Math Jams

## 2007 AIME Math Jam

Go back to the Math Jam Archive

AoPS instructors will lead a discussion on all 15 problems from the 2007 AIME.

#### Facilitator: Dave Patrick

DPatrick (17:59:12)
Hello and welcome to the 2007 AIME I Math Jam!

DPatrick (17:59:26)
Before we get started, please note that this classroom is moderated, meaning that students can type into the classroom, but only the moderator can choose a comment to drop into the classroom. This helps keep the discussion organized and on track.

DPatrick (18:00:01)
In general, only [b]well-written[/b] comments will be dropped into the classroom, so please take time to write responses that are complete and easy to read.

DPatrick (18:00:14)
Due to the large size of this Math Jam, I probably won't be able to respond to individual questions, nor post a lot of responses. Please do not complain if your comments are not responded to.

DPatrick (18:00:33)
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 (18:00:49)
Also, these solutions are not necessarily the fastest or slickest, but these solutions 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 (18:01:01)
We'll do all 15 problems, in order.

DPatrick (18:01:05)
Let's get started!

DPatrick (18:01:11)

DPatrick (18:01:14)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-170570223.gif

DPatrick (18:01:24)
(you can click on the link to pop up the problem in a separate window)

tjhance (18:01:43)
find the prime factorization of 24

longwan (18:01:46)
24=2*2*2*3

DPatrick (18:01:54)
Note that 24 = 2^3 * 3. How does this help?

ra5249 (18:01:46)
To be a perfect square and a multiple of 24, it must also be a multiple of 144 (the smallest mult. of 24 also a perfect square)

EunuchOmerta (18:02:02)
multiply by 2 and 3; we need paired factors

DPatrick (18:02:14)
A perfect square must have even exponents for all its prime factors.

Therefore, any multiple of 2^3 * 3 that is a perfect square must in fact be a multiple of 2^4 * 3^2 = 144.

tjhance (18:02:46)
The numbers are of the form 144x where x is also a perfect square

DPatrick (18:02:56)
So all are of the form 144*n^2 for some positive integer n.

DPatrick (18:03:02)

cincodemayo5590 (18:03:09)
So we can take the square root of both sides.

pianoforte (18:03:16)
square root, 12n<1000.

DPatrick (18:03:28)
Take the square root to get 12n < 1000.

redcomet46 (18:02:54)
1000/12=83.33333 which means there are 83 multiples of 12 between 1 and 1000.

DPatrick (18:03:51)
We see that 83 < 1000/12 < 84.

DPatrick (18:03:56)

DPatrick (18:04:22)
As is typical for #1, this was fairly straightforward.

DPatrick (18:04:31)

DPatrick (18:04:37)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-122792126.gif

DPatrick (18:05:00)
For a long-worded problem like this...

13375P34K43V312 (18:04:54)
read it CAREFULLY would be the first step

DPatrick (18:05:14)
Absolutely. I would read it *twice*.

mysmartmouth (18:04:54)
start writing equations, knowing rate times time = distance

guyinPA (18:04:57)
make equations for each person

DPatrick (18:05:39)
Yes. We can let t be the number of seconds that have elapsed since the start (when Al steps on).

DPatrick (18:05:58)
What are the equations for the positions of the three people?

davidyko (18:05:48)
6t, 10(t-2), 8(t-4)

cincodemayo5590 (18:06:10)
A=6t , B=(6+4)(t-2), C= (8)(t-4)

13375P34K43V312 (18:06:10)
A=6t, B=10t-20, C=8t-32

HoratioK (18:06:16)
A = 6t
B = 10t - 20
C = 8t - 32

DPatrick (18:06:29)

DPatrick (18:06:53)
Note that A's rate is 6 (the speed of the walkway), B's rate is 10 (walkway + 4), and C's rate is 8 (his own rate since he's not on the walkway).

DPatrick (18:07:10)
B's ""time"" is t-2 since he starts 2 seconds late.

DPatrick (18:07:17)
C's ""time"" is t-4 since he starts 4 seconds late.

DPatrick (18:07:24)
Now what?

guyinPA (18:07:00)
find average of two of the equations

davidyko (18:07:32)
One is the average of the other two

HoratioK (18:07:56)
three cases: a in the middle, b in the middle, c in the middle

DPatrick (18:08:10)
One person being exactly between the other two means that one of these is the average of the other two.

In other words, the sum of two of these should equal twice the third.

cincodemayo5590 (18:08:13)
Set up the three averages, then see which gives a t that works

DPatrick (18:08:34)
We would need to check 2A = B+C, 2B = A+C, and 2C = A+B and solve for t in each.

ra5249 (18:08:07)
assume A is the average then B then C

Andrew Kim (18:08:33)
It turns out that Al is it in this problem...

DPatrick (18:08:53)
A little thought (or experimentation) will show that it must be the first one.

DPatrick (18:09:00)
2A = B + C

DPatrick (18:09:10)

DPatrick (18:09:27)
So 6t = 52.

meenamathgirl (18:09:21)
Why can the other two be eliminated?

DPatrick (18:09:52)
You can try them and see: they both give negative values for t.

hyang (18:09:45)

Assassin of Death (18:10:05)
one of them is never true and the other results in a negative

DPatrick (18:10:32)
Actually, you're right. One has no solutions for t, the other one gives a negative t.

DPatrick (18:10:41)
Since 6t -- the position of the middle person A -- is what the problem asked for, that's our answer!

DPatrick (18:10:50)

DPatrick (18:11:04)
(Note we don't need to solve for t, though we easily could of course.)

Andrew Kim (18:11:05)
I think many (too many) people missed the Cy not being on the walkway part.

DPatrick (18:11:24)
I know that a lot of students here when I proctored did.

DPatrick (18:11:33)
Reading the problem carefully is always the first step.

DPatrick (18:11:48)

DPatrick (18:11:53)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-102727262.gif

PenguinIntegral (18:11:51)
Equate z^2's and z^3's imaginary parts and solve for b?

mysmartmouth (18:11:53)
Just square and cube (9+bi), then equate the imaginary parts of each.

phoenixstrike (18:12:02)
straightforward i think. just brute force it

tjhance (18:12:08)
expand (9+bi)^2 and (9+bi)^3 and set the imaginary parts equal

davidyko (18:12:09)
Expand both, and compare.

DPatrick (18:12:20)
There's nothing really clever to do here; we just compute!

DPatrick (18:12:29)

DPatrick (18:12:46)

DPatrick (18:13:01)
Notice that I just wrote ""whatever"" for the real part of z^3. We don't need to bother computing it, since we won't need it.

DPatrick (18:13:23)

DPatrick (18:13:43)

mburg (18:13:40)
solve for b

IntrepidMath (18:13:41)

bubble (18:13:43)
divide by b

Andrew Kim (18:13:49)
b can be dived from both sides since it cannot be zero

DPatrick (18:14:00)

DPatrick (18:14:24)
(I thought that this was the easiest problem on the test.)

