Alternate AIME Math Jam

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rrusczyk (19:01:38)


rrusczyk (19:01:48)
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arjunlandes (19:01:44)
examine a diagram

rrusczyk (19:01:56)
We can start by sketching a picture:

rrusczyk (19:02:02)


rrusczyk (19:02:06)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/48df4d91619d3015ef902f528890a649.png

rrusczyk (19:02:42)
What do we know about a hexagon that will help us?

tamwynheir (19:02:54)
the two right angles?

arjunlandes (19:02:56)
right angles A and D, all sides equal

Zmastr (19:03:02)
All their angles add up to 720 degrees

chesspro (19:03:03)
sum of the angles

rrusczyk (19:03:18)
Their angles sum to 720. What does this tell us about the non-right angles?

tongchen1226 (19:03:25)
B, C, E, F are all 135

siphor (19:03:25)
135

dark_sunburst (19:03:30)
135 degrees

rrusczyk (19:03:38)
Therefore, if we let x be the measure of angle B, C, E, or F, we know that 4x + 180 = 720, hence x = 135.

rrusczyk (19:03:44)
Now what?

Phelpedo (19:02:16)
The hexagon can be divided into two iscoceles right triangles and a rectangle.

tongchen1226 (19:02:32)
Connect BF and CE

perfectnumber628 (19:03:01)
draw CE and BF and you get right triangles

chesspro (19:03:49)
BCEF is a rectangle

rrusczyk (19:04:24)
That's one route we can take. Did anyone find a solution such that the hexagon is part of a larger figure?

joezfang (19:04:21)
the hexagon is a square with two corners cut out.

rrusczyk (19:04:57)
It's relatively hard to work with the area of a hexagon, but areas of squares and right triangles are easy! We could cut the hexagon into pieces, or...

rrusczyk (19:05:03)
Enclose ABCDEF inside a square:

tongchen1226 (19:04:50)
extend DC,AB, DE, AF

rrusczyk (19:05:15)


rrusczyk (19:05:17)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/d0abb5836e711d6da87477e01ecb4e46.png

rrusczyk (19:05:22)
Let s be the side length of the hexagon.

rrusczyk (19:05:35)
What is the area of the big square?

chesspro (19:05:49)
s^2 = (AB + AB/sqrt(2))^2

perfectnumber628 (19:05:54)
(s+s/sqrt2)^2

arjunlandes (19:05:55)
(s+s/sqrt(2))^2

rrusczyk (19:06:10)
Then the two outside triangles are each 45-45-90, and the side length is s/sqrt(2).

rrusczyk (19:06:15)


rrusczyk (19:06:33)
So, what is the area of the hexagon in terms of s?

Mr.Ocax (19:07:09)
s^2+s^2*sqrt(2)

tongchen1226 (19:07:21)
(1+sqrt(2))s^2

chesspro (19:07:28)
s^2(1 + \sqrt(2))

rrusczyk (19:07:36)
The area of the hexagon is the difference of these, which is

rrusczyk (19:07:40)


rrusczyk (19:07:45)


rrusczyk (19:07:53)
And what do we get for our answer?

mkkool64 (19:07:59)
46

arjunlandes (19:08:00)
thus s^2=2116

joezfang (19:08:01)
46

rrusczyk (19:08:10)
So we set this equal to 2116(sqrt(2)+1), and we see that s = sqrt(2116) = 46

rrusczyk (19:08:18)
Next problem!

rrusczyk (19:08:26)


rrusczyk (19:08:35)
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rrusczyk (19:08:38)
What determines whether three numbers can be the sides of a triangle?

jen7 (19:08:34)
triangle inequality

arjunlandes (19:08:38)
triangle inequality!

chesspro (19:08:45)
triangle inequality

chess64 (19:08:47)
Triangle inequality

rrusczyk (19:08:55)
They must satisfy the Triangle Inequality: each side must be smaller than the sum of the other two.

rrusczyk (19:09:00)
Therefore, the following equations must be satisfied:

rrusczyk (19:09:03)


rrusczyk (19:09:06)
(Note that there is a third equation with log_10 12 on the left side, but this is automatically satisfied since log_10 12 < log_10 75.)

rrusczyk (19:09:11)
Now what?

chess64 (19:09:20)
log x + log y = log (xy)
log x - log y = log (x/y)

chesspro (19:09:25)
equivalently
12n > 75
12*75 > n

mkkool64 (19:09:26)
use log properties to combine logs

Phelpedo (19:09:28)
Addition property of logs

perfectnumber628 (19:09:32)
when you add logs, you multiply the arguments

Vihang (19:09:32)
thus 75<12*n and n<12*75

arjunlandes (19:09:33)
combine logs with loga+logb=logab

rrusczyk (19:09:42)
We can use the log product rule to get 75 < 12n and n < 12*75.

rrusczyk (19:09:49)
Hence 75/12 < n < 12*75 = 900.

rrusczyk (19:09:55)
So, what is our answer?

dark_sunburst (19:10:08)
893

tongchen1226 (19:10:14)
893

ramdude (19:10:19)
893

Phelpedo (19:10:26)
Maximum possible value is 899 and minimum is 7, for 893 possible values

chess64 (19:10:06)
6<n<900 -> 893

rrusczyk (19:10:36)
So n can be any integer between 7 and 899 (inclusive), and hence there are 899-7+1 = 893 possible values for n.

rrusczyk (19:10:51)
Next problem!

rrusczyk (19:10:56)


rrusczyk (19:11:18)
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rrusczyk (19:11:23)
How should we proceed?

jen7 (19:11:16)
P contains odd #'s from 1-199

rrusczyk (19:11:53)
Indeed; how will we count the powers of 3 in the product?

Phelpedo (19:11:47)
The 100th positive odd integer is 199. Find the number of multiples of 3 in 199! and subtract out the even ones.

jen7 (19:12:09)
find multiples of 3 and subtract even multiples of 3

perfectnumber628 (19:12:22)
find how many are mult of 3, how many are mult of 9, etc, and add

rrusczyk (19:13:05)
We can definitely do it this way. Can anyone give me a mathematical expression that equals P without using ... or product notation (use only factorials, exponents, and mult/division)

rrusczyk (19:13:52)
(Notice that the answer isn't just counting the multiples of 3 in 199 factorial; we can't include the even numbers!)

mkkool64 (19:13:51)
200!/(2^100 * 100!)

chess64 (19:14:00)


Phelpedo (19:14:15)
199!/(2^99*99!(

rrusczyk (19:14:23)
We can count the powers of 3 in P directly, but we can also write

rrusczyk (19:14:26)


rrusczyk (19:14:40)
(Note that this is the same as chess64 &Phelpedo's expression)

rrusczyk (19:14:57)
So, how can we count the powers of 3 in P?

chesspro (19:15:22)
total power of 3 in 200! - total power of 3 in 100!

Phelpedo (19:15:32)
Count the number of 3's in 200! and subtract the number in 100!.

topaz (19:15:33)
find the number of factors of 3 in 200! and subtract the number of 3's in 100!

rrusczyk (19:16:05)
Then the number of powers of 3 in P is just the number of powers of 3 in 200! minus the number of powers of 3 in 100!.

rrusczyk (19:16:10)
How do we determine the number of powers of 3 in n! for any n?

rrusczyk (19:16:54)
(For those who aren't sure why we are doing this subtraction, consider:

Elemennop (19:15:17)
count all in the top, count all in bottom...subtract!

chess64 (19:15:20)
#(Numerator)-#(denominator)

mkkool64 (19:15:23)
#in numerator - #in denominator

rrusczyk (19:17:11)
Here's how to count the number of powers of 3 in n!:

mkkool64 (19:16:44)
floor[n/3] + floor[n/9] + floor[n/27] etc.

