2006 AIME Math Jam
Go back to the Math Jam ArchiveAoPS instructors will lead a discussion on all 15 problems from the 2006 AIME.
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Facilitator: Mathew Crawford
MCrawford (18:59:12)
2006 AIME Math Jam
Before we get started, please note that this classroom is moderated, meaning that students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. 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 to write responses that are complete and easy to read.
MCrawford (18:59:41)
MCrawford (18:59:48)
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MCrawford (18:59:50)
(By clicking on the link, you can bring up a copy of the problem in a separate window, to use to follow along as we work through the solution. You may need to hold down the Ctrl key while you click on the link. If that doesn't work, you many need to disable your popup-blocker.)
MCrawford (19:00:03)
What's our first step?
RC-7th (19:00:09)
Draw a diagram
robinhe (19:00:10)
draw it
uclabb (19:00:10)
diagram
Chocoboom (19:00:11)
diagram!
hockeyadi23 (19:00:13)
Start with a picture
ohhoonkwon (19:00:17)
Draw a neat picture?
MCrawford (19:00:32)
We start with a quick sketch:
MCrawford (19:00:35)
Yes, a neat one!
MCrawford (19:00:41)
MCrawford (19:00:48)
As chesspro pointed out after taking the test, you could draw this precisely and measure AD with your ruler if you were so inclined. But if you weren't so inclined, what's the quick route to the solution?
lasagnaman (19:00:10)
to find AC
mathwiz2588 (19:00:18)
Pythagorean
DavidL (19:00:10)
Use two right triangles, and Pythagorean Theorem to get AD = 31
DavidL (19:00:57)
Pythag twice
RC-7th (19:01:02)
Pythagorean Theorem!
hockeyadi23 (19:01:05)
Now it becomes clear (once you label out the sides). Use the pythagorean theorem twice to find AD.
uclabb (19:01:06)
pythag a couple times
robinhe (19:01:09)
ABC is a right traingle and AC^2=765
E^(pi*i)=-1 (19:01:10)
Pythag theorem
lasagnaman (19:01:12)
find AC using pythag, then find AD using pythag
MCrawford (19:01:28)
All we need is AD, and a couple applications of the Pythagorean Theorem gives it to us:
MCrawford (19:01:31)
MCrawford (19:01:33)
MCrawford (19:01:40)
Combining these, we have
MCrawford (19:01:42)
DavidL (19:01:39)
So The Perimeter = 84
MCrawford (19:02:10)
Therefore, AD = 31, so our perimeter is 18+21+14+31 = 084.
MCrawford (19:02:20)
MCrawford (19:02:32)
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hockeyadi23 (19:02:29)
Find the min and max
E^(pi*i)=-1 (19:02:42)
Find the min (first 90) and max (last 90)
Chocoboom (19:02:43)
you just find the minimum possible sum and the maximum possible sum
mathclass (19:02:44)
all sums from the minimum sum to the maximum sum can be achieved bc it's 100 consecutive numbers.
robinhe (19:02:45)
the smallest is 1+2+...90 and the largest is 11+12+...100
MCrawford (19:03:14)
We can quickly find the highest and the lowest sums. The lowest is 1 + 2 + . . . + 90 and the highest is 11 + 12 + . . . + 100.
MCrawford (19:03:20)
We wonder if we can attain every sum in between. Can we?
Hamster1800_2 (19:03:03)
So find the min and the max, and just "see" that you can form any integer value between them.
Chocoboom (19:03:11)
And everything else inbetween can be possible as well.
MCrawford (19:04:32)
It would be nice if we could prove our intuition.
DavidL (19:03:45)
Yes, consec ints
robinhe (19:03:56)
yes because just add 1 to one of the sums
uni412 (19:04:11)
Yes, because you can always reduce one of the numers in the set by 1
E^(pi*i)=-1 (19:04:11)
Yes, if we have a certain sum, just increase an element by one to get a bigger sum by one.
rachel (19:04:23)
yes-start with the lowest and increase the highest to lowest terms (induction) to get everything in between
MCrawford (19:05:09)
Yes! Let our lowest sum be T: T = 1 + 2 + . . . + 90. We add 1 to the highest term of T: T = 1 + 2 + . . . + 89 + 91.
MCrawford (19:05:11)
We can keep adding 1 to the highest term of T that isn't 'maxed out' (i.e. after 1 + 2 + . . . + 89 + 100 comes 1 + 2 + . . . + 88 + 90 + 100) until we reach our largest possible sum. Therefore, our problem is finding the number of integers from 1 + 2 + . . . + 90 to 11 + 12 + . . . + 100, inclusive.
hockeyadi23 (19:02:37)
Take the difference and add 1
Hexahedron (19:03:27)
difference between those is (11-1)*90=900, and have to add one
hockeyadi23 (19:03:52)
10*90
MCrawford (19:05:31)
We subtract the two numbers and add 1. Subtraction cancels most of the terms (and points out that we could have solved this by considering sums of 10-member subsets) and leaves us with:
(91 + 92 + . . . + 100) - (1 + 2 + . . . + 10) +1
MCrawford (19:05:37)
This equals
(91 - 1) + (92 - 2) + (93 - 3) + . . . + (100 - 10) + 1 = 10(90) + 1 = 901.
MCrawford (19:05:43)
MCrawford (19:05:50)
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Elemennop (19:05:52)
Set it up algebraicly
MCrawford (19:06:00)
Letting A be our original number and B being the new number, what information can we get about A and B from the problem?
hockeyadi23 (19:06:00)
I like this one: call n the digit removed, and k the two digit number that's left over
JRechel (19:06:01)
you know it has to end in 5
robinhe (19:06:07)
list all 3-digit multiples of 29?
uclabb (19:06:10)
29x=100+x
MCrawford (19:06:23)
Not 100, but 100n -- some multiple of 100.
mathwiz2588 (19:06:16)
29A =B
DavidL (19:06:14)
Use an equation 100a + 10b +c = 290 b + 29c
tiredepartment (19:06:22)
100a+10b+c
fubu (19:06:10)
if we try two digits, we can immediately see that it doesn't work. such as 10x + y
hockeyadi23 (19:06:13)
A= 100n+k, b=k
epi14159 (19:06:16)
since the last digit remain the same, only 0 and 5 can be the last digit of the number
lasagnaman (19:06:24)
so, A must be a multiple of 25, also, quikc trial and error shows that A must be 3 digits
MCrawford (19:06:42)
Lot's of ideas here.
MCrawford (19:06:46)
We know that 29B = A, and from the given information, we know that the last digits of B and A are the same.
MCrawford (19:06:55)
MCrawford (19:07:04)
Now what?
uclabb (19:07:24)
that makes it a mult of 25
Hexahedron (19:07:31)
A=0(mod 25)
DavidL (19:07:34)
Why put it in Modular Arithmetic when we can avoid it altogether?
MCrawford (19:07:52)
I think it's simplest.
MCrawford (19:07:56)
MCrawford (19:08:01)
Now we simply try the smallest positive multiple of 25:
25 x 29 = 725.
We have a winner, and it is 725!
MCrawford (19:08:03)
And that's all there is to it.
Elemennop (19:07:30)
There's no reason to split up the two digit part into two variables. Let N be a digit, and let X be a two digit number. 100N + X = 29X, or 100N-28X = 0, which has solution (N,X)=(7,25)
MCrawford (19:08:22)
There are many ways to work this one.
MCrawford (19:08:40)
Modular arithmetic has the advantage of helping us quickly isolate one of the variables to find its value.
MCrawford (19:08:49)
Solutions using modular arithmetic, divisibility, and a bit of algebra are covered in the AoPS Introductory Number Theory class as well as the upcoming Introductory Number Theory book.
MCrawford (19:08:56)
MCrawford (19:09:05)
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MCrawford (19:09:17)
In order to count the number of 0's at the end, what do we have to count?
hockeyadi23 (19:09:09)
Basically count the number of fives
Sunday Math-er_2 (19:09:10)
count # of 5s
mathwiz2588 (19:09:13)
well 100! has 24 zeros
E^(pi*i)=-1 (19:09:15)
We need the number of 5's in the product
ohhoonkwon (19:09:17)
Find all the fives
MCrawford (19:09:40)
We have to count the number of factors of 5 in this gigantic product. How are we going to do it in an organized way? Obviously we can forget about the first 4 factorials. Are we then going to count all the 5's in 5!, then all the 5's in 6!, etc., all the way up to 100!?
We could do this, but we can organize our counting more effectively - how?
llamallad (19:09:34)
100 has 24 zeros and 99-95 has 22 0's because they lose 2 factors of 5. just continue this down to 5. and add up all zeros
perfectnumber628 (19:09:39)
to find the amount of 0s at the end of x!, you figure out how many 5s or powers of 5 are less than it
robinhe (19:09:41)
1-4 factorials don't have 0's
5-9 each have 1
10-14 each have 2
...
MCrawford (19:10:14)
We can note that the number 5 shows up in 96 factorials (5! through 100!, inclusive). The number 10 shows up in 91 factorials (10! through 100!), the number 15 in 86 factorials, etc.
Following in this vein, we have how many multiples of 5?
Hexahedron (19:10:02)
5-9 is same number, 10-14 is same....
DavidL (19:10:06)
Group them by five
rachel (19:10:08)
count all 5s, then all 25s separately
E^(pi*i)=-1 (19:10:13)
5 appears 96 times, 10 appears 91 times....
MCrawford (19:11:17)
What's 96 + 91 + 86 + ... + 1?
E^(pi*i)=-1 (19:11:11)
970 multiples of 5, but some have two
MCrawford (19:12:07)
We have 96 + 91 + . . . + 6 + 1 = (97)(10) = 970 multiples of 5 in our list.
MCrawford (19:12:14)
We have to remember the multiples of 25, which add a factor of 5 to our product. How many multiples of 25 are there?
perfectnumber628 (19:12:01)
but then don't you have to also count multiples of 25?
Sunday Math-er_2 (19:12:20)
then count 5^2*k again
hockeyadi23 (19:12:29)
154
matt276eagles (19:12:34)
76+51+26+1=2(77)=154
MCrawford (19:13:04)
We follow similar reasoning as before. The number 25 shows up in 76 factorials (25! through 100!), 50 shows up in 51 factorials, 75 in 26 factorials, and 100 in 1 factorial, for a total of 76 + 51 + 25 + 1 = 154.
E^(pi*i)=-1 (19:12:50)
76 25's, 51 50's, 26 75's, 1 100
MCrawford (19:13:14)
So, what is our final answer?
robinhe (19:13:04)
970+154=1124
so the answer is 124
MCrawford (19:13:33)
Our final answer is the remainder when 970+154 = 1124 is divided by 1000, or 124.
MCrawford (19:13:44)
MCrawford (19:13:54)
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MCrawford (19:13:56)
MCrawford (19:13:58)
What do we do with that?
tiredepartment (19:14:00)
set them equal and square both sides
robinhe (19:14:01)
set it up as an equation and then square both sides
Elemennop (19:14:02)
Square the variable expression, and set it equal to the square.
alan (19:14:02)
square it
MCrawford (19:14:33)
The big square root on the right is intimidating. We can get rid of it by squaring both sides:
MCrawford (19:14:39)
matt276eagles (19:14:24)
square and match terms
fubu (19:14:28)
the coefficients of the roots have to be the same
MCrawford (19:14:59)
We can match up coefficients of the radicals:
2ab = 104
2ac = 468
2bc = 144
mel (19:14:46)
Then equate corresponding parts to find ab, bc, and ca.
Kelly_F (19:14:59)
then you can try to simplify the numbers in the square roots on the right side to come with the a comparatable equation on the left.
robinhe (19:15:00)
so then 2ab=104, 2ac=468...
MCrawford (19:16:03)
Divide by 2:
ab = 52
ac = 234
bc = 72
DavidL (19:15:07)
Multiply tme all up
Elemennop (19:15:09)
a,b,c are rational, so ab=52, ac=234, bc=72. Multiply all together and take the square root to get abc!
DavidL (19:15:22)
and then you get a2b2c2 = somebignumber
hockeyadi23 (19:15:29)
Then prime factorize, and take the square root of the product
fubu (19:15:36)
find ab, ac, bc, multiply them altogether and then take the square root ... prime factorization will help with the square rooting
MCrawford (19:16:55)
The product of our three equations gives us (abc)^2:
MCrawford (19:17:06)
(ab)(ac)(bc) = (52)(234)(72) = (4)(13)(13)(18)(8)(9).
MCrawford (19:17:29)
What's our answer?
hockeyadi23 (19:16:50)
Thus the answer is 72*13=936
robinhe (19:16:56)
936
Kelly_F (19:17:03)
abc=976
Chocoboom (19:17:20)
But it's too complicated squaring without a calculator.
