shrinking up and to the left.
ThinkFlow21:11:08
2^2, 1^2
Meta_Knight21:11:08
then 2^2
DPatrick21:11:17
Continuing in this vein, (12,1) will give us 2^2 and (13,1) will give us 1^2.
DPatrick21:11:23
We can update our table:
DPatrick21:11:27
Aeroalon21:11:37
Will the same pattern hold for b=2,3,4?
DPatrick21:11:53
Indeed, we would be surprised if it doesn't!
ArtofMath21:12:00
b=2 | 0 0 1 4 9 4 1 0 0
joeislittle21:12:00
It isn't too much of a stretch to reason that it does...
ThinkFlow21:12:00
next row: 1^2, 2^3, 3^2, 2^2, 1^2
DPatrick21:12:20
I'd probably go with the pattern at this point (after all, on the AMC contest, the clock is ticking!).
DPatrick21:12:30
But let's just look at a couple of pictures for the b=2 case.
DPatrick21:12:36
We can see that there is only one square for (8,2):
DPatrick21:12:41
DPatrick21:12:54
We can see that there are 2^2 squares for (9,2): one is shown below, and the green dots are the allowed upper-left corners:
DPatrick21:13:01
DPatrick21:13:17
We'll basically play the same game for (8,2), (9,2), (10,2), etc., up through (12,2), and we'll find the pattern continues:
DPatrick21:13:23
DPatrick21:13:39
At this point, it's probably obvious how the pattern finishes:
DPatrick21:13:46
DPatrick21:13:59
So, by careful casework, we've listed all the squares! How do we finish?
Yongyi78121:14:13
But wait! For b=1,2,3,4, we have to multiply by 2!
rubik21:14:13
sum everything
Aeroalon21:14:13
Add everything.
worthawholebean21:14:13
Sum this monster.
DPatrick21:14:29
We add them up, but we remember that all the b>0 cases have to be multiplied by 2!:
DPatrick21:14:37
...because the diagonal squares come in pairs:
DPatrick21:14:42
DPatrick21:15:06
So, we add the top row of our chart, and double the sums of the rest of the rows, and we get:
DPatrick21:15:17
schnook21:15:33
the answer is 225, E
Complex_Ninja21:15:33
225
Just4Math21:15:33
85+ 70*2=225
DPatrick21:15:40
This comes out to 225, so that's our answer. (E)
DPatrick21:15:57
There are (at least) a couple of other ways you can come to this result.
Denny103821:16:02
is this the fastest way to solve this problem?
DPatrick21:16:13
I don't know if it's the fastest, but it's the most "straightforward" I think.
DPatrick21:16:40
The official AMC solution (that was distributed to teachers) contains what I would essentially call magic.
DPatrick21:16:55
The solution I just presented was the way Richard Rusczyk solved the problem.
DPatrick21:17:02
I solved it myself slightly differently...
DPatrick21:17:21
...but I also started by counting the "easy" squares, then looked at this picture:
DPatrick21:17:26
DPatrick21:17:36
And fill in the rest of the green triangles:
DPatrick21:17:40
DPatrick21:17:49
We can now view ABCD as "bounded" by the little green square and the bigger blue square.
DPatrick21:18:00
Every choice of two concentric squares (parallel to the axes) like this gives us exactly 2 tilted squares (one in each direction).
DPatrick21:18:09
And conversely every possible ABCD is bounded by a pair of squares.
DPatrick21:18:27
So we need to count the pairs of concentric squares, multiply by 2, and add to our previous count of the first 85 squares (with integer side lengths).
DPatrick21:18:42
On your own, you can see if you can develop this observation into a full solution.
darksigma21:19:31
how about the AMC official solution?
Yongyi78121:19:31
What was the official AMC solution?
Quickster9421:19:31
what is in the AMC solution?
DPatrick21:19:40
Ask your teacher for a copy, or perhaps someone will post it.
DPatrick21:19:58
I don't find it useful since I have no idea how anyone could have possibly thought of it in real-time while taking the test.
DPatrick21:20:28
We've already been here for 2 hours and 20 minutes, but I'll do 1 or 2 more problems by request from among 16-20 on either contest.