Vihang (19:16:56)
|n/3| + |n/9| + ...
Connsider | | to be the floor function

chess64 (19:16:57)



rrusczyk (19:18:05)


rrusczyk (19:18:18)
So we need to compute [200/3] + [200/9] + [200/27] + [200/81] - [100/3] - [100/9] - [100/27] - [100/81].

rrusczyk (19:18:21)
What do we get?

joezfang (19:18:36)
49

LordoftheMorons (19:18:39)
49

rrusczyk (19:18:50)
This is 66+22+7+2-33-11-3-1 = 49.

rrusczyk (19:19:03)
Next problem!

rrusczyk (19:19:08)


rrusczyk (19:19:18)
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rrusczyk (19:19:21)
Before diving in the calculations, are there any observations we can make about such permutations?

arjunlandes (19:19:13)
a_6 must be 1

tongchen1226 (19:19:17)
1 is fixed

Go Beyond (19:19:22)
a6 is 1

chess64 (19:19:23)
a_6 = 1 for all permutations.

rrusczyk (19:19:38)
a_6 is smaller than every other element, so a_6 must be 1.

rrusczyk (19:19:44)
Now what?

LordoftheMorons (19:19:48)
Since a6 is clearly 1, We cleverly note that if we choose numbers for one side, the other side is predetermined, as it is the remaining numbers in increasing order.

Phelpedo (19:19:49)
For every permutation of 5 numbers, there is exactly 1 way to arrange them from least to greatest

arjunlandes (19:19:59)
also, for any set of 5 numbers a_1 to a_5, there is one unique way to arrrange them in ascending order

chess64 (19:20:18)
So, if we just pick any 5 integers from 2 - 12, put them in decreasing order, and put them on the left side of a_6, or permutation is determined.

tongchen1226 (19:20:22)
choose 5 elements to put into the first 5 blanks

arjunlandes (19:20:24)
and, if you select the first 5, the other five are already determined

topaz (19:20:37)
you can choose any five numbers and arrange them in decreasing order and arrange the rest in decreasing order too

rrusczyk (19:21:00)
Then, as soon as we decide which 5 (of the 11) remaining elements are going to make up a_1 through a_5, their order is fixed. Similarly the ordering of a_7 through a_12 is fixed.

rrusczyk (19:21:08)
So to construct such a permutation, all we need to do is choose 5 of the elements (2,3,...,12) to form a_1 through a_5.

rrusczyk (19:21:13)
So, what is our answer?

camathkid (19:21:04)
butm a6 has to be 5 a's less that a1

mkkool64 (19:21:22)
11C5 = 462

arjunlandes (19:21:25)
11 choose 5=462

rrusczyk (19:21:42)


rrusczyk (19:21:51)
This is an example of what I like to call 'constructive counting'. We think about how we would make one list (1 in the middle, pick 5 numbers for the first five and their order is determined, then the other 6 have their order determined), and this guides us in counting how many lists there are total.

rrusczyk (19:22:05)
Next problem!

rrusczyk (19:22:11)


rrusczyk (19:22:16)
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rrusczyk (19:22:21)
Yikes, this question seems complicated. Make sure you read it carefully before proceeding.

kops723 (19:22:33)
it's not so bad

joezfang (19:23:01)
it's less complicated than it seems

rrusczyk (19:23:13)
Indeed; often on the AIME, the long, complicated problems aren't so bad. The short, simple looking ones near the end are often the nightmares.

rrusczyk (19:23:26)
Let's set the probability of rolling face F to be p (which is what we want to find to solve the problem).

rrusczyk (19:23:32)
What's the probability of the face opposite from F?

tongchen1226 (19:23:40)
1/3-p

chesspro (19:23:43)
1/3 - p

Phelpedo (19:23:45)
1/3-p

TimReyn (19:23:49)
1/3-p

rrusczyk (19:23:57)
It's 1 - p - 4(1/6) = 1/3 - p, since the sum of the probabilities of the six faces must sum to 1.

rrusczyk (19:24:09)
Now what's the probability of rolling a 7 on the two dice (in terms of p)?

kops723 (19:24:26)
1/9 +2p(p-1/3)

jen7 (19:24:53)
4(1/6)(1/6)+2(p)(1/3-P)=47/288 considering the # of ways to roll each pair

rrusczyk (19:25:08)
There are 2 ways to roll a 7 using F and the face opposite to F. Each of these occur with probability p(1/3 - p).

rrusczyk (19:25:14)
There are 4 ways to roll a 7 using two of the "normal" faces. Each of these occur with probability (1/6)^2 = 1/36.

rrusczyk (19:25:23)


rrusczyk (19:25:33)
But we know that this is 47/288, so we can set it equal and solve for p.

rrusczyk (19:25:39)


rrusczyk (19:26:01)
What do we find for p?

LordoftheMorons (19:26:07)
5/24

kops723 (19:26:08)
5/24

Phelpedo (19:26:11)
p = 5/24

joezfang (19:26:26)
5/24

rrusczyk (19:27:18)
Clear the denominators: 192p(1-3p) + 32 = 47.

rrusczyk (19:27:22)
This gives -576p^2 + 192p - 15 = 0.

rrusczyk (19:27:24)
Dividing by -3 gives 192p - 64p + 5 = 0.

rrusczyk (19:27:33)
Now for the quadratic equation (if we don't see how to factor):

rrusczyk (19:27:37)


rrusczyk (19:27:46)
Why do we want the '+' root?

jen7 (19:27:53)
>1/6

Elemennop (19:27:57)
p is greater than 1/6.

dark_sunburst (19:27:58)
p is greater than 1/6

jli (19:28:01)
because it is greater thatn 1/6\

joezfang (19:28:02)
p is greater than 1/6

rrusczyk (19:28:09)
We want the "+" root, since we know that p > 1/6, hence p = 80/384 = 5/24.

rrusczyk (19:28:11)
So the answer is 5+24 = 29.

rrusczyk (19:28:24)
Next problem!

rrusczyk (19:28:26)


rrusczyk (19:28:30)
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kops723 (19:28:35)
diagram

rrusczyk (19:28:43)
We clearly need to start by drawing a diagram. Let's first just draw the square and the triangle AEF.

rrusczyk (19:28:52)


rrusczyk (19:28:54)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/43f646c5361dfc48655e49631340c0a0.png

rrusczyk (19:28:58)
What do we know about the angles in this picture?

the1337master (19:29:10)
aef fea fae is 60

Phelpedo (19:29:33)
ECF is an iscoceles right triange; the BEA and ADF are 15-75-90 triangles

the1337master (19:29:41)
cef is 45 and cfe is 45

tongchen1226 (19:30:00)
15, 60, 15 for A

perfectnumber628 (19:30:04)
AC is a line of symmetry

rrusczyk (19:30:12)
We know since EAF = 60 and since the picture is symmetric with respect to the diagonal AC, we must have BAE = FAD = 15.

rrusczyk (19:30:18)
Now let's add the little square.

rrusczyk (19:30:23)


rrusczyk (19:30:25)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/2d44e47261c09ab6010c3360ef7ac146.png

rrusczyk (19:30:43)
I've labeled the vertices of the little square as shown.

rrusczyk (19:30:47)
Call the side of the little square x. What equations can we write for x?

perfectnumber628 (19:30:47)
similar triangles

Vihang (19:30:54)
APQ is similar to QRE

rrusczyk (19:31:09)
Indeed, what else?

rrusczyk (19:31:32)
(We can get to the answer with similar triangles; are there other ways?)

jen7 (19:31:17)
PA=1-x, PQ=x, so tan(15)=x/(1-x)

rrusczyk (19:31:46)
We know PQ/PA = x/(1-x) = tan 15.

rrusczyk (19:31:52)
What's tan 15?

perfectnumber628 (19:32:03)
2-sqrt3

Elemennop (19:32:06)
tan 15 = 2-rt3

rrusczyk (19:32:54)
How can we figure this out if we don't have it memorized?