DPatrick (18:14:41)

DPatrick (18:14:43)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-57766270.gif

DPatrick (18:15:41)
There are probably lots of ways to do this problem. A lot of you are suggesting GCD and LCM solutions. But here's how I did it.

DPatrick (18:15:59)
First, let's get the rates of the three planets, meaning: how many degrees does each planet rotate each year?

EunuchOmerta (18:16:13)
6, 360/84, 360/140

DPatrick (18:16:27)
These simplify to 6, 30/7, and 18/7.

Let's write these as 42/7, 30/7, and 18/7 degrees/year.

DPatrick (18:16:44)
But there's a clever way we can think about it to make the calculation slightly easier.

Imagine the slowest planet is fixed (or, alternatively, imagine that you're rotating your head at the same rate as the slow planet, so that it appears never to move). How do the other planets rotate relative to the 3rd planet?

chessgeek (18:16:55)
subtract rates

IntrepidMath (18:17:01)

Xyteron (18:17:03)

DPatrick (18:17:47)
Aha! Now we see when the next collinearity will take place. It'll be when the first planet has rotated all the way around (relative to the 3rd planet), and the 2nd planet has rotated half-way around, since the 1st planet moves twice as fast (relative to the 3rd planet) as the 2nd.

DPatrick (18:18:18)
The first planet takes 360/(24/7) = 105 years to rotate all the way back to the 3rd planet.

DPatrick (18:18:23)

shubidubab (18:18:32)
How do we know what ""currently collinear"" means? I mean it there are many positions that satisfy that the four objects are collinear.

DPatrick (18:18:52)
True, and this is a valid point: we don't know that all 3 planets start in the same place (meaning that they all lie on the same ray with vertex at the star). But it turns out that it doesn't matter. Think about why.

chessgeek (18:18:57)
can you explain the LCM and GCF solution?

DPatrick (18:19:18)
Check the message board -- a lot of solutions are posted there.

DPatrick (18:19:37)
As I said at the front, we don't have time to do every solution to every problem. I'm just going to present one for each.

DPatrick (18:19:54)

DPatrick (18:20:01)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-183207727.gif

IntrepidMath (18:20:16)
Find C to F formula?

DPatrick (18:20:32)

DPatrick (18:20:41)

probability1.01_4 (18:20:55)
How about substituting X = F - 32 first...

DPatrick (18:21:17)
That's what I did next: the 32 is more or less irrelevant, so let's get rid of it.

DPatrick (18:21:29)
Let's set G = F-32. (So it matches my notes :) )

DPatrick (18:21:34)

EunuchOmerta (18:20:23)
calculate a few values; 9 sounds good because it should repeat every 9 because we're dividing by 9

cincodemayo5590 (18:21:45)
mod 9, or mod 45 ?

indianamath (18:22:11)
mod 9

DPatrick (18:22:30)
We see that G mod 9 is what's important, so let's make a chart of the different residue classes mod 9:

DPatrick (18:22:43)

Assassin of Death (18:22:51)
residue classes?

DPatrick (18:23:15)
That just means the different possible remainders for G when we divide by 9.

DPatrick (18:23:39)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-175920676.gif

DPatrick (18:23:47)
(that's a link for the table if you didn't get it on your screen)

exponent (18:23:27)
So it works 5 out of 9 times.

apusey (18:22:32)
5 out of the first 9 work

DPatrick (18:24:17)
Right: we conclude that if G is equivalent to one of {0,2,4,5,7} mod 9, then it will work.

DPatrick (18:24:35)
How many such G are there with 0 <= G <= 968?

davidyko (18:24:34)
Divide 968 by 9.

kshweh (18:24:37)
I think 9k+8 is screwed up

DPatrick (18:24:53)
You're right -- my table has a typo.

DPatrick (18:25:11)
The last row should be 9k+8 5k+4 9k+7

DPatrick (18:25:17)

DPatrick (18:25:32)
This is why you should always check your work. :)

exponent (18:25:25)
Then we just need to divide 968 by 9, multiply the result by 5, and use the table to work out the remainder

Mmew (18:25:38)
968/9 paying attention to remainder(s)

DPatrick (18:25:50)
Right.

DPatrick (18:26:04)
The largest multiple of 9 below 968 is 963.

So for 0 <= G <= 963 we have (963/9)*5 = 535 numbers.

DPatrick (18:26:27)
Actually, that should be for 0 <= G < 963.

DPatrick (18:26:40)
We have to check the top numbers by hand.

EunuchOmerta (18:26:43)
963 = 9k, 964 = 9k+1... 968=9k+5

DPatrick (18:26:56)
The remaining possible values are 963, 964, 965, 966, 967, and 968. 963, 965, 966, and 968 all work, since they're 0, 2, 4, 5 (respectively) mod 9.

apusey (18:26:57)
plus 963, 965, 967, and 968

DPatrick (18:27:27)
Right (I have a lot of typos in this problem [img id=em-1])

DPatrick (18:27:37)
The remaining possible values are 963, 964, 965, 966, 967, and 968. 963, 965, 967, and 968 all work, since they're 0, 2, 4, 5 (respectively) mod 9.

EunuchOmerta (18:27:22)
add these 4 to 535 = 539

DPatrick (18:27:48)

DPatrick (18:28:11)
This was a problem that you might have done all the work correctly, but miscounted at the end, so be careful all the way through!

jamieding51 (18:28:04)
Is there a method for people who don't know much number theory?

DPatrick (18:28:35)
You don't need to know any really advanced number theory.

DPatrick (18:28:41)
The key step was the chart.

DPatrick (18:29:00)
...knowing to look at the different possible remainders when dividing by 9.

DPatrick (18:29:13)

DPatrick (18:29:18)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-165165039.gif

cincodemayo5590 (18:29:24)
A number-line is a good place to start.

DPatrick (18:29:50)
Right, I started by just drawing a line with all the possible stopping points.

DPatrick (18:29:56)
[b]0[/b] 3 6 9 12 [b]13[/b] 15 18 21 24 [b]26[/b] 27 30 33 36 [b]39[/b]

DPatrick (18:30:00)
The multiples of 13 are in bold.

cincodemayo5590 (18:30:17)
Then, we can count the number of ways to get to the first multiple of 13, and the one way to jump over it.

mburg (18:30:33)
but from 3,6,9,12 the frog can go directly to 13

DPatrick (18:30:52)
We can just start listing, from left to right, the number of ways to get to each point.

chessgeek (18:31:01)
5 ways to get to 13

DPatrick (18:31:18)
There's only one path through each of 3, 6, 9, 12.

DPatrick (18:31:24)

DPatrick (18:31:44)
We can jump there straight to 13 from any of 0,3,6,9,12. So there are 5 ways.

