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2010 AIME I Math Jam

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AoPS instructors will lead a discussion on all 15 problems from the 2010 AIME I.

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Facilitator: David Patrick

DPatrick 2010-03-18 19:00:36
Hello and welcome to the 2010 AIME I Math Jam!
DPatrick 2010-03-18 19:00:47
My name is Dave Patrick, and I'll be leading tonight's discussion.
DPatrick 2010-03-18 19:00:54
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick 2010-03-18 19:01:05
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick 2010-03-18 19:01:17
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read. Only write comments that are relevant to the discussion.
DPatrick 2010-03-18 19:01:37
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!
DPatrick 2010-03-18 19:02:07
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. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick 2010-03-18 19:02:37
We do have an assistant tonight who can help answer some of your questions. She is Wendy Hou (wobster109). Wendy is a college sophomore majoring in computer programming. She has taken the USAMO four times and competed in a broad spectrum of high school competitions, and she had the honor of representing the U.S. in China as a member of the first U.S.A. CGMO team. She has extensive experience from summer programs and with Art of Problem Solving, both as a student and as a mentor. She spends her days frantically battling down a monstrous courseload. She haunts a bell tower. At night, she transforms into a velociraptress and howls at the moon.
DPatrick 2010-03-18 19:02:53
(so I am told)
wobster109 2010-03-18 19:03:02
Hi everyone!
DPatrick 2010-03-18 19:03:07
Wendy can answer questions by whispering to you or by opening a window with you to chat 1-on-1.
DPatrick 2010-03-18 19:03:34
Please also remember that the purpose of this Math Jam is to work through the solutions to the AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. 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.
DPatrick 2010-03-18 19:04:20
I don't claim that the solutions that we'll present today are the "definitive" or "best" solutions to each problem. Many of the problems have more than one plausible approach. We don't have time to consider every possible approach to every problem.
DPatrick 2010-03-18 19:04:35
Also, these solutions are not necessarily the fastest or slickest, but these solution are the ones that (to me) are the most logical or the ones that you'd be most likely to think of on your own.
DPatrick 2010-03-18 19:05:07
Our agenda is to work through all 15 problems on the 2010 AIME I, in order.
DPatrick 2010-03-18 19:05:27
So let's get started!
DPatrick 2010-03-18 19:05:38
DPatrick 2010-03-18 19:05:48
(Notice I will also put the current problem at the top of the window, for us to refer to as we discuss it. You can resize the region at the top by dragging the horizontal gray bar.)
likemath 2010-03-18 19:06:10
First prime factorize 2010
prezcoin 2010-03-18 19:06:10
2010= 2*3*5*67
maraks 2010-03-18 19:06:10
prime factorize 2010 to get 2*3*5*67
MathTwo 2010-03-18 19:06:10
write prime factorization of 2010^2 first
DPatrick 2010-03-18 19:06:18
To work with divisors, we probably need the prime factorization of 2010.
DPatrick 2010-03-18 19:06:28
This is 2010 = 2 * 3 * 5 * 67.
DPatrick 2010-03-18 19:06:48
superbob 2010-03-18 19:07:02
First, you may want to find the number of factors 2010^2 has
theone142857 2010-03-18 19:07:03
Find how many divisors
DPatrick 2010-03-18 19:07:13
So now we can count the total number of divisors of 2010^2.
karatemagic7 2010-03-18 19:07:38
The number of factors of 2010 is (2+1)^4=81.
natiator 2010-03-18 19:07:38
(2+1)^4
GeorgiaTechMan 2010-03-18 19:07:38
The number of factors is (2+1)(2+1)(2+1)(2+1)=81
gh625 2010-03-18 19:07:38
(2+1)(2+1)(2+1)(2+1)=81
mathwiz314 2010-03-18 19:07:38
(2+1)*(2+1)*(2+1)*(2+1)
DPatrick 2010-03-18 19:07:43
DPatrick 2010-03-18 19:07:55
So there are 3 choices for each prime power, and thus there are 3*3*3*3 = 81 divisors.
DPatrick 2010-03-18 19:08:04
In how many ways can we randomly choose two distinct divisors, without regard to order?
Ihatepie 2010-03-18 19:08:30
thus, there are 81C2 ways=3240
victorzhou 2010-03-18 19:08:30
81C2 = 81*40
1=2 2010-03-18 19:08:30
karatemagic7 2010-03-18 19:08:30
81*80/2=3240
DPatrick 2010-03-18 19:08:37
DPatrick 2010-03-18 19:08:48
(I won't multiply it out because we'll likely cancel some factors later.)
DPatrick 2010-03-18 19:09:00
How do we pick two such that exactly one of them is a perfect square?
victorzhou 2010-03-18 19:09:31
pick 1 square and 1 non-square.
Goldey 2010-03-18 19:09:32
Number of perfect squares times number of not perfect squares
DPatrick 2010-03-18 19:09:37
Right, we pick one perfect square and one non-square.
DPatrick 2010-03-18 19:09:46
How many of the 81 divisors are perfect squares?
Ihatepie 2010-03-18 19:10:17
well for each factor, you can choose two or zero of it, so there are 2^4=16 perfect squares
Goldey 2010-03-18 19:10:17
likemath 2010-03-18 19:10:17
Each power must be a 0 or a 2 so 2^4=16 divisors
karatemagic7 2010-03-18 19:10:17
16 because there are 16 factors of 2010.
v_Enhance 2010-03-18 19:10:17
For a number to be a perfect square, each of a,b,c,d must be even; 0 or 2. Thus we have 2^4 = 16
DPatrick 2010-03-18 19:10:33
Right. One way is that each prime must occur an even number of times. So each exponent must be 0 or 2. Thus, there are 2*2*2*2 = 16 perfect square divisors.
DPatrick 2010-03-18 19:11:03
Or, we could note that a perfect square divisor of 2010^2 is the square of a divisor of 2010, so again there are 16.
DPatrick 2010-03-18 19:11:12
Hence, there are 81 - 16 = 65 non-square divisors.
prezcoin 2010-03-18 19:11:23
there are 16*(81-16)= 16*65 ways to choose
GeorgiaTechMan 2010-03-18 19:11:27
16*65 possible ways
DPatrick 2010-03-18 19:11:43
Yes, there are 16 * 65 ways to pick a square and a non-square.
likemath 2010-03-18 19:11:55
mathwiz314 2010-03-18 19:11:55
16*65/(81*40)
karatemagic7 2010-03-18 19:11:55
16*65/(81*40)=26/81=m/n
DPatrick 2010-03-18 19:12:01
DPatrick 2010-03-18 19:12:11
Not multiplying this out will make simplification easier.
DPatrick 2010-03-18 19:12:26
DPatrick 2010-03-18 19:12:55
Ihatepie 2010-03-18 19:13:08
easiest one on test
DPatrick 2010-03-18 19:13:15
I agree -- this was a lot easier than #1.
-$45! 2010-03-18 19:13:28
put it in mod 1000!
dinoboy 2010-03-18 19:13:28
Everything except 9,99 is just -1 mod 1000
SuitcaseAsian 2010-03-18 19:13:28
999, 9999 ,.... = -1 mod 1000
gh625 2010-03-18 19:13:28
Everything after 999 is congruent to -1.
Goldey 2010-03-18 19:13:28
All of the numbers 999 and after are -1 mod 1000
DPatrick 2010-03-18 19:13:40
Exactly. The key to this problem is that we can do everything mod 1000 (meaning we just keep track of remainders when we divide by 1000).
DPatrick 2010-03-18 19:13:54
DPatrick 2010-03-18 19:14:09
nikeballa96 2010-03-18 19:14:41
891(-1)=-891(mod 1000)
prezcoin 2010-03-18 19:14:41
this is -891 mod 1000, which is 109 mod 1000
karatemagic7 2010-03-18 19:14:41
=-891=109 (mod 1000)
DPatrick 2010-03-18 19:14:50
DPatrick 2010-03-18 19:15:28
I don't really have anything else to say about this one -- if you know modular arithmetic, it was really easy.
DPatrick 2010-03-18 19:15:37
DPatrick 2010-03-18 19:16:16
Many people are suggesting substituting right way, but I slightly prefer to get rid of the exponents first.
nikeballa96 2010-03-18 19:16:36
logarithms.
v_Enhance 2010-03-18 19:16:37
Logarithms on second equation.
nikeballa96 2010-03-18 19:16:37
y log x=x log y
-$45! 2010-03-18 19:16:37
use logs
DPatrick 2010-03-18 19:16:43
Right, we can take logs!
DPatrick 2010-03-18 19:16:56
DPatrick 2010-03-18 19:17:05
(It doesn't matter what base we choose, so let's not worry about the base until we have to, then we can choose it conveniently if necessary.)
2012 2010-03-18 19:17:31
Logs make life so much easier!
nikeballa96 2010-03-18 19:17:31
substitute now and we'll find that x's cancel.
DPatrick 2010-03-18 19:17:45
Yes, now it's not so intimidating to plug in y = 3/4 x and see what happens:
DPatrick 2010-03-18 19:17:52
prezcoin 2010-03-18 19:18:08
divide x
DPatrick 2010-03-18 19:18:09
Happily, x cancels (since it has to be nonzero). Thus we have:
DPatrick 2010-03-18 19:18:14
1=2 2010-03-18 19:18:47
Separate the second log, and then let's make the base 3/4.
DPatrick 2010-03-18 19:19:04
Well, let's not pick a base yet, but I definitely like using log(ab) = log a + log b on the right side:
DPatrick 2010-03-18 19:19:09
DPatrick 2010-03-18 19:19:25
This lets us solve for log x!
panjia123 2010-03-18 19:19:53
log x=-4log(3/4)
prezcoin 2010-03-18 19:19:53
(-1/4)log(x)=log(3/4)
v_Enhance 2010-03-18 19:19:53
1/4 log x = - log 3/4
dinoboy 2010-03-18 19:19:53
log x = -4log(3/4)
DPatrick 2010-03-18 19:20:00
dinoboy 2010-03-18 19:20:15
Now remove the logs!