MCrawford (19:18:38)
MCrawford (19:18:57)
Those of you who suggested using prime factorization are correct.
MCrawford (19:19:03)
You say an enormous amount of time multiplying by factors instead of large numbers.
MCrawford (19:19:13)
MCrawford (19:19:23)
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MCrawford (19:19:27)
Hexahedron (19:19:30)
it needs to be of the form (abc/999)
hockeyadi23 (19:19:26)
9*8*7=720
perfectnumber628 (19:19:35)
there are (10)(9)(8)=720 numbers
Hexahedron (19:19:36)
abc, is distinct
E^(pi*i)=-1 (19:19:37)
Fractions abc/999
robinhe (19:19:36)
12/999
Elemennop (19:19:29)
Gauss!! Gaussian Pairing = easy solution.
DavidL (19:19:31)
10x9x8 ways
fubu (19:19:34)
that decimal is equal to (100a+10b+c)/999
matt276eagles (19:19:50)
we have 10*9*8=720 fractions, each pair sums to 1
MCrawford (19:20:09)
Finding every fraction is going to be a nightmare: there are (10)(9)(8) = 720 of them! What a pain!
MCrawford (19:20:16)
We can look at how many times each digit shows up in each place. For example, if we stick a 0 in the first place, we have 9 ways to choose the next digit, and 8 ways to choose the third digit, for a total of 72 ways. (This shouldn't be surprising: 1/10 of our 720 numbers start with 0, by symmetry.)
MCrawford (19:20:30)
Similarly, there are 72 ways to start with 1, 72 ways to start with 2, etc. From here there are many ways to solve the problem. We can add across each digit, or we could add up all the 1's (then 2's, then 3's) in all digits, etc.
pilot (19:20:26)
Each digit occurs 9*8=72 times in each of the 3 places
Mathgod (19:20:22)
they pair up
Mathgod (19:20:32)
so it's 360
Hamster1800_2 (19:20:17)
Since there are 720 numbers and 10 digits for each place,each digit should appear 72 times in each spot.
MCrawford (19:20:53)
We can consider the average of each digit. The 'average' first digit is the average of the numbers 0 through 9, or 4.5. The average second digit is 4.5, as is the average third digit, and so on.
MCrawford (19:20:57)
Therefore, our average number is:
MCrawford (19:21:01)
MCrawford (19:21:04)
The sum in the parentheses is a geometric series with sum (1/10)/(1-1/10) = 1/9, so our average number is (4.5)(1/9) = 1/2.
MCrawford (19:21:21)
Though pairing in sums of 1 is certainly faster!
MCrawford (19:21:36)
The average number is 1/2, and there are 720 of them, so the sum of the numbers is (720)(1/2)=360.
MCrawford (19:22:06)
MCrawford (19:22:09)
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MCrawford (19:22:12)
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MCrawford (19:22:19)
MCrawford (19:22:31)
How can we proceed?
Hexahedron (19:22:34)
so all the triangles are similiar
Hamster1800_2 (19:22:50)
All of the triangles formed are similar, right?
dhuang614 (19:22:54)
let 1 equal the distance between two lines
hockeyadi23 (19:22:55)
so then solve for the base of a, calling the distance between parallel lines 1
Elemennop (19:22:57)
Heights are all equal except for A...+ Parallel lines, so the triangles are similar. Then find the ratios of heights, whose square is the ratio of area.
MCrawford (19:23:37)
Tons of parallel lines should make you think of similar triangles.
MCrawford (19:23:41)
Let's assume (for simplicity) that the distance between the parallel lines is 1.
Then we can express the height of the small triangle A to be x, where 0 < x < 1.
MCrawford (19:23:44)
So the heights of the successive triangles are x, 1+x, 2+x, 3+x, ....
How can we write the ratio of the shaded areas C and B?
DavidL (19:24:04)
As sums and differences of squares of heights
Elemennop (19:24:11)
(4+x)^2-(3+x)^2 / (2+x)^2 - (1+x)^2
hockeyadi23 (19:23:59)
11/5
MCrawford (19:24:35)
C is the area of a triangle with height 4+x minus the area of a triangle with height 3+x.
B is the area of a triangle with height 2+x minus the area of a triangle with height 1+x.
MCrawford (19:24:38)
MCrawford (19:25:09)
So 11/5 = (2x+7)/(2x+3).
This gives 22x+33 = 10x + 35, so 12x = 2, hence x = 1/6.
Then what is the ratio we want to find?
hockeyadi23 (19:25:29)
areas
Elemennop (19:25:31)
(6+x)^2 - (5+x)^2 / x^2
pilot (19:25:53)
((6+x)^2-(5+x)^2)/x^2
MCrawford (19:26:40)
Hamster1800_2 (19:26:17)
The area of D to the area of A, which would be (6+1/6)^2 - (5+1/6)^2 to (1/6)^2
MCrawford (19:27:09)
This is (2x+11)/x^2.
Elemennop (19:26:47)
Plugging in 1/6 yields the ratio as 408, which is what we want.
MCrawford (19:27:38)
Plugging in x=1/6 gives us (34/3)*36 = 34*12 = 408.
MCrawford (19:27:46)
Remember, parallel lines mean similar triangles!
Problems like this are covered in the new AoPS Introductory Geometry book.
MCrawford (19:28:01)
MCrawford (19:28:08)
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MCrawford (19:28:15)
MCrawford (19:28:20)
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MCrawford (19:28:33)
Any intuition as to what's going on here?
krsattack (19:28:27)
figure out max and min areas of T
d343seven (19:28:38)
call AFE=2theta
Elemennop (19:29:09)
Well, we can squeeze the hexagon until T becomes zero, so if we find the max area of T we can achieve every integer less than max T because it's function of area is a smooth function .
dhuang614 (19:29:10)
define an angle
MCrawford (19:29:33)
You might see "intuitively" that we can make T arbitrarily small by making the outer rhombi P, Q, R, and S look more like squares:
MCrawford (19:29:36)
epi14159 (19:29:27)
the min areas of T is 0
Optimosis (19:29:23)
area of T based on angles of the other 4 rhombuses
MCrawford (19:30:04)
Let's label the other four points:
MCrawford (19:30:14)
MCrawford (19:30:56)
And T also appears to grow bigger as we "flatten" the outer rhombi:
MCrawford (19:31:01)
Elemennop (19:30:28)
We can use angles or we can use the diagonals of T (one of which is twice the height of the other rhombi) to find the side of the rhombus in terms of the diagonal
mel (19:29:51)
MCrawford (19:31:46)
Let d be the side length of the rhombi, and let theta be angle FWX.
MCrawford (19:31:57)
(Note: you can also work with angle AWX -- it'll come out essentially the same.)
MCrawford (19:32:05)
MCrawford (19:32:13)
Angle XWZ is 360 - 2*theta, so:
MCrawford (19:32:18)
MCrawford (19:32:24)
But sin(360-x) = sin(-x) = -sin(x), so the area of T is just d^2 * -sin(2*theta).
MCrawford (19:32:38)
Now we can use the double-angle formula!
MCrawford (19:32:47)
MCrawford (19:33:08)
How do we use what we've learned to identify the answer?
perfectnumber628 (19:33:17)
the cos is negative
Elemennop (19:33:25)
Cos ranges from 90 to 180, so the area ranges from 0 to just above 89.
MCrawford (19:33:50)
Hence, as theta varies from 90 degrees up through 180 degrees, cos(theta) varies from 0 down to -1, so the area of T varies from 0 up to 2*sqrt(2006) (as we expected).
Thus the answer is the largest integer less than 2*sqrt(2006) = sqrt(8024).
perfectnumber628 (19:33:42)
do you have to extract sqrt2006?
E^(pi*i)=-1 (19:33:41)
89 possibilities for theta
mel (19:33:53)
MCrawford (19:34:14)
Note that 8024 < 8100 = 90^2, so we check that 89^2 = (90-1)^2 = 8100 - 180 + 1 = 7921 < 8024.
MCrawford (19:34:23)
We don't need to extract the exact square root.
MCrawford (19:34:25)
So 89 < sqrt(8024) < 90, and the answer is 089.
Chocoboom (19:33:04)
Whats the double-angle formula?
MCrawford (19:34:58)
MCrawford (19:35:20)
I recommend asking how it's derived in the Intermediate forum or trying it yourself.
MCrawford (19:35:38)
Trig identities like that become essential at this level of problem solving.
DavidL (19:35:37)
Hint: use the unit circle!
MCrawford (19:35:54)
At this time, Dave Patrick will take over the Math Jam.
DPatrick (19:36:00)
DPatrick (19:36:05)
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DPatrick (19:36:13)
What should we do to make this equation easier to work with?
Hamster1800_2 (19:36:17)
write the logs as one log
quantum leap (19:36:22)
combine them into one log
DavidL (19:36:23)
Well, it's just log base 8 of (a1a2...)
DPatrick (19:36:35)
We can use the identity log_8 (a) + log_8 (b) = log_8 (ab) to rewrite it as
DPatrick (19:36:40)
tiredepartment (19:36:29)
def of geo seq: a1=a
a2=ar
a3=ar^2
weiyan (19:36:28)
write a1, a2, etc in terms of a and r, such as a, ar, ar^2...
DPatrick (19:37:04)
We know that a_1 = a, a_2 = ar, a_3 = ar^2, and so on up to a_12 = ar^11.
DPatrick (19:37:09)
So what is the product of a_1...a_12?
Elemennop (19:36:58)
a_1a_2...a_{12} = a^12r^66
DPatrick (19:37:27)
We have 12 copies of a, and (1+2+3+...+11) copies of r.
DPatrick (19:37:43)
DPatrick (19:37:49)
Now what?
hockeyadi23 (19:37:29)
a^12r^66=8^2006
fubu (19:37:54)
change it to exponential form
DPatrick (19:38:17)
We can get rid of the log.
DPatrick (19:38:25)
anniesez (19:38:12)
a^12r^66=8^2006=2^6018
Hexahedron (19:38:17)
= 2^6012
perfectnumber628 (19:38:21)
8^2006=2^6018
HuAreYou? (19:38:22)
or 2^6018
DPatrick (19:38:54)
Yes, better to write this as a power of a prime:
DPatrick (19:38:57)
b-flat (19:38:38)
you know both a and r must be powers of 2
DPatrick (19:39:12)
Since a and r are positive integers, we know that they must both be powers of 2 (the only prime factor).
E^(pi*i)=-1 (19:39:13)
we know a and r are powers of 2, so 12x+66y=6018
DPatrick (19:39:31)
So let a = 2^x and r = 2^y, where x and y are nonnegative integers.
Then our equation becomes 12x + 66y = 6018.
DavidL (19:39:31)
Diophantine Time!
mathwiz2588 (19:39:35)
diophantine eq.
DPatrick (19:39:55)
Yes, this is what is known as a Diophantine equation, since we are looking for integer solutions.
alan (19:39:51)
epi14159 (19:39:53)
2x+11y=1003
Elemennop (19:40:02)
2x + 11y = 1003 is a nicer form.
DPatrick (19:40:15)
Yes, it's nicer if we divide by 6 to get 2x + 11y = 1003.
DPatrick (19:40:23)
How do we solve this?
Hamster1800_2 (19:40:28)
so y must be odd.
dhuang614 (19:40:35)
then take remainders modulo 2
Hexahedron (19:40:40)
y must be odd
DPatrick (19:41:07)
This will have a solution for any odd y.
DPatrick (19:41:16)
If y is odd, then x = 1003-11y / 2.
DPatrick (19:41:21)
(If y is even, then 1003-11y is odd, so we can't divide by 2 and get an integer.)
DPatrick (19:41:37)
How many values of y make x nonnegative?
5alive (19:41:26)
y can be 1,3,5,...91
epi14159 (19:41:33)
(496,1)...(1,91)
lasagnaman (19:41:38)
also ,11y < 1003
DPatrick (19:42:15)
Exactly -- we need y < 91.
DPatrick (19:42:21)
So y = 1,3,5,...91.
bthings_2000 (19:42:10)
91= 45+46, 46 odds since 1 and 91 are odd, so it's 46
5alive (19:42:25)
or equal to 91 **
DPatrick (19:42:43)
Right -- I should have written y <= 91, thanks.
DPatrick (19:42:47)
So there are 046 possible solutions.
DPatrick (19:43:01)
DPatrick (19:43:05)
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DPatrick (19:43:16)
DPatrick (19:43:19)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/8bf79dfef92bfbba1af268bb14c04190.png
krsattack (19:43:17)
look at where the circles intersect. Connecting these lines creates symmetrical patterns which should help us get lucky
DPatrick (19:43:46)
Indeed, that will work.
DPatrick (19:44:02)
Just to warm up to the problem, let's first modify it to make it much easier. First, imagine that there were nine circles in the plane instead of eight.
