AMC 10/12 Contest A Math Jam
Go back to the Math Jam ArchiveWe will review the problems from the 2005 AMC-10 A and AMC-12 A.
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Facilitator: Mathew Crawford
AMC 2005 A MATH JAM
MCrawford (19:28:20)
Before we begin, we have a seminar class at the end of this month that some of you might be interested in taking. The special AIME seminar involves two three hours classes, including a half hour break each day, plus a mock AIME designed to help students prepare for the AIME. Time will be spent on AIME problems covering algebra, geometry, number theory, counting, and probability. The cost of this seminar is $40 and you can find details here:
MCrawford (19:28:26)
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MCrawford (19:28:33)
At this time Dave Patrick will walk through the last 5 problems from the AMC 10 A.
DPatrick (19:28:50)
21.
DPatrick (19:28:52)
DPatrick (19:29:00)
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DPatrick (19:29:15)
(If you click on the link, you'll get a copy of the problem in a separate window, which is easier to follow along with)
DPatrick (19:29:25)
What seems like a reasonable first step with this problem?
NightFlarer (19:29:13)
write it as (n(n+1))/2
pika (19:29:18)
n(n+1)/2
everyday847 (19:29:19)
Well, 1+2+...+n = n(n+1)/2
mna851 (19:29:29)
n(n+1)/2
DPatrick (19:29:49)
DPatrick (19:30:02)
If this is not a sum that you are familiar with, it would be a good idea to discuss this in
DPatrick (19:30:10)
the Basics or Getting Started forums.
DPatrick (19:30:21)
Now, we have rewritten the given problem in a simpler form -- this is already a great leap in the challenge of problem solving. We like easier problems. We are now asking for how many positive integers n(n+1)/2 evenly divides 6n. Where can we go from here?
pika (19:29:51)
When is 6n/(n(n+1)/2) an integer?
RC-7th_? (19:30:00)
Then 12/(n+1) must be an integer
NightFlarer (19:30:11)
so (n(n+1)/2 has to divide 6n, cancel out the n and multiply to get 12/(n+1) = some integer k
mna851 (19:30:36)
n+1 must divide 12
everyday847 (19:30:47)
Thus 6n / (n(n+1)/2) is an integer.
DPatrick (19:30:58)
DPatrick (19:31:10)
How can we count the number of positive integers n for which this is true?
mathenthusiast8 (19:31:21)
factors of 12
zabelman (19:31:22)
look at divisors of 12
everyday847 (19:31:23)
Factors of 12 are 1,2,3,4,6,12
aznarchmage (19:31:33)
find the factors of 12
DPatrick (19:31:48)
everyday847 (19:31:47)
And n-values can be 0 - not valid, because it's not positive, 1,2,3,5,11
isaacchao (19:31:49)
omit 1, 0 is not possitive integer
Scrambled (19:32:00)
0 cannot be n
sam3.14 (19:32:03)
n+1 cannot equal 1
DPatrick (19:32:17)
We know that n + 1 is a positive divisor of 12, but n must be a positive integer. This means that we count all of the positive divisors of 12 except 1, so our answer is 5. (B).
Scrambled (19:32:08)
wait a minute; does 1+2+...+n mean that n cannot be 1 or 2? (because there was an answer choice for people who assumed that n cant be 1 or 2and for those who assumed it could be)
DPatrick (19:33:18)
No: 1 + 2 + ... + n is a notational convenience for the sum of the positive integers from 1 through n.
DPatrick (19:33:41)
So even though it may look a little strange, n=1 or n=2 is valid.
DPatrick (19:34:00)
22.
DPatrick (19:34:13)
DPatrick (19:34:15)
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DPatrick (19:34:25)
What can we first do with this problem?
everyday847 (19:34:44)
To be common, they'd have to be divisible by 12, the LCM.
RC-7th_? (19:34:48)
Look at the LCM of 4 and 6
RC-7th_? (19:35:40)
The common elements are divisible by 4 and 6, so 12
mathenthusiast8 (19:35:41)
so let me rephrase: find the LCM of 4 and 6
kostya (19:35:45)
#'s that divide by 12 are in both sets
everyday847 (19:35:31)
I'd say that thus, every other member of T would be in S, and every third member of S would be in T.