DPatrick 2010-03-18 19:20:29
-$45! 2010-03-18 19:20:43
now find y
mathelete 2010-03-18 19:20:50
But we want x+y
DPatrick 2010-03-18 19:20:55
Aha -- don't fall into the trap that the answer is 256 + 81 = 337. That's not what was asked for!
DPatrick 2010-03-18 19:21:10
yankeesrule007 2010-03-18 19:21:19
256/81(7/4) gives you the answer
DPatrick 2010-03-18 19:21:24
DPatrick 2010-03-18 19:21:42
DPatrick 2010-03-18 19:22:03
As I mentioned at the top, there are often lots of ways to solve a problem, and certainly here there were lots of ways to do the algebra.
DPatrick 2010-03-18 19:22:29
But I usually like making the equation as pretty as possible to start with, and to do that I made taking logs my first step, to get rid of ugly exponents early.
DPatrick 2010-03-18 19:22:54
mgao 2010-03-18 19:23:16
casework: 0 heads, 1 head, 2 heads, 3 heads
theone142857 2010-03-18 19:23:16
Both of them get 0H, 1H, 2H, or 3H
prezcoin 2010-03-18 19:23:16
cases by number of heads?
DPatrick 2010-03-18 19:23:26
When I read it, I was hoping there was a clever way to solve this, but it's essentially brute-force casework computation.
1=2 2010-03-18 19:23:47
Let's find the probability that one of them gets 0/1/2/3 heads.
DPatrick 2010-03-18 19:23:56
First, let's just tabulate the probabilities of any person flipping 0, 1, 2, or 3 heads.
DPatrick 2010-03-18 19:24:04
0 heads is easy:
dinoboy 2010-03-18 19:24:21
0 is simple: 1/2 * 1/2 * 3/7 = 3/28
GeorgiaTechMan 2010-03-18 19:24:21
1/2*1/2*3/7=3/28 for 0
v_Enhance 2010-03-18 19:24:21
0 heads is easy: 1/2*1/2*3/7 = 3/28
AsimoIsFuture 2010-03-18 19:24:21
0 = (1/2)(1/2)(3/7)
DPatrick 2010-03-18 19:24:33
DPatrick 2010-03-18 19:24:40
3 heads is also easy:
dinoboy 2010-03-18 19:24:57
3 is simple as well: 1/2 * 1/2 * 4/7 = 1/7
GeorgiaTechMan 2010-03-18 19:24:57
3 is also easy1/2*1/2*4/7=1/7 for 3 heads
v_Enhance 2010-03-18 19:24:57
1/2*1/2*4/7 = 1/7
jxl28 2010-03-18 19:24:57
1/2*1/2*4/7=1/7
DPatrick 2010-03-18 19:25:06
DPatrick 2010-03-18 19:25:13
(Tip: leave them all over 28 since we'll have to do a little more arithmetic later and it's probably convenient to have everything over a common denominator.)
DPatrick 2010-03-18 19:25:41
Now for the slightly messier cases. How about 1 head?
1=2 2010-03-18 19:26:23
1 head: (1/2)(1/2)(3/7)+(1/2)(1/2)(3/7)+(1/2)(1/2)(4/7)
yankeesrule007 2010-03-18 19:26:23
for one you have three cases, HTT, THT and TTH which together add to 2(1/2)^2(3/7)+1/2^2(4/7)=5/14=10/28 (to get common denominators)
megahertz 2010-03-18 19:26:23
1: 1/2 * 1/2 * 3/7 + 1/2 * 1/2 * 3/7 + 1/2 * 1/2 * 4/7 = 10/28
jxl28 2010-03-18 19:26:23
(1/2)(1/2)(3/7)(2)+(1/2)(1/2)(4/7)=10/28
DPatrick 2010-03-18 19:26:28
I think these are correct.
DPatrick 2010-03-18 19:26:34
Either we get 1 head from the fair coins and a tail from the unfair coin (with probability 1/2 * 3/7)... ...or we get 0 heads from the fair coins and a head from the unfair coin (with probability 1/4 * 4/7).
DPatrick 2010-03-18 19:26:52
DPatrick 2010-03-18 19:27:10
How about 2 heads?
GeorgiaTechMan 2010-03-18 19:27:15
For 2 is is 28-4-3-10/28=11/28
DPatrick 2010-03-18 19:27:37
Right, that's a bit of a time-saver. Since they have to sum to 1, we know it must be 11/28. But let's check nonetheless.
theone142857 2010-03-18 19:27:58
(1/2)(1/2)(4/7)2+(1/2)(1/2)(3/7)
v_Enhance 2010-03-18 19:27:58
2 heads is similarly 1/2*4/7 + 1/4*3/7 = 11/28
karatemagic7 2010-03-18 19:27:58
1/2*4/7+1/4*3/7=11/28
mathwiz314 2010-03-18 19:27:58
1/2*1/2*3/7+2*1/2*1/2*4/7
DPatrick 2010-03-18 19:28:06
I think these are all correct too.
DPatrick 2010-03-18 19:28:11
Either we get 1 head from the fair coins and a head from the unfair coin (with probability 1/2 * 4/7)... ...or we get 2 heads from the fair coins and a tail from the unfair coin (with probability 1/4 * 3/7).
DPatrick 2010-03-18 19:28:30
DPatrick 2010-03-18 19:28:38
DPatrick 2010-03-18 19:28:47
(And again, as a sanity check, they sum to 1.)
DPatrick 2010-03-18 19:28:55
Using this table, what's the probability that they both get the same number of heads?
prezcoin 2010-03-18 19:29:06
square these values and add
karatemagic7 2010-03-18 19:29:06
Now, let's square the possibilities and add them together.
Goldey 2010-03-18 19:29:06
Square them all and add the squares for having the same
KingMax 2010-03-18 19:29:06
Square each probability and add them up
megahertz 2010-03-18 19:29:06
Probability of them getting the same amount is each case squared
EHSTutor 2010-03-18 19:29:06
square valuves and add
rd5493 2010-03-18 19:29:06
So to find the probability that the get the same, shouldn't we square each case and then find the sum of the squares.
DPatrick 2010-03-18 19:29:18
DPatrick 2010-03-18 19:29:38
DPatrick 2010-03-18 19:29:48
Leaving them all over 28 was helpful!
-$45! 2010-03-18 19:29:51
so (9+100+121+16)/28^2
DPatrick 2010-03-18 19:30:08
DPatrick 2010-03-18 19:30:24
DPatrick 2010-03-18 19:30:53
Frankly, I thought this problem was kinda boring.
DPatrick 2010-03-18 19:31:11
DPatrick 2010-03-18 19:31:29
What do we notice here?
mathwiz314 2010-03-18 19:31:43
difference of squares on the second equation
QuincyCello 2010-03-18 19:31:43
diff. of two squares on the second equation
BarbieRocks 2010-03-18 19:31:43
flamingmath 2010-03-18 19:31:43
we can use difference of squares
professordad 2010-03-18 19:31:51
you could start out by factoring so (a-b)(a+b) + (c+d)(c-d) = 2010
GeorgiaTechMan 2010-03-18 19:31:51
a^2-b^2+c^2-d^2 can be written as (a+b)(a-b)+(c+d)(c-d)
DPatrick 2010-03-18 19:32:00
Right, we see differences of squares in the quadratic expression, so we can factor it:
DPatrick 2010-03-18 19:32:07
DPatrick 2010-03-18 19:32:25
How does this help?
prezcoin 2010-03-18 19:33:00
since a-b and c-d are positive, both values equal 1
KingMax 2010-03-18 19:33:00
a-b and c-d are both 1
nikeballa96 2010-03-18 19:33:00
hence, a-b and c-d must equal 1.
Jadesymdragon 2010-03-18 19:33:00
a-b=1 and c-d = 1
DPatrick 2010-03-18 19:33:21
This quadratic equals 2010. But a+b+c+d also equals 2010.
DPatrick 2010-03-18 19:33:32
DPatrick 2010-03-18 19:33:48
Since everything here is a positive integer, and moreover a-b > 0 and c-d > 0, we must have a-b = 1 and c-d = 1.
GeorgiaTechMan 2010-03-18 19:34:14
so b can be a-1 and d can be c-1
mathelete 2010-03-18 19:34:14
nikeballa96 2010-03-18 19:34:14
hence, we can express b in terms of and d in terms of c
DPatrick 2010-03-18 19:34:22
So we can let b = a-1 and d = c-1 and simplify things quite a bit.
DPatrick 2010-03-18 19:34:36
theone142857 2010-03-18 19:34:40
a+c=1006
rd5493 2010-03-18 19:34:40
a+c=1006
DPatrick 2010-03-18 19:34:49
Thus a + c = 1006.
DPatrick 2010-03-18 19:34:54
How do we count the possible a's?
-$45! 2010-03-18 19:35:08
but a>c
flamingmath 2010-03-18 19:35:08
a>c
prezcoin 2010-03-18 19:35:08
a must be greater than c
DPatrick 2010-03-18 19:35:19
In fact, a > a-1 > c > c-1 > 0, so we must have a > c+1, where c > 1.
yankeesrule007 2010-03-18 19:35:35
then your extreme cases are (504,502) and (1004,2) since a can't equal c and c can't equal 1 since then d would be 0
karatemagic7 2010-03-18 19:35:35
So we just need to count the number of positive integers a and c such that a-1>c >1and a+c=1006. a goes from 504 to 1004 so there are 501 possibilities.
SuitcaseAsian 2010-03-18 19:35:35
504<a<1004. C>2
DPatrick 2010-03-18 19:35:43
Right. The pair (a,c) can be anything from (1004,2), (1003,3), (1002,4),... down to (504,502).
DPatrick 2010-03-18 19:36:03
(I like listing them out like this just to be safe.)
DPatrick 2010-03-18 19:36:08
So a is anything in the set {504,505,...,1004}.
yankeesrule007 2010-03-18 19:36:15
then your answer is 1004-504+1=501
flamingmath 2010-03-18 19:36:15
1004-504+1
DPatrick 2010-03-18 19:36:24
DPatrick 2010-03-18 19:36:55
DPatrick 2010-03-18 19:37:06
This problem was my favorite of those on the first page.