DPatrick (19:44:11)
DPatrick (19:44:16)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/8cdddc9d9df932f5557b3570f25c3074.png
DPatrick (19:44:24)
Where would the line go in this case?
dhuang614 (19:44:27)
all lines must pass through the center circle
tiredepartment (19:44:29)
If there were 9, then the line must pass through the center of the center circle
mathwiz2588 (19:44:32)
through thte center of the cetner circle
DPatrick (19:44:42)
Right, it would go through the center of the center circle.
DPatrick (19:44:49)
DPatrick (19:44:52)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/242b36f07d573d43775d177f36c59ad1.png
DPatrick (19:45:02)
I want to point out that the case with nine circles is very easy, because there is a lot of symmetry in our diagram. Any line that goes through the center of a circle is automatically an axis of symmetry of that circle, and we want to divide a region into two regions of equal area. Both of these points strongly suggest that we want to use symmetry.
DPatrick (19:45:08)
Unfortunately, the original problem with eight circles does not quite have as much symmetry. What other lines can we try drawing?
nsato (19:45:18)
DPatrick (19:45:32)
(ignore that!)
happydude (19:45:27)
Transform the line to the left
dhuang614 (19:45:27)
lines through tangent points
DPatrick (19:45:52)
After removing the ninth circle in the corner, we need to move the line a little to the left to balance this missing circle.
barce11 (19:45:51)
draw lines where they are tangent
DPatrick (19:46:25)
We could try the line passing through the point of tangency between circles 5 and 8. That's the first obvious place to try as we move to the left.
DPatrick (19:46:31)
I'll also label the circles for our convenience:
DPatrick (19:46:42)
DPatrick (19:46:53)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/832a06a6006311e6cd23f3b0a0f17e21.png
DPatrick (19:47:03)
One thing should immediately leap out of the diagram at us. What is that?
Hamster1800_2 (19:46:47)
Why are points of tangency "obvious choices?"
DPatrick (19:47:20)
Good question. We're really hoping to get lucky and find a "nice" line that works.
DPatrick (19:47:27)
The only points we have to work with are the centers of the circles and the points where they're tangent.
pilot (19:47:25)
they divide the two circles equally
dhuang614 (19:47:26)
the points of tangency split two circles into equal area
b-flat (19:47:36)
the cut off portions in circles 5 and 8 are equal...
lasagnaman (19:47:31)
there are 2 whole circles on each side, and 1 and 2 are evenly split bewteen the 2 sides, as are 5 and 8
perfectnumber628 (19:47:41)
you only have to worry about circles 1,2,5,8
DPatrick (19:48:09)
Right, it should be pretty clear by symmetry that the line I drew above splits circles 5 and 8 evenly.
DPatrick (19:48:12)
Why does it also split circles 1 and 2 evenly?
DPatrick (19:48:30)
The line appears to pass through the point of tangency of circles 1 and 2. Is this true?
DavidL (19:48:26)
Because it happens to pass through the tangent point
dasherm (19:48:29)
because it hits their tangent as well
mathwiz2588 (19:48:31)
it passes thru their tangency also
DPatrick (19:49:05)
Yes, because the line has slope 3, which is also the slope between the two points of tangency. (The horizontal part is one radius, and the vertical part is three radii.) A simple diagram wiill hopefully make this clear:
DPatrick (19:49:08)
DPatrick (19:49:13)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/bc78521e15e4ba4fdaccf547de08a3ff.png
DPatrick (19:49:23)
So the line splits circles 1 and 2 evenly as well.
DPatrick (19:49:56)
So this is the line that we want!! It splits 5&8 evenly, 1&2 evenly, and has two other whole circles on each side.
krsattack (19:49:54)
so all we have left is to find the equation of the line
DPatrick (19:50:14)
Yes -- the rest is now a very easy computation. The line passes through the point (1.5, 2) and has slope 3.
DavidL (19:50:12)
So we use point slope form
bthings_2000 (19:50:23)
point slope time!
krsattack (19:50:24)
point slope formula
DPatrick (19:50:39)
The equation of the line is then y - 2 = 3(x - 1.5), which becomes 2y = 6x - 5.
DPatrick (19:50:46)
Therefore, the final answer is 2^2 + 6^2 + 5^2 = 065.
DPatrick (19:50:56)
It's very easy to make a problem like this too complicated. We might search for a formula that would give areas in terms of lines and circles, but this would get very unwieldy, very quickly. Problems on the AIME tend to have simple solutions, and this problem is no exception; you just have to look for them. Here, we solved the problem by checking key points and using symmetry - the simplest approach is often the best.
DPatrick (19:51:21)
DPatrick (19:51:28)
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DPatrick (19:51:46)
How might we start to get a handle on this problem?
barce11 (19:51:50)
use simpler cases
pilot (19:51:56)
smaller cases
dhuang614 (19:51:57)
solve a simpler problem
lasagnaman (19:52:02)
try working with 2, then 3 then 4 cubes
DPatrick (19:52:20)
OK, how many stacks with 2 cubes?
alpha123 (19:52:26)
2
perfectnumber628 (19:52:26)
2
DPatrick (19:52:40)
For 2 blocks, there are clearly 2 ways: (1 2) and (2 1). (Let the first number be the 'bottom'.)
DPatrick (19:52:46)
How about 3 cubes?
E^(pi*i)=-1 (19:52:49)
6 stacks
alpha123 (19:52:50)
3! = 6
Hannes (19:52:51)
6
sirorange (19:52:51)
2*3=6
uni412 (19:52:51)
6
DPatrick (19:53:10)
Again, all orders work, for a total of 3! = 6.
DPatrick (19:53:18)
How about 4?
Elemennop (19:52:58)
The same with 3 cubes, 6. For four cubes, only 18 of them work (anything with 14 together in that orer doesn't work).
E^(pi*i)=-1 (19:53:19)
18 stacks
cobylu (19:53:25)
18
alpha123 (19:53:25)
18
hockeyadi23 (19:53:25)
18
DPatrick (19:53:52)
For 4, we have to omit the orders where 4 comes right after 1: (1 4 2 3) (2 1 4 3), and so on. All the rest work.
perfectnumber628 (19:53:42)
4! but subtract cases where 1 is right under 4
DPatrick (19:54:36)
We could consider 1-4 together a single block, and see that there are 3! ways to order 1-4, 2, and 3.
DPatrick (19:54:47)
So there are 6 bad cases, hence 4! - 6 = 18 ways to stack them.
DPatrick (19:54:52)
So, we have a bit of a feel for the problem. Stepping up to 5 blocks, we would have to throw out all the 5-2's, the 5-1's, and 4-1's from the 5! orders, which is starting to get complicated.
DPatrick (19:55:05)
We wonder if just as we build our illegal 4-block towers from the 3-block towers, can we also build all the legal ones?
krsattack (19:54:49)
put the nth cube on bottom, over n-1th cube, or over n-2th cube
Elemennop (19:54:59)
2,6,18...multiply by 3 ?
lasagnaman (19:55:12)
now, for 5 cubes, you can take all possible orders for 4 cubes, then insert the 5th block in any of 5 positions in between blocks, then rule out situations where 5 comes right after 1 or 2
Hamster1800_2 (19:55:29)
so we can put the 5 at the beginning, after 3, or after 4 (3 places). For 4 blocks, you could put it at the beginning, after 2, or after 3.
hockeyadi23 (19:55:32)
RECURSION!!!
DPatrick (19:56:13)
Right. For example, to go from a 3-block tower to a 4-block tower, we can either put block 4 on the bottom, or just above the 2, or just above the 3, for a total of 3 legal 4-block towers for each 3-block tower.
DPatrick (19:56:24)
Clearly doing this over all 3-block towers won't create any tower more than once (i.e. it doesn't overcount).
DPatrick (19:56:33)
We must also make sure that we haven't missed any legal 4-block towers; i.e. that there aren't any 4-block towers which cannot be formed by just adding block 4 to a 3-block tower.
DPatrick (19:57:00)
But this works too -- if we have a 4-block tower, and remove block 4, we get a legal 3-block tower.
DPatrick (19:57:08)
So what's the conclusion?
Hamster1800_2 (19:57:26)
To get from the number of n block towers to the number of n+1 block towers, multiply by 3.
DPatrick (19:58:08)
Right. Every tower with n blocks gives exactly 3 of the towers of n+1 blocks. And every tower with n+1 blocks comes from an n-block tower.
krsattack (19:57:56)
formula time : 2 * 3^(k-2)
hockeyadi23 (19:58:00)
So the closed form is 2*3^n-2
DPatrick (19:58:50)
Right. There are 2 = 2*3^0 2-block towers, 6 = 2*3^1 3-block towers, and since we multiply by 3 everytime we add a block, we get 2*3^(n-2) n-block towers.
alpha123 (19:58:29)
so for 8 cubes stacks = 2*3^6 = 1458
kostya (19:58:36)
so we get 2 * 3^6
E^(pi*i)=-1 (19:58:43)
2*3^6=1458
DPatrick (19:59:05)
Letting n=8, we have 2(3^6) = 1458 legal 8-block towers. The remainder when this is divided by 1000 is 458.
DPatrick (19:59:18)
DPatrick (19:59:25)
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DPatrick (19:59:35)
How can we start working with this equation?
Hamster1800_2 (19:59:37)
put everything in terms of 4x and x
alpha123 (19:59:41)
say cos3x = cos(4x-x)
DPatrick (19:59:58)
The key observation to notice is that 3x = 4x - x and 5x = 4x + x.
DPatrick (20:00:13)
So we can use the cosine angle-sum formulas.
DPatrick (20:00:21)
DPatrick (20:00:35)
So our original equation becomes:
DPatrick (20:00:38)
DPatrick (20:00:55)
When we expand the left side, what happens?
5alive (20:01:00)
much cancellation on the left side
Hamster1800_2 (20:01:05)
When we cube them, most of the sines cancel, but not all.
krustyteklown_2 (20:01:05)
2 terms cancel
DPatrick (20:01:27)
Two of the four terms from each cubic cancel.
DPatrick (20:01:37)
We're left with:
DPatrick (20:01:40)
DPatrick (20:02:08)
Now what?
alpha123 (20:01:53)
and divide by 6
Hamster1800_2 (20:02:14)
move everything to one side so we can use 0 to help us.
DPatrick (20:02:57)
Yes, but I would also use sin^2 4x = 1 - cos^2 4x and sin^2 x = 1 - cos^2 x to write everything in terms of cos 4x and cos x.
DPatrick (20:03:02)
For ease of writing, let's set a = cos 4x and b = cos x.
DPatrick (20:03:24)
DPatrick (20:03:40)
When we expand this out, we see that the a^3b^3 term cancels, and we're left with
DPatrick (20:03:44)
DPatrick (20:04:05)
So the solutions are either a=0, b=0, or a^2+b^2 = 1.
DPatrick (20:04:17)
We can eliminate b=0, since cos x is never 0 for 100 < x < 200.
DPatrick (20:04:31)
When is cos 4x = 0 in this range?
krsattack (20:04:49)
4x = 90 + 180k
perfectnumber628 (20:04:56)
if a=0, 4x=90 or 270 or 450, etc
DPatrick (20:05:16)
Clearly 400 < 4x < 800, so the angles in this range which have cos 4x = 0 are 4x = 450 and 4x = 630. (cos is 0 at odd multiples of 90.)
DPatrick (20:05:22)
So x = 450/4 and x=630/4 are two solutions. These solutions sum to 1080/4 = 270.
DPatrick (20:05:41)
Finally we need the solutions to a^2+b^2 = 1.
These are solutions to cos^2(x) + cos^2(4x) = 1.
How can we find these?
E^(pi*i)=-1 (20:06:12)
use double angle formula?
DPatrick (20:06:40)
We can use the half-angle formula to rewrite cos^2(x) + cos^2(4x) = 1 as (1+cos(2x))+(1+cos(8x)) = 2, so we have cos(2x) + cos(8x) = 0.
DPatrick (20:06:58)
Then we can use the sum-to-product formula to rewrite this as 2cos(5x)cos(3x) = 0.
DPatrick (20:07:16)
So we must have cos(5x)=0 or cos(3x)=0.
DPatrick (20:07:35)
What values of 100 < x < 200 give cos(5x)=0 or cos(3x)=0?
perfectnumber628 (20:07:39)
so 5x=90, 270, etc
platinumnerd (20:08:10)
same for 3x
Hamster1800_2 (20:08:24)
we want 500 < 5x < 1000 and 300 < 3x < 600 where the middle is 90+180k for each
DPatrick (20:09:00)
Right. cos(5x)=0 gives 126, 162, and 198 in the correct range, and cos(3x)=0 gives 150.
DPatrick (20:09:22)
So these four solutions sum to 126+162+198+150 = 636.