Go Beyond (19:32:48)
half-angle formula

kops723 (19:33:01)
half angle formula

jli (19:33:01)
half angle

rrusczyk (19:33:12)


rrusczyk (19:33:16)
We can also use:

chess64 (19:33:05)
tan (45-30)

rrusczyk (19:33:23)
(If you don't know this formula, it's pretty easy to derive using cos 2x = 1 - 2sin^2(x) = 2cos^2(x) - 1.)

rrusczyk (19:33:34)


rrusczyk (19:33:43)


rrusczyk (19:34:01)
What do we find for x?

mkkool64 (19:35:09)
x= (3-sqr(3))/6

rrusczyk (19:35:26)


rrusczyk (19:35:39)
So the answer is 3+3+6 = 12

rrusczyk (19:36:10)
Next problem!

rrusczyk (19:36:22)


rrusczyk (19:36:26)
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rrusczyk (19:36:31)
How can we count these?

amy tsao! (19:36:54)
by counting and subtracting

perfectnumber628 (19:36:59)
find the pairs that add up to 1000, subtract ones with 0

hqin2007 (19:37:03)
Count all subtract ones with zero digit.

rrusczyk (19:37:23)
That's one approach. Can we do it directly?

rrusczyk (19:37:46)
(One of you who took this indirect approach, write me up a full solution I can post at the end)

Phelpedo (19:37:13)
Divide it into cases where (1) both a and b are 3 digit numbers, (2) one of them is a 2 digit number, and (3) one is a 1 digit number

rrusczyk (19:38:08)
Let's start by looking at the case where a and b are both three-digit numbers without any 0's.

chess64 (19:37:25)
Casework based on # of digits

rrusczyk (19:38:23)
What next?

rrusczyk (19:38:28)
How will we count these?

kops723 (19:36:55)
we know that the 'ones' digits add to 10, the other two add to 9 each

Vihang (19:38:54)
last digit of a is x then last digit of b is 10-x

rrusczyk (19:39:04)
When we add them, the last digits must sum to 10 so that we get a 0 for the units digit of the sum.

rrusczyk (19:39:11)
Then, since we're carrying a 1, the middle digits of a and b must sum to 9 so that we get a 0 for the middle digit of the sum.

rrusczyk (19:39:16)
Finally, since we're again carrying a 1, the first digits of a and b must sum to 9.

rrusczyk (19:39:20)
So, how many for this case?

sirorange (19:39:03)
hmm, we can take 8 (for the first digits) * 8(for the second digit)* 9, for any of the last digits=

kops723 (19:39:35)
9*8*8=576

rrusczyk (19:39:53)
Thus there are 8*8*9 = 576 possibilities: we choose the digits of a to be 1-8 in the hundreds position, 1-8 in the tens position, and 1-9 in the units position. (We can't put a 9 in the hundreds or tens positions, since then b would have a 0 in that position).

rrusczyk (19:40:01)
Now what if one of a or b is less than 100? We can assume that a is, and just multiply by 2 to account for those cases where b is.

Vihang (19:40:37)
first digit of b is 9

kops723 (19:40:56)
b has a 9, last digit of a and b sum to 10, "tens" digits sum to 9

rrusczyk (19:41:29)
If a < 100, then the first digit of b must be 9.

rrusczyk (19:41:34)
Now what?

tongchen1226 (19:41:30)
011~089= 8*9= 72

rrusczyk (19:41:56)
If 9 < a < 100, then as above, the tens digits of a and b must sum to 9 and the units digits must sum to 10. So there are 8*9 = 72 possibilities.

rrusczyk (19:42:00)
What's left?

tongchen1226 (19:42:06)
001~009

Hokkage (19:42:06)
Ones.

rrusczyk (19:42:20)
If a < 10, then the middle digit of b must be 9. Any choice of a 1 through 9 works, so there are 9 possibilities in this case (a = 1, b = 999; a = 2, b = 998; etc. up to a = 9, b = 991).

rrusczyk (19:42:27)
Hence the number of possibilities with a < 100 is 9+72 = 81. Similarly there are 81 with b < 100.

rrusczyk (19:42:32)
So the final answer is 576 + 2(81) = 738

Here's another look at the problem:

Phelpedo (19:43:25)
There are 1000 total pairs (a = 1..999 b = 999...1). There are 0 1-digit Method II-There are 999 total pairs (a = 1..999, b = 999...1).
There are: 0 1-digit numbers with a 0.
9 2-digit numbers with a 0 (in the forum x0, where x = 1..9),
81 3-digit numbers with a 0 in the form xx0
9 3-digit numbers with a 0 in the form x00.
81 3-digit numbers with a 0 in the form x0x

Since the complement of all the numbers in the first 3 cases has a 0, they only need to be counted once. However, numbers of the form x0x are need to be counted twice because their complements have no 0's. Therefore, the answer is 999-9-81-9-(2*81)=738
numbers with

rrusczyk (19:43:03)


rrusczyk (19:43:56)
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rrusczyk (19:44:13)
Once again, a lengthy problem statement that needs to be read carefully.

rrusczyk (19:44:18)
We'll start with a picture.

rrusczyk (19:44:23)


rrusczyk (19:46:17)
What's the first observation we might make?

sirorange (19:43:46)
Let's start with the center triangle

kops723 (19:44:42)
we need to find the number of ways the outer three triangles can be colored and multiply by 6

jen7 (19:46:27)
center can take all 6 colors

chesspro (19:46:27)
assume center is a single color, multiply result by 6

rrusczyk (19:46:48)
The center color is unique in the sense that two large triangles can only match if their center colors are equal. So there are 6 choices for the center color, and we have to determine how many outside color choices are possible up to rotation and reflection.

rrusczyk (19:46:58)
How might we approach counting these?

jen7 (19:47:12)
consider 3,2,and 1 different colors for outer 3

Hokkage (19:47:26)
They are all the same, two are the same, or none are the same (of the three outers).

mkkool64 (19:47:46)
3 cases: all same color, two the same color, all different colors

rrusczyk (19:47:59)
The easiest way seems to use cases based on the number of different outside colors that we have.

rrusczyk (19:48:06)
What if all three are the same color?

Vihang (19:48:14)
6 ways

ramdude (19:48:16)
6 ways

kops723 (19:48:17)
6 possibilities

Hokkage (19:48:18)
6 possibilities

rrusczyk (19:48:27)
If all three outside triangles have the same color, these are all the same under rotation and reflection. So there are 6 possibilities.

rrusczyk (19:48:37)
How about two of one and one of another?