DPatrick 2010-03-18 19:37:22
Some of you might see right away what the best thing to do is.
DPatrick 2010-03-18 19:37:37
I'll come back to that...for now, let's pretend you don't see it. (I didn't when I first solved it!)
DPatrick 2010-03-18 19:38:13
What's a more naive thing to do, or let me ask it a different way: what's often a good tactic with inequalities?
sparkle123 2010-03-18 19:38:33
move terms to one side
Goldey 2010-03-18 19:38:33
Make the smaller end 0?
DPatrick 2010-03-18 19:38:49
Right. What I did first is subtract the smallest term so that it had a 0 on one end.
DPatrick 2010-03-18 19:39:08
Again, this is not the most clever thing you can do, but it is (to me) the most natural way to attack the problem if you don't notice the clever method.
DPatrick 2010-03-18 19:39:21
(I'll mention the more clever way at the end; many of you have pointed it out.)
DPatrick 2010-03-18 19:39:34
DPatrick 2010-03-18 19:39:52
jxl28 2010-03-18 19:40:01
Perfect square!
prezcoin 2010-03-18 19:40:01
the right side is (x-1)^2
yankeesrule007 2010-03-18 19:40:01
factor that as (x-1)^2
theone142857 2010-03-18 19:40:01
(x-1)^2
fortenforge 2010-03-18 19:40:01
hey the thing on the right can be factored!
DPatrick 2010-03-18 19:40:18
Right! Now we ought to notice that the right term is just (x-1)^2.
DPatrick 2010-03-18 19:40:26
DPatrick 2010-03-18 19:40:28
What does that tell us about Q(x)?
BarbieRocks 2010-03-18 19:41:02
It's in the form k(x-1)^2?
abcak 2010-03-18 19:41:03
it must be a(x-1)^2 where 0<a<1?
sparkle123 2010-03-18 19:41:03
it's "flatter" than y=(x-1)^2
ytrewq 2010-03-18 19:41:03
It must have a vertex at x=1
DPatrick 2010-03-18 19:41:13
DPatrick 2010-03-18 19:41:26
There are a couple of ways we can show this. One is to just sketch a picture.
DPatrick 2010-03-18 19:41:31
DPatrick 2010-03-18 19:41:46
It is clear from the picture that a parabola (y = Q(x)) that lies above the red line (0) but below the blue parabola ((x-1)^2) must be a parabola that has its vertex at (1,0).
DPatrick 2010-03-18 19:42:13
Algebraically, we can see this by noting that Q(1) = 0 and that Q(x) >= 0 means that the parabola has a double root at x=1.
DPatrick 2010-03-18 19:42:25
abcak 2010-03-18 19:42:37
then plug in x=11 and solve for c
DPatrick 2010-03-18 19:42:49
Right, we plug in x=11 and use P(11) = 181 to solve for c:
DPatrick 2010-03-18 19:43:01
professordad 2010-03-18 19:43:16
so c = 4/5
karatemagic7 2010-03-18 19:43:16
c=80/100=4/5
rd5493 2010-03-18 19:43:16
c=.8
DPatrick 2010-03-18 19:43:21
Thus c = 0.8.
v_Enhance 2010-03-18 19:43:41
Now just plug in x=16 to find the answer!
DPatrick 2010-03-18 19:43:46
DPatrick 2010-03-18 19:44:10
There is a simpler solution that starts with a slightly different first step and makes the algebra a little quicker.
sparkle123 2010-03-18 19:44:24
complete square?
abcak 2010-03-18 19:44:24
COMPLETE THE SQUARE!
lightning 2010-03-18 19:44:24
complete the square on both sides of the equation?
DPatrick 2010-03-18 19:44:36
Right: we can start by completing the squares of the original quadratics.
DPatrick 2010-03-18 19:44:44
DPatrick 2010-03-18 19:45:15
So now we've got P(x) squeezed between two parabolas with vertices at (1,1), and thus by essentially the same reasoning, P(x) must be a parabola with vertex at (1,1) too.
mathwiz314 2010-03-18 19:45:29
P(x)=a(x-1)^2+1
abcak 2010-03-18 19:45:35
then you can say P(x)=a(x-1)^2+1 where 1<=a<=2
DPatrick 2010-03-18 19:45:43
DPatrick 2010-03-18 19:45:58
jxl28 2010-03-18 19:46:13
Then substitute x=11 and P(x)=181
megahertz 2010-03-18 19:46:14
then plug in 11 and solve for k?
y2ucc 2010-03-18 19:46:14
plug in x = 11
1=2 2010-03-18 19:46:19
k would be 1.8 here
DPatrick 2010-03-18 19:46:24
DPatrick 2010-03-18 19:46:50
It's the same idea as our original solution, but completing the square first allowed us to skip an algebraic step in the middle.
DPatrick 2010-03-18 19:47:18
mathwiz314 2010-03-18 19:47:54
make a venn diagram, as a visual aid
DPatrick 2010-03-18 19:48:03
I agree: because we have an intersection of sets, I naturally thought of drawing a Venn diagram:
DPatrick 2010-03-18 19:48:11
sparkle123 2010-03-18 19:48:28
nothing in the middle
Wesmaster160 2010-03-18 19:48:28
the center is empty
y2ucc 2010-03-18 19:48:35
the complete intersection is blank and the individual intersections are 1
DPatrick 2010-03-18 19:48:41
We know the sizes of some of these regions...there are no elements in the center and exactly 1 element in the double-intersection regions:
DPatrick 2010-03-18 19:48:46
DPatrick 2010-03-18 19:49:05
(Those numbers are sizes of the regions, not actual numbers in those sets.)
Jason27603 2010-03-18 19:49:16
7*6*5=210 ways to pick one number to be in A and B, one to be in B and C, and one to be in A and C
mz94 2010-03-18 19:49:16
there are 7*6*5 ways to choose those 3 1s in the intersections
DPatrick 2010-03-18 19:49:49
Exactly: we can choose numbers to put in those regions in 7 * 6 * 5 ways (7 choices for the number in A&B, then 6 choices remaining for the number in B&C, then 5 choices remaining for A&C).
fortenforge 2010-03-18 19:49:56
4 elements left to place
PhireKaLk6781 2010-03-18 19:50:03
there are 4 numbers left to put into one of the 4 remaining regions, so 4^4
1=2 2010-03-18 19:50:03
Then there are 4 regions to put the other four numbers, so we multiply that by 4^4
mz94 2010-03-18 19:50:03
and each other number can go in set A, set B, set C, or none. so we multioply by 4^4
mathwiz314 2010-03-18 19:50:10
you have 4 numbers left, and 4 places to put them (A, B, C, or none), so 4^4
DPatrick 2010-03-18 19:50:25
Each of the remaining 4 numbers has 4 choices: in A, in B, in C, or in none.
DPatrick 2010-03-18 19:50:40
So that's 4^4 = 256 more choices, after the choices we've already made.
DPatrick 2010-03-18 19:50:55
Thus there are 210 * 256 = 53760 triples of sets.
fortenforge 2010-03-18 19:51:07
4^4 * 7 * 6 * 4 = 53760 = 760 (mod 1000)
DPatrick 2010-03-18 19:51:13
DPatrick 2010-03-18 19:52:01
This was a classic example of "constructive counting": when we figure out how to construct the objects we want, and keep track of the number of choices we make along the way.
rd5493 2010-03-18 19:52:07
seems way easier than it looks
DPatrick 2010-03-18 19:52:24
Lots of AIME problems are like that: they look scary but are not really are difficult as they first appear to be.
DPatrick 2010-03-18 19:52:56
I'm going to take a slight break to give my typing fingers a slight rest, but my colleague Naoki Sato will step in and do the next two problems. I'll see you in a few minutes for #10. :)
nsato 2010-03-18 19:53:08
nsato 2010-03-18 19:53:28
This is a wordy problem. Where can we begin?
Jason27603 2010-03-18 19:54:11
graph it
-$45! 2010-03-18 19:54:11
Draw a diagram
alligator112 2010-03-18 19:54:11
graph it
flamingmath 2010-03-18 19:54:11
we could start by making a graph
1=2 2010-03-18 19:54:11
Let's try graphing R
richardpan 2010-03-18 19:54:11
graphing R
nsato 2010-03-18 19:54:16
Obviously, we want to get a sense of what R looks like. But how can we do that?
likemath 2010-03-18 19:55:09
find the ordered pairs of integers that satisfy x^2+y^2=25
alligator112 2010-03-18 19:55:09
first find integral points of x^2 + y^2 = 25
Goldey 2010-03-18 19:55:10
[x] and [y] are always integers, so we should look at the critical points on the circle x^2+y^2=25 that have integer coordinates ((3,4) (4,3) (5,0) etc.)
Wesmaster160 2010-03-18 19:55:20
find integer squares that add up to 25
nsato 2010-03-18 19:55:24
First, we want to find the solutions of the equation m^2 + n^2 = 25, where m is the floor of x, and n is the floor of y.
nsato 2010-03-18 19:55:41
How can we express 25 as the sum of two perfect squares?
BarbieRocks 2010-03-18 19:56:27
3^2+4^2, 5^2+0^2
-$45! 2010-03-18 19:56:27
3^2+4^2 or 5^2 + 0^2
to94as 2010-03-18 19:56:27
3^2+4^2 or 0^2+5^2
fortenforge 2010-03-18 19:56:27
3^2 + 4^2 = 25 and 0^2 + 5^2 = 25
yankeesrule007 2010-03-18 19:56:27
0+25 or 9+16
nsato 2010-03-18 19:56:38
25 = 0^2 + 5^2 or 25 = 3^2 + 4^2.
nsato 2010-03-18 19:56:47
nsato 2010-03-18 19:57:03
We want to count the negative solutions, and the permutations.
nsato 2010-03-18 19:57:24
(Note that I've used set notation above, not ordered pairs, so we have all of them.)
nsato 2010-03-18 19:57:48
This gives 12 solutions for [x] and [y].
nsato 2010-03-18 19:58:02
How does each solution give a part of the region R?