DPatrick (20:09:25)
Finally, adding this to the 270 that we found before, we get our final answer of 636+270 = 906.
DPatrick (20:09:53)
Note: there are other ways you can attack this problem -- you could try using the sum-to-product or product-to-sum at the very beginning.
DPatrick (20:10:02)
But this is how I did it. :)
DPatrick (20:10:20)
DPatrick (20:10:24)
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DPatrick (20:10:33)
Where can we start with this problem?
krustyteklown_2 (20:10:45)
try a first few couple n's
5alive (20:10:49)
write out some sn
DPatrick (20:11:05)
Looking at small n's might give a pattern.
DPatrick (20:11:14)
For example, take n = 5.
DPatrick (20:11:20)
Then we can generate the following table.
DPatrick (20:11:26)
DPatrick (20:11:41)
(you'd probably do even smaller n's first, but I'm trying to speed things up a bit here!)
jiangwei (20:11:17)
2 shows up every other term, 4 shows up every 4 terms, .....
perfectnumber628 (20:11:52)
half of them are 2, 1/4 are 4, etc
DPatrick (20:12:07)
Right. The pattern seems clear.
DPatrick (20:12:11)
DPatrick (20:12:24)
The number 2 appears eight times, the number 4 appears four times, the number 8 appears twice, the number 16 appears once, and the number 32 appears once.
DPatrick (20:12:35)
DPatrick (20:12:45)
This way of writing the sum makes the general formula much more accessible.
DPatrick (20:13:14)
How many times does the number 2^m appear among these numbers, where 1 <= m <= n?
E^(pi*i)=-1 (20:13:40)
2^(n-m)
DPatrick (20:14:06)
Close...it's actually 2^(n-m-1) times.
DPatrick (20:14:30)
...except for 2^n, which appears once.
DPatrick (20:14:41)
DPatrick (20:15:04)
We don't have to prove it to get an answer for the AIME (and it would probably waste valuable time), but it's a good exercise to go through the proof on your own.
DPatrick (20:15:17)
What can we do with this expression?
krustyteklown_2 (20:15:01)
=(n+1)*2^(n-1)
DavidL (20:15:06)
(n+1) 2^(n-1)
dhuang614 (20:15:16)
factor into (2^(n-1))(n+1)
DPatrick (20:15:33)
DPatrick (20:15:45)
We now have two nice factors to work with.
DPatrick (20:15:48)
When is this a perfect square?
alpha123 (20:15:56)
when n is odd
krustyteklown_2 (20:15:46)
so we need n+1 to be a perfect square, but n-1 has to be even, so the power of 2 is a perfect square
hockeyadi23 (20:15:57)
now all we have to look for is an odd n that's 1 less than a perfect square
DPatrick (20:16:27)
Right. If n is odd, then 2^(n-1) is a perfect square, so we'd need n+1 to be a perfect square as well.
DPatrick (20:16:40)
What's the largest 3-digit odd n so that n+1 is a perfect square?
hockeyadi23 (20:15:58)
899
DavidL (20:16:25)
899 is the biggest.
alpha123 (20:16:48)
899
DPatrick (20:17:11)
Right. 900 = 30^2 but 32^2 = 1024 is too bug.
DPatrick (20:17:16)
So the largest n is 899.
DPatrick (20:17:19)
Could n be even?>
alpha123 (20:17:29)
no
pilot (20:17:30)
no
maverick417 (20:17:32)
no
DavidL (20:17:32)
certainly not
lasagnaman (20:17:40)
no, beacuse then you need a factor of 2 in n +1
DPatrick (20:18:02)
Right. We'd have an odd power of 2 times an odd number, which can't be a perfect square.
DPatrick (20:18:13)
Hence, the final answer is n = 899.
DPatrick (20:18:25)
DPatrick (20:18:37)
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DPatrick (20:18:43)
Scary 3-D geometry.
hockeyadi23 (20:18:35)
Diagram!
Chocoboom (20:18:46)
draw a tetrahedron?
DPatrick (20:19:06)
Do we need to draw two different diagrams, one for before and for after?
DPatrick (20:19:38)
I'd be inclined to try to get all our information into one (large) diagram.
DPatrick (20:19:49)
Let's set A to be the top of the tripod, and let B, C, D be the feet.
DPatrick (20:20:00)
Let E be the point on AB such that AE = 4. The 'after' tripod has base ECD and top still at A.
DPatrick (20:20:14)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/3c24aa8131f86987c61168956eaabdf4.png
DPatrick (20:20:19)
In terms of this diagram, what distance is the answer to the problem?
Elemennop (20:20:48)
perpendicular from A to ECD.
Hamster1800_2 (20:20:48)
The distance from A to the plane defined by E, C, and D
DPatrick (20:21:12)
We are looking for the distance from the point A to a plane in which we have a triangle (ECD). What general geometric tool might we consider?
krustyteklown_2 (20:21:48)
or volumes
DPatrick (20:22:01)
We consider volume, since our desired height is simply the altitude from A to base ECD of tetrahedron AECD. If we can find the volume of AECD and the area of ECD, we can find the altitude.
DPatrick (20:22:28)
How does the volume of AECD relate to the volume of ABCD?
Chocoboom (20:22:54)
4/5?
DPatrick (20:23:11)
Because E is AE = (4/5)AB, the altitude from E to face ACD is 4/5 the altitude from B to ACD.
DPatrick (20:23:28)
ACD is a base of both the 'before' and 'after' tetrahedra (ABCD and AECD), so the volume of AECD is 4/5 that of ABCD.
DPatrick (20:23:33)
To see this clearly, we can draw altitudes from E and B to face ACD:
DPatrick (20:23:39)
DPatrick (20:23:44)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/5ed4dd33b59a0c8653a595598e21961c.png
DPatrick (20:23:48)
Triangle EFA and BGA are similar, so EF/BG = AE/AB = 4/5.
DPatrick (20:24:10)
So vol(AECD) = (4/5)*vol(ABCD). How do we find the volume of ABCD?
lasagnaman (20:24:26)
1/3 bh
fubu (20:24:29)
is the base of ABCD a equilateral triangle?
perfectnumber628 (20:24:30)
find the base area, and you know the height is 4
DPatrick (20:24:48)
Yes! Because AB = AC = AD and the angles between the legs are equal, BCD is equilateral. Furthermore, the height from A to BCD meets the plane of BCD at the circumcenter of BCD (let the foot of the altitude be X; triangles AXB, AXC, and AXD are congruent by HL, so XB=XC=XD).
DPatrick (20:24:57)
DPatrick (20:25:03)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/9dc2294c31b881148c8d166eaa0aee9e.png
5alive (20:24:59)
the base of abcd is equilateral triangle w/ side length 3 root 3
DPatrick (20:25:21)
Good. Since AX = 4, AB = 5, and AXB is a right triangle, XB = 3. Since X is the circumcenter of equilateral BCD, it is also the centroid and the orthocenter. Therefore, altitude BM of BCD is also a median, and it has length (3/2)(BX) = 4.5. From 30-60-90 triangle BMC, we see that each side has length (4.5)(2/sqrt(3)) = 9/sqrt(3) = 3sqrt(3).
DPatrick (20:26:10)
So what is the area of BCD?
b-flat (20:26:31)
alan (20:26:32)
DPatrick (20:26:50)
DPatrick (20:27:05)
DPatrick (20:27:27)
So, now we have the volume of AECD!
DPatrick (20:27:38)
DPatrick (20:28:01)
So if we can find the area of ECD, we'll be done.
DPatrick (20:28:10)
We have CD = 3sqrt(3), so we just need the altitude from E to CD.
DPatrick (20:28:15)
DPatrick (20:28:17)
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DPatrick (20:28:46)
In general, when hunting for a length, one powerful tactic is to build right triangles and use the Pythagorean Theorem. Where can we build a right triangle here to find altitude EM of triangle ECD?
maverick417 (20:29:00)
triangle EMC
Chocoboom (20:29:03)
EMC?
DPatrick (20:29:34)
Yeah, we know that EMC is right, but where else can we find one?
lasagnaman (20:29:15)
altitude from E in EBC
Chocoboom (20:29:41)
CEB
DPatrick (20:30:08)
Not a bad choice.
kostya (20:29:57)
EBM?
platinumnerd (20:29:58)
EMB
DPatrick (20:30:13)
Even better.
DPatrick (20:30:27)
We can draw an altitude from E to triangle BCD.
DPatrick (20:30:32)
DPatrick (20:30:36)
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DPatrick (20:30:54)
How can we find EY and YM easily?
Chocoboom (20:30:13)
3-4-5 triangle!
b-flat (20:31:04)
you have similar right triangles
alan (20:31:05)
DPatrick (20:31:37)
Yes, EBY is a 3-4-5 right triangle! So EY = 4/5 (since EB = 1).
DPatrick (20:31:50)
And BY = 3/5, so what's YM?
DPatrick (20:32:31)
(this is where the calculations start to get a little ugly!)
maverick417 (20:32:25)
4.5-3/5
DPatrick (20:33:03)
Right...we know that BM = 9/2 (since BM is a median of the equilateral triangle BCD that we found before), so YM = BM - BY = 9/2 - 3/5 = 39/10.
lasagnaman (20:31:51)
how do we nkow it is 3-4-5?
Sunday Math-er_2 (20:32:59)
Why does BEY must be a 3-4-5 triangle?
DPatrick (20:33:19)
A couple people have asked this -- perhaps this picture makes it clearer:
DPatrick (20:33:27)
DPatrick (20:33:29)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/b17fa1b1ce5b3eba8d3d243282ffe881.png
DPatrick (20:33:46)
BEY and BAX are similar, and we knew that BAX is 3-4-5, so BEY is too.
DPatrick (20:34:12)
So we can use the Pythagorean Theorem on EYM to find EM.
DPatrick (20:34:47)
It works out to EM = sqrt((4/5)^2 + (39/10)^2) = sqrt(1585)/10.
DPatrick (20:35:07)
Now it's just crunching all the numbers.
DPatrick (20:35:11)
CD = 3*sqrt(3) from before.
DPatrick (20:35:36)
We also know that the volume of AECD is 36*sqrt(3) / 5.
DPatrick (20:35:44)
So we're ready to solve for the height from A to ECD.
DPatrick (20:35:56)
Since volume of AECD = (h)([ECD])/3 = (h)(EM)(CD)/6, we have h = 6(Volume)/(EM)(CD). Solving, we find h = 144/sqrt(1585). Since 39 < sqrt(1585) < 40, our desired answer is 144+39 = 183.
DPatrick (20:36:05)
(You can check the details on your own later if you like!)
DPatrick (20:36:26)
DPatrick (20:36:31)
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Chocoboom (20:36:46)
Absolute values scare me.
DPatrick (20:37:09)
Indeed, they make things really tough to work with -- what can we do to make them go away?
DPatrick (20:37:24)
(This is a common technique -- if you see something ugly, try to make it go away)
perfectnumber628 (20:37:26)
square?
DPatrick (20:37:37)
Aha!
DPatrick (20:37:58)
But before getting too far, it's not hard to see that every x_k must be a multiple of 3. What can we do to simplify things?
DavidL (20:38:09)
Factor out a 3?
Hamster1800_2 (20:38:09)
divide them all by 3?
DPatrick (20:38:17)
Sure, we can make the substitution y_k = x_k/3.
DPatrick (20:38:23)
DPatrick (20:38:43)
DPatrick (20:39:10)
DPatrick (20:39:25)
We can get rid of the ugly absolute values by squaring everything!
DPatrick (20:39:32)
DPatrick (20:39:37)
Now what?
DavidL (20:39:46)
Sum thme
mna851 (20:39:47)
add?
DPatrick (20:40:08)
Sure! Since we want y_1+y_2+... anyway, let's add them!
DPatrick (20:40:25)
Something very, very cool happens!
DPatrick (20:40:31)
DPatrick (20:40:49)
(We have also used the fact that y_0 = 0.)
DPatrick (20:41:00)
So to minimize the sum y_1 + y_2 + ... + y_{2006}, what should we do?
DPatrick (20:41:52)
We should make y_{2007} as close to the square root of 2007 as possible.
DPatrick (20:42:29)
What value of y_2007 gives (y_2007)^2 as close to 2007 as possible?
E^(pi*i)=-1 (20:42:19)
45?
kostya (20:42:30)
45
perfectnumber628 (20:42:32)
45^2=2025
DPatrick (20:42:57)
44^2 = 1936 and 45^2 = 2025.
maverick417 (20:43:45)
so make y_{2007}=45
Hamster1800_2 (20:44:06)
so plug in 45 for y_2007
DPatrick (20:44:19)
How do we know that we can do this?
kostya (20:44:39)
parity
DPatrick (20:45:26)
Yeah, we know that all of the y_k's where k is odd are odd, and all of the y_k's where k is even are even, by looking at the parity (the parity alternates).