Vihang (19:48:45)
6*5 ways

LordoftheMorons (19:48:48)
6*5=30 ways

jen7 (19:48:52)
(6 C 2)*2

rrusczyk (19:49:00)
If we have two of one color and a third of other, there are also all the same under rotation and reflection. So there are 6*5 = 30 possibilities (6 choices for the color on 2 triangles, and 5 remaining choices for the color on the 3rd triangle.)

rrusczyk (19:49:08)
How about all three different?

jen7 (19:49:13)
(6 C 3)

Phelpedo (19:49:26)
6c3 = 20

mkkool64 (19:49:35)
6C3=20

LordoftheMorons (19:49:37)
6C3 =20

rrusczyk (19:49:45)
If they are all different colors, these are also all the same under rotation and reflection. So there are C(6,3) = 20 choices for the 3 colors, since all we must do is choose the three colors.

rrusczyk (19:49:58)
So, what is our total number of colorings?

Vihang (19:50:03)
so 6*(20+30+6)=6*56=336

LordoftheMorons (19:50:14)
6*56=336

Hokkage (19:50:16)
6*(20+30+6)

rrusczyk (19:50:27)
Thus the answer is 6*(6+30+20) = 6*56 = 336.

rrusczyk (19:50:35)
All right - next problem

rrusczyk (19:50:40)


rrusczyk (19:50:45)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-42001662.gif

perfectnumber628 (19:51:03)
draw them

rrusczyk (19:51:12)
If you brought graph paper, a compass, and a ruler with you to the AIME (and didn't just decide to skip this problem :) ), you should be able to draw a fairly accurate picture.

rrusczyk (19:51:20)


rrusczyk (19:51:25)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/d7330042954e04d8023a844af54a8d50.png

rrusczyk (19:51:30)
What can we determine from this picture?

Go Beyond (19:51:32)
similar triangle

dark_sunburst_2 (19:51:44)
tangents form right angles with radius

rrusczyk (19:51:59)
Let's label some points so we can talk about this more simply

rrusczyk (19:52:06)
Let's label the points in our picture. Call the centers of the three circles A, B, C, the four tangent points (from left to right) S, T, U, V, the points where t_1 and t_2 hit the x-axis X and Y, and the point we're trying to find Z:

rrusczyk (19:52:11)


rrusczyk (19:52:14)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/7b98e954efd45f6dd2fb3af7d8567b1b.png

LordoftheMorons (19:51:57)
similar triangles

rrusczyk (19:52:44)
Right angles often mean similar triangles. Where are the similar triangles here?

kops723 (19:52:39)
SXA ~ TXB

Hokkage (19:52:57)
ASX ~ BTX and CYV ~ UYB

Vihang (19:52:59)
ASX BTX

kops723 (19:52:59)
UYB ~ VYC

rrusczyk (19:53:34)
Triangles ASX and BTX are similar right triangles.

rrusczyk (19:53:39)
How does this help?

chess64 (19:53:22)
tan (AXS) = 1/SX = 2/XT, so XT = 2SX, which implies that XB = 2AX, so AX=4 and XB=8, so X=(4,0).

jen7 (19:53:29)
AS=1 and TB=2, so by similar triangles, AX=4 and XB=8. Thus, X has coordinates (4,0).

Vihang (19:54:01)
AB=12
AS=1 BT=2
Thus AX=4 XB=8

rrusczyk (19:54:29)
Therefore, XA/XB = SA/BT = 1/2.

rrusczyk (19:54:34)
So since AB = 12, we have XA = 4 and XB = 8. Hence X is the point (4,0).

rrusczyk (19:54:39)


rrusczyk (19:54:42)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/7b98e954efd45f6dd2fb3af7d8567b1b.png

rrusczyk (19:54:57)
What do we get for BY and BC using our other similarity?

chess64 (19:55:10)
Y(16,0)

dark_sunburst_2 (19:55:51)
BY = 4, YC = 8, so Y (16,0)

Someone (19:56:12)
4,8 The point Y is (16,0)

Phelpedo (19:56:16)
4 and 8 again

rrusczyk (19:56:44)
Similarly (no pun intended!), BUY and CVY are similar.

rrusczyk (19:56:46)
So BY/CY = BU/CV = 1/2, and hence BY = 4 and YC = 8. Hence Y is the point (16,0).

rrusczyk (19:56:51)
So now we can focus our picture on just the middle circle!

rrusczyk (19:56:56)


rrusczyk (19:56:58)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/09f958472d3f942d9a8a1fe6bde067a5.png

rrusczyk (19:57:03)
We could start pounding through this with a whole lot of analytic geometry, but maybe we can find a slicker approach!

rrusczyk (19:57:13)
What length do we seek in order to answer the problem?

jen7 (19:57:32)
x coordinate of Z

rrusczyk (19:58:06)
OK -> in terms of this diagram, can we define a length that, if we had it, would give us our answer quickly?

rrusczyk (19:59:01)
(You can define other points if you want; in this diagram, what would give us the 'x coordinate of Z'?)

chess64 (19:59:13)
Drop a perpendicular ZH to XY.

perfectnumber628 (19:59:20)
drop an altitude to XY?

kops723 (19:59:26)
construct a perpendicular from Z to the axis, call the point of intersection M, find BM

rrusczyk (19:59:33)
We seek the distance from X to the foot of the altitude from Z to XY. Let this foot be point P.

rrusczyk (19:59:37)


rrusczyk (19:59:40)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/4317e9bb5f04c343243a8af1657c638e.png

rrusczyk (19:59:47)
We now have a bunch of right angles in the diagram; what tool might we use to solve the problem?

chess64 (19:59:45)
Area

rrusczyk (20:00:19)
With all those altitudes, we reach for area to solve the problem.

rrusczyk (20:00:26)
Area can be a very powerful problem solving tool.

rrusczyk (20:00:28)
Let's give it a try.

rrusczyk (20:00:43)
Let ZP have length y. Do we know any other lengths in terms of y?

rrusczyk (20:00:52)
(Don't forget the lengths we have:

rrusczyk (20:00:57)


rrusczyk (20:01:00)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/09f958472d3f942d9a8a1fe6bde067a5.png

rrusczyk (20:01:51)
Do we know any angles?

rrusczyk (20:02:15)
(Aside from the right angles)

Someone (20:02:11)
UBY = 60

perfectnumber628 (20:02:13)
ZYB=30 degrees

kops723 (20:02:13)
BYU = 30

rrusczyk (20:02:25)
Since UB = 2 and hypotenuse YB = 4 in right triangle BUY, we have <BYU = 30. How does this help?

rrusczyk (20:02:41)


rrusczyk (20:02:44)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/4317e9bb5f04c343243a8af1657c638e.png

kops723 (20:02:44)
we know that ZPY is 30-60-90

rrusczyk (20:02:55)
ZP = y; what other lengths do we now have?