BarbieRocks 2010-03-18 19:59:02
The region will be the unit square with the integer points above in the bottom left corner
mgao 2010-03-18 19:59:29
each solution serves as a lower left corner for a unit square containing soutions (x,y)
dinoboy 2010-03-18 19:59:29
You "color in" the square which has the point as its bottom left corner
nsato 2010-03-18 19:59:36
Each solution is the lower-left corner of a unit square in R.
stinstin 2010-03-18 19:59:42
R will consist of 12 unit squares.
nsato 2010-03-18 20:00:01
For example, if [x] = 3 and [y] = 4, then the portion of R that corresponds to this solution is the unit square whose lower-left corner is (3,4).
nsato 2010-03-18 20:00:09
Now we can confidently graph R:
-$45! 2010-03-18 20:00:12
We can now graph R!
flamingmath 2010-03-18 20:00:15
now we can make a graph
nsato 2010-03-18 20:00:19
nsato 2010-03-18 20:00:22
(By the way, this problem is a strong endorsement of having graph paper with you when you take the AIME! You are permitted graph paper, a rule, a compass, and a protractor. Make sure you have them all!)
nsato 2010-03-18 20:00:39
rule -> ruler
nsato 2010-03-18 20:00:45
What do you notice about this graph?
gh625 2010-03-18 20:01:28
symmetric around (1/2, 1/2)
pii 2010-03-18 20:01:28
centered around (1/2,1/2)
1=2 2010-03-18 20:01:28
It's "centered" at (1/2,1/2)
yankeesrule007 2010-03-18 20:01:28
its symmetrical about (1/2,1/2) NOT the origin
ftong 2010-03-18 20:01:28
it is symmetric about (1/2, 1/2)
nsato 2010-03-18 20:01:37
We notice that this picture is symmetric about the point (0.5,0.5).
nsato 2010-03-18 20:01:54
Hence, the smallest disc containing R must also have this center.
nsato 2010-03-18 20:02:10
How do we determine the radius of this disc?
1=2 2010-03-18 20:03:33
We need to find the point that's the farthest away from our designated centre.
prezcoin 2010-03-18 20:03:34
find the point that is the furthest from (1/2, 1/2)
flamingmath 2010-03-18 20:03:34
we find the distance between (1/2,1/2) and the farthest part of R
remy1140 2010-03-18 20:03:34
find the distance from (1/2,1/2) to the farthest point in one of the squares
Just_Beginner 2010-03-18 20:03:34
find the point furthest away from (1/2,1/2)
nsato 2010-03-18 20:04:19
We find the point (or points) in R that are furthest from (0.5, 0.5). It's not hard to see that this is going to be a vertex of one of the squares.
nsato 2010-03-18 20:05:02
That means we can check the vertices. One such vertex is (4,5).
nsato 2010-03-18 20:05:24
So what is the minimum radius?
1=2 2010-03-18 20:07:34
sqrt(130)/2
TheCopleyIndian 2010-03-18 20:07:34
sqrt130/2
GeorgiaTechMan 2010-03-18 20:07:34
root 130/2
panjia123 2010-03-18 20:07:34
sqrt(32.5)=sqrt(130)/2
theone142857 2010-03-18 20:07:34
root130/2
jxl28 2010-03-18 20:07:34
sqrt{130}/2
gh625 2010-03-18 20:07:34
sqrt{130}/2
nsato 2010-03-18 20:07:35
The minimum radius is the square root of (4 - 1/2)^2 + (5 - 1/2)^2 = 49/4 + 81/4 = 130/4.
nsato 2010-03-18 20:07:49
This simplifies as sqrt(130/4) = sqrt(130)/2.
mathelete 2010-03-18 20:08:27
AwesomeToad 2010-03-18 20:08:27
flamingmath 2010-03-18 20:08:27
130+2=132 which is our answer
victorzhou 2010-03-18 20:08:27
so the answer is 132?
birdofsorrow 2010-03-18 20:08:31
answer is 132
abcak 2010-03-18 20:08:32
so our answer is 132
nsato 2010-03-18 20:08:34
nsato 2010-03-18 20:08:48
Once you draw the right diagram, the problem becomes quite easy.
nsato 2010-03-18 20:08:54
mgao 2010-03-18 20:09:31
Look at that symmetry. Can we exploit it somehow?
nsato 2010-03-18 20:09:34
How can we take advantage of some of the similarities of these equations?
-$45! 2010-03-18 20:10:00
add the equations?
megahertz 2010-03-18 20:10:00
Add the 3 equations?
birdofsorrow 2010-03-18 20:10:00
add the three equations together
professordad 2010-03-18 20:10:00
maybe you could add them all up
mathwiz314 2010-03-18 20:10:08
add the three equations together to get x^3+y^3+z^3=28+3xyz
birdofsorrow 2010-03-18 20:10:08
x^3+y^3+z^3-3xyz=28 ?
nsato 2010-03-18 20:10:12
nsato 2010-03-18 20:10:19
nsato 2010-03-18 20:10:57
What else can we do with the equations, to determine xyz?
mgao 2010-03-18 20:11:45
x^3=2+xyz, etc
nsato 2010-03-18 20:11:57
We can isolate x^3, y^3, and z^3:
nsato 2010-03-18 20:12:00
nsato 2010-03-18 20:12:07
What can we do with these equations?
abcak 2010-03-18 20:12:36
multiply!
Mij 2010-03-18 20:12:36
multiply them together
mathcire 2010-03-18 20:12:36
Multiply these equations.
sparkle123 2010-03-18 20:12:36
multiply together
GoldenFrog1618 2010-03-18 20:12:36
multiply
nsato 2010-03-18 20:12:41
gh625 2010-03-18 20:13:03
multiply and let u=xyz
anaverageaopser 2010-03-18 20:13:03
substitute xyz as one variable?
nsato 2010-03-18 20:13:08
Since we can express everything in terms of xyz, it makes sense to make a substitution.
1=2 2010-03-18 20:13:11
Let xyz=t. We now have a cubic in t, which can reduce to a quadratic easily.
nsato 2010-03-18 20:13:15
mgao 2010-03-18 20:13:28
then solve the quadratic in xyz
abcak 2010-03-18 20:13:28
expand, (xyz)^3 cancels
Mij 2010-03-18 20:13:37
expand and p^3 cancels
nsato 2010-03-18 20:13:40
nsato 2010-03-18 20:13:50
What are the solutions to this quadratic equation?
jxl28 2010-03-18 20:14:30
p=-4 or -15/7
mathwiz314 2010-03-18 20:14:30
(7p+15)(p+4)=0, p={-15/7,-4}
panjia123 2010-03-18 20:14:30
-4,-15/7 rather
AwesomeToad 2010-03-18 20:14:30
-15/7 and -4
mgao 2010-03-18 20:14:34
-15/7,-4
nsato 2010-03-18 20:14:40
nsato 2010-03-18 20:14:48
Which solution do we want?
jxl28 2010-03-18 20:15:20
But we want the maximum, so p=-15/7
1=2 2010-03-18 20:15:20
The bigger one; -15/7.
gh625 2010-03-18 20:15:20
-15/7, because it's greater
AdithyaGanesh 2010-03-18 20:15:20
The greater one, namely -15/7.
Mij 2010-03-18 20:15:20
-15/7 because it is greater
-$45! 2010-03-18 20:15:20
the greater one, or -15/7
mathcire 2010-03-18 20:15:20
p=-15/7 is greater.
nsato 2010-03-18 20:15:26
We want the larger solution, so we take p = -15/7.
nsato 2010-03-18 20:15:42
sparkle123 2010-03-18 20:16:04
158
professordad 2010-03-18 20:16:04
wow so answer is 158
anaverageaopser 2010-03-18 20:16:04
158 is answer.
nsato 2010-03-18 20:16:07
nsato 2010-03-18 20:16:16
To be completely correct, we should verify that any solution for p gives us a solution (x,y,z) to the original system. But since any number has a cube root, this is not a problem.
nsato 2010-03-18 20:16:34
I now turn things over back to Dave Patrick.
DPatrick 2010-03-18 20:16:52
My fingers are now rested and I'm ready to go to the finish!
DPatrick 2010-03-18 20:17:02
DPatrick 2010-03-18 20:17:21
One could do this by brute force, but let's see a somewhat more clever solution.
DPatrick 2010-03-18 20:17:34
A key bit of data is 0 <= a_i <= 99. How can we use that data?
y2ucc 2010-03-18 20:18:06
place value
mgao 2010-03-18 20:18:24
so you let a_i = 10b_i + c_i
DPatrick 2010-03-18 20:18:34
DPatrick 2010-03-18 20:19:01
The fact that a_i is between 0 and 99 means every possible choice of (t_i,u_i) corresponds to exactly one a_i and vice versa.
DPatrick 2010-03-18 20:19:11
QuantumTiger 2010-03-18 20:19:29
Factor.
DPatrick 2010-03-18 20:19:42
Sure, now we just collect the t's and the u's:
DPatrick 2010-03-18 20:19:48
DPatrick 2010-03-18 20:20:04
What's a much easier way to write this?
mathcire 2010-03-18 20:20:44
2010=10*T+U with 0<=T,U<=9999
-$45! 2010-03-18 20:20:50
combine them
DPatrick 2010-03-18 20:20:56
DPatrick 2010-03-18 20:21:12
DPatrick 2010-03-18 20:21:26
And again, each choice of t and u gives us a unique choice for the a's, and vice versa.
DPatrick 2010-03-18 20:21:40
But now it's really easy to count. How many possible (t,u) pairs are there?
BarbieRocks 2010-03-18 20:21:58
for every t from 0 to 201, there is only one u... so the answer is 202
1=2 2010-03-18 20:21:58
202, t can be anything from 0 to 201.
RunpengFAILS 2010-03-18 20:21:58
so like t=0,...,201 which is like 202 pairs
DPatrick 2010-03-18 20:22:03
t can be anything from 0 to 201 (inclusive), and then u = 2010 - 10t.