DPatrick (20:45:50)
DPatrick (20:46:09)
DPatrick (20:46:40)
That's it! Thanks for coming!
2006 AIME Math Jam
Before we get started, please note that this classroom is moderated, meaning that students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. 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 to write responses that are complete and easy to read.
MCrawford (18:59:41)
MCrawford (18:59:48)
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MCrawford (18:59:50)
(By clicking on the link, you can bring up a copy of the problem in a separate window, to use to follow along as we work through the solution. You may need to hold down the Ctrl key while you click on the link. If that doesn't work, you many need to disable your popup-blocker.)
MCrawford (19:00:03)
What's our first step?
RC-7th (19:00:09)
Draw a diagram
robinhe (19:00:10)
draw it
uclabb (19:00:10)
diagram
Chocoboom (19:00:11)
diagram!
hockeyadi23 (19:00:13)
Start with a picture
ohhoonkwon (19:00:17)
Draw a neat picture?
MCrawford (19:00:32)
We start with a quick sketch:
MCrawford (19:00:35)
Yes, a neat one!
MCrawford (19:00:41)
MCrawford (19:00:48)
As chesspro pointed out after taking the test, you could draw this precisely and measure AD with your ruler if you were so inclined. But if you weren't so inclined, what's the quick route to the solution?
lasagnaman (19:00:10)
to find AC
mathwiz2588 (19:00:18)
Pythagorean
DavidL (19:00:10)
Use two right triangles, and Pythagorean Theorem to get AD = 31
DavidL (19:00:57)
Pythag twice
RC-7th (19:01:02)
Pythagorean Theorem!
hockeyadi23 (19:01:05)
Now it becomes clear (once you label out the sides). Use the pythagorean theorem twice to find AD.
uclabb (19:01:06)
pythag a couple times
robinhe (19:01:09)
ABC is a right traingle and AC^2=765
E^(pi*i)=-1 (19:01:10)
Pythag theorem
lasagnaman (19:01:12)
find AC using pythag, then find AD using pythag
MCrawford (19:01:28)
All we need is AD, and a couple applications of the Pythagorean Theorem gives it to us:
MCrawford (19:01:31)
MCrawford (19:01:33)
MCrawford (19:01:40)
Combining these, we have
MCrawford (19:01:42)
DavidL (19:01:39)
So The Perimeter = 84
MCrawford (19:02:10)
Therefore, AD = 31, so our perimeter is 18+21+14+31 = 084.
MCrawford (19:02:20)
MCrawford (19:02:32)
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hockeyadi23 (19:02:29)
Find the min and max
E^(pi*i)=-1 (19:02:42)
Find the min (first 90) and max (last 90)
Chocoboom (19:02:43)
you just find the minimum possible sum and the maximum possible sum
mathclass (19:02:44)
all sums from the minimum sum to the maximum sum can be achieved bc it's 100 consecutive numbers.
robinhe (19:02:45)
the smallest is 1+2+...90 and the largest is 11+12+...100
MCrawford (19:03:14)
We can quickly find the highest and the lowest sums. The lowest is 1 + 2 + . . . + 90 and the highest is 11 + 12 + . . . + 100.
MCrawford (19:03:20)
We wonder if we can attain every sum in between. Can we?
Hamster1800_2 (19:03:03)
So find the min and the max, and just "see" that you can form any integer value between them.
Chocoboom (19:03:11)
And everything else inbetween can be possible as well.
MCrawford (19:04:32)
It would be nice if we could prove our intuition.
DavidL (19:03:45)
Yes, consec ints
robinhe (19:03:56)
yes because just add 1 to one of the sums
uni412 (19:04:11)
Yes, because you can always reduce one of the numers in the set by 1
E^(pi*i)=-1 (19:04:11)
Yes, if we have a certain sum, just increase an element by one to get a bigger sum by one.
rachel (19:04:23)
yes-start with the lowest and increase the highest to lowest terms (induction) to get everything in between
MCrawford (19:05:09)
Yes! Let our lowest sum be T: T = 1 + 2 + . . . + 90. We add 1 to the highest term of T: T = 1 + 2 + . . . + 89 + 91.
MCrawford (19:05:11)
We can keep adding 1 to the highest term of T that isn't 'maxed out' (i.e. after 1 + 2 + . . . + 89 + 100 comes 1 + 2 + . . . + 88 + 90 + 100) until we reach our largest possible sum. Therefore, our problem is finding the number of integers from 1 + 2 + . . . + 90 to 11 + 12 + . . . + 100, inclusive.
hockeyadi23 (19:02:37)
Take the difference and add 1
Hexahedron (19:03:27)
difference between those is (11-1)*90=900, and have to add one
hockeyadi23 (19:03:52)
10*90
MCrawford (19:05:31)
We subtract the two numbers and add 1. Subtraction cancels most of the terms (and points out that we could have solved this by considering sums of 10-member subsets) and leaves us with:
(91 + 92 + . . . + 100) - (1 + 2 + . . . + 10) +1
MCrawford (19:05:37)
This equals
(91 - 1) + (92 - 2) + (93 - 3) + . . . + (100 - 10) + 1 = 10(90) + 1 = 901.
MCrawford (19:05:43)
MCrawford (19:05:50)
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Elemennop (19:05:52)
Set it up algebraicly
MCrawford (19:06:00)
Letting A be our original number and B being the new number, what information can we get about A and B from the problem?
hockeyadi23 (19:06:00)
I like this one: call n the digit removed, and k the two digit number that's left over
JRechel (19:06:01)
you know it has to end in 5
robinhe (19:06:07)
list all 3-digit multiples of 29?
uclabb (19:06:10)
29x=100+x
MCrawford (19:06:23)
Not 100, but 100n -- some multiple of 100.
mathwiz2588 (19:06:16)
29A =B
DavidL (19:06:14)
Use an equation 100a + 10b +c = 290 b + 29c
tiredepartment (19:06:22)
100a+10b+c
fubu (19:06:10)
if we try two digits, we can immediately see that it doesn't work. such as 10x + y
hockeyadi23 (19:06:13)
A= 100n+k, b=k
epi14159 (19:06:16)
since the last digit remain the same, only 0 and 5 can be the last digit of the number
lasagnaman (19:06:24)
so, A must be a multiple of 25, also, quikc trial and error shows that A must be 3 digits
MCrawford (19:06:42)
Lot's of ideas here.
MCrawford (19:06:46)
We know that 29B = A, and from the given information, we know that the last digits of B and A are the same.
MCrawford (19:06:55)
MCrawford (19:07:04)
Now what?
uclabb (19:07:24)
that makes it a mult of 25
Hexahedron (19:07:31)
A=0(mod 25)
DavidL (19:07:34)
Why put it in Modular Arithmetic when we can avoid it altogether?
MCrawford (19:07:52)
I think it's simplest.
MCrawford (19:07:56)
MCrawford (19:08:01)
Now we simply try the smallest positive multiple of 25:
25 x 29 = 725.
We have a winner, and it is 725!
MCrawford (19:08:03)
And that's all there is to it.
Elemennop (19:07:30)
There's no reason to split up the two digit part into two variables. Let N be a digit, and let X be a two digit number. 100N + X = 29X, or 100N-28X = 0, which has solution (N,X)=(7,25)
MCrawford (19:08:22)
There are many ways to work this one.
MCrawford (19:08:40)
Modular arithmetic has the advantage of helping us quickly isolate one of the variables to find its value.
MCrawford (19:08:49)
Solutions using modular arithmetic, divisibility, and a bit of algebra are covered in the AoPS Introductory Number Theory class as well as the upcoming Introductory Number Theory book.
MCrawford (19:08:56)
MCrawford (19:09:05)
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MCrawford (19:09:17)
In order to count the number of 0's at the end, what do we have to count?
hockeyadi23 (19:09:09)
Basically count the number of fives
Sunday Math-er_2 (19:09:10)
count # of 5s
mathwiz2588 (19:09:13)
well 100! has 24 zeros
E^(pi*i)=-1 (19:09:15)
We need the number of 5's in the product
ohhoonkwon (19:09:17)
Find all the fives
MCrawford (19:09:40)
We have to count the number of factors of 5 in this gigantic product. How are we going to do it in an organized way? Obviously we can forget about the first 4 factorials. Are we then going to count all the 5's in 5!, then all the 5's in 6!, etc., all the way up to 100!?
We could do this, but we can organize our counting more effectively - how?
llamallad (19:09:34)
100 has 24 zeros and 99-95 has 22 0's because they lose 2 factors of 5. just continue this down to 5. and add up all zeros
perfectnumber628 (19:09:39)
to find the amount of 0s at the end of x!, you figure out how many 5s or powers of 5 are less than it
robinhe (19:09:41)
1-4 factorials don't have 0's
5-9 each have 1
10-14 each have 2
...
MCrawford (19:10:14)
We can note that the number 5 shows up in 96 factorials (5! through 100!, inclusive). The number 10 shows up in 91 factorials (10! through 100!), the number 15 in 86 factorials, etc.
Following in this vein, we have how many multiples of 5?
Hexahedron (19:10:02)
5-9 is same number, 10-14 is same....
DavidL (19:10:06)
Group them by five
rachel (19:10:08)
count all 5s, then all 25s separately
E^(pi*i)=-1 (19:10:13)
5 appears 96 times, 10 appears 91 times....
MCrawford (19:11:17)
What's 96 + 91 + 86 + ... + 1?
E^(pi*i)=-1 (19:11:11)
970 multiples of 5, but some have two
MCrawford (19:12:07)
We have 96 + 91 + . . . + 6 + 1 = (97)(10) = 970 multiples of 5 in our list.
MCrawford (19:12:14)
We have to remember the multiples of 25, which add a factor of 5 to our product. How many multiples of 25 are there?
perfectnumber628 (19:12:01)
but then don't you have to also count multiples of 25?
Sunday Math-er_2 (19:12:20)
then count 5^2*k again
hockeyadi23 (19:12:29)
154
matt276eagles (19:12:34)
76+51+26+1=2(77)=154
MCrawford (19:13:04)
We follow similar reasoning as before. The number 25 shows up in 76 factorials (25! through 100!), 50 shows up in 51 factorials, 75 in 26 factorials, and 100 in 1 factorial, for a total of 76 + 51 + 25 + 1 = 154.
E^(pi*i)=-1 (19:12:50)
76 25's, 51 50's, 26 75's, 1 100
MCrawford (19:13:14)
So, what is our final answer?
robinhe (19:13:04)
970+154=1124
so the answer is 124
MCrawford (19:13:33)
Our final answer is the remainder when 970+154 = 1124 is divided by 1000, or 124.
MCrawford (19:13:44)
MCrawford (19:13:54)
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MCrawford (19:13:56)
MCrawford (19:13:58)
What do we do with that?
tiredepartment (19:14:00)
set them equal and square both sides
robinhe (19:14:01)
set it up as an equation and then square both sides
Elemennop (19:14:02)
Square the variable expression, and set it equal to the square.
alan (19:14:02)
square it
MCrawford (19:14:33)
The big square root on the right is intimidating. We can get rid of it by squaring both sides:
MCrawford (19:14:39)
matt276eagles (19:14:24)
square and match terms
fubu (19:14:28)
the coefficients of the roots have to be the same
MCrawford (19:14:59)
We can match up coefficients of the radicals:
2ab = 104
2ac = 468
2bc = 144
mel (19:14:46)
Then equate corresponding parts to find ab, bc, and ca.
Kelly_F (19:14:59)
then you can try to simplify the numbers in the square roots on the right side to come with the a comparatable equation on the left.
robinhe (19:15:00)
so then 2ab=104, 2ac=468...
MCrawford (19:16:03)
Divide by 2:
ab = 52
ac = 234
bc = 72
DavidL (19:15:07)
Multiply tme all up
Elemennop (19:15:09)
a,b,c are rational, so ab=52, ac=234, bc=72. Multiply all together and take the square root to get abc!
DavidL (19:15:22)
and then you get a2b2c2 = somebignumber
hockeyadi23 (19:15:29)
Then prime factorize, and take the square root of the product
fubu (19:15:36)
find ab, ac, bc, multiply them altogether and then take the square root ... prime factorization will help with the square rooting
MCrawford (19:16:55)
The product of our three equations gives us (abc)^2:
MCrawford (19:17:06)
(ab)(ac)(bc) = (52)(234)(72) = (4)(13)(13)(18)(8)(9).
MCrawford (19:17:29)
What's our answer?
hockeyadi23 (19:16:50)
Thus the answer is 72*13=936
robinhe (19:16:56)
936
Kelly_F (19:17:03)
abc=976
Chocoboom (19:17:20)
But it's too complicated squaring without a calculator.
MCrawford (19:18:38)
MCrawford (19:18:57)
Those of you who suggested using prime factorization are correct.