DPatrick (18:31:48)

frodo (18:31:23)
We don't have to get through 13 right?

HoratioK (18:31:42)
13: 5, 15: 6

apusey (18:31:55)
6 ways to 15

The Zuton Force (18:31:59)
1+5 = 6 ways to get to 15

ra5249 (18:32:03)
we can go to 15 in 6 ways then?

DPatrick (18:32:28)
Right. We can get to 15 either from 12 or from 13. So there are 6 ways to 15.

ra5249 (18:32:23)
18 in 6 ways also, same for 21, 24

DPatrick (18:32:43)
For each of 15,18,21,24, there are 6 ways there: we can jump there either from 12 skipping 13, or we can jump there from 13.

DPatrick (18:32:49)

apusey (18:32:37)
29 wayst to get to 26

xelaw07 (18:32:57)
29 ways to 26

guyinPA (18:33:00)
29 to 26

DPatrick (18:33:11)
This makes 5+6+6+6+6 = 29 paths to 26 (from 13, 15, 18, 21, 24)

DPatrick (18:33:18)

The Zuton Force (18:33:26)
And then 6+29=35 ways to get to 27

hyang (18:33:41)
6+29=35 for 27

DPatrick (18:34:02)
Right: to get to 27, we can come either from 24 or 26. So there are 6+29 = 35 paths.

DPatrick (18:34:07)
Same for 30, 33, 36.

DPatrick (18:34:13)

apusey (18:34:26)
169 to 39 becasue 35 to each of 27 30 33 and 36 and 29 to 26

chessgeek (18:34:30)
so 29+35+35+35+35=169 ways to get to 39

DPatrick (18:34:44)

DPatrick (18:35:01)

jamieding51 (18:35:09)
Is it just a coincidence that 13^2 = 169?

HoratioK (18:35:25)
is it a coincidence that 169 = 13 ^ (3 - 1)?

DPatrick (18:35:48)
I'm not sure...I hadn't even noticed until you pointed it out!

DPatrick (18:36:06)

DPatrick (18:36:12)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-74920521.gif

davidyko (18:36:20)
Note that the floor and ceiling functions are equal iff k is a power of 2.

calc rulz (18:36:28)
Think about what the thing inside the parentheses can be.

daermon (18:36:29)
is it true that it will be 0 if k is a power of 2?

DPatrick (18:36:54)
Right. We notice that the term in parentheses is either 0 or 1.

And, it's only 0 if k is a power of 2.

monkeymatt (18:37:25)
So we add all of the numbers that are not powers of 2.

IntrepidMath (18:37:32)
so then we have 1000*1001/2 - (1+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9)

daermon (18:37:38)
so its 1000*1001/-#of powers of 2?

DPatrick (18:38:25)
That's a good approach: we can sum all 1000 terms, and subtract the ones that aren't there (where the parentheses term is 0) by subtracting all the powers of 2.

DPatrick (18:38:36)
I did something slightly different (but which amounts to the same thing).

DPatrick (18:38:49)
We can pair the terms: the k term pairs with the 1000-k term. If the parentheses part is 1 for both terms, then these terms will add to 1000, so they won't contribute anything to the final answer.

DPatrick (18:39:06)
(But beware: the k=500 term doesn't have a pair!)

DPatrick (18:39:55)
So all we need to add is the sum of all the (1000-k)'s, where k is a power of 2. This is the same as subtracting all the powers of 2.

DPatrick (18:40:21)
In any event, we get 500-(1+2+4+...+512).

guyinPA (18:40:43)
2^0+...2^9is 2^10 - 1

DPatrick (18:41:19)
Right, the term in parenthesis is 1024-1 = 1023, so since we only care about the last 3 digits, we can pretend that it's 023.

DPatrick (18:41:33)

DPatrick (18:42:18)

DPatrick (18:42:21)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-169799631.gif

Note: this problem was presented incorrectly in the original Math Jam. What follows is an edited, corrected solution.

DPatrick (18:42:50)
If two polynomials are each factors of a third, then what else is a factor?

PenguinIntegral (18:42:46)
GCD.

HoratioK (18:43:10)
their gcf

DPatrick (18:43:29)
Right.

DPatrick (18:43:44)
In particular, we can use the Euclidean Algorithm to combine Q_1 and Q_2 to get rid of the quadratic term.

DPatrick (18:43:53)

DPatrick (18:44:02)
What do we conclude from this?

calc rulz (18:44:36)
15x+3k?

DPatrick (18:45:02)
Yeah, lost an important x there.

DPatrick (18:45:14)

calc rulz (18:45:01)

frodo (18:45:19)
x+k/5 is a factor.

DPatrick (18:45:57)
Right. We get that this polynomial -- x + k/5 (after we divide by 15) -- must be a factor of P(x). Why? Because the GCD of Q1 and Q2 is a divisor of 15x+3k, so it's either a linear factor or a constant. But if the GCD would be a constant, then each of the two quadratics would countribute 4 roots to P(x), which is impossible as P(x) is a cubic polynomial.

DPatrick (18:46:07)
So -k/5 is a root.

calc rulz (18:46:16)
Plug it back into one of the equations!

DPatrick (18:46:50)
Indeed, we can plug in x = -k/5 back into the Q's, and solve for k.

DPatrick (18:47:04)
If -k/5 is a root of Q_1, then we have the following equation:

DPatrick (18:47:07)

Assassin of Death (18:47:19)
why are they equal to 0?

DPatrick (18:47:42)
because -k/5 must be a root of Q_1(x), so if we substitute x = -k/5 we must get 0.

DPatrick (18:48:08)

IntrepidMath (18:48:09)
k=30

xelaw07 (18:48:21)
k=0,30

DPatrick (18:48:43)
This has solutions k=0 and k=30.

calc rulz (18:48:30)

HoratioK (18:48:41)
30 largest

DPatrick (18:49:00)

DPatrick (18:49:11)
We can check it in Q_2 if we like as well:

DPatrick (18:49:16)

DPatrick (18:49:25)

DPatrick (18:49:38)
Same quadratic, same roots: k=0 and k=30.

meenamathgirl (18:48:57)

jnApp66 (18:49:09)
why does q2(x)-2q1(x) have to be a root

DPatrick (18:50:43)
Because any common root of Q1 and Q2 is also a root of Q2-2Q1.

DPatrick (18:51:05)
It's basically the Euclidean Algorithm for polynomials.

DPatrick (18:51:24)
Think about it some more on your own, and/or post to the message board thread for this problem.

DPatrick (18:51:30)
Let's move on:

DPatrick (18:51:35)

DPatrick (18:51:41)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-118431214.gif

mburg (18:51:45)
draw a picture

davidyko (18:51:52)
Draw a diagram.

DPatrick (18:52:28)
Certainly we'll want a picture.