DPatrick (19:36:45)
In light of this last comment -- the first thing that we can do is rephrase the problem such that we are asking a more simple question. Of the two sets S and T, S has a smaller range. This means we could just search for the number of multiples of 6 in set S.
DPatrick (19:37:04)
How can we easily count the number of multiples of 6 in set S?
gleb (19:36:43)
Rewrite the sets as {2^2, 2^2n,...2^2n^2004} and {2*3, 2*3n,...2*3n^2004}
pika (19:36:56)
int(2005/3)=668
mna851 (19:37:17)
They follow a pattern of every third number.
Ars (19:37:25)
divide by 3
mo (19:37:36)
They occur for once for every 3 multiples of 4.
sam3.14 (19:37:37)
it's every three
everyday847 (19:37:39)
Every third multiple of 4 is a multiple of 6.
zabelman (19:37:46)
every third element is a multiple of 6
DPatrick (19:37:53)
Now, in order for an integer to be a multiple of 6, it must be both a multiple of 2 and 3. Clearly ever member of S is a multiple of 2, so we are left counting the members of S that are multiples of 3.
DPatrick (19:38:02)
pokey (19:38:18)
Why is there no E choice?
DPatrick (19:38:45)
Sorry, it got chopped off. (E) was 1001.
DPatrick (19:39:06)
Let's move on to #23
DPatrick (19:39:11)
DPatrick (19:39:14)
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DPatrick (19:39:19)
DPatrick (19:39:22)
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DPatrick (19:39:42)
Let's start by labeling the points that are left unlabeled. Are there any points that we might give a particular label (that might be helpful)?
zabelman (19:39:50)
O
Scrambled (19:39:56)
the point opposite to d
mathenthusiast8 (19:39:58)
F, as in chord DF
joml88 (19:39:59)
center...O
pika (19:39:59)
Intersection of DC and the circdle
DPatrick (19:40:34)
DPatrick (19:40:39)
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pilot (19:40:02)
I would set r=3 AC=2 and BC=4 since we're dealing with a ratio
kostya (19:40:17)
AC=2
DPatrick (19:40:48)
I agree. We can go about hunting down relative lengths, though I would personally start by assigning a length to some segment and working from there. I personally find it makes my work neater whether or not it actually comes in handy in the problem. In this case I'll let AC = 2. This means BC = 4.
DPatrick (19:41:00)
Now, keeping in mind that trig isn't required for the AMC10, what can we do to compare the areas of DCE and ABD?
MGriffin487 (19:41:30)
compare the parts that you multiply to find the area
MGriffin487 (19:41:34)
aka base and height.
fedwinri (19:41:36)
Compare the lengths of the bases (since the hights are the same).
DPatrick (19:42:02)
We see several triangles that share bases, parts of bases, or heights.
pika (19:41:14)
Altitudes and bases
zabelman (19:41:15)
calculate them :)
gleb (19:41:57)
find the ratio of their altitudes
DPatrick (19:42:43)
Let's compare DCO and ACB. They have the same height. What is the ratio of their bases?
DPatrick (19:42:54)
Sorry...I meant DCO and ADB.
pilot (19:43:05)
1:6
RC-7th_? (19:43:07)
1:6
SuperNova (19:43:09)
1:6
MaThWhIz2004 (19:43:15)
1/6
zabelman (19:43:16)
1/6
DPatrick (19:43:37)
Right: CO = 1 and AB = 6, so the bases have ratio 1:6. This means the area of DCO is 1/6 of the area of ABD.
DPatrick (19:43:49)
What other triangles can we compare?