MCrawford (19:19:03)
You say an enormous amount of time multiplying by factors instead of large numbers.
MCrawford (19:19:13)
MCrawford (19:19:23)
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MCrawford (19:19:27)
Hexahedron (19:19:30)
it needs to be of the form (abc/999)
hockeyadi23 (19:19:26)
9*8*7=720
perfectnumber628 (19:19:35)
there are (10)(9)(8)=720 numbers
Hexahedron (19:19:36)
abc, is distinct
E^(pi*i)=-1 (19:19:37)
Fractions abc/999
robinhe (19:19:36)
12/999
Elemennop (19:19:29)
Gauss!! Gaussian Pairing = easy solution.
DavidL (19:19:31)
10x9x8 ways
fubu (19:19:34)
that decimal is equal to (100a+10b+c)/999
matt276eagles (19:19:50)
we have 10*9*8=720 fractions, each pair sums to 1
MCrawford (19:20:09)
Finding every fraction is going to be a nightmare: there are (10)(9)(8) = 720 of them! What a pain!
MCrawford (19:20:16)
We can look at how many times each digit shows up in each place. For example, if we stick a 0 in the first place, we have 9 ways to choose the next digit, and 8 ways to choose the third digit, for a total of 72 ways. (This shouldn't be surprising: 1/10 of our 720 numbers start with 0, by symmetry.)
MCrawford (19:20:30)
Similarly, there are 72 ways to start with 1, 72 ways to start with 2, etc. From here there are many ways to solve the problem. We can add across each digit, or we could add up all the 1's (then 2's, then 3's) in all digits, etc.
pilot (19:20:26)
Each digit occurs 9*8=72 times in each of the 3 places
Mathgod (19:20:22)
they pair up
Mathgod (19:20:32)
so it's 360
Hamster1800_2 (19:20:17)
Since there are 720 numbers and 10 digits for each place,each digit should appear 72 times in each spot.
MCrawford (19:20:53)
We can consider the average of each digit. The 'average' first digit is the average of the numbers 0 through 9, or 4.5. The average second digit is 4.5, as is the average third digit, and so on.
MCrawford (19:20:57)
Therefore, our average number is:
MCrawford (19:21:01)
MCrawford (19:21:04)
The sum in the parentheses is a geometric series with sum (1/10)/(1-1/10) = 1/9, so our average number is (4.5)(1/9) = 1/2.
MCrawford (19:21:21)
Though pairing in sums of 1 is certainly faster!
MCrawford (19:21:36)
The average number is 1/2, and there are 720 of them, so the sum of the numbers is (720)(1/2)=360.
MCrawford (19:22:06)
MCrawford (19:22:09)
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MCrawford (19:22:12)
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MCrawford (19:22:19)
MCrawford (19:22:31)
How can we proceed?
Hexahedron (19:22:34)
so all the triangles are similiar
Hamster1800_2 (19:22:50)
All of the triangles formed are similar, right?
dhuang614 (19:22:54)
let 1 equal the distance between two lines
hockeyadi23 (19:22:55)
so then solve for the base of a, calling the distance between parallel lines 1
Elemennop (19:22:57)
Heights are all equal except for A...+ Parallel lines, so the triangles are similar. Then find the ratios of heights, whose square is the ratio of area.
MCrawford (19:23:37)
Tons of parallel lines should make you think of similar triangles.
MCrawford (19:23:41)
Let's assume (for simplicity) that the distance between the parallel lines is 1.
Then we can express the height of the small triangle A to be x, where 0 < x < 1.
MCrawford (19:23:44)
So the heights of the successive triangles are x, 1+x, 2+x, 3+x, ....
How can we write the ratio of the shaded areas C and B?
DavidL (19:24:04)
As sums and differences of squares of heights
Elemennop (19:24:11)
(4+x)^2-(3+x)^2 / (2+x)^2 - (1+x)^2
hockeyadi23 (19:23:59)
11/5
MCrawford (19:24:35)
C is the area of a triangle with height 4+x minus the area of a triangle with height 3+x.
B is the area of a triangle with height 2+x minus the area of a triangle with height 1+x.
MCrawford (19:24:38)
MCrawford (19:25:09)
So 11/5 = (2x+7)/(2x+3).
This gives 22x+33 = 10x + 35, so 12x = 2, hence x = 1/6.
Then what is the ratio we want to find?
hockeyadi23 (19:25:29)
areas
Elemennop (19:25:31)
(6+x)^2 - (5+x)^2 / x^2
pilot (19:25:53)
((6+x)^2-(5+x)^2)/x^2
MCrawford (19:26:40)
Hamster1800_2 (19:26:17)
The area of D to the area of A, which would be (6+1/6)^2 - (5+1/6)^2 to (1/6)^2
MCrawford (19:27:09)
This is (2x+11)/x^2.
Elemennop (19:26:47)
Plugging in 1/6 yields the ratio as 408, which is what we want.
MCrawford (19:27:38)
Plugging in x=1/6 gives us (34/3)*36 = 34*12 = 408.
MCrawford (19:27:46)
Remember, parallel lines mean similar triangles!
Problems like this are covered in the new AoPS Introductory Geometry book.
MCrawford (19:28:01)
MCrawford (19:28:08)
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MCrawford (19:28:15)
MCrawford (19:28:20)
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MCrawford (19:28:33)
Any intuition as to what's going on here?
krsattack (19:28:27)
figure out max and min areas of T
d343seven (19:28:38)
call AFE=2theta
Elemennop (19:29:09)
Well, we can squeeze the hexagon until T becomes zero, so if we find the max area of T we can achieve every integer less than max T because it's function of area is a smooth function .
dhuang614 (19:29:10)
define an angle
MCrawford (19:29:33)
You might see "intuitively" that we can make T arbitrarily small by making the outer rhombi P, Q, R, and S look more like squares:
MCrawford (19:29:36)
epi14159 (19:29:27)
the min areas of T is 0
Optimosis (19:29:23)
area of T based on angles of the other 4 rhombuses
MCrawford (19:30:04)
Let's label the other four points:
MCrawford (19:30:14)
MCrawford (19:30:56)
And T also appears to grow bigger as we "flatten" the outer rhombi:
MCrawford (19:31:01)
Elemennop (19:30:28)
We can use angles or we can use the diagonals of T (one of which is twice the height of the other rhombi) to find the side of the rhombus in terms of the diagonal
mel (19:29:51)
MCrawford (19:31:46)
Let d be the side length of the rhombi, and let theta be angle FWX.
MCrawford (19:31:57)
(Note: you can also work with angle AWX -- it'll come out essentially the same.)
MCrawford (19:32:05)
MCrawford (19:32:13)
Angle XWZ is 360 - 2*theta, so:
MCrawford (19:32:18)
MCrawford (19:32:24)
But sin(360-x) = sin(-x) = -sin(x), so the area of T is just d^2 * -sin(2*theta).
MCrawford (19:32:38)
Now we can use the double-angle formula!
MCrawford (19:32:47)
MCrawford (19:33:08)
How do we use what we've learned to identify the answer?
perfectnumber628 (19:33:17)
the cos is negative
Elemennop (19:33:25)
Cos ranges from 90 to 180, so the area ranges from 0 to just above 89.
MCrawford (19:33:50)
Hence, as theta varies from 90 degrees up through 180 degrees, cos(theta) varies from 0 down to -1, so the area of T varies from 0 up to 2*sqrt(2006) (as we expected).
Thus the answer is the largest integer less than 2*sqrt(2006) = sqrt(8024).
perfectnumber628 (19:33:42)
do you have to extract sqrt2006?
E^(pi*i)=-1 (19:33:41)
89 possibilities for theta
mel (19:33:53)
MCrawford (19:34:14)
Note that 8024 < 8100 = 90^2, so we check that 89^2 = (90-1)^2 = 8100 - 180 + 1 = 7921 < 8024.
MCrawford (19:34:23)
We don't need to extract the exact square root.
MCrawford (19:34:25)
So 89 < sqrt(8024) < 90, and the answer is 089.
Chocoboom (19:33:04)
Whats the double-angle formula?
MCrawford (19:34:58)
MCrawford (19:35:20)
I recommend asking how it's derived in the Intermediate forum or trying it yourself.
MCrawford (19:35:38)
Trig identities like that become essential at this level of problem solving.
DavidL (19:35:37)
Hint: use the unit circle!
MCrawford (19:35:54)
At this time, Dave Patrick will take over the Math Jam.
DPatrick (19:36:00)
DPatrick (19:36:05)
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DPatrick (19:36:13)
What should we do to make this equation easier to work with?
Hamster1800_2 (19:36:17)
write the logs as one log
quantum leap (19:36:22)
combine them into one log
DavidL (19:36:23)
Well, it's just log base 8 of (a1a2...)
DPatrick (19:36:35)
We can use the identity log_8 (a) + log_8 (b) = log_8 (ab) to rewrite it as
DPatrick (19:36:40)
tiredepartment (19:36:29)
def of geo seq: a1=a
a2=ar
a3=ar^2
weiyan (19:36:28)
write a1, a2, etc in terms of a and r, such as a, ar, ar^2...
DPatrick (19:37:04)
We know that a_1 = a, a_2 = ar, a_3 = ar^2, and so on up to a_12 = ar^11.
DPatrick (19:37:09)
So what is the product of a_1...a_12?
Elemennop (19:36:58)
a_1a_2...a_{12} = a^12r^66
DPatrick (19:37:27)
We have 12 copies of a, and (1+2+3+...+11) copies of r.
DPatrick (19:37:43)
DPatrick (19:37:49)
Now what?
hockeyadi23 (19:37:29)
a^12r^66=8^2006
fubu (19:37:54)
change it to exponential form
DPatrick (19:38:17)
We can get rid of the log.
DPatrick (19:38:25)
anniesez (19:38:12)
a^12r^66=8^2006=2^6018
Hexahedron (19:38:17)
= 2^6012
perfectnumber628 (19:38:21)
8^2006=2^6018
HuAreYou? (19:38:22)
or 2^6018
DPatrick (19:38:54)
Yes, better to write this as a power of a prime:
DPatrick (19:38:57)
b-flat (19:38:38)
you know both a and r must be powers of 2
DPatrick (19:39:12)
Since a and r are positive integers, we know that they must both be powers of 2 (the only prime factor).
E^(pi*i)=-1 (19:39:13)
we know a and r are powers of 2, so 12x+66y=6018
DPatrick (19:39:31)
So let a = 2^x and r = 2^y, where x and y are nonnegative integers.
Then our equation becomes 12x + 66y = 6018.
DavidL (19:39:31)
Diophantine Time!
mathwiz2588 (19:39:35)
diophantine eq.
DPatrick (19:39:55)
Yes, this is what is known as a Diophantine equation, since we are looking for integer solutions.
alan (19:39:51)
epi14159 (19:39:53)
2x+11y=1003
Elemennop (19:40:02)
2x + 11y = 1003 is a nicer form.
DPatrick (19:40:15)
Yes, it's nicer if we divide by 6 to get 2x + 11y = 1003.
DPatrick (19:40:23)
How do we solve this?
Hamster1800_2 (19:40:28)
so y must be odd.
dhuang614 (19:40:35)
then take remainders modulo 2
Hexahedron (19:40:40)
y must be odd
DPatrick (19:41:07)
This will have a solution for any odd y.
DPatrick (19:41:16)
If y is odd, then x = 1003-11y / 2.
DPatrick (19:41:21)
(If y is even, then 1003-11y is odd, so we can't divide by 2 and get an integer.)
DPatrick (19:41:37)
How many values of y make x nonnegative?
5alive (19:41:26)
y can be 1,3,5,...91
epi14159 (19:41:33)
(496,1)...(1,91)
lasagnaman (19:41:38)
also ,11y < 1003
DPatrick (19:42:15)
Exactly -- we need y < 91.
DPatrick (19:42:21)
So y = 1,3,5,...91.
bthings_2000 (19:42:10)
91= 45+46, 46 odds since 1 and 91 are odd, so it's 46
5alive (19:42:25)
or equal to 91 **
DPatrick (19:42:43)
Right -- I should have written y <= 91, thanks.
DPatrick (19:42:47)
So there are 046 possible solutions.
DPatrick (19:43:01)
DPatrick (19:43:05)
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DPatrick (19:43:16)
DPatrick (19:43:19)
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krsattack (19:43:17)
look at where the circles intersect. Connecting these lines creates symmetrical patterns which should help us get lucky
DPatrick (19:43:46)
Indeed, that will work.
DPatrick (19:44:02)
Just to warm up to the problem, let's first modify it to make it much easier. First, imagine that there were nine circles in the plane instead of eight.