HoratioK (18:52:07)
AB = 34

guyinPA (18:52:27)
the hopetnuse is 34

DPatrick (18:52:47)
I would notice that off the bat too: ABC is a 8-15-17 triangle.

DPatrick (18:52:58)
Note that O1 and O2 lie on a line parallel to AB. Let's draw the picture.

DPatrick (18:53:03)

DPatrick (18:53:05)

knexpert (18:52:59)
let the radius of each circle be ;$r$

DPatrick (18:53:20)
Sure, letLet the unknown radius be r.

DPatrick (18:53:41)
Naturally, we'll also want to draw all the relevant radii from the two circles to their points of tangencies.

DPatrick (18:53:52)

DPatrick (18:53:54)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/704900b59ce0a12d6eaf039e9e90b6ae.png

DPatrick (18:54:15)
I've seen many ways to proceed from here. I'm not sure which is best, but there's one thing I noticed right away from this diagram.

daermon (18:54:35)
seems like we have a bunch of similar triangles

mysmartmouth (18:54:39)
a bunch of similar triangles

davidyko (18:54:40)
Similar triangles?

DPatrick (18:55:09)
Parallel lines give similar triangles!

DPatrick (18:55:17)

DPatrick (18:55:42)

DPatrick (18:55:57)
So side HJ is (17/8 + 2 + 17/15)r.

DPatrick (18:56:12)
What now?

daermon (18:56:13)
so maybe should we find CJ and CH and use pythag?

DPatrick (18:56:43)
We can find CJ and CH, since the big triangle CHJ is also similar to all the small ones.

DPatrick (18:56:56)
We know that CJ = (8/17)HJ = (8/17)(17/8+2+17/15)r = (1+16/17+8/15)r.

DPatrick (18:57:08)
Similarly (pun intended), CH = (15/17)HJ = (15/17)(17/8+2+17/15)r = (15/8+30/17+1)r.

DPatrick (18:57:25)
(Notice how I'm not combining these fractions; don't combine them until we need to, leave them in this simpler form for now.)

daermon (18:57:11)
hmm won't this all cancel out because we know theyre all 8-15-17s?

DPatrick (18:57:44)
A lot of it will cancel out!

DPatrick (18:58:00)
For example, we can find AF and BG now.

DPatrick (18:58:26)
We also that CB = 16 and GJ = (8/15)r, so BG = (1+16/17+8/15)r - (8/15)r - 16 = (33/17)r - 16.

DPatrick (18:58:59)
(You should probably be labeling all these lengths on your diagram, as I did when I first did this problem. It looked messy on the computer, though, so I didn't include them.)

DPatrick (18:59:07)
Also AF = (15/8+30/17+1)r - (15/8)r - 30 = (47/17)r - 30.

bubble (18:59:28)
what was the problem asking for again?

DPatrick (18:59:57)
We're trying to find r. (This is important -- when doing a geometry problem, don't lose track of what we're trying to do!)

jamieding51 (18:59:57)
Could you put the picture up again?

DPatrick (19:00:05)
Yeah:

DPatrick (19:00:13)

DPatrick (19:00:15)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/704900b59ce0a12d6eaf039e9e90b6ae.png

DPatrick (19:00:25)
So we've got all the lengths of all the outside segments.

DPatrick (19:00:29)
How do we finish up?

gstsclq (19:00:24)
AF+BG+2r=34

Andrew Kim (18:59:50)

DPatrick (19:00:52)

DPatrick (19:00:56)
Same way we know BE.

DPatrick (19:01:07)
This means that AB = AD + DE + EB = AF + DE + EB = (47/17)r -30 + 2r + (33/17)r - 16.

This simplifies to (114/17)r - 46.

But we know that AB = 34! So we can equate and solve for r!

DPatrick (19:01:34)
(114/17)r - 46 = 34

frodo (19:01:35)
17*80/114=r

DPatrick (19:01:54)
r = (80*17)/114 = (40*17)/57 = 680/57.

DPatrick (19:02:00)

DPatrick (19:02:19)
Like a lot of geometry problems, this didn't require any heavy machinery -- just similar triangles!

Ars (19:01:24)
I used trig identities on EB and AD immediately it was easier

DPatrick (19:02:43)
There were certainly other solutions: two that I know of are using trig and using coordinates.

DPatrick (19:02:51)
But this is the most ""low-tech"" solution.

DPatrick (19:03:10)

DPatrick (19:03:14)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-186260094.gif

DPatrick (19:03:25)

DPatrick (19:03:27)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/326521a8b7536ccc34def612ccb55508.png

cincodemayo5590 (19:03:27)
Symmetry and case-work?

PenguinIntegral (19:03:35)
Anything beside casework?

DPatrick (19:04:23)
I've seen basically two solutions for this: one that focuses on the columns, and one that focuses on the rows. I happened to do it focusing on the rows.

DPatrick (19:04:39)
What are the possibilities for any given row?

frodo (19:04:46)
4C2 ways to color the top row.

cincodemayo5590 (19:04:47)
c(4,2) = 6

irisxu0 (19:04:51)
6C2

daermon (19:04:52)
hmm there are 6 ways to scramble each row

DPatrick (19:05:28)
There are C(4,2) = 6 choices for each row (since we have to choose 2 squares to be black), but the way I thought about it is that there are 3 patterns:

DPatrick (19:05:41)

DPatrick (19:05:46)

DPatrick (19:05:54)

DPatrick (19:06:09)
Note that we use two shades of grey, rather than white and black, to distinguish the two colors.

k8reindeer (19:06:05)
and all are reversible

DPatrick (19:06:19)
Right.

DPatrick (19:06:26)
And how many of each pattern must be use?

alexluh (19:06:36)
two

jamieding51 (19:06:40)
2?

Andrew Kim (19:06:44)
2 2 2?

DPatrick (19:07:29)
If we have any row with one pattern, then another row must have the same pattern with the colors reversed.

daermon (19:07:36)
why?

k8reindeer (19:07:40)
why?

PenguinIntegral (19:07:43)
Proof?

DPatrick (19:07:59)
Let me just sketch the argument.

DPatrick (19:08:06)
Suppose we had an odd number of these:

DPatrick (19:08:14)

DPatrick (19:08:43)
Each of the other two patterns gives an equal number of white and black squares to the dark-shaded columns and the light-shaded columns.

DPatrick (19:09:07)
But if we have an odd number of these, two of the columns are going to have more black than white, and vice versa.

DPatrick (19:09:28)
This is obviously a pretty rough sketch of the proof -- you might think about it more on your own.

DPatrick (19:09:47)
Anyway, the point is that each pattern needs to appear an even number of times.

cincodemayo5590 (19:10:15)
So we're picking three pairs, with replacement, then we can worry about the ordering.

DPatrick (19:10:24)
Right.

DPatrick (19:10:43)
The three cases are: all 3 pairs the same, 2 of one type and one of another, and all 3 different.