Hokkage (20:02:52)
PY = y*sqrt3

kops723 (20:03:10)
PY = y*sqrt3

Hokkage (20:03:11)
PY = y*sqrt3

kops723 (20:03:15)
ZY = 2y

Vihang (20:03:16)
ZY=2p

rrusczyk (20:03:27)
Since ZP = y and <ZYP = 30 in triangle ZYP, we have PY=y*sqrt{3} and ZY=2y.

rrusczyk (20:03:36)
Keep going! What other lengths do we have in terms of y?

tongchen1226 (20:03:50)
XP=sqrt(15)y

jen7 (20:04:52)
XP=12-y*sqrt(3)?

chesspro (20:04:54)
XP = 12 - sqrt(3) y

rrusczyk (20:05:33)
Interesting. We have two different expressions for XP. Are they both correct?

rrusczyk (20:05:47)


rrusczyk (20:05:50)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/4317e9bb5f04c343243a8af1657c638e.png

jli (20:05:43)
yes

Someone (20:05:58)
yep, why not set them equal and solve

rrusczyk (20:06:12)
Let's see why they are both correct!

rrusczyk (20:06:17)


rrusczyk (20:06:24)
That one is pretty easy to see.

rrusczyk (20:06:41)
How about XP = y*sqrt(15)? Where did that come from?

Bictor717 (20:07:20)
XPZ similar to XTB

rrusczyk (20:07:41)
The Pythagorean Theorem gives us XT = 2sqrt(15), and triangles BTX and ZPX are similar, so XP/ZP = XT/BT = sqrt(15)

rrusczyk (20:07:48)
So, XP = y*sqrt(15).

rrusczyk (20:08:15)
We set out to use areas (and if we hadn't seen XP = y*sqrt(15), we still could have), but we found another path!

rrusczyk (20:08:29)
We have two expressions for XP, so what can we do?

Hokkage (20:08:16)
y*sqrt(15)=12-y*sqrt(3)

perfectnumber628 (20:08:36)
solve for y

jli (20:08:39)
set them equal

kops723 (20:08:39)
set them equal

Hokkage (20:08:47)
Set them equal and solve for y.

rrusczyk (20:08:58)
And what do we get?

Hokkage (20:09:31)
y=sqrt(15)-sqrt(3)

Bictor717 (20:09:33)
12/(sqrt15+sqrt3)

rrusczyk (20:09:57)
Setting these expressions equal gives y = sqrt(15) - sqrt(3).

rrusczyk (20:10:11)
So, what is our desired x-coordinate?

rrusczyk (20:12:44)


rrusczyk (20:12:56)
(A reminder of the diagram, ZP is y)

mkkool64 (20:11:30)
19-3sqr(5)

Mr.Ocax (20:12:42)
19-3sxqrt(5)

tongchen1226 (20:12:58)
19-3sqrt(5)

rrusczyk (20:13:13)
Our desired x-coordinate is 4 + XP = 4 + y*sqrt(15) = 19 -3sqrt(5).

rrusczyk (20:13:16)
So the answer is 19+3+5 =27

rrusczyk (20:13:37)
When the question is in terms of analytic geometry, that doesn't mean we have to use analyt to solve it!

rrusczyk (20:13:47)
Notice, no nasty equations to deal with here!

rrusczyk (20:14:05)
Next problem

rrusczyk (20:14:10)


rrusczyk (20:14:18)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-87318478.gif

rrusczyk (20:14:30)
What observations can we make?

Vihang (20:14:59)
seven teams so 6 matches per team

rrusczyk (20:15:11)
And A has beaten B, so

tongchen1226 (20:15:06)
A has 1 point now, B has 0

rrusczyk (20:15:21)
And how many games are left?

rrusczyk (20:15:23)
(For them)

perfectnumber628 (20:15:26)
5

Hokkage (20:15:27)
Five.

mkkool64 (20:15:28)
5 each

rrusczyk (20:15:40)
So our problem is:

kops723 (20:15:29)
b needs to win strictly more games than A in 5 games

rrusczyk (20:15:52)
Each team plays 5 other games.

rrusczyk (20:15:57)
Do we have to wade through all possible cases? A scores 6 and B scores 0. A scores 6 and B scores 1. A scores 6 and B scores 2.

noneoftheabove (20:16:11)
no

rrusczyk (20:16:19)
We hope not. This could take forever.

rrusczyk (20:16:23)
How can we define our cases to work through this more quickly?

noneoftheabove (20:16:49)
b is less than a, b is equal to a, and b is greater than a

rrusczyk (20:17:44)
We can consider three cases: B scores more than A in the last 5 games, A scores more than B in the last 5, and A and B score the same in the last 5.

rrusczyk (20:18:01)
Do we have to break down into zillions of cases to evaluate these? Or can we make a simplifying observation?

kops723 (20:18:11)
we know that b < a is the same probability as b > a

rrusczyk (20:18:37)
We can use symmetry to simplify the calculation enormously, since "P(Team A scores more points in 5 games)" and "P(Team B scores more points in 5 games)" are equal, and these are both half of 1 - "P(Team A and Team B score the same number of points)".

rrusczyk (20:19:11)
So let's compute P(Team A and Team B score the same number of points in 5 games).

rrusczyk (20:19:23)
How do we do it?

perfectnumber628 (20:19:51)
there are 6 possible scores for each team, so 6 ways they can be equal

rrusczyk (20:20:34)
All right, and how many different outcomes are there for the games for each team?

rrusczyk (20:21:23)
(Notice that 'both teams win 1' is not as likely as 'both teams win 2', so we can't just say '6 ways for them to be the same out of however many possibilities)

jen7 (20:20:48)
2^5

rrusczyk (20:21:32)
For each team, there are 2^5 = 32 possible outcomes of the games.

rrusczyk (20:21:42)
How many ways can each team win 0 games?

Hokkage (20:21:48)
1

kops723 (20:21:50)
1

Vihang (20:21:54)
1

noneoftheabove (20:21:54)
1

rrusczyk (20:22:06)
How many ways can each team win 1 game?

jen7 (20:22:08)
5

Hokkage (20:22:12)
5

kops723 (20:22:14)
5

Vihang (20:22:18)
5

noneoftheabove (20:22:18)
5

Phelpedo (20:22:20)
5

rrusczyk (20:22:52)
That's per team; we want both teams to win exactly 1 game.

kops723 (20:22:28)
5^2 = 25

rrusczyk (20:23:17)
So, what is the probability they both win one game of their last 5?

kops723 (20:23:32)
25/1024

kops723 (20:23:37)
25/1024

Hokkage (20:23:37)
25/1024

rrusczyk (20:24:08)
Probability A wins 1 out 5 = 5/32. Same for B. Probability they both win 1 out of 5 = (5/32)^2

rrusczyk (20:24:28)
Continuing in this manner, what is the probability A and B win the same number of games in their last 5 games?

Hokkage (20:24:49)
252/1024

Bictor717 (20:25:33)
252/1024

LordoftheMorons (20:25:43)
252/1024

jen7 (20:25:44)
2(100/1024) + 2(25/1024) + 2(1/1024)

noneoftheabove (20:25:49)
1/1024+25/1024+100/1024+100/1024+25/1024/1/1024=252/1024

Vihang (20:25:56)
(1+25+100+100+25+1)/1024

rrusczyk (20:26:04)
There are 1, 5, 10, 10, 5, 1 ways to score 0, 1, 2, 3, 4, and 5 points (respectively) for each team: these are just the binomial coefficients C(5,0), C(5,1), etc.

rrusczyk (20:26:12)


rrusczyk (20:26:16)
(Notice that 252 = C(10,5). This is not a coincidence.)

rrusczyk (20:26:41)
So, what is the probability that A scores more than B in the last 5 games?