DPatrick 2010-03-18 20:22:11
DPatrick 2010-03-18 20:22:40
Again, you could definitely have brute-force counted the a's using messy casework (for example, a_3 must be 2, 1, or 0, etc...)
DPatrick 2010-03-18 20:22:47
But it's slow.
DPatrick 2010-03-18 20:23:15
Using the clue that 0 <= a_i <= 99 means a_i is a 2-digit number (with a possible leading zero) and breaking up into digits basically solved the problem.
DPatrick 2010-03-18 20:23:42
DPatrick 2010-03-18 20:23:55
I really hope you had graph paper when you tried to solve this!
mz94 2010-03-18 20:24:15
yay for graph paper
flamingmath 2010-03-18 20:24:20
graph it and find the intersections
DPatrick 2010-03-18 20:24:32
Right, we start by sketching the two curves that are used in the inequalities.
DPatrick 2010-03-18 20:24:44
3y - x = 15 is easier: it's just a line with slope 1/3 passing through (0,5). I'll start with that.
DPatrick 2010-03-18 20:24:50
DPatrick 2010-03-18 20:25:01
The region R lies above this blue line. (We'll only need the first quadrant, as we shall see.)
DPatrick 2010-03-18 20:25:18
What about |8-x| + y = 10?
alligator112 2010-03-18 20:25:44
(8,10) is the vertex
theone142857 2010-03-18 20:25:44
2 lines
DPatrick 2010-03-18 20:25:50
It clearly passes through (8,10). Then, as x gets away from 8, y must decrease at the same rate.
Goldey 2010-03-18 20:25:57
It has a vertex at x=8, and two lines with slopes of 1 and -1 going from it
prezcoin 2010-03-18 20:25:57
2 rays
DPatrick 2010-03-18 20:26:04
So it is two rays starting at (8,10) with slopes 1 and -1, going downwards.
DPatrick 2010-03-18 20:26:17
I'll add this curve in red:
DPatrick 2010-03-18 20:26:23
DPatrick 2010-03-18 20:26:44
The region R lies below this curve, so it is the region below the red but above the blue.
karatemagic7 2010-03-18 20:26:47
So R is a triangle!
DPatrick 2010-03-18 20:26:53
DPatrick 2010-03-18 20:27:11
Before going to the 3D, knowing the intersection points is probably helpful.
prezcoin 2010-03-18 20:27:33
set the equations equal
-$45! 2010-03-18 20:27:33
Use a 2x2 system of equations to find them
mgao 2010-03-18 20:27:33
the first one's (9/2, 13/2)
DPatrick 2010-03-18 20:27:38
The left point is the intersection of 3y - x = 15 and 8 - x + y = 10.
DPatrick 2010-03-18 20:27:51
Solving gives (x,y) = (9/2,13/2). Let's call this point A.
DPatrick 2010-03-18 20:28:03
The right point is the intersection of 3y - x = 15 and x - 8 + y = 10.
DPatrick 2010-03-18 20:28:09
Solving gives (x,y) = (39/4, 33/4). Let's call this point B.
DPatrick 2010-03-18 20:28:20
(I won't waste our time going through the computations for the above!)
BarbieRocks 2010-03-18 20:28:31
Does the fact that it's a right triangle make the problem any easier?
DPatrick 2010-03-18 20:28:43
It sure did for me, although you can solve it without noticing this.
DPatrick 2010-03-18 20:28:59
Indeed, our triangle is a right triangle with the blue line as its hypotenuse.
DPatrick 2010-03-18 20:29:16
What is the solid that we get when we rotate the gray region around the blue line?
flamingmath 2010-03-18 20:29:38
two right cones
gh625 2010-03-18 20:29:39
two cones with their bases attached
Jadesymdragon 2010-03-18 20:29:39
2 cones attached to each other
mathwiz314 2010-03-18 20:29:39
it makes 2 cones with the same base
jxl28 2010-03-18 20:29:39
two cones joined at the base?
DPatrick 2010-03-18 20:29:54
We get two circular cones. Each has a base that's a circle, with readius equal to the height of the triangle from the red intersection point to the blue line.
DPatrick 2010-03-18 20:30:10
Let me put the pic up again:
DPatrick 2010-03-18 20:30:17
alligator112 2010-03-18 20:30:26
since 2 cones with same base, you can use the sum of the heights as the height
DPatrick 2010-03-18 20:30:50
DPatrick 2010-03-18 20:31:08
The total heights of the cones is the length of the blue segment, which is the hypotenuse of the right triangle.
DPatrick 2010-03-18 20:31:19
DPatrick 2010-03-18 20:31:35
But h looks like a pain to calculate. How can we avoid it?
theone142857 2010-03-18 20:32:01
use area
theone142857 2010-03-18 20:32:01
and right triangles
panjia123 2010-03-18 20:32:05
You could get the area with two different bases
DPatrick 2010-03-18 20:32:21
DPatrick 2010-03-18 20:32:29
DPatrick 2010-03-18 20:32:45
sparkle123 2010-03-18 20:32:54
but you need a and b
DPatrick 2010-03-18 20:33:01
Yeah: now we have to compute, but there's one more minor thing we can do (if we want) to make the computation a little nicer.
PhireKaLk6781 2010-03-18 20:33:20
dilation by 4x
DPatrick 2010-03-18 20:33:30
That's what I did. We can scale everything by a factor of 4 to get rid of the denominators.
DPatrick 2010-03-18 20:33:41
So the right angle is now at (32,40) and the other two points are (18,26) and (39,33).
DPatrick 2010-03-18 20:33:52
This will multiply the volume by 4^3, so we'll have to divide by 4^3 at the end.
QuantumTiger 2010-03-18 20:34:07
But the numbers are now bigger.
DPatrick 2010-03-18 20:34:17
They are, but I think they're a lot easier to work with, without denominators.
DPatrick 2010-03-18 20:34:37
DPatrick 2010-03-18 20:35:04
AwesomeToad 2010-03-18 20:35:37
DPatrick 2010-03-18 20:36:10
DPatrick 2010-03-18 20:37:02
So we plug these side lengths into our volume formula from earlier, and don't forget to divide by the 4^3 to cancel out the scaling we did.
DPatrick 2010-03-18 20:37:08
DPatrick 2010-03-18 20:37:45
Now we simplify.
DPatrick 2010-03-18 20:38:08
DPatrick 2010-03-18 20:38:29
We can then cancel 8 out of 56 and 2 out of 98, and 4^2 in the denominator:
DPatrick 2010-03-18 20:38:34
magixter 2010-03-18 20:38:44
m + n + p = 365
master123 6 2010-03-18 20:38:44
365 is m+n+p
DPatrick 2010-03-18 20:38:48
It won't simplify any further.
DPatrick 2010-03-18 20:38:52
Yongyi781 2010-03-18 20:39:12
You forgot a pi in the final expression, but that doesn't matter in obtaining m+n+p.
DPatrick 2010-03-18 20:39:16
Right, thanks.
DPatrick 2010-03-18 20:39:54
You could certainly do the calculation differently once you found the coordinates of the triangle, but noticing it was a right triangle allowed me to simplify things a little bit.
DPatrick 2010-03-18 20:40:07
In particular, I didn't have to compute the radius of the base of the cone.
DPatrick 2010-03-18 20:40:20
(Not that it's particuarly hard, but it's slightly messy.)
DPatrick 2010-03-18 20:40:36
DPatrick 2010-03-18 20:41:01
I thought that this was the most interesting problem of the final third (11-15). How can we attack it?
maraks 2010-03-18 20:41:22
I guessed and checked this question during the AIME
megahertz 2010-03-18 20:41:22
Proof by contradiction + guess and check?
RunpengFAILS 2010-03-18 20:41:22
Guess the answer and prove
flamingmath 2010-03-18 20:41:22
brute force?
DPatrick 2010-03-18 20:41:34
More or less. We can try to construct a "bad" partition, in which both subsets don't have ab=c inside of them.
DPatrick 2010-03-18 20:42:08
Our idea will be to see how big we can make m before we can no longer construct a bad partition. That will at least be an upper bound on the answer.
karatemagic7 2010-03-18 20:42:18
Well, let's just look at 3 and its powers.
anchenyao 2010-03-18 20:42:20
Try 3, 9, 27, 81, and 243
DPatrick 2010-03-18 20:42:33
The simplest place to start is with powers of 3.
DPatrick 2010-03-18 20:42:46
For example, suppose A and B are a bad partition (for some m), with 3 in A.
DPatrick 2010-03-18 20:42:57
Where must 9 go?
mathwiz314 2010-03-18 20:43:11
B
1=2 2010-03-18 20:43:11
B
Yongyi781 2010-03-18 20:43:11
B
QuincyCello 2010-03-18 20:43:11
B
gh625 2010-03-18 20:43:11
B
DPatrick 2010-03-18 20:43:27
Since 3*3=9, and we want our partition to be bad, 9 must be in B.
DPatrick 2010-03-18 20:43:33
What about 27?
Yongyi781 2010-03-18 20:43:58
Either
prezcoin 2010-03-18 20:43:58
either
SuitcaseAsian 2010-03-18 20:43:58
Either?
karatemagic7 2010-03-18 20:43:58
We don't know (yet).
DPatrick 2010-03-18 20:44:03
It doesn't matter (at first), so let's try 27 in A.
mgao 2010-03-18 20:44:21
if 27 goes in A nowhere for 81. then 27 is in B
gh625 2010-03-18 20:44:21
27 has to go in B, because putting it in A means that 81 can't go in either.
Wesmaster160 2010-03-18 20:44:21
81 cant go in either
-$45! 2010-03-18 20:44:21
81 can't go anywhere!
DPatrick 2010-03-18 20:44:33
Right. Now we can't place 81, since 81 = 3*27 (in A) = 9*9 (in B).
DPatrick 2010-03-18 20:44:42
So let's try 27 in B instead.