DPatrick (19:11:06)
How many if all 3 pairs are the same?

HoratioK (19:11:12)
3

guyinPA (19:11:13)
3

daermon (19:11:13)
3?

DPatrick (19:11:47)
Yeah, there's 3 choices for which pattern, but then we have to choose which rows start with a white square and which rows start with a black square.

DPatrick (19:11:58)
There are 3 choices for the type, and then C(6,3) = 20 choices for which 3 rows start with black, for a total of 3*20 = 60 possibilities.

DPatrick (19:12:18)
I'm going to skip ahead to the other two cases...

DPatrick (19:12:30)
Case 2: 4 rows of one type, 2 rows of another. There are 3*2 choices for the types, then C(6,2) choices for which 2 rows get the 2nd type, then C(4,2) choices for which of the 4 rows of the first type start with black, then 2 choices for which of the 2 rows of the second type starts with black.

DPatrick (19:12:52)
This is a total of 6*C(6,2)*C(4,2)*2 = 6*15*6*2 = 1080 possibilities in this case.

DPatrick (19:13:07)
Case 3: 2 rows of each of the 3 types. There are C(6,2)*C(4,2) choices for assigning the rows to types, then 2^3 choices for which rows (in each type) starts with black.

DPatrick (19:13:21)
This is a total of C(6,2)*C(4,2)*8 = 15*6*8 = 720 possibilities in this case.

DPatrick (19:13:38)

DPatrick (19:14:11)
Like I said, you could also do this problem by focusing on the columns instead of the rows; check the message board for this alternate solution.

DPatrick (19:14:24)
And there are other ways too.

k8reindeer (19:14:37)
would the columns be easier because there are only 4, not 6?

DPatrick (19:15:02)
Not necessarily. My train of thought was that there were only 3 patterns for the rows, and I went from there.

DPatrick (19:15:25)

DPatrick (19:15:29)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-169167038.gif

DPatrick (19:15:48)

Umm the rounded square root

SaifHakim (19:16:13)
round(sqrt(p))

xelaw07 (19:16:15)
sqrt(p) rounded to the nearest integer

Someone (19:16:16)
take the square root, round to the nearest number

DPatrick (19:16:30)

DPatrick (19:16:38)

DPatrick (19:16:59)
You could do a few small values of p and see the pattern pretty quickly I think.

frodo (19:16:47)
(p-1/2)^2+1<=k<=(p+1/2)^2

DPatrick (19:17:34)
Right...if you wanted to be more rigorous, we can examine (k+1/2)^2.

DPatrick (19:17:47)

DPatrick (19:18:02)

DPatrick (19:18:38)
Right...though you may have meant to say 2k values for every k.

DPatrick (19:18:50)

cincodemayo5590 (19:18:29)
and for the remaining values will be k+1. So we can set up a summation!

DPatrick (19:19:13)
Right...how high does k go?

Ars (19:19:24)
44

HoratioK (19:19:34)
45

DPatrick (19:19:55)
(My question was a little ambiguous.) We see that 44*45 = 1980 and 45*46 = 2070, so all values of p above 1980 will have b(p) = 45. There are 2007-1980 = 27 such numbers.

Sum it

DPatrick (19:20:27)

DPatrick (19:20:51)
The sum of the first n squares is given by the formula n(n+1)(2n+1)/6.

DPatrick (19:20:59)

DPatrick (19:21:26)
Now you have to be careful in computing this!

calc rulz (19:21:30)
15(81+44*89)

DPatrick (19:21:54)
Yeah, I'd factor out 15 to get 15((44)(89)+81) = 15(3997).

DPatrick (19:22:15)
We can replace 3997 with -3, since we only care about the remainder upon division by 1000.

DPatrick (19:22:30)

DPatrick (19:23:19)
Even though it's still me, I'm going to turn thing over to Naoki Sato, who wrote our solutions for #12 and #13...

DPatrick (19:23:38)

DPatrick (19:23:48)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-23764942.gif

DPatrick (19:23:55)
First, we draw a diagram. We must draw a very accurate diagram, otherwise we may not notice that the region common to both triangles is a quadrilateral, not a triangle. This is one problem for which a ruler and a protractor would really come in handy to draw a very accurate diagram!

DPatrick (19:24:06)
Let B' and C' be the images of B and C under the rotation.

DPatrick (19:24:19)

DPatrick (19:24:22)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/ac1b790a7cbec3a6e443c3144813cd40.png

DPatrick (19:24:27)
Where can we start?

calc rulz (19:24:38)
How do we know B' is outside ABC?

DPatrick (19:25:21)
That's why we draw an accurate diagram. Ruler, compass, and protractor are permitted, and will probably come in handy here.

cincodemayo5590 (19:24:33)
Angle chasing?

PenguinIntegral (19:25:16)
Find some angles

DPatrick (19:26:08)
Let's start by filling in as many angles as we can.

IntrepidMath (19:25:22)
C'AC=15 degrees

frodo (19:26:11)
b'ac=60 degrees

knexpert (19:26:13)
CAB'=60

mburg (19:26:14)
we know <bac is 75 so it was rotated 15 degrees

DPatrick (19:26:38)

DPatrick (19:26:40)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/e1f3a6943e8419a3687e349ec6ecdcaf.png

irisxu0 (19:26:17)
angle chasing shows AB' perpindicular to CB

DPatrick (19:27:02)
Note that AB' is perpendicular to BC.

DPatrick (19:27:19)
Let D be the intersection of BC and B'C', and let E be the intersection of AB' and BC.

DPatrick (19:27:30)

DPatrick (19:27:32)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/544d2c7e2b1a20c2ae6989a8f3f04576.png

DPatrick (19:27:41)
We have that B'C' is the image of BC under the rotation around A, and that D is the intersection of BC and B'C'. So what is AD with respect to lines BC and B'C'?

cincodemayo5590 (19:27:56)
angle bisector?

cincodemayo5590 (19:28:09)
oooh, that's cool :)

k8reindeer (19:28:17)
it bisects the obtuse angle there!

knexpert (19:28:28)
angle bisector

DPatrick (19:28:40)
AD must be an angle bisector of BDC'.

DPatrick (19:28:45)
To make this fact more clear, let E' be the image of E under the rotation.

DPatrick (19:28:53)

DPatrick (19:28:56)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/11a8ab243e9ed8df70ea754743911503.png

DPatrick (19:29:01)
Since AE is perpendicular to BC, AE' is perpendicular to B'C'. Also, AE' = AE. What can we conclude?

Elemennop_2 (19:29:34)
HL congruency

k8reindeer (19:29:41)
the triangles are congruent. HL.