Hokkage (20:26:56)
(1-(252/1024))/2

perfectnumber628 (20:27:01)
(1-252/1024)/2

joezfang (20:27:08)
386/1024

jen7 (20:27:15)
386/1024?

rrusczyk (20:27:25)
Hence the probability that Team A scores more points is (1/2)(1-252/1024) = 386/1024.

rrusczyk (20:27:32)
Thus our probability of success is (252/1024) + (386/1024) = 638/1024 = 319/512, and our answer is 319+512 = 831

rrusczyk (20:27:47)
All right. We're 2/3 of the way there.

rrusczyk (20:28:19)
Now for the next problem (and a note to those of you who are at this Math Jam to see what our classes are like: most of our classes have a somewhat less frenetic pace!)

rrusczyk (20:28:24)


rrusczyk (20:28:34)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-199741246.gif

rrusczyk (20:28:49)
One way of doing this problem is to calculate all the terms up to a_{28}, and then the given sum manually. We could do the calculations mod 1000, which would help, but this is still extremely tedious; we would only want to do this as a last resort.

rrusczyk (20:28:56)
What might we do instead?

mkkool64 (20:29:25)
look for a pattern

Phelpedo (20:29:30)
List some terms and look for a pattern.

joezfang (20:29:35)
find a pattern

rrusczyk (20:30:02)
If anyone played around and found a simple pattern, please write up the solution and post it.

rrusczyk (20:30:15)
How might we use our recursion with our summation?

rrusczyk (20:30:25)
Note that we are given the values of a_{28}, a_{29}, and a_{30}. Is this a clue? How can we use them?

rrusczyk (20:31:40)
How might we use our recursion to rewrite our summation

Vihang (20:32:04)
a5=a4+a3+a2=2a4-a1
a6=a5+a4+a3=2a5-a2...

rrusczyk (20:32:27)
Let's see what happens if we write a whole stack of these:

rrusczyk (20:32:32)


rrusczyk (20:32:37)
(We go up to a_{30}, because that is the last term we are given.)

rrusczyk (20:32:42)
What can we do with these equations?

perfectnumber628 (20:32:55)
add?

kops723 (20:33:11)
add them

rrusczyk (20:33:23)


rrusczyk (20:33:43)


rrusczyk (20:34:00)
We are now very close to finding the sum we want. What adjustments do we need to make?

chesspro (20:34:27)
2S + a_3 - a_1 - a_28 = a_30

jli (20:34:33)
add a_1, -a_3 and a_28

Bictor717 (20:34:40)
add a28 to both sides and since a1=a3, divide by 2

rrusczyk (20:35:00)


rrusczyk (20:35:22)


rrusczyk (20:35:30)
And what do we get for our final answer?

ramdude (20:35:47)
834

joezfang (20:36:00)
834

rrusczyk (20:36:29)


rrusczyk (20:36:34)
Therefore, the answer is [b]834[/b].

rrusczyk (20:37:13)
Notice that our strategy here was not to mess around trying to cancel and find this or that term. We swung for the fences and tried to find a way to get an expression that included the summation we wanted.

rrusczyk (20:37:43)
Next problem!

joezfang (20:37:43)
that's so much better than brute force

rrusczyk (20:37:53)
:)

rrusczyk (20:37:58)


rrusczyk (20:38:06)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-250752062.gif

rrusczyk (20:38:11)
Our first step is to draw a diagram.

rrusczyk (20:38:19)


rrusczyk (20:38:24)
http://www.artofproblemsolving.com/MathJams/aime2006-2-11.JPG

yuhehe (20:38:41)
and then what?

rrusczyk (20:39:15)
Good question. We focus on our target triangle CGB. Do we know anything about it? Anything about its angles or sides?

perfectnumber628 (20:39:07)
you can find a side of the equilateral triangle

rrusczyk (20:39:25)
Indeed! We can find BC. What is BC?

ramdude (20:39:32)
2sqrt3

Hokkage (20:39:53)
2sqrt3

yuhehe (20:40:02)
2root 3

mkkool64 (20:40:02)
2sqr(3)

rrusczyk (20:40:15)
Since triangle ABC is equilateral and inscribed in a circle of radius 2, BC = 2 sqrt(3).

rrusczyk (20:40:29)
OK. Do we know anything else about the angles or sides of BGC?

rrusczyk (20:40:34)


rrusczyk (20:40:37)
http://www.artofproblemsolving.com/MathJams/aime2006-2-11.JPG

yuhehe (20:39:37)
CGB is 120 degrees

perfectnumber628 (20:40:22)
angle BGC=120

jen7 (20:40:44)
<GBC=<CAG and BGC=180-60=120 by cyclic quad., so triangles BGC and AEF are similar. Thus, BC/AF=CG/EF and BC/AF=BG/AE.

chesspro (20:41:07)
CGB = 120, due to cyclic quads

rrusczyk (20:42:10)
Having pegged <CGB as 120 (since ACGB is cyclic), we look at the other angles of CGB. Why is CGB similar to FEA and ADF?

chesspro (20:42:29)
AEFD is a parallelogram

rrusczyk (20:43:01)
Since <A = 60, this tells us that <D = <E = 120.

rrusczyk (20:43:11)
We need another angle. Where will we get it?

Vihang (20:42:50)
<BAG=<BCG

perfectnumber628 (20:43:20)
BAG and BCG intercept the same arc

rrusczyk (20:43:30)


rrusczyk (20:43:32)
http://www.artofproblemsolving.com/MathJams/aime2006-2-11.JPG

rrusczyk (20:43:44)
BAG and BCG are inscribed in the same arc, so they are equal!

rrusczyk (20:44:13)
(Don't just stare at geometry problems. Find the angles and lengths you can find, hunt down and mark equal angles. Then the similar triangles will pop out.)

rrusczyk (20:44:25)
It says that triangles CBG and AFD are similar. Now the problem reduces to comparing the sides of these triangles.

rrusczyk (20:45:10)
We know AD = 13. What else can we find?

jen7 (20:44:53)
law of cosines on AEF (or ADF)

chesspro (20:45:03)
you could find AF via LoC and then find area of either ADF or GBC, scaling sides/areas if necessary

Vihang (20:45:16)
Use cosine rule for AEF to get AF is apossible way

chesspro (20:45:22)
DF = AE = 11

rrusczyk (20:45:34)
. Since ADFE is a parallelogram, DF = AE = 11, and angle ADF is 120 degrees.

rrusczyk (20:45:42)


rrusczyk (20:45:48)
How does this help?

Vihang (20:46:03)
BC=2root3

perfectnumber628 (20:46:07)
that is in proportion to BC

Hokkage (20:46:10)
Ratios. Find BG and GC.

joezfang (20:46:11)
find BG and CG

rrusczyk (20:46:16)


rrusczyk (20:46:25)
Now what?

chesspro (20:46:42)
we don't need BG and CG, find the area of ADF and multiply by appropriate scaling factor

perfectnumber628 (20:46:44)
so you can find the area of the big triangle and then scale it down

rrusczyk (20:47:12)
How do we find the area of the big triangle?

perfectnumber628 (20:47:29)
use trig A=1/2 absinC

Vihang (20:47:35)
1/2*11*13*sin120

chesspro (20:47:38)
sine area formula or heron's, i prefer the first in this case :)

rrusczyk (20:47:53)
Me too.

rrusczyk (20:47:55)


rrusczyk (20:48:00)
So what is the area of triangle CBG?