DPatrick (19:44:11)
DPatrick (19:44:16)
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DPatrick (19:44:24)
Where would the line go in this case?
dhuang614 (19:44:27)
all lines must pass through the center circle
tiredepartment (19:44:29)
If there were 9, then the line must pass through the center of the center circle
mathwiz2588 (19:44:32)
through thte center of the cetner circle
DPatrick (19:44:42)
Right, it would go through the center of the center circle.
DPatrick (19:44:49)
DPatrick (19:44:52)
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DPatrick (19:45:02)
I want to point out that the case with nine circles is very easy, because there is a lot of symmetry in our diagram. Any line that goes through the center of a circle is automatically an axis of symmetry of that circle, and we want to divide a region into two regions of equal area. Both of these points strongly suggest that we want to use symmetry.
DPatrick (19:45:08)
Unfortunately, the original problem with eight circles does not quite have as much symmetry. What other lines can we try drawing?
nsato (19:45:18)
DPatrick (19:45:32)
(ignore that!)
happydude (19:45:27)
Transform the line to the left
dhuang614 (19:45:27)
lines through tangent points
DPatrick (19:45:52)
After removing the ninth circle in the corner, we need to move the line a little to the left to balance this missing circle.
barce11 (19:45:51)
draw lines where they are tangent
DPatrick (19:46:25)
We could try the line passing through the point of tangency between circles 5 and 8. That's the first obvious place to try as we move to the left.
DPatrick (19:46:31)
I'll also label the circles for our convenience:
DPatrick (19:46:42)
DPatrick (19:46:53)
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DPatrick (19:47:03)
One thing should immediately leap out of the diagram at us. What is that?
Hamster1800_2 (19:46:47)
Why are points of tangency "obvious choices?"
DPatrick (19:47:20)
Good question. We're really hoping to get lucky and find a "nice" line that works.
DPatrick (19:47:27)
The only points we have to work with are the centers of the circles and the points where they're tangent.
pilot (19:47:25)
they divide the two circles equally
dhuang614 (19:47:26)
the points of tangency split two circles into equal area
b-flat (19:47:36)
the cut off portions in circles 5 and 8 are equal...
lasagnaman (19:47:31)
there are 2 whole circles on each side, and 1 and 2 are evenly split bewteen the 2 sides, as are 5 and 8
perfectnumber628 (19:47:41)
you only have to worry about circles 1,2,5,8
DPatrick (19:48:09)
Right, it should be pretty clear by symmetry that the line I drew above splits circles 5 and 8 evenly.
DPatrick (19:48:12)
Why does it also split circles 1 and 2 evenly?
DPatrick (19:48:30)
The line appears to pass through the point of tangency of circles 1 and 2. Is this true?
DavidL (19:48:26)
Because it happens to pass through the tangent point
dasherm (19:48:29)
because it hits their tangent as well
mathwiz2588 (19:48:31)
it passes thru their tangency also
DPatrick (19:49:05)
Yes, because the line has slope 3, which is also the slope between the two points of tangency. (The horizontal part is one radius, and the vertical part is three radii.) A simple diagram wiill hopefully make this clear:
DPatrick (19:49:08)
DPatrick (19:49:13)
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DPatrick (19:49:23)
So the line splits circles 1 and 2 evenly as well.
DPatrick (19:49:56)
So this is the line that we want!! It splits 5&8 evenly, 1&2 evenly, and has two other whole circles on each side.
krsattack (19:49:54)
so all we have left is to find the equation of the line
DPatrick (19:50:14)
Yes -- the rest is now a very easy computation. The line passes through the point (1.5, 2) and has slope 3.
DavidL (19:50:12)
So we use point slope form
bthings_2000 (19:50:23)
point slope time!
krsattack (19:50:24)
point slope formula
DPatrick (19:50:39)
The equation of the line is then y - 2 = 3(x - 1.5), which becomes 2y = 6x - 5.
DPatrick (19:50:46)
Therefore, the final answer is 2^2 + 6^2 + 5^2 = 065.
DPatrick (19:50:56)
It's very easy to make a problem like this too complicated. We might search for a formula that would give areas in terms of lines and circles, but this would get very unwieldy, very quickly. Problems on the AIME tend to have simple solutions, and this problem is no exception; you just have to look for them. Here, we solved the problem by checking key points and using symmetry - the simplest approach is often the best.
DPatrick (19:51:21)
DPatrick (19:51:28)
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DPatrick (19:51:46)
How might we start to get a handle on this problem?
barce11 (19:51:50)
use simpler cases
pilot (19:51:56)
smaller cases
dhuang614 (19:51:57)
solve a simpler problem
lasagnaman (19:52:02)
try working with 2, then 3 then 4 cubes
DPatrick (19:52:20)
OK, how many stacks with 2 cubes?
alpha123 (19:52:26)
2
perfectnumber628 (19:52:26)
2
DPatrick (19:52:40)
For 2 blocks, there are clearly 2 ways: (1 2) and (2 1). (Let the first number be the 'bottom'.)
DPatrick (19:52:46)
How about 3 cubes?
E^(pi*i)=-1 (19:52:49)
6 stacks
alpha123 (19:52:50)
3! = 6
Hannes (19:52:51)
6
sirorange (19:52:51)
2*3=6
uni412 (19:52:51)
6
DPatrick (19:53:10)
Again, all orders work, for a total of 3! = 6.
DPatrick (19:53:18)
How about 4?
Elemennop (19:52:58)
The same with 3 cubes, 6. For four cubes, only 18 of them work (anything with 14 together in that orer doesn't work).
E^(pi*i)=-1 (19:53:19)
18 stacks
cobylu (19:53:25)
18
alpha123 (19:53:25)
18
hockeyadi23 (19:53:25)
18
DPatrick (19:53:52)
For 4, we have to omit the orders where 4 comes right after 1: (1 4 2 3) (2 1 4 3), and so on. All the rest work.
perfectnumber628 (19:53:42)
4! but subtract cases where 1 is right under 4
DPatrick (19:54:36)
We could consider 1-4 together a single block, and see that there are 3! ways to order 1-4, 2, and 3.
DPatrick (19:54:47)
So there are 6 bad cases, hence 4! - 6 = 18 ways to stack them.
DPatrick (19:54:52)
So, we have a bit of a feel for the problem. Stepping up to 5 blocks, we would have to throw out all the 5-2's, the 5-1's, and 4-1's from the 5! orders, which is starting to get complicated.
DPatrick (19:55:05)
We wonder if just as we build our illegal 4-block towers from the 3-block towers, can we also build all the legal ones?
krsattack (19:54:49)
put the nth cube on bottom, over n-1th cube, or over n-2th cube
Elemennop (19:54:59)
2,6,18...multiply by 3 ?
lasagnaman (19:55:12)
now, for 5 cubes, you can take all possible orders for 4 cubes, then insert the 5th block in any of 5 positions in between blocks, then rule out situations where 5 comes right after 1 or 2
Hamster1800_2 (19:55:29)
so we can put the 5 at the beginning, after 3, or after 4 (3 places). For 4 blocks, you could put it at the beginning, after 2, or after 3.
hockeyadi23 (19:55:32)
RECURSION!!!
DPatrick (19:56:13)
Right. For example, to go from a 3-block tower to a 4-block tower, we can either put block 4 on the bottom, or just above the 2, or just above the 3, for a total of 3 legal 4-block towers for each 3-block tower.
DPatrick (19:56:24)
Clearly doing this over all 3-block towers won't create any tower more than once (i.e. it doesn't overcount).
DPatrick (19:56:33)
We must also make sure that we haven't missed any legal 4-block towers; i.e. that there aren't any 4-block towers which cannot be formed by just adding block 4 to a 3-block tower.
DPatrick (19:57:00)
But this works too -- if we have a 4-block tower, and remove block 4, we get a legal 3-block tower.
DPatrick (19:57:08)
So what's the conclusion?
Hamster1800_2 (19:57:26)
To get from the number of n block towers to the number of n+1 block towers, multiply by 3.
DPatrick (19:58:08)
Right. Every tower with n blocks gives exactly 3 of the towers of n+1 blocks. And every tower with n+1 blocks comes from an n-block tower.
krsattack (19:57:56)
formula time : 2 * 3^(k-2)
hockeyadi23 (19:58:00)
So the closed form is 2*3^n-2
DPatrick (19:58:50)
Right. There are 2 = 2*3^0 2-block towers, 6 = 2*3^1 3-block towers, and since we multiply by 3 everytime we add a block, we get 2*3^(n-2) n-block towers.
alpha123 (19:58:29)
so for 8 cubes stacks = 2*3^6 = 1458
kostya (19:58:36)
so we get 2 * 3^6
E^(pi*i)=-1 (19:58:43)
2*3^6=1458
DPatrick (19:59:05)
Letting n=8, we have 2(3^6) = 1458 legal 8-block towers. The remainder when this is divided by 1000 is 458.
DPatrick (19:59:18)
DPatrick (19:59:25)
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DPatrick (19:59:35)
How can we start working with this equation?
Hamster1800_2 (19:59:37)
put everything in terms of 4x and x
alpha123 (19:59:41)
say cos3x = cos(4x-x)
DPatrick (19:59:58)
The key observation to notice is that 3x = 4x - x and 5x = 4x + x.
DPatrick (20:00:13)
So we can use the cosine angle-sum formulas.
DPatrick (20:00:21)
DPatrick (20:00:35)
So our original equation becomes:
DPatrick (20:00:38)
DPatrick (20:00:55)
When we expand the left side, what happens?
5alive (20:01:00)
much cancellation on the left side
Hamster1800_2 (20:01:05)
When we cube them, most of the sines cancel, but not all.
krustyteklown_2 (20:01:05)
2 terms cancel
DPatrick (20:01:27)
Two of the four terms from each cubic cancel.
DPatrick (20:01:37)
We're left with:
DPatrick (20:01:40)
DPatrick (20:02:08)
Now what?
alpha123 (20:01:53)
and divide by 6
Hamster1800_2 (20:02:14)
move everything to one side so we can use 0 to help us.
DPatrick (20:02:57)
Yes, but I would also use sin^2 4x = 1 - cos^2 4x and sin^2 x = 1 - cos^2 x to write everything in terms of cos 4x and cos x.
DPatrick (20:03:02)
For ease of writing, let's set a = cos 4x and b = cos x.
DPatrick (20:03:24)
DPatrick (20:03:40)
When we expand this out, we see that the a^3b^3 term cancels, and we're left with
DPatrick (20:03:44)
DPatrick (20:04:05)
So the solutions are either a=0, b=0, or a^2+b^2 = 1.
DPatrick (20:04:17)
We can eliminate b=0, since cos x is never 0 for 100 < x < 200.
DPatrick (20:04:31)
When is cos 4x = 0 in this range?
krsattack (20:04:49)
4x = 90 + 180k
perfectnumber628 (20:04:56)
if a=0, 4x=90 or 270 or 450, etc
DPatrick (20:05:16)
Clearly 400 < 4x < 800, so the angles in this range which have cos 4x = 0 are 4x = 450 and 4x = 630. (cos is 0 at odd multiples of 90.)
DPatrick (20:05:22)
So x = 450/4 and x=630/4 are two solutions. These solutions sum to 1080/4 = 270.
DPatrick (20:05:41)
Finally we need the solutions to a^2+b^2 = 1.
These are solutions to cos^2(x) + cos^2(4x) = 1.
How can we find these?
E^(pi*i)=-1 (20:06:12)
use double angle formula?
DPatrick (20:06:40)
We can use the half-angle formula to rewrite cos^2(x) + cos^2(4x) = 1 as (1+cos(2x))+(1+cos(8x)) = 2, so we have cos(2x) + cos(8x) = 0.
DPatrick (20:06:58)
Then we can use the sum-to-product formula to rewrite this as 2cos(5x)cos(3x) = 0.
DPatrick (20:07:16)
So we must have cos(5x)=0 or cos(3x)=0.
DPatrick (20:07:35)
What values of 100 < x < 200 give cos(5x)=0 or cos(3x)=0?
perfectnumber628 (20:07:39)
so 5x=90, 270, etc
platinumnerd (20:08:10)
same for 3x
Hamster1800_2 (20:08:24)
we want 500 < 5x < 1000 and 300 < 3x < 600 where the middle is 90+180k for each
DPatrick (20:09:00)
Right. cos(5x)=0 gives 126, 162, and 198 in the correct range, and cos(3x)=0 gives 150.
DPatrick (20:09:22)
So these four solutions sum to 126+162+198+150 = 636.
DPatrick (20:09:25)
Finally, adding this to the 270 that we found before, we get our final answer of 636+270 = 906.
DPatrick (20:09:53)
Note: there are other ways you can attack this problem -- you could try using the sum-to-product or product-to-sum at the very beginning.