DPatrick (19:29:53)

DPatrick (19:30:01)
In particular, angles <DAE and <DAE' are equal. What are they?

k8reindeer (19:30:33)
7.5 degrees

b-flat (19:30:40)
7.5

guyinPA (19:30:45)
7.5

15/2

DPatrick (19:30:58)
Since <E'AE is 15 degrees (because this is the angle of rotation), <DAE = <DAE' = 7.5 degrees.

DPatrick (19:31:04)
Let F be the intersection of AC and B'C'.

DPatrick (19:31:12)

DPatrick (19:31:15)

DPatrick (19:31:19)
What can we say about triangle AE'F?

frodo (19:31:27)
45-45-90

Elemennop_2 (19:31:28)
Right Isosceles.

k8reindeer (19:31:32)
it's a 45-45-90 triangle

Assassin of Death (19:31:32)
45-45-90

xelaw07 (19:31:34)
Isosceles right triangle

DPatrick (19:32:03)
It's a 45-45-90 triangle!

DPatrick (19:32:13)

DPatrick (19:32:16)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/cd1dbc301c1f7f51548d7c05ff2e6153.png

cincodemayo5590 (19:31:45)
AE' = E'F

DPatrick (19:32:26)
Then FE' = AE' = AE.

k8reindeer (19:31:57)
now we need to find some sides...

DPatrick (19:32:48)
Now we are in a position to start calculating the area of the common region, which is quadrilateral DEAF.

DPatrick (19:33:04)
The diagram already tells us to split this quadrilateral into triangles AE'F, AE'D, and AED.

DPatrick (19:33:15)
We know AB is 20. What is AE?

Elemennop_2 (19:33:23)
20cos 15.

frodo (19:33:43)
20cos15

DPatrick (19:34:10)
AE = AB cos 15 = 20 cos 15.

DPatrick (19:34:17)
Then FE' = AE' = AE = 20 cos 15.

DPatrick (19:34:23)
What is the area of triangle AE'F?

cincodemayo5590 (19:34:42)
1/2 AE^2

bubble (19:34:44)
(20cos15)^2/2

guyinPA (19:34:48)
200cos15^2

Elemennop_2 (19:34:49)
200(cos15)^2

calc rulz (19:34:52)
400cos^2(15)/2 = 200*cos(15)*cos(15)

shubidubab (19:34:58)
200 cos^2 15

DPatrick (19:35:17)

DPatrick (19:35:25)
How can we calculate cos^2 15?

davidyko (19:35:32)
Half-angle?

knexpert (19:36:01)
double angle\

DPatrick (19:36:15)
We can use the double angle formula:

DPatrick (19:36:21)

Elemennop_2 (19:35:32)
1/2 (1+cos 30)

Assassin of Death (19:35:36)
(1+cos30)/2

Mmew (19:35:48)
(1+cos(30))/2

calc rulz (19:36:15)
(1/2+cos(30)/2)

DPatrick (19:37:03)
Then

DPatrick (19:37:14)

frodo (19:37:05)
100+50rt3

DPatrick (19:37:23)
What about the areas of triangles AE'D and AED?

guyinPA (19:37:31)
equal

knexpert (19:37:35)
they are the same

DPatrick (19:38:03)
These are both right triangles, so

DPatrick (19:38:09)

frodo (19:38:06)
AE=20cos15

xelaw07 (19:37:45)
We need to find DE

DPatrick (19:38:20)
We know AE. What is DE?

frodo (19:38:50)
20cos15sin7.5

knexpert (19:38:58)
20tan 7.5 cos 15

DPatrick (19:39:05)

DPatrick (19:39:13)
Therefore,

DPatrick (19:39:17)

k8reindeer (19:38:34)
do we need to find the tan7.5? that's icky

DPatrick (19:39:37)
We know cos^2 15 already. We just need to compute tan 7.5. How can we do that?

Elemennop_2 (19:39:47)
half angles

k8reindeer (19:39:50)
half angle formula again...

knexpert (19:39:55)
half angle for tangent

DPatrick (19:40:22)
Since 7.5 is half of an integer, we can use the half-angle formula:

DPatrick (19:40:30)

calc rulz (19:39:53)

xelaw07 (19:40:11)
tan7.5=cos15/(1+sin15)

DPatrick (19:40:43)
Therefore,

DPatrick (19:40:47)

b-flat (19:35:51)

DPatrick (19:41:05)
We also know

DPatrick (19:41:10)

Someone (19:35:44)
cos15 = cos(45-30)

eryaman (19:35:57)
cos(45-30)

Andrew Kim (19:36:15)
you can use the angle addition formula cos(45-30)=cos45cos30+sin45sin30

DPatrick (19:41:24)
(These can be derived from trigonometric identities, and using the fact that 15 = 45 - 30. These are good identities to have in your pocket.)

DPatrick (19:41:29)
So in the formulas for tan 7.5 above, which is easier to use?

cincodemayo5590 (19:41:57)
the second.

cincodemayo5590 (19:42:09)
Less rationalizing. :)

DPatrick (19:42:19)
Probably the second one, as the denominator is less complicated.

DPatrick (19:42:22)
So,

DPatrick (19:42:25)

DPatrick (19:42:33)
Now what?

knexpert (19:42:40)
rationalize

cincodemayo5590 (19:42:47)
multiply by the conjugate. and have fun! :P

multiply by the conjugate

SaifHakim (19:42:53)
multiply by (sqrt(6)+sqrt(2))/(sqrt(6)+sqrt(2)

guyinPA (19:43:11)
multiply by 6^.5+2^.5

DPatrick (19:43:21)
Now we rationalize the denominator:

DPatrick (19:43:23)

mburg (19:43:31)
put it all together - we are looking at the coeficients for the answers anyway

DPatrick (19:43:55)
The rest is now just a calculation:

DPatrick (19:44:00)

DPatrick (19:44:16)
Therefore,

DPatrick (19:44:21)

longwan (19:44:17)
we need to be careful on calculations

DPatrick (19:44:32)
Always!

DPatrick (19:44:37)

knexpert (19:44:36)

Elemennop_2 (19:44:42)
.5 (500+350+300+600) = .5 (1750) = 875.

davidyko (19:44:47)
1750/2 = 875

guyinPA (19:44:54)
875

cincodemayo5590 (19:45:01)
875 ?

DPatrick (19:45:19)

Mmew (19:45:17)
That was quite the lengthy problem.

DPatrick (19:45:53)
Yes it was. There doesn't seem to be any way of avoiding long calculations.

DPatrick (19:46:06)

DPatrick (19:46:14)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-174007358.gif

knexpert (19:46:15)
diagram

DPatrick (19:46:24)
First, let's draw a diagram. Let P, Q, and R be the midpoints of AE, BC, and CD, respectively.

DPatrick (19:46:37)

DPatrick (19:46:40)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/dfc96e2b16ee61f0c2cf251c60fb8ceb.png

frodo (19:46:34)
Make it on an 3-D coordinate system.