Vihang (20:48:56)
429/433*root3

jen7 (20:48:59)
429sqrt(3)/433

mkkool64 (20:49:09)
429sqr(3)/433

rrusczyk (20:49:16)


rrusczyk (20:49:23)
Therefore, the answer is 429 + 3 + 433 = 865.

rrusczyk (20:49:40)
Next problem!

rrusczyk (20:49:47)


rrusczyk (20:49:54)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-90715535.gif

rrusczyk (20:50:23)
Where do we start?

rrusczyk (20:51:27)
How can we get from the words of this problem to something mathematical?

mkkool64 (20:51:56)
write N = 2k+1 + (2k+3) ... 2k+9

rrusczyk (20:52:36)
We have to turn the words of the problem into mathematics, so we write a mathematical expression for our sum of j consecutive positive odd integers.

rrusczyk (20:52:51)
Above, we have a sum of 5 consecutive integers.

rrusczyk (20:52:56)
What do we have for j of them?

mkkool64 (20:52:43)
N=2k+1 + 2k+3 + ... 2k + (2j-1)

Vihang (20:53:12)
N=2jk+j^2

perfectnumber628 (20:53:23)
go up to 2k+2j-1

rrusczyk (20:53:40)
Our sum is (2k+1) + (2k+3) + ... + (2k+2j-1).

rrusczyk (20:53:54)
As Vihang notes, the closed form of this sum is:

rrusczyk (20:53:59)
(2k+1+2k+2j-1)(j/2) = (4k+2j)(j/2) = j(2k+j).

rrusczyk (20:54:04)
What do we observe from this?

Vihang (20:51:26)
j must have the same parity for a fixed N

Vihang (20:52:00)
j and N have same parity

rrusczyk (20:54:21)
If N is odd, then j is an odd factor, and if N is even, then j is an even factor.

jli (20:50:59)
split into cases where j is even and when j is odd

rrusczyk (20:54:38)
Let's tackle the "N is odd" case first. What do we know from N = j(2k+j)?

davidyko (20:54:27)
what's parity?

rrusczyk (20:54:55)
Oddness or evenness (1 and 33 have the same parity)

Vihang (20:54:52)
j divides N

rrusczyk (20:55:12)
What else? What fact have we not used yet?

perfectnumber628 (20:55:27)
there must be exactly 5 values of j

jen7 (20:55:33)
5 values of j

rrusczyk (20:56:03)
Using that with the fact N = j(2k+j) tells us what about N (when N is odd)?

perfectnumber628 (20:56:55)
N has at least 5 odd factors

rrusczyk (20:57:25)
What are the possibilities for the number of factors when N is odd?

joezfang (20:57:56)
10, 9

rrusczyk (20:58:23)
If N is odd, then N must have exactly 9 or 10 factors: the possible values of j will be the smaller half of those factors (including sqrt(N) if N has 9 factors). (Remember, we are told there are exactly 5 possible values of j!)

rrusczyk (20:58:37)
(Note that N = j(2k+j), and j < 2k + j)

rrusczyk (20:58:53)
In terms of the prime factorization of N, what does this mean?

rrusczyk (20:59:12)
What are the possible forms of the prime factorization of N?

mkkool64 (20:59:40)
N=p^8 or p^2*q^2 or p^9 or p * q^4

rrusczyk (21:00:05)
We can have N = p^9, or N = pq^4, or N = p^8, or N = p^2q^2, where p and q are odd primes.

rrusczyk (21:00:10)
p^8 and p^9 are too big (since N < 1000, note 3^8 = 65641).

rrusczyk (21:00:23)
How about N=pq^4?

Bictor717 (21:00:52)
only q=3

mkkool64 (21:00:53)
81*5, 81*7, 81*11

rrusczyk (21:01:10)
If N=pq^4, then q must be 3 (any other q is too big), and then p = 5, 7, or 11. This gives 3 possibilities (namely N=405, 567, and 891)

rrusczyk (21:01:18)
N = (pq)^2?

jen7 (21:01:49)
225, 441

Vihang (21:01:59)
pq<31

dyaka (21:02:01)
15^2, 21^2

rrusczyk (21:02:07)
If N = (pq)^2, then pq < sqrt(1000) = 31.6..., so pq < 31. The only possibilities are pq = 15 or 21, so this gives 2 possibilities (namely N=225 and 441).

rrusczyk (21:02:13)
Hence there are 5 odd N.

rrusczyk (21:02:21)
Now we consider N = j(2k+j) even.

rrusczyk (21:02:25)
Since both j and 2k+j are even, N must be a multiple of 4.

rrusczyk (21:02:33)
So we are looking for N such that there are exactly 5 values of N with j and N/j even.

rrusczyk (21:03:45)
How do we deal with this?

Vihang (21:04:28)
divide j and N/j by 2 to reduce the problem

rrusczyk (21:05:44)
Alternatively, we can write N = 2^b * M, where M is odd.

rrusczyk (21:06:20)
(Though keep in mind that b is at least 2, and at least one factor of 2 goes with j and one with (2k+j) in N = j(2k+j). )

rrusczyk (21:06:50)
So, how many pairs of factors of N, in terms of b and factors of M, have both factors even?

rrusczyk (21:08:09)
The number of pairs of factors with both factors even is (b-1)*(# of factors of M)/2.

rrusczyk (21:08:33)
(We give one of the 2's to j and one to 2k+j)

rrusczyk (21:08:50)
Hence we need (b-1)*(# of factors of M) to be either 9 or 10 (just as in the "N odd" case).

rrusczyk (21:09:20)
So, we have a bunch of nasty casework to do

rrusczyk (21:09:26)
This means that we have the following cases, since we have 9 = 1*9 = 3*3 and 10 = 1*10 = 2*5.

rrusczyk (21:09:29)
Cases with 9 factors:
b=2, # of factors of M = 9
b=4, # of factors of M = 3
b=10, # of factors of M = 1

rrusczyk (21:09:32)
Cases with 10 factors:
b = 2, # of factors of M = 10
b = 3, # of factors of M = 5
b = 6, # of factors of M = 2
b = 11, # of factors of M = 1

rrusczyk (21:09:36)
The b=10 and b=11 cases are too big.

rrusczyk (21:09:42)
So we are left with N being one of the following forms:
2^2 * p^8
2^2 * p^2 * q^2
2^4 * p^2
2^2 * p^9
2^2 * p * q^4
2^3 * p^4
2^6 * p
(where p and q are distinct odd primes)

rrusczyk (21:09:54)
2^2 * p^8 is too big.

rrusczyk (21:10:07)
What about 2^2 * (pq)^2 ?

perfectnumber628 (21:10:28)
p=3, q=5

Bictor717 (21:10:32)
p=3, q=5

rrusczyk (21:10:39)
2^2 * (pq)^2 means pq < sqrt(250) = 15.8..., so we must have p = 3 and q = 5. This gives 1 possibility (namely N = 900).

rrusczyk (21:10:47)
How about 2^4 p^2?

jen7 (21:11:56)
p=3,5,7?

rrusczyk (21:12:06)
2^4 * p^2 means p^2 < 1000/16 = 62.5, so we must have p=3, 5, or 7. This gives 3 possibilities (namely N = 144, 400, 784).

rrusczyk (21:12:09)
2^2 * p^9 is too big.

rrusczyk (21:12:13)
2^2 * p * q^4 means q=3, but then it's too big.

rrusczyk (21:12:19)
2^3 * p^4 means p^4 < 1000/8 = 125, so p=3. This gives 1 possibility (namely N=648).