DPatrick21:20:58
Type in your request and I'll see which are most popular...
DPatrick21:21:08
(you have to say which contest too)
DPatrick21:21:46
I see a lot of requests for #19 on the 12, so I'll do that one. It's an interesting algebra problem.
DPatrick21:21:53
applepie21:22:18
factor
rubik21:22:19
factor?
DPatrick21:22:42
Since we want to find primes, it makes sense to try to factor this thing, and in particular find the value(s) of n for which it can only be factored trivially.
DPatrick21:22:59
The tricky part is figuring out how to factor this thing.
Paboga21:23:20
Difference of two squares
worthawholebean21:23:20
Try and find a difference of squares'
DPatrick21:23:31
Difference of squares is the easiest factorization if we can find it.
DPatrick21:23:39
We might also notice that n^4 and 400 are already perfect squares.
gf484821:23:55
completing the square
stupidityismygam21:23:55
complete the square
DPatrick21:24:16
This suggests a sort of "reverse" completing the square, where we try to make n^4 and 400 two of our terms.
DPatrick21:24:23
What binomial squares to give terms n^4 and 400?
Meta_Knight21:24:45
n^2+20
webmath21:24:45
N^2 + 20
rubik21:24:45
n^2+20
kthxbai21:24:45
(n^2+20)
DPatrick21:24:47
Right.
DPatrick21:25:04
DPatrick21:25:30
DPatrick21:25:33
Aha!
pieater21:25:52
(n^2+20)^2-400n^2
DPatrick21:25:57
even better..
bluebear21:26:00
differnce of squares
DPatrick21:26:23
Yongyi78121:26:53
(n^2+20n+20)(n^2-20n+20)
DPatrick21:27:09
DPatrick21:27:29
Now, let's not forget why we were doing this...when will this be prime?
Smartguy21:27:54
when one of the factors is 1
webmath21:27:54
one of the factors is one
worthawholebean21:27:54
it is necessary that one of them be +/- 1
Complex_Ninja21:27:54
when 1 of the factors = 1
applepie21:27:54
one of them is 1
gh62521:27:54
When one of the factors is 1
DPatrick21:28:30
Right. We only have a hope for this being prime if one of the factors is 1 or -1. (If both factors are neither 1 or -1 then the number is either composite or 0.)
stupidityismygam21:28:41
fortunately n^2+20n+20 is always >1 so that simplifies the process
DPatrick21:29:06
Indeed, the second factor n^2 + 20n + 20 is always positive (and always more than 20 for that matter) if n is a positive integer.
DPatrick21:29:16
Just4Math21:29:24
just because one of them is 1 does that mean that the other one will produce a prime number?
DPatrick21:29:36
Good point: you are right. We will have to verify that any number(s) we get are indeed prime.
rubik21:30:05
n = 19, 1
tinytim21:30:05
which gives n=1,19
pieater21:30:05
(n-19)(n-1)
james4l21:30:05
n=1 or 19
DPatrick21:30:25
We solve n^2 - 20n + 20 = 1 by writing it as n^2 - 20n + 19 = (n-19)(n-1) = 0.
DPatrick21:30:29
So n must be 1 or 19.
DPatrick21:30:46
n=1 gives f(n) = 41, which is prime.
DPatrick21:31:27
I don't want to plug n=19 into f(n), but remember that we've already factored it, and the first factor was 1, so we only need to plug it into the second factor.
DPatrick21:31:43
n=19 gives 19^2 + 20(19) + 20 = 39(19) + 20 = 40(20)-39.
funtwo21:31:55
actually we know they both yeild prime results otherwise the correct answer choice wouldn't be there
tinytim21:31:55
f(19)=461. Since none of the answers are 41, we cheat and assume 761 is prime.
DPatrick21:32:06
Right. 41 is not an answer choice, so this number must be prime too.
DPatrick21:32:16
The sum of them is 40(20) - 39 + 41 = 802.
DPatrick21:32:19
Answer (E).
DPatrick21:32:48
(There's no need to compute 40(20) - 39, since we need to add 41 to it anyway to get the final answer; easier to leave it in its current form.)
DPatrick21:33:13
Please type in a request for one more problem, I'll take whichever is most popular...