DPatrick (19:46:53)
The easiest way to tackle this problem seems to be to use three-dimensional coordinates. (There may also be some clever vector-based solutions, but I found that the 3-D coordinate solution was the most straightforward.)

DPatrick (19:46:58)
First, we need the height of the pyramid. How can we calculate this?

DPatrick (19:47:09)
Let O be the midpoint of the square base ABCD. We want OE.

DPatrick (19:47:16)
Since O is the projection of E onto the base, angle COE is 90 degrees. How can we use this to calculate OE?

DPatrick (19:47:24)

DPatrick (19:47:27)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/0f74495b4b940862802b1983dd99df4e.png

Assassin of Death (19:47:16)
distance formula in three vars

maniacman384 (19:47:26)
right triangles?

knexpert (19:47:34)
pythagorean theorem

DPatrick (19:47:51)
By Pythagoras, CO^2 + OE^2 = CE^2.

DPatrick (19:47:56)
What is CO?

frodo (19:47:23)
2rt2

Kamil Witek (19:47:40)
its 2 root 2

frodo (19:47:42)
OC=2rt2

Someone (19:48:07)
2 rt.2

knexpert (19:48:08)
2rt2

mburg (19:48:17)
1/2 the diagonal

DPatrick (19:48:33)
CO is half the diagonal of a square of side length 4, so CO = 1/2 * 4 * sqrt(2) = 2 * sqrt(2).

DPatrick (19:48:38)
What is CE?

cincodemayo5590 (19:48:44)
4

SaifHakim (19:48:45)
4

Someone (19:48:47)
4

bubble_2 (19:48:51)
4?

DPatrick (19:49:04)
CE = 4.

Elemennop_2 (19:47:35)
EC = 4, OC = 2rt2, so OE = 2rt2.

Kamil Witek (19:47:51)
its 2 root 2

maniacman384 (19:48:08)
2root2

knexpert (19:48:45)
EO=2rt2

irisxu0 (19:48:45)
EO=2sqrt(2)

DPatrick (19:49:27)
Therefore, OE^2 = CE^2 - CO^2 = 16 - 8 = 8, so OE = 2 sqrt(2).

DPatrick (19:49:43)
We place the pyramid in coordinate space, so that the square base ABCD lies in the yz-plane, and its center coincides with the origin.

DPatrick (19:49:55)

DPatrick (19:49:58)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/2ef92a91f70915d099447b07dfb5cbd6.png

DPatrick (19:50:03)
We want to find the area of the intersection of plane PQR with the pyramid, so let's try to find the equation of plane PQR.

DPatrick (19:50:11)
The equation of plane PQR has the form ax + by + cz = d, for some constants a, b, c, and d. When we plug in the coordinates of P, Q, and R, we obtain the equations

DPatrick (19:50:18)

DPatrick (19:50:26)
Solving, we get a = d/2 and b = -d/2. What is c?

Elemennop_2 (19:50:49)
drt2.

IntrepidMath (19:50:52)
c=d(root2)

knexpert (19:50:53)
d\rt2

mthiyaga (19:50:56)
c=d(sqrt2)

DPatrick (19:51:10)

DPatrick (19:51:15)
so c = sqrt(2) * d.

DPatrick (19:51:20)
Therefore, the equation of plane PQR is

DPatrick (19:51:23)

DPatrick (19:51:25)
Dividing both sides by d/2, we get

DPatrick (19:51:30)

DPatrick (19:51:33)
What should we do next?

Elemennop_2 (19:51:43)
There are two more points it hits the pyramid at.

The Zuton Force (19:51:53)
Find points of intersection with the edges

yjneb (19:52:09)
Find intersection pts w/ other edges

DPatrick (19:52:28)
We should find where the plane intersects BE and DE.

DPatrick (19:52:35)
Since B = (2,2,0) and E = (0,0, 2sqrt(2)), a point on the line BE is given parametrically by

DPatrick (19:52:41)

DPatrick (19:52:56)
Plugging this into the equation of plane PQR, we get

DPatrick (19:53:01)

DPatrick (19:53:09)
so t = 3/4.

DPatrick (19:53:21)
Let S be the intersection of plane PQR and line BE. Then S = (3/2, 3/2, sqrt(2)/2).

DPatrick (19:53:29)

DPatrick (19:53:32)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/3c4092499f9f3bac479c05019daba985.png

DPatrick (19:53:37)
Let T be intersection of plane PQR and DE.

DPatrick (19:53:52)
To find the coordinates of T, we could go through all the calculations, but it easier to say that by symmetry, T = (-3/2, -3/2, sqrt(2)/2).

DPatrick (19:54:05)

DPatrick (19:54:08)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/2fa7ca2d8b8ae7561038d8e60c5566da.png

DPatrick (19:54:23)
Now we can try to start finding the area of pentagon PSQRT. How can we do that?

Elemennop_2 (19:54:41)
Triangle PST and Trapezoid RTSQ, I think.

frodo (19:54:33)
It is an isosceles triangle on top of an isosceles trapezoid.

frodo (19:55:04)
Draw TS

DPatrick (19:55:59)
We can split pentagon PSQRT into triangles PST and trapezoid SQRT.

DPatrick (19:56:12)

DPatrick (19:56:14)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/0368894c72a86682377b72881119552f.png

DPatrick (19:56:20)
What are the relevant dimensions we need?

The Zuton Force (19:54:49)
We need the length from P to midpoint of R and Q, as well as ST

guyinPA (19:56:28)
ts is 3rt2

k8reindeer (19:56:47)
TS, RQ, and total height

shubidubab (19:56:56)
TS RQ and P to the midpoint of RQ

guyinPA (19:57:09)
rq is 2 rt2

IntrepidMath (19:57:14)
RQ=2rt2

knexpert (19:56:44)
TS, RQ, and the distance from P to RQ

knexpert (19:57:45)
TS=3rt2

DPatrick (19:58:09)
We need base ST, base QR, the height of triangle PST, and the height of trapezoid SQRT. All of these are easy to calculate from the three-dimensional coordinates.

irisxu0 (19:58:13)
you could use distance formula to find everything

DPatrick (19:58:24)

DPatrick (19:58:26)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/f9b1795ffd8e094523bd779509f28244.png

DPatrick (19:58:34)
So what is the area of pentagon PSQRT?

trap + triangle whatever that is

yjneb (19:59:06)
3rt5/2+5rt5/2

irisxu0 (19:59:16)
4sqrt(5)?

mthiyaga (19:59:23)
4sqrt5

guyinPA (19:59:25)
4rt5

Andrew Kim (19:59:28)
4sqrt (5)!!! 5rt(5)/2+3rt(5)/2

jnApp66 (19:59:29)
sqrt(80)

frodo (19:59:31)
rt80

yjneb (19:59:32)
4rt5

DPatrick (19:59:50)

IntrepidMath (19:59:53)

yjneb (20:00:00)

DPatrick (20:00:13)

DPatrick (20:00:38)
Now we'll turn things over to Valentin Vornicu, who will solve the last two problems.