rrusczyk (21:12:24)
2^6 * p means p < 1000/2^6 = 15.625. This gives p =3,5,7,11,13. This gives 5 possibilities: 192, 320, 448, 704, 832.

rrusczyk (21:12:41)
(I'm just running through the cases in the interest of time - it's getting late!)

rrusczyk (21:12:44)
So there are 1+3+1+5 = 10 even N.

rrusczyk (21:12:51)
Combining this with the 5 odd N from before gives a final answer of 15

rrusczyk (21:12:59)
That was relatively un-fun, eh?

rrusczyk (21:13:24)
Let's hope for a slicker solution to #14

rrusczyk (21:13:26)


rrusczyk (21:13:39)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-37925326.gif

rrusczyk (21:13:50)
It seems hard to get a handle on this at first. What might we compute as a first step?

mkkool64 (21:13:59)
S_1

perfectnumber628 (21:14:00)
n=1

rrusczyk (21:14:25)
Most likely knowing the sum of the reciprocals from 1 to 9 will be useful at some point.

rrusczyk (21:14:42)
Note that the LCM of 1 through 9 is 2^3 * 3^2 * 5 * 7 = 8 *9 * 35 = 2520.

rrusczyk (21:14:47)


rrusczyk (21:14:55)
(Actually, the numerator's not important.)

rrusczyk (21:15:02)
When we sum reciprocals from 1 to 10^n, how many times is this sum counted?

calc rulz (21:15:39)
n*10^(n-1)

Phelpedo (21:16:06)
n*10(n-1)

rrusczyk (21:16:18)
Each digit appears 10^{n-1} times in each slot, so it gets counted 10^{n-1}* n times.

rrusczyk (21:16:35)
So, what does this make our problem?

perfectnumber628 (21:17:05)
n10^(n-1)/2520 is an integer

Bictor717 (21:17:11)
smallest n that makes n*10^(n-1) a multiple of 2520

Hokkage (21:17:13)
10^(n-1)*n is a multiple of 2520

rrusczyk (21:17:27)
Thus we need the minimal n such that 10^{n-1} * n is a multiple of 2520.

rrusczyk (21:17:41)
And what is that n (and why?)

mkkool64 (21:18:15)
63, because factors of 2 and 5 cancel and we need to cancel the 9 and 7

perfectnumber628 (21:18:34)
n=63 because it cancels with the 2520 to get 40

rrusczyk (21:18:50)
The power of 10 will take care of all the factors of 2 and 5 in 2520, so this leaves n = 3^2 * 7 = 63 as our answer.

rrusczyk (21:18:57)
Much nicer than #13, in my opinion.

rrusczyk (21:19:03)
But that leaves #15.

rrusczyk (21:19:08)


rrusczyk (21:19:12)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-246807102.gif

rrusczyk (21:19:46)
We might try bashing with algebra (have fun - it will work eventually), but instead, let's see if we can find a more clever way to approach the problem.

rrusczyk (21:20:01)
What might the square root expressions remind us of?

calc rulz (21:19:56)
Square roots of differences of squares, hmm I smell geometry (pythag theorem)...

perfectnumber628 (21:20:12)
pythagorean theorem?

mkkool64 (21:20:14)
pythagorean theorem

rrusczyk (21:20:30)
The square roots of y^2 - 1/16 and z^2 - 1/16 might remind us of Pythagoras.

rrusczyk (21:20:35)


Someone (21:21:04)
y = hypotenuse, 1/4 = length of one leg

jen7 (21:21:18)
rt. triangle of hypotenuse y and leg of 1/4

rrusczyk (21:21:29)
By building a right triangle with hypotenuse y and one leg 1/4.

rrusczyk (21:21:35)


rrusczyk (21:21:36)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/d2dc4fa1d0618f93928768aa41fb4d41.png

rrusczyk (21:21:45)
And what might we use that 1/4 to build next?

topaz (21:22:15)
build a triangle off of 1/4

perfectnumber628 (21:22:16)
make another one with hypoteneuse z

jen7 (21:22:13)
the triangle with hypotenuse z

rrusczyk (21:22:33)
To get the square root of z^2 - 1/16, we can build a right triangle on the other side.

rrusczyk (21:22:38)


rrusczyk (21:22:42)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/4d5cac7c2ea5a58876d9a7a3fc2f6217.png

rrusczyk (21:22:52)
Why does this make us smile?

calc rulz (21:22:01)
Z on the other side...so x y and z are sides of the triangle!

Bictor717 (21:22:49)
base is x, yay

topaz (21:22:58)
the sum of the bases is x

rrusczyk (21:23:16)


rrusczyk (21:23:25)
In other words, in the triangle with side lengths x, y, and z, the altitude to the side of length x is 1/4.

rrusczyk (21:23:28)
It is not hard to see that we can do the same with the other equations, to build the other altitudes.

rrusczyk (21:23:34)


rrusczyk (21:23:37)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/1f3b3d62a2fbfb25344916f782290617.png

rrusczyk (21:23:49)
Now we think we're barking at the right tree.

rrusczyk (21:23:52)
What now?

mkkool64 (21:24:10)
use multiple expressions for area

Hokkage (21:24:11)
Areas are the same?

Someone (21:24:18)
so by using area you can get all of them in terms of one variable, then resubstitute in one of the equations and solve

rrusczyk (21:24:33)


rrusczyk (21:24:40)
What does that say about the sides x, y, and z?

jen7 (21:24:32)
(1/6)z=(1/4)x=(1/5)y

mkkool64 (21:25:06)
ratio of 4:5:6

joezfang (21:25:12)
they are in the progression 4:5:6

rrusczyk (21:25:22)
It says that x, y, and z are in the ratio 4:5:6.

rrusczyk (21:25:25)
So how can we find x, y, and z?

Someone (21:26:01)
you have x, 5/4 x, and 3x/2 then resubstitute in the equations shown above and solve

Bictor717 (21:26:19)
substitute y and z in first equation in terms of x

rrusczyk (21:26:42)
We definitely can bash from here. If we want to keep going, what can we do with geometry?

mkkool64 (21:25:40)
maybe heron's formula

calc rulz (21:26:52)
heron's formula

rrusczyk (21:27:15)
Probably the easiest way is to find the altitudes of a 4-5-6 triangle, and then scale appropriately. We can find the area using Heron.

rrusczyk (21:27:47)


rrusczyk (21:28:21)
By what do we have to scale our 4-5-6 triangle to get an altitude of 1/4?

jen7 (21:28:53)
2/(15sqrt7)

Someone (21:29:10)
2 rt. 7/105

Hokkage (21:29:17)
2/(15sqrt7)

rrusczyk (21:29:31)


rrusczyk (21:29:38)


rrusczyk (21:29:40)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/3e9c695bfef6dc1bd1750b84bc8675cd.png

rrusczyk (21:30:02)
So, what's our x+y+z?

mkkool64 (21:30:18)
2/sqr(7)

noneoftheabove (21:30:22)
2/sqrt(7)

Hokkage (21:30:25)
2/sqrt7

perfectnumber628 (21:30:35)
x+y+z=30/(15sqrt7)=2/sqrt7

rrusczyk (21:30:43)


rrusczyk (21:30:51)


rrusczyk (21:31:15)
Thank you all for coming to the Math Jam. Good luck to those of you who qualified for the USAMO.