DPatrick21:34:03
I'll do two short ones that we both requested by a lot of people: #19 and #20 on the AMC 10B.
DPatrick21:34:12
Mewto5555521:35:00
it will only be right when there are no 1's
DPatrick21:35:07
Right.
DPatrick21:35:16
How many (out of the 12 hours) will have no 1's?
DPatrick21:35:38
(We only need to look at 12 hours, the other 12 hours will be exactly the same)
rubik21:35:55
8
smokewisps21:35:55
8
kelleyzhao21:35:55
8
pieater21:35:55
8
mathtyro21:35:55
1,10,11,12 have 1's so 8 hours do NOT have 1
crazypianist111621:35:55
8
joeislittle21:35:55
and 8
DPatrick21:36:07
Right. We need the hour to be from 2 through 9, so there are 8 (out of 12) hours with no 1.
DPatrick21:36:15
How many tens digits for the minutes have no 1?
Meta_Knight21:36:29
5
joeislittle21:36:29
5
pieater21:36:29
5
skaterpigsusa21:36:29
5
DPatrick21:36:40
Sure, any valid digit 0,2,3,4,5 other than 1. So 5 (out of 6) tens digits work.
DPatrick21:36:56
And of course 9 (of the 10) units digits work. (I won't insult your intelligence by asking)
DPatrick21:37:17
So we multiply these fraction (since the numbers are essentially independent)
DPatrick21:37:25
(8/12) * (5/6) * (9/10)
schnook21:37:30
The answer is 1/2, A
Meta_Knight21:37:30
(2/3)*(5/6)*(9/10)=1/2 (A)
DPatrick21:37:38
This simplifies to 1/2, so the answer is (A).
DPatrick21:37:55
I'll finish tonight with AMC 10B #20:
DPatrick21:38:00
DPatrick21:38:05
DPatrick21:38:32
I wasn't particular fond of this problem, since if you know the right trick, it can be solved in essentially one line.
Twin Prime Conjecture21:38:38
angle bisector theorem
crazypianist111621:38:38
angle bisector theorem!
Mewto5555521:38:38
angle bisector theorem
megamath21:38:38
angle bisector theorem
yankeesrule00721:38:38
angle bisector theorem?
DPatrick21:38:47
Yes, the Angle Bisector Theorem.
DPatrick21:39:17
...which states that the angle bisector of a triangle splits the opposite side in the same proportion as the two adjacent sides.
boringguy21:39:26
ab/ac = bd/dc
DPatrick21:39:35
Which, in this problem, means AB/AC = BD/BC.
DPatrick21:39:43
What's AC?
Yongyi78121:39:52
sqrt(5)
Meta_Knight21:39:52
sqrt5
crazypianist111621:39:52
sqrt5
DPatrick21:40:02
AB = 1, BC = 2, and the triangle is right, so AC = sqrt(5).
DPatrick21:40:12
So BD/DC = 1/sqrt(5).
Twin Prime Conjecture21:40:21
dc=2-bd
DPatrick21:40:25
Indeed, DC = 2 - BD.
DPatrick21:40:44
So we have BD/(2-BD) = 1/sqrt(5), and we just solve for BD from here.
DPatrick21:40:58
It works out to answer (B).
DPatrick21:41:27
Well, I'm tired of typing after 2 hours and 40 minutes, so I'll just finish up with some general AMC info and announcements.
DPatrick21:41:49
In case you hadn't heard, the AMC has announced that the AIME qualifying scores for the A-date AMC tests are:
97.5 or higher on the AMC 12A
120 or higher on the AMC 10A
DPatrick21:42:18
The AMC will announce the B-date qualifying scores next week or the week after. They will be at most 100 (for the 12B) and 120 (for the 10B), but they might go slightly lower (as the 12A score did).
DPatrick21:42:44
please hold your questions until the end
DPatrick21:43:08
Also, the USAMTS has announced that students scoring 45 or higher on Rounds 1 and 3 of the USAMTS have qualified for the AIME. Log on to your account on the USAMTS website and access your profile for instructions if you have been participating in the USAMTS.