-$45! 2010-03-18 20:45:01
now 81 must go in A
gh625 2010-03-18 20:45:01
81 goes in A
mathwiz314 2010-03-18 20:45:01
81 must go in A
jxl28 2010-03-18 20:45:01
81 goes in A then.
DPatrick 2010-03-18 20:45:07
We can't have 81 in B (since 9*9 = 81), so 81 must be in A.
DPatrick 2010-03-18 20:45:34
To recap: so far we have 3 and 81 in A, and 9 and 27 in B, and everything is cool. This partition is still "bad".
magixter 2010-03-18 20:45:49
But then 243 can't go anywhere!
GeorgiaTechMan 2010-03-18 20:45:49
243 can't go anywhere
-$45! 2010-03-18 20:45:49
And now 243 can't go in either!
Mewto55555 2010-03-18 20:45:49
nowhere for 243
anchenyao 2010-03-18 20:45:49
Then 243 has no place to go, so 243 would be the lowest.
Luminescence 2010-03-18 20:45:49
but then 243 cant go into either
DPatrick 2010-03-18 20:45:54
But now there's nowhere for 243 = 3^5 to go, since 243 = 3*81 (in A) = 9*27 (in B).
DPatrick 2010-03-18 20:46:18
What we have shown is that if m >= 243, we can't construct a bad partition, because there's nowhere to place 243. Thus the answer is at most 243.
DPatrick 2010-03-18 20:46:56
We might be tempted to guess 243 is the answer.
DPatrick 2010-03-18 20:47:09
(And if low on time I would indeed be very tempted.0
DPatrick 2010-03-18 20:47:21
To prove it: can we always construct a bad partition of {3,4,...,m} for m < 243?
DPatrick 2010-03-18 20:47:31
So far, we've only worried about the powers of 3.
-$45! 2010-03-18 20:47:45
yes!
Bachukas 2010-03-18 20:47:45
YEs
theone142857 2010-03-18 20:47:45
Yes
mathelete 2010-03-18 20:47:45
we can always construct it
DPatrick 2010-03-18 20:47:47
How?
theone142857 2010-03-18 20:48:20
If n is less than 243, Consider A={3 through 8} U {81 through n}, B={9 through 80} doen't satisfy the condition(bad)! So it is 243.
-$45! 2010-03-18 20:48:20
{3,4,5,6,7,8,81,82,...,242} and {9,10,11,...80}
karatemagic7 2010-03-18 20:48:20
Yes, let A={3,4,...8,81,82,...242} and B={9,10,...80} and subtract terms when m<242.
mathelete 2010-03-18 20:48:26
the same way we assigned the first 4 powers of 3 into sets
DPatrick 2010-03-18 20:48:32
Right. We can mimic a bit what we did above.
DPatrick 2010-03-18 20:48:50
We put {3,4,...,8} in A, then {9,10,...,80} in B, and then 81 and everything above back in A. (If m < 81 then abridge this accordingly.)
DPatrick 2010-03-18 20:49:17
We can quickly check that this works. Multiplying two elements of A less than 9 will land you in B. Multiplying one or two elements of A >=81 will be at least 243, which is larger than m so not in the set. Multiplying two elements of B will be at least 81, so not in B. Thus this partition is bad.
DPatrick 2010-03-18 20:49:46
DPatrick 2010-03-18 20:50:12
What's slightly annoying about this problem is that you can relatively quickly figure out that 243 works, and then make an educated guess that 243 is the answer, without doing the work of showing that no m < 243 works too. So it's not ideal as an AIME problem.
DPatrick 2010-03-18 20:50:33
It's a really interesting problem, but I would have prefered it as a USAJMO problem where you have to show your work.
DPatrick 2010-03-18 20:50:56
-$45! 2010-03-18 20:51:17
Make a diagram!!!!
birdofsorrow 2010-03-18 20:51:17
draw a diagram
DPatrick 2010-03-18 20:51:23
We can start by drawing a picture that shows the given data:
DPatrick 2010-03-18 20:51:29
Yongyi781 2010-03-18 20:51:47
Scale everything by 42 (or 21), much computationally easier!
-$45! 2010-03-18 20:51:47
I guess you could scale it down by 14 to make it cleaner
DPatrick 2010-03-18 20:52:04
You could definitely scale it down by some convenient factor.
DPatrick 2010-03-18 20:52:15
Let's press on with these numbers, though -- it's not too bad since they're all integers.
DPatrick 2010-03-18 20:52:29
Also, I don't claim that this picture is to scale.
DPatrick 2010-03-18 20:52:37
The only data not explicitly in the diagram is the 1:2 ratio of the areas. It certainly appears that the left region is the smaller one.
KingMax 2010-03-18 20:52:47
The radius of the semicircle is 126
Maximilliann 2010-03-18 20:53:01
We can figure out that the radius is 126
DPatrick 2010-03-18 20:53:11
One thing that we hopefully notice is that the diameter of the semicircle is 252, which is twice AN. What does that mean?
mgao 2010-03-18 20:53:34
first observe that BAN = 60 degrees
theone142857 2010-03-18 20:53:34
angle AON=60 degrees
karatemagic7 2010-03-18 20:53:34
If O is the center of the semicircle then ANO is an equilateral triangle.
1=2 2010-03-18 20:53:34
Let O be the midpoint of AB. ANO is equilateral.
prezcoin 2010-03-18 20:53:34
we can draw in the radius and center to N
anchenyao 2010-03-18 20:53:34
<NAU is 60 degrees
DPatrick 2010-03-18 20:53:41
If we let O be the center of the semicircle, then AON is an equilateral triangle:
DPatrick 2010-03-18 20:53:51
DPatrick 2010-03-18 20:54:08
So what?
Wesmaster160 2010-03-18 20:54:51
sector NOA is 1/3 the are of the semicircle
1=2 2010-03-18 20:54:51
We can find the area of the "sector" AUN by finding the area of sector AON and subtracting the area of OUN.
DPatrick 2010-03-18 20:55:01
Yes. Angle AON is 60 degrees, and angle BON is 120 degrees.
DPatrick 2010-03-18 20:55:21
This means that the line NUT splits the semicircle into 2 pieces with area 1:2 EXCEPT for the triangle UON which is "extra" on the right side.
DPatrick 2010-03-18 20:55:39
What does that tell us about NUT and the rectangle?
DPatrick 2010-03-18 20:56:09
There are a bunch of different ways you could go about this -- one way is to set up coordinates and just bash it -- but here's a way to think about it:
DPatrick 2010-03-18 20:56:30
If I drop a perpendicular from U up to DC, what ratio does that split the rectangle's area into?
Maximilliann 2010-03-18 20:56:53
1:2
skyhog 2010-03-18 20:56:53
1:2
prezcoin 2010-03-18 20:56:53
1:2
DPatrick 2010-03-18 20:57:15
Right: since AU : UB = 1 : 2, the vertical line from U would split ABCD in ratio 1 : 2.
prezcoin 2010-03-18 20:57:24
triangle OUN has the same area as the right triangle formed by perps
DPatrick 2010-03-18 20:57:27
Exactly.
DPatrick 2010-03-18 20:57:36
This is very clear when I draw the picture:
DPatrick 2010-03-18 20:57:41
Maximilliann 2010-03-18 20:57:45
So the extra area of triangle UON is equal to the extra area of that new triangle
DPatrick 2010-03-18 20:58:09
Exactly. Below, the right side is 2/3 of the semicircle, plus the extra blue area UON.
DPatrick 2010-03-18 20:58:22
So above, I need the extra area on the left side to "cancel out" the extra area below.
DPatrick 2010-03-18 20:58:34
UP splits the rectangle's area in the ratio 1:2. So we need the "extra" red region PTU on the left side to cancel out the extra blue region on the right side, for the regions overall to be in ratio 1:2.
DPatrick 2010-03-18 20:59:04
So we need [UON] = [UTP].
DPatrick 2010-03-18 20:59:15
Now it's just calculation. What's the blue area UON?
mz94 2010-03-18 21:00:10
42*63sqrt3/2=1323sqrt3
anchenyao 2010-03-18 21:00:10
63sqrt(3)*21
Maximilliann 2010-03-18 21:00:10
The base is 126 - 84 = 42, and the height is 63 sqrt(3), so the area is 21*63 sqrt(3)
DPatrick 2010-03-18 21:00:18
DPatrick 2010-03-18 21:00:27
DPatrick 2010-03-18 21:00:45
skyhog 2010-03-18 21:01:14
draw perp from N to AO and use similar triangles
anchenyao 2010-03-18 21:01:20
similar triangle
DPatrick 2010-03-18 21:01:38
Basically, yes, although to be honest I didn't do this explicitly.
DPatrick 2010-03-18 21:01:46
Here's the pic again:
DPatrick 2010-03-18 21:01:51
DPatrick 2010-03-18 21:02:02
Let x = AD be the height. (Note this is what we are trying to find.) What is the length of PT in terms of x?
skyhog 2010-03-18 21:03:02
x/3sqrt3
DPatrick 2010-03-18 21:03:35
DPatrick 2010-03-18 21:03:45
Maximilliann 2010-03-18 21:04:06
so the area of PTU is x^2 / 6sqrt(3)
DPatrick 2010-03-18 21:04:23
Right, and we can compare this to what we know the area must be, to solve for x.
DPatrick 2010-03-18 21:04:30
skyhog 2010-03-18 21:04:59
x=63sqrt6
anchenyao 2010-03-18 21:04:59
63sqrt(6)
DPatrick 2010-03-18 21:05:08
DPatrick 2010-03-18 21:05:17
DPatrick 2010-03-18 21:05:50
I thought this was a nice problem, if perhaps a little on the easy side for a #13.
DPatrick 2010-03-18 21:06:35
The next problem, as has been widely discussed, was way too easy for a #14. In my opinion the difficulty level of the next problem was pretty badly misjudged.
DPatrick 2010-03-18 21:06:40
DPatrick 2010-03-18 21:07:22
It looks really scary, but it's not as scary as it looks.