Valentin Vornicu (20:01:00)
Hi everybody. Thanks Dave. Actually I solved them yesterday :)

Valentin Vornicu (20:01:18)

Valentin Vornicu (20:01:25)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-104109566.gif

Valentin Vornicu (20:01:53)
This solution is an amalgam of solutions I saw on the message board, and my important idea (one-liner punch line).

Valentin Vornicu (20:01:58)
First of all, note that all terms of the sequence are positive.

frodo (20:01:49)
a_2007/a_2006+a_2006/a_2007

Valentin Vornicu (20:02:18)

Valentin Vornicu (20:02:43)
(I actually solved the problem without ever noticing this fact, but it's shorter this way).

Valentin Vornicu (20:02:50)
The other thing to note is that the 2007 is probably a red herring, meaning that it probably doesn't play a significant role in the solution. So anything we can do to make it go away will probably help.

Valentin Vornicu (20:02:58)
One idea, then, is to try to cancel it out.

Valentin Vornicu (20:03:08)
Any ways to do that?

cincodemayo5590 (20:03:11)
we can move everything else to the side and use the n+1 expression

Valentin Vornicu (20:03:37)

Elemennop_2 (20:03:25)
Take the definition for n = k and n = k-1, then take the difference of the two equations.

Valentin Vornicu (20:03:54)

Valentin Vornicu (20:04:04)
What now?

Valentin Vornicu (20:04:58)
Here comes the punch line that solves the problem: the above expression can be rearranged to look more ""symmetric"":

Valentin Vornicu (20:05:05)

Valentin Vornicu (20:05:35)
What have we obtained?

xelaw07 (20:05:42)
A constant value

Elemennop_2 (20:05:45)
An invariant.

IntrepidMath (20:06:01)
constant formula?

cincodemayo5590 (20:06:01)
a way to compare a_1, a_2 to a_2006, a_2007 !

The Zuton Force (20:06:05)
An invariant.

Valentin Vornicu (20:06:15)

Valentin Vornicu (20:06:25)

Valentin Vornicu (20:07:01)
This is indeed a much nicer recurrence relationship, so let's use it a little bit.

Valentin Vornicu (20:07:06)
Since $a_3>a_2$, by induction we can easily check that $a_n >a_{n-1}$ for all $n\geq 3$, so in fact $a_{n+1} > 224 a_n$, for $n\geq 3$, so
$$a_{2007} > 224^{2004} \cdot 672 . \eqno (5)$$

Valentin Vornicu (20:07:11)

Valentin Vornicu (20:07:40)
One idea at this point is to use an approximation argument. It's clear that the sequence grows very rapidly.

Valentin Vornicu (20:07:53)

Valentin Vornicu (20:08:18)
What do we want and what do we know about b_n?

cincodemayo5590 (20:08:29)
and b_n is approximately 224 ?

cincodemayo5590 (20:08:48)
*a little higher than 224 actually

Valentin Vornicu (20:09:10)

Valentin Vornicu (20:09:22)

Valentin Vornicu (20:10:06)
The whole trick in this problem was discovering the constant sequence y_n.

Valentin Vornicu (20:11:09)
Another similar problem you might want to give a try is the following http://www.artofproblemsolving.com/Forum/viewtopic.php?p=492872#p492872

Valentin Vornicu (20:11:32)
which I proposed for the Romanian IMO Team Selection Tests last year. It's based on a very similar idea.

Valentin Vornicu (20:11:40)

Valentin Vornicu (20:11:47)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-24641006.gif

Valentin Vornicu (20:12:09)
Let's sketch a picture. Warning: this picture is not to scale!!

Valentin Vornicu (20:12:14)

Valentin Vornicu (20:12:19)

Valentin Vornicu (20:12:32)

Valentin Vornicu (20:12:47)

Valentin Vornicu (20:12:53)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/e26e30e11b89e3812c4c849e37319b48.png

Valentin Vornicu (20:13:00)
This picture is easier to follow.

Valentin Vornicu (20:13:08)
Since we're given an area, we might try an approach involving areas...

Valentin Vornicu (20:13:13)
Any first ideas?

cincodemayo5590 (20:13:21)
can we just use sine area formula to get a system going?

Altheman (20:13:23)
Sine area formula

Valentin Vornicu (20:13:48)
How would you use it more specifically?

cincodemayo5590 (20:14:04)
Sum the four and it's equal to the larger triangle

Altheman (20:14:08)
[ABC]=[AEF]+[BFD]+[DCE]+[DEF]

Valentin Vornicu (20:14:26)
So let's write up the sum of all the areas of the 4 smaller triangles, which must add up to the area of the triangle $ABC$. We'll use the 1/2*ab*sin(C) formula for most of the triangles.

Valentin Vornicu (20:14:31)

Valentin Vornicu (20:14:38)

Valentin Vornicu (20:14:56)

Valentin Vornicu (20:15:06)
We set this equal to the area of ABC. The sqrt(3)/4 terms cancels (as does the x^2 term), and after simplification, we get:

Valentin Vornicu (20:15:11)

Valentin Vornicu (20:15:28)
But what else do we have?

Altheman (20:15:42)
a 60 degree angle

Valentin Vornicu (20:15:52)
And how do we use it?

cincodemayo5590 (20:15:48)
use law of cosines to find teh side lengths of the smaller triangle?

Valentin Vornicu (20:16:44)
That crossed my mind yesterday, but it involves huge computations. As I only allowed a few minutes to solve this problem I tried to think about something different (read easier).

mburg (20:16:11)
weknow the area and one angle in edf so we can find length df and then have similar triangles

Valentin Vornicu (20:17:14)
I like the similar triangles idea. What triangles are on our picture that are similar?

Valentin Vornicu (20:18:07)
Triangles AEF and CDE are similar! Try to see why (hint: angle chasing).

Valentin Vornicu (20:18:25)

Valentin Vornicu (20:18:41)
Therefore we can set up a ratio of sides:

Elemennop_2 (20:18:23)
2 / y = x-y / 5

Valentin Vornicu (20:18:52)

Valentin Vornicu (20:18:57)
We substitute this into our earlier equation relating x and y, and simplify to get:

Valentin Vornicu (20:19:01)

Valentin Vornicu (20:19:10)

Valentin Vornicu (20:19:19)
We don't have to fully simplify this, as we only care about the squarefree part of the discriminant (under the square root).

Valentin Vornicu (20:19:35)

Valentin Vornicu (20:19:49)

Valentin Vornicu (20:20:09)
That's it for the Math Jam.

Valentin Vornicu (20:20:18)
The AIME II is on March 28, and the AIME II Math Jam will be on March 30.