DPatrick21:43:21
As you may know, the regular date for the AIME is Tuesday, March 17, and the alternate date is Wednesday, April 1. You can take the AIME only if you qualify, and only on one date, not both. You are supposed to take it on the first date unless you have something (illness, school holiday or closure, academic conflict, etc.) that makes it necessary to take it on the second date.
DPatrick21:43:45
There is a rumor that the second AIME is harder to discourage people from taking it. This rumor is totally false.
DPatrick21:43:54
Our AIME Math Jams are scheduled for 2 days after each contest: Thursday March 19 and Friday April 3. We will cover all 15 problems on each AIME.
DPatrick21:44:14
Also, we have a special weekend AIME seminar coming up! It's Saturday and Sunday, March 7 and 8, and will last from 3:30 - 6:30 Eastern (12:30 - 3:30 Pacific) each day. Each day consists of general approaches and important facts needed for problems within a given subject area, followed by a discussion of specific problems from past AIME competitions, or from other contests of a similar difficulty level. Day 1 is Counting & Algebra, Day 2 is Number Theory & Geometry. The cost is $65.
DPatrick21:44:36
Note: if you are just beginning at the AIME level and only expect to get 1 or 2 (out of 15), then it is probably too advanced for you. The target audience for the seminar is students scoring in the 3-9 range trying to get up into the 10+ range.
DPatrick21:44:54
Finally, I will not speculate on the USAMO qualifying standards, so please do not ask me about them.
DPatrick21:45:21
I will stick around until 10 PM (ET) / 7 PM (PT) to answer questions if anybody has any.
brightzhu21:45:31
how do you participate in the USAMTS?
DPatrick21:45:34
Go to www.usamts.org
rubik21:45:39
is there a chance that the cutoff for the 10 and 12 will be greater than 100 and 120 points respectively?
DPatrick21:45:40
No.
darksigma21:45:47
Did you think that the AMC12 and AMC 10 were harder or easier than usual?
DPatrick21:46:17
That depends on each individual's taste. By my taste, I though the 10s were slightly easier and the 12s were slightly harder. But others will have different opinions to be sure.
crazypianist111621:46:24
if you participated in usamts, how will they calculate your usamo index?
DPatrick21:46:31
I don't know...they haven't told me.
Complex_Ninja21:46:36
Umm when will our school in Canada get our results on who has qualified?
DPatrick21:46:42
You should ask the AMC this. I have no idea.
DPatrick21:47:13
In general, administrative questions such as this are best directed to the AMC.
Meta_Knight21:47:20
what's the max score for each round of USAMTS?
DPatrick21:47:22
25
megamath21:47:26
What is a good way to practice?
DPatrick21:47:37
For the AIME? Practice on old problems.
DPatrick21:48:05
My absolute best piece of advice: read each problem carefully, and try to cut down on "silly" mistakes. Check your work -- a necessary step to do this is to keep your work organized.
crazypianist111621:48:11
is it possible to get through aops vol. 2 in 2 weeks to prepare for aime?
DPatrick21:48:17
Unlikely, sorry.
DPatrick21:48:29
At this point it's hard to learn "new" math. You want to practice problem solving.
Meta_Knight21:48:33
if our school doesn't do AMC then how do we get the tests
DPatrick21:49:01
The AMC must be done through a school. However, "school" can mean "college": many colleges now offer the AMC as a public service to their communities.
DPatrick21:49:03
Check the AMC website.
DPatrick21:49:08
Of course, it is too late for 2009.
megamath21:49:12
Is AIME harder than AMC10 and AMC12?
DPatrick21:49:17
Yes.
mathtyro21:49:27
What are the general average scores for the AIME?
DPatrick21:49:38
Anywhere from 2 to 5 is usually the average (out of 15).
rubik21:49:57
so would AIME be like #s 20-25 of the AMC-12?
DPatrick21:50:09
I think any of 21-25 on this year's AMC 12 would have been easy-to-medium AIME problems.
DPatrick21:50:32
AMC 12B #25 would have made a nice mid-level AIME problem I think.
DPatrick21:51:24
Well, the questions are frankly starting to get a little silly, so I'm going to end it here.
DPatrick21:51:35
Thanks for coming and good luck on the AIME if you're taking it!