DPatrick 2010-03-18 21:07:28
AceOfDiamonds 2010-03-18 21:07:56
digits-1
AsimoIsFuture 2010-03-18 21:07:56
how many digits x has - 1
MathTwo 2010-03-18 21:07:56
one less than its number of digits
master123 6 2010-03-18 21:07:56
number of digits minus one
karatemagic7 2010-03-18 21:08:00
1 less than the number of digits of x.
DPatrick 2010-03-18 21:08:05
It is one less than the number of digits in the decimal representation of x. (Any number between 10 and 99 will have a log of 1.something, anything between 100 and 999 will have a log of 2.something, and so on.)
DPatrick 2010-03-18 21:08:22
So what (in English) is the interpretation of the sum in the problem?
mathwiz314 2010-03-18 21:08:42
number of digits -100
master123 6 2010-03-18 21:08:43
total number of digits minus 100
DPatrick 2010-03-18 21:08:48
It's 100 less than the total number of digits when we write out n, 2n, 3n, ..., up to 100n.
DPatrick 2010-03-18 21:09:04
So we're looking for the largest number n such that writing out n, 2n, 3n, ..., up to 100n uses no more than 400 total digits.
DPatrick 2010-03-18 21:09:14
How do we find it?
AsimoIsFuture 2010-03-18 21:09:36
guess and check!!!
MathTwo 2010-03-18 21:09:36
I guessed and checked this problem
mathwiz314 2010-03-18 21:09:36
guess and check/brute force
Jadesymdragon 2010-03-18 21:09:36
guess and check
DPatrick 2010-03-18 21:09:41
That's certainly what I did.
skyhog 2010-03-18 21:10:04
is there a better way?
DPatrick 2010-03-18 21:10:09
Not that I know of.
magixter 2010-03-18 21:10:17
Try n = 100
DPatrick 2010-03-18 21:10:39
n=100 is a good place to start since it's easy to calculate and should be about right, since 100k is a 4-digit number for most 1 <= k <= 100.
DPatrick 2010-03-18 21:11:10
n...9n each use 3 digits, 10n...99n each use 4 digits, and 100n uses 5 digits. So that's a total of 9*3 + 90*4 + 1*5 = 392 digits.
DPatrick 2010-03-18 21:11:34
(oops, I got by n's and k's backwards. That should be k in the last posting.)
prezcoin 2010-03-18 21:11:39
really close
Yoshi 2010-03-18 21:11:44
Then try n=110
DPatrick 2010-03-18 21:12:01
We're already really close. I tried n=110 next, since it's also easy to calculate.
DPatrick 2010-03-18 21:12:25
k...9k each use 3, 10k...90k each use 4, 91k...100k each use 5. That gives 9*3 + 81*4 + 10*5 = 401 digits. Oops, 1 too many.
1=2 2010-03-18 21:12:35
1 off! Try n=109!
prezcoin 2010-03-18 21:12:35
try 109
Jadesymdragon 2010-03-18 21:12:35
109
QuantumTiger 2010-03-18 21:12:37
What about n=109?
DPatrick 2010-03-18 21:12:46
Since we're only off by 1, clearly let's check 109.
DPatrick 2010-03-18 21:13:05
k...9k each use 3, 10k...91k each use 4, 92k...100k each use 5.
DPatrick 2010-03-18 21:13:14
Indeed, this is one fewer digit than n=110 (since 91*109 is 4 digits but 91*110 is 5 digits, and that's the only one that's changed).
DPatrick 2010-03-18 21:13:31
(Oops, got n and k backwards again I think. Hopefully you get the idea!)
DPatrick 2010-03-18 21:13:38
So n=109 uses exactly 400 digits.
DPatrick 2010-03-18 21:13:54
DPatrick 2010-03-18 21:14:56
This was very, very easy for a #14. I have to assume that the AIME problem committee didn't realize that a guess-and-check solution was so straightforward. Or perhaps they thought that the logs and the floors and the summation symbol would be too scary.
DPatrick 2010-03-18 21:15:26
On the other hand, I wouldn't call #15 "very easy" by any means. :)
DPatrick 2010-03-18 21:15:30
EHSTutor 2010-03-18 21:15:57
diagram!
mathelete 2010-03-18 21:15:57
First and foremost...DIAGRAM!!!
DPatrick 2010-03-18 21:16:03
DPatrick 2010-03-18 21:16:18
(Note that the picture is probably not to scale. I also won't draw the incircles.)
DPatrick 2010-03-18 21:16:34
We'll probably want variable names for the other three lengths. Let's set a = AM and c = CM.
DPatrick 2010-03-18 21:16:39
Let's also set m = BM. I'll add these labels:
DPatrick 2010-03-18 21:16:43
DPatrick 2010-03-18 21:17:13
There are a number of ways to proceed. I'll first present the solution that requires the least amount of "cleverness", but which also has the messiest algebra. Then I'll present a nice solution for which you have to be really clever at the beginning, but the resulting algebra is much simpler.
DPatrick 2010-03-18 21:17:34
First, let's keep track of what we want to find. Let x = a/c. We're trying to find x.
MathTwo 2010-03-18 21:17:41
c=15-a
DPatrick 2010-03-18 21:17:51
Certainly, we also have a+c = 15. That's one equation.
DPatrick 2010-03-18 21:18:22
Ideally I'd like everything in terms of x, so I can end up with an equation (in x) that we can solve.
DPatrick 2010-03-18 21:18:29
So let's write a and c in terms of x.
jxl28 2010-03-18 21:18:56
a=cx, c=a/x
QuantumTiger 2010-03-18 21:18:56
a = 15-c
DPatrick 2010-03-18 21:19:10
Right, all I want to do now is combine these to write a and c in terms of x.
DPatrick 2010-03-18 21:19:23
For instance, c(1+x) = 15, so c = 15/(1+x).
DPatrick 2010-03-18 21:19:41
Then also a = cx = 15x/(1+x).
DPatrick 2010-03-18 21:19:53
I haven't really done anything yet...this is all bookkeeping.
DPatrick 2010-03-18 21:20:13
In particular, I haven't used the data about the incircles yet. How does that apply?
GeorgiaTechMan 2010-03-18 21:20:34
Use the property that the inradius of a triangle*1/2 the perimeter is the area of the triangle? It worked for me.
to94as 2010-03-18 21:20:34
inradius=2*area/semiperimeter
sparkle123 2010-03-18 21:20:34
find total area using incircle radii
megahertz 2010-03-18 21:20:34
Area is equal to radius times semiperimeter
Luke-Mathwalker 2010-03-18 21:20:34
Area = rs
skyhog 2010-03-18 21:20:37
use A=rs for each and set r equal?
DPatrick 2010-03-18 21:20:43
We have the formula A = rs for the area of a triangle, where r is the radius and s is the semiperimeter (half the perimeter). (If you don't know this formula, it's a fun exercise to try to prove it!)
mathwiz314 2010-03-18 21:21:14
x= (12+m+a)/(13+m+c)
DPatrick 2010-03-18 21:21:20
Right.
DPatrick 2010-03-18 21:21:35
We look at the ratio of the areas [ABM] / [ACM].
DPatrick 2010-03-18 21:21:50
They have the same height, so the ratio of their areas is a/c = x.
DPatrick 2010-03-18 21:21:56
But they also have the same inradius, so the ratio of their areas is also the ratio of their semiperimeters.
DPatrick 2010-03-18 21:22:18
Oops, I meant the ration [ABM] / [CBM].
DPatrick 2010-03-18 21:22:21
(ratio)
DPatrick 2010-03-18 21:22:30
mgao 2010-03-18 21:22:41
but notice that x = a/c so we can "eliminate" a/c from the ratio and get (12 + m)/(13 + m)
DPatrick 2010-03-18 21:22:55
DPatrick 2010-03-18 21:23:07
mathwiz314 2010-03-18 21:23:36
How do you simpilfy it?
gh625 2010-03-18 21:23:36
how do you just remove the a from the top and the c from the bottom
DPatrick 2010-03-18 21:23:40
Think about it. :)
DPatrick 2010-03-18 21:23:48
It's a useful algebraic "trick" to know.
DPatrick 2010-03-18 21:23:59
Anyway, the nice thing is that now I can solve for m in terms of x.
DPatrick 2010-03-18 21:24:12
DPatrick 2010-03-18 21:24:48
At this stage, I have all of a, c, and m in terms of x. If I can find an equation relating them, I can probably solve for x (which is what we want!).
magixter 2010-03-18 21:25:00
Stewart's theorem?
skyhog 2010-03-18 21:25:00
stewarts theorem
DPatrick 2010-03-18 21:25:28
That's one possibility, but I'm not smart enough to remember Stewart's Theorem.
DPatrick 2010-03-18 21:26:00
But I found a slightly better approach anyway.
skyhog 2010-03-18 21:26:04
law of cosines + law of sines?
DPatrick 2010-03-18 21:26:11
We have all three sides of ABC, so we can use the Law of Cosines to get the cosines of the three angles. This may help.
DPatrick 2010-03-18 21:26:21
DPatrick 2010-03-18 21:26:49
Angle B seems useless, so we'll want to try A or C. (That way we can use the same angle on the big triangle and one of the little ones.)
DPatrick 2010-03-18 21:26:57
Let's try angle A.
DPatrick 2010-03-18 21:27:02
skyhog 2010-03-18 21:27:21
cos = 5/9
DPatrick 2010-03-18 21:27:28
DPatrick 2010-03-18 21:27:56
5/9 is pretty simple, so that's good enough for me. Let's use it again but this time on BAM instead of BAC.
DPatrick 2010-03-18 21:28:07
DPatrick 2010-03-18 21:28:31
DPatrick 2010-03-18 21:28:43
DPatrick 2010-03-18 21:28:57
Aha, now we have an equation for x. :) Hopefully we can solve this.
BarbieRocks 2010-03-18 21:29:22
That looks like a very annoying quartic.
DPatrick 2010-03-18 21:29:30
This is icky, but is certainly accessible.
DPatrick 2010-03-18 21:29:43
You just have to do it step-by-step and be careful!