DPatrick (20:10:02)
But this is how I did it. :)
DPatrick (20:10:20)
DPatrick (20:10:24)
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DPatrick (20:10:33)
Where can we start with this problem?
krustyteklown_2 (20:10:45)
try a first few couple n's
5alive (20:10:49)
write out some sn
DPatrick (20:11:05)
Looking at small n's might give a pattern.
DPatrick (20:11:14)
For example, take n = 5.
DPatrick (20:11:20)
Then we can generate the following table.
DPatrick (20:11:26)
DPatrick (20:11:41)
(you'd probably do even smaller n's first, but I'm trying to speed things up a bit here!)
jiangwei (20:11:17)
2 shows up every other term, 4 shows up every 4 terms, .....
perfectnumber628 (20:11:52)
half of them are 2, 1/4 are 4, etc
DPatrick (20:12:07)
Right. The pattern seems clear.
DPatrick (20:12:11)
DPatrick (20:12:24)
The number 2 appears eight times, the number 4 appears four times, the number 8 appears twice, the number 16 appears once, and the number 32 appears once.
DPatrick (20:12:35)
DPatrick (20:12:45)
This way of writing the sum makes the general formula much more accessible.
DPatrick (20:13:14)
How many times does the number 2^m appear among these numbers, where 1 <= m <= n?
E^(pi*i)=-1 (20:13:40)
2^(n-m)
DPatrick (20:14:06)
Close...it's actually 2^(n-m-1) times.
DPatrick (20:14:30)
...except for 2^n, which appears once.
DPatrick (20:14:41)
DPatrick (20:15:04)
We don't have to prove it to get an answer for the AIME (and it would probably waste valuable time), but it's a good exercise to go through the proof on your own.
DPatrick (20:15:17)
What can we do with this expression?
krustyteklown_2 (20:15:01)
=(n+1)*2^(n-1)
DavidL (20:15:06)
(n+1) 2^(n-1)
dhuang614 (20:15:16)
factor into (2^(n-1))(n+1)
DPatrick (20:15:33)
DPatrick (20:15:45)
We now have two nice factors to work with.
DPatrick (20:15:48)
When is this a perfect square?
alpha123 (20:15:56)
when n is odd
krustyteklown_2 (20:15:46)
so we need n+1 to be a perfect square, but n-1 has to be even, so the power of 2 is a perfect square
hockeyadi23 (20:15:57)
now all we have to look for is an odd n that's 1 less than a perfect square
DPatrick (20:16:27)
Right. If n is odd, then 2^(n-1) is a perfect square, so we'd need n+1 to be a perfect square as well.
DPatrick (20:16:40)
What's the largest 3-digit odd n so that n+1 is a perfect square?
hockeyadi23 (20:15:58)
899
DavidL (20:16:25)
899 is the biggest.
alpha123 (20:16:48)
899
DPatrick (20:17:11)
Right. 900 = 30^2 but 32^2 = 1024 is too bug.
DPatrick (20:17:16)
So the largest n is 899.
DPatrick (20:17:19)
Could n be even?>
alpha123 (20:17:29)
no
pilot (20:17:30)
no
maverick417 (20:17:32)
no
DavidL (20:17:32)
certainly not
lasagnaman (20:17:40)
no, beacuse then you need a factor of 2 in n +1
DPatrick (20:18:02)
Right. We'd have an odd power of 2 times an odd number, which can't be a perfect square.
DPatrick (20:18:13)
Hence, the final answer is n = 899.
DPatrick (20:18:25)
DPatrick (20:18:37)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-94294281.gif
DPatrick (20:18:43)
Scary 3-D geometry.
hockeyadi23 (20:18:35)
Diagram!
Chocoboom (20:18:46)
draw a tetrahedron?
DPatrick (20:19:06)
Do we need to draw two different diagrams, one for before and for after?
DPatrick (20:19:38)
I'd be inclined to try to get all our information into one (large) diagram.
DPatrick (20:19:49)
Let's set A to be the top of the tripod, and let B, C, D be the feet.
DPatrick (20:20:00)
Let E be the point on AB such that AE = 4. The 'after' tripod has base ECD and top still at A.
DPatrick (20:20:14)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/3c24aa8131f86987c61168956eaabdf4.png
DPatrick (20:20:19)
In terms of this diagram, what distance is the answer to the problem?
Elemennop (20:20:48)
perpendicular from A to ECD.
Hamster1800_2 (20:20:48)
The distance from A to the plane defined by E, C, and D
DPatrick (20:21:12)
We are looking for the distance from the point A to a plane in which we have a triangle (ECD). What general geometric tool might we consider?
krustyteklown_2 (20:21:48)
or volumes
DPatrick (20:22:01)
We consider volume, since our desired height is simply the altitude from A to base ECD of tetrahedron AECD. If we can find the volume of AECD and the area of ECD, we can find the altitude.
DPatrick (20:22:28)
How does the volume of AECD relate to the volume of ABCD?
Chocoboom (20:22:54)
4/5?
DPatrick (20:23:11)
Because E is AE = (4/5)AB, the altitude from E to face ACD is 4/5 the altitude from B to ACD.
DPatrick (20:23:28)
ACD is a base of both the 'before' and 'after' tetrahedra (ABCD and AECD), so the volume of AECD is 4/5 that of ABCD.
DPatrick (20:23:33)
To see this clearly, we can draw altitudes from E and B to face ACD:
DPatrick (20:23:39)
DPatrick (20:23:44)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/5ed4dd33b59a0c8653a595598e21961c.png
DPatrick (20:23:48)
Triangle EFA and BGA are similar, so EF/BG = AE/AB = 4/5.
DPatrick (20:24:10)
So vol(AECD) = (4/5)*vol(ABCD). How do we find the volume of ABCD?
lasagnaman (20:24:26)
1/3 bh
fubu (20:24:29)
is the base of ABCD a equilateral triangle?
perfectnumber628 (20:24:30)
find the base area, and you know the height is 4
DPatrick (20:24:48)
Yes! Because AB = AC = AD and the angles between the legs are equal, BCD is equilateral. Furthermore, the height from A to BCD meets the plane of BCD at the circumcenter of BCD (let the foot of the altitude be X; triangles AXB, AXC, and AXD are congruent by HL, so XB=XC=XD).
DPatrick (20:24:57)
DPatrick (20:25:03)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/9dc2294c31b881148c8d166eaa0aee9e.png
5alive (20:24:59)
the base of abcd is equilateral triangle w/ side length 3 root 3
DPatrick (20:25:21)
Good. Since AX = 4, AB = 5, and AXB is a right triangle, XB = 3. Since X is the circumcenter of equilateral BCD, it is also the centroid and the orthocenter. Therefore, altitude BM of BCD is also a median, and it has length (3/2)(BX) = 4.5. From 30-60-90 triangle BMC, we see that each side has length (4.5)(2/sqrt(3)) = 9/sqrt(3) = 3sqrt(3).
DPatrick (20:26:10)
So what is the area of BCD?
b-flat (20:26:31)
alan (20:26:32)
DPatrick (20:26:50)
DPatrick (20:27:05)
DPatrick (20:27:27)
So, now we have the volume of AECD!
DPatrick (20:27:38)
DPatrick (20:28:01)
So if we can find the area of ECD, we'll be done.
DPatrick (20:28:10)
We have CD = 3sqrt(3), so we just need the altitude from E to CD.
DPatrick (20:28:15)
DPatrick (20:28:17)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/85493687c44585c8d8ea253c22039ead.png
DPatrick (20:28:46)
In general, when hunting for a length, one powerful tactic is to build right triangles and use the Pythagorean Theorem. Where can we build a right triangle here to find altitude EM of triangle ECD?
maverick417 (20:29:00)
triangle EMC
Chocoboom (20:29:03)
EMC?
DPatrick (20:29:34)
Yeah, we know that EMC is right, but where else can we find one?
lasagnaman (20:29:15)
altitude from E in EBC
Chocoboom (20:29:41)
CEB
DPatrick (20:30:08)
Not a bad choice.
kostya (20:29:57)
EBM?
platinumnerd (20:29:58)
EMB
DPatrick (20:30:13)
Even better.
DPatrick (20:30:27)
We can draw an altitude from E to triangle BCD.
DPatrick (20:30:32)
DPatrick (20:30:36)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/fa08f0a266e780fe92a27a27eb893219.png
DPatrick (20:30:54)
How can we find EY and YM easily?
Chocoboom (20:30:13)
3-4-5 triangle!
b-flat (20:31:04)
you have similar right triangles
alan (20:31:05)
DPatrick (20:31:37)
Yes, EBY is a 3-4-5 right triangle! So EY = 4/5 (since EB = 1).
DPatrick (20:31:50)
And BY = 3/5, so what's YM?
DPatrick (20:32:31)
(this is where the calculations start to get a little ugly!)
maverick417 (20:32:25)
4.5-3/5
DPatrick (20:33:03)
Right...we know that BM = 9/2 (since BM is a median of the equilateral triangle BCD that we found before), so YM = BM - BY = 9/2 - 3/5 = 39/10.
lasagnaman (20:31:51)
how do we nkow it is 3-4-5?
Sunday Math-er_2 (20:32:59)
Why does BEY must be a 3-4-5 triangle?
DPatrick (20:33:19)
A couple people have asked this -- perhaps this picture makes it clearer:
DPatrick (20:33:27)
DPatrick (20:33:29)
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/b17fa1b1ce5b3eba8d3d243282ffe881.png
DPatrick (20:33:46)
BEY and BAX are similar, and we knew that BAX is 3-4-5, so BEY is too.
DPatrick (20:34:12)
So we can use the Pythagorean Theorem on EYM to find EM.
DPatrick (20:34:47)
It works out to EM = sqrt((4/5)^2 + (39/10)^2) = sqrt(1585)/10.
DPatrick (20:35:07)
Now it's just crunching all the numbers.
DPatrick (20:35:11)
CD = 3*sqrt(3) from before.
DPatrick (20:35:36)
We also know that the volume of AECD is 36*sqrt(3) / 5.
DPatrick (20:35:44)
So we're ready to solve for the height from A to ECD.
DPatrick (20:35:56)
Since volume of AECD = (h)([ECD])/3 = (h)(EM)(CD)/6, we have h = 6(Volume)/(EM)(CD). Solving, we find h = 144/sqrt(1585). Since 39 < sqrt(1585) < 40, our desired answer is 144+39 = 183.
DPatrick (20:36:05)
(You can check the details on your own later if you like!)
DPatrick (20:36:26)
DPatrick (20:36:31)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-231416222.gif
Chocoboom (20:36:46)
Absolute values scare me.
DPatrick (20:37:09)
Indeed, they make things really tough to work with -- what can we do to make them go away?
DPatrick (20:37:24)
(This is a common technique -- if you see something ugly, try to make it go away)
perfectnumber628 (20:37:26)
square?
DPatrick (20:37:37)
Aha!
DPatrick (20:37:58)
But before getting too far, it's not hard to see that every x_k must be a multiple of 3. What can we do to simplify things?
DavidL (20:38:09)
Factor out a 3?
Hamster1800_2 (20:38:09)
divide them all by 3?
DPatrick (20:38:17)
Sure, we can make the substitution y_k = x_k/3.
DPatrick (20:38:23)
DPatrick (20:38:43)
DPatrick (20:39:10)
DPatrick (20:39:25)
We can get rid of the ugly absolute values by squaring everything!
DPatrick (20:39:32)
DPatrick (20:39:37)
Now what?
DavidL (20:39:46)
Sum thme
mna851 (20:39:47)
add?
DPatrick (20:40:08)
Sure! Since we want y_1+y_2+... anyway, let's add them!
DPatrick (20:40:25)
Something very, very cool happens!
DPatrick (20:40:31)
DPatrick (20:40:49)
(We have also used the fact that y_0 = 0.)
DPatrick (20:41:00)
So to minimize the sum y_1 + y_2 + ... + y_{2006}, what should we do?
DPatrick (20:41:52)
We should make y_{2007} as close to the square root of 2007 as possible.
DPatrick (20:42:29)
What value of y_2007 gives (y_2007)^2 as close to 2007 as possible?
E^(pi*i)=-1 (20:42:19)
45?
kostya (20:42:30)
45
perfectnumber628 (20:42:32)
45^2=2025
DPatrick (20:42:57)
44^2 = 1936 and 45^2 = 2025.
maverick417 (20:43:45)
so make y_{2007}=45
Hamster1800_2 (20:44:06)
so plug in 45 for y_2007
DPatrick (20:44:19)
How do we know that we can do this?
kostya (20:44:39)
parity
DPatrick (20:45:26)
Yeah, we know that all of the y_k's where k is odd are odd, and all of the y_k's where k is even are even, by looking at the parity (the parity alternates).
DPatrick (20:45:50)
DPatrick (20:46:09)
DPatrick (20:46:40)
That's it! Thanks for coming!
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