DPatrick 2010-03-18 21:29:51
DPatrick 2010-03-18 21:30:11
DPatrick 2010-03-18 21:30:30
(You can check these at home later if you care!)
DPatrick 2010-03-18 21:30:36
skyhog 2010-03-18 21:30:52
so it's not quartic after all! kind of
DPatrick 2010-03-18 21:31:00
Yeah, happily, the x^4 and constant terms cancel, so we can divide by x and get a quadratic:
DPatrick 2010-03-18 21:31:13
RunpengFAILS 2010-03-18 21:31:23
we can basically cheat using the fact that x is rational...
MathTwo 2010-03-18 21:31:23
well, this is the AIME so it must give a rational root by RRT
DPatrick 2010-03-18 21:31:40
Right, the problem tells us the answer is rational, so we should be able to factor this.
DPatrick 2010-03-18 21:31:50
DPatrick 2010-03-18 21:32:20
Which is it?
skyhog 2010-03-18 21:32:29
i imagine x=2/3 breaks something... but i don't know what
Yongyi781 2010-03-18 21:32:59
x=2/3 makes m negative, ew.
Ihatepie 2010-03-18 21:32:59
it makes m negative
DPatrick 2010-03-18 21:33:02
In fact x = 2/3 doesn't work, since then m < 0.
DPatrick 2010-03-18 21:33:09
DPatrick 2010-03-18 21:33:39
If you had time, it would not be unreasonable to reverse-engineer from this point to check your answer. Note we get a = 22/3 and b = 23/3. We also get m = 10. You could then check Stewart's Theorem and/or verify that the ratio of the semiperimeters works out to 22/23 as it should.
DPatrick 2010-03-18 21:33:55
The algebra got a bit hairy in this solution. Let me now present a more geometrically-elegant solution for which the algebra is much simpler.
DPatrick 2010-03-18 21:34:10
I wish I had thought of this solution, but I didn't. :(
DPatrick 2010-03-18 21:34:24
This solution was posted on our message board by AoPS user CatalystOfNostalgia. I only added some pictures to make it prettier.
DPatrick 2010-03-18 21:34:41
There is a magic word you have to know to make the pretty solution work.
DPatrick 2010-03-18 21:34:44
That word is...
ytrewq 2010-03-18 21:34:54
homothety
DPatrick 2010-03-18 21:34:55
...homothety.
DPatrick 2010-03-18 21:35:08
A circle with a pair of tangents, or, better yet, two circles sandwiched between two common tangents, is often a clue to try homothety. (Homothety is a fancy word for "dilation".)
DPatrick 2010-03-18 21:35:23
Here, we have two circles with a pair of tangents. They aren't sandwiched between two common tangents, but they do share one tangent, side AC. However, is there a natural third circle we should consider?
panjia123 2010-03-18 21:35:46
Incircle of ABC?
KingMax 2010-03-18 21:35:46
incircle of the whole thing
DPatrick 2010-03-18 21:35:51
We can compare each to the incircle of ABC:
DPatrick 2010-03-18 21:35:57
DPatrick 2010-03-18 21:36:12
(The scaling makes it a bit hard to see---the blue and green circles are NOT tangent to BM at the same point.)
DPatrick 2010-03-18 21:36:22
Now, we have two pairs of circles that are sandwiched between common tangents.
DPatrick 2010-03-18 21:36:44
We say that the incircle of ABM (blue) and the incircle of ABC (red) are "homothetic about A" because there is a dilation with center A that maps the incircle of ABM to the incircle of ABC.
DPatrick 2010-03-18 21:36:58
Similarly, the incircle CBM and the incircle of ABC are homothetic about B.
DPatrick 2010-03-18 21:37:03
How do these homotheties help us?
megahertz 2010-03-18 21:37:44
wait so is homothety with 2 tangents an application of similar triangles by drawing lines connecting the two points of tangency?
ytrewq 2010-03-18 21:38:02
they have the same scale factor
DPatrick 2010-03-18 21:38:04
More or less. The key is that the blue and green circles are the same size.
DPatrick 2010-03-18 21:38:13
That is, the factor we use to scale the incircle of ABM up to get the incircle of ABC is the same as the factor we use to scale the incircle of BCM up to get the incircle of ABC.
DPatrick 2010-03-18 21:38:25
So when we scale from blue up to red, we use the same factor as green up to red.
DPatrick 2010-03-18 21:38:41
We're actually going to go in the opposite direction, and scale down from red to blue and green.
DPatrick 2010-03-18 21:38:51
That's because we can compute a lot of distances using the red circle.
QuantumTiger 2010-03-18 21:38:57
Are the blue and green circles touching? (I can't really tell.)
DPatrick 2010-03-18 21:39:15
They're not, but you can't really tell with this scale. It'll be a little more obvious in a moment that they're not.
DPatrick 2010-03-18 21:39:28
Let's look at the picture with just the red circle:
DPatrick 2010-03-18 21:39:34
DPatrick 2010-03-18 21:39:41
Can we compute some of the distances in this picture?
AsimoIsFuture 2010-03-18 21:40:38
ay = az zb = bx cy = cx
GeorgiaTechMan 2010-03-18 21:40:40
we can figure out ZB, BX, XC, CY, AY, and AZ
DPatrick 2010-03-18 21:41:00
Right! AY = AZ because the distance to the two tangent points are equal. Same for BX = BZ and CX = CY.
DPatrick 2010-03-18 21:41:41
But we also know AZ + BZ = 12, and the other two sides give a sum too, so that let's us solve for all the lengths in the picture.
birdofsorrow 2010-03-18 21:42:10
BX=BZ=5, AX=AY=7 CY=CX=8
DPatrick 2010-03-18 21:42:20
You meant AZ = AY = 7, but otherwise that's right.
DPatrick 2010-03-18 21:42:29
The two lengths from A are 7 and the two lengths from C are 8.
DPatrick 2010-03-18 21:42:49
Now we can look at the picture with the smaller incircles, and the corresponding points labeled:
DPatrick 2010-03-18 21:42:54
DPatrick 2010-03-18 21:43:14
Again, our key fact here is that the homotheties shrink the red circle to the blue and green circles by a common factor. Let's call this factor k.
DPatrick 2010-03-18 21:43:37
What lengths do we now know (in terms of k)?
skyhog 2010-03-18 21:43:59
AY1=7k and CY2=8k
DPatrick 2010-03-18 21:44:26
Exactly. We have AY_1 = 7k and CY_2 = 8k. This is just the shrinking of AY and CY each by the same factor k.
DPatrick 2010-03-18 21:44:43
So now we can "length chase" to get the rest of the lengths in terms of k.
DPatrick 2010-03-18 21:45:08
DPatrick 2010-03-18 21:45:37
oops, that first one is AZ_1 = AY_1 = 7k.
DPatrick 2010-03-18 21:46:00
DPatrick 2010-03-18 21:46:32
What about the lengths at point M?
ytrewq 2010-03-18 21:47:25
they add to 15-15k and differ by 1-k
DPatrick 2010-03-18 21:47:29
Right: we have to work a little harder, but we can get them too.
DPatrick 2010-03-18 21:47:37
DPatrick 2010-03-18 21:48:54
(Basically, adding MY_1 and MY_2 gives Y_1Y_2 along the bottom, which is 15 - 15k. But going up, their difference is also MX_1 - MZ_2, and we already computed this difference as the difference of the lengths coming down from B. This is the hardest step so you may need to try to reconstruct it later if you're interested.)
DPatrick 2010-03-18 21:49:12
The point is, we have all of our lengths now in terms of k.
DPatrick 2010-03-18 21:49:17
DPatrick 2010-03-18 21:49:37
How do we finish?
1=2 2010-03-18 21:50:01
Stewarts theorem bash! :D
skyhog 2010-03-18 21:50:01
stewarts theorem!
dinoboy 2010-03-18 21:50:01
Law of cosines again?
MathTwo 2010-03-18 21:50:01
stewarts theorem
1=2 2010-03-18 21:50:01
Or law of cosines/sines...
DPatrick 2010-03-18 21:50:24
No way. The point of all this was to save ourselves that nastiness! Let's do something much simpler.
ytrewq 2010-03-18 21:50:28
ratio of perimiters and areas
DPatrick 2010-03-18 21:50:41
Right: we just go back to the ratio of the areas of the two small triangles.
jxl28 2010-03-18 21:50:49
AM/MC=(AB+BM+AM)/(BC+AM+MC)?
DPatrick 2010-03-18 21:50:51
Exactly.
DPatrick 2010-03-18 21:50:59
DPatrick 2010-03-18 21:51:09
DPatrick 2010-03-18 21:51:23
DPatrick 2010-03-18 21:51:42
GeorgiaTechMan 2010-03-18 21:52:00
k is 2 or 2/3
Yongyi781 2010-03-18 21:52:08
k = 2/3 is the only solution between 0 and 1
DPatrick 2010-03-18 21:52:29
Right: clearly the geometry of the problem requires 0 < k < 1 (since k is our "shrinking" factor), so it must be k = 2;/3.
DPatrick 2010-03-18 21:52:42
And to finish, we have AM/CM = (8-k)/(7+k) = (24-2)/(21 + 2) = 22/23.
DPatrick 2010-03-18 21:52:49
Ihatepie 2010-03-18 21:52:56
Catalyst is a beast
DPatrick 2010-03-18 21:53:06
Yes, this was a really nice solution that he posted.
DPatrick 2010-03-18 21:53:22
It requires a bit more geometry to set up, but the resulting algebra is much, much simpler. And you don't need Law of Cosines. :)
DPatrick 2010-03-18 21:53:53
Reminder: the complete transcript of tonight's session will be posted on the website as soon as we finish up here.
DPatrick 2010-03-18 21:54:07
(Click on the "Math Jams" link, then the "Transcripts" tab.)
DPatrick 2010-03-18 21:54:17
Also, mark your calendar! The 2010 AIME II is Wednesday, March 31, and we'll be having our Math Jam on Friday, April 2 at 7 PM ET / 4 PM PT.
DPatrick 2010-03-18 21:54:32
Thanks for coming!

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