DPatrick (19:44:04)
Remember that we want DCE in the end...
zabelman (19:44:01)
DCO and COE
Bictor717 (19:44:05)
DCO and OCE
fedwinri (19:44:11)
And ECO = DCO
RC-7th_? (19:44:12)
[DOE]=[DCO]
riposte (19:44:13)
DCO and COE
pilot (19:44:15)
DCO=COE
DPatrick (19:44:26)
By symmetry, if we drop a perpendicular from E to AB, it would be the same length as DC. This means triangles DCO and COE share a base and have equal heights. The area of DCE is twice the area of either of these triangles.
DPatrick (19:44:36)
What is our final answer?
RC-7th_? (19:44:23)
So the answer is 1/3
zabelman (19:44:43)
1/3
d343seven (19:44:44)
1/3
fedwinri (19:44:46)
1/3
Monkey (19:44:47)
1/3
pokey (19:44:48)
1/3
pilot (19:44:49)
1/3
joyofpi (19:44:49)
1/3
riposte (19:44:49)
1/3
DPatrick (19:45:11)
Since DCE has twice the area of DCO, the ratio of the area of DCE to the area of ABD is 1/3, (C).
DPatrick (19:45:14)
Here we used our understanding of symmetry and the formula bh/2 for calculating the area of a triangle. Remember that this formula allows us to compare areas by comparing lengths of sides of triangle. That often comes in handy.
DPatrick (19:45:22)
Now for #24:
DPatrick (19:45:30)
DPatrick (19:45:33)
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DPatrick (19:46:01)
This problem is going to require a bit of exploration. It's important to begin mathematical exploration on solid footing, so let's collect some facts. What facts can we derive from the problem statement?
NightFlarer (19:45:52)
n must be the square of a prime number
RC-7th_? (19:46:13)
n and n+48 are both squares
solafidefarms_2 (19:46:15)
n is a square
pika (19:46:16)
and n+48 is the square of a prime
DPatrick (19:46:41)
We know that n must be a perfect square in order for the square root of n to be a prime (integer).
RC-7th_? (19:46:32)
n+48 is also the square of a prime number
aznarchmage (19:46:33)
n+48 must also be the square of a prime number
DPatrick (19:46:54)
The functional equations we are given tell us that n and n + 48 are each square of not just integers, but of prime integers. How can we use this fact?
isaacchao (19:47:20)
p^2 + 48 = q^2
RC-7th_? (19:47:22)
(x+y)(x-y)=48, where x and y are primes?
NightFlarer (19:47:30)
call n p^2 and let n+ 48 = q^2
DPatrick (19:47:37)
Many number theory problems require us to step into the realm of algebra. This is one of them. We can help ourselves by labeling the primes that we intend to discuss.
DPatrick (19:47:44)
DPatrick (19:47:55)
mna851 (19:48:05)
Difference of squares
everyday847 (19:48:05)
Well, q^2 - p^2 = 48
zabelman (19:48:14)
(q-p)(q+p)=48, and look at divisors
DPatrick (19:48:23)
DPatrick (19:48:40)
DPatrick (19:48:56)
What does this equation tell us?
mathenthusiast8 (19:48:15)
their sum and difference are factors of 48
mandy_pal (19:49:03)
find factors of 48
Scrambled (19:49:07)
we can find factors of 48
pika (19:49:10)
factors of 48
DPatrick (19:49:21)
The equation now tells us that both (q + p) and (q - p) are divisors of 48 that pair up so that their product is 48. How can we work with that information?
Jimmy (19:49:36)
decompose 48
zabelman (19:49:38)
list factor pairs of 48 and solve for p and q in each case
ekrem (19:49:50)
try the factors
DPatrick (19:50:09)
DPatrick (19:50:12)
For which of these pairs of divisors are there prime numbers p and q such that the sum and difference of the primes is equal to each divisor in the pair?
mathenthusiast8 (19:49:34)
the factors must be of the same parity
mna851 (19:49:50)
q and p must also have the same parity
DPatrick (19:50:52)
Experienced students of number theory might quicken their search knowing that q + p and q - p must both be even or both be odd. This is because their difference is 2p -- an even number. Since the divisors can't both be odd, we are left with only three pairs to examine.
DPatrick (19:51:11)
DPatrick (19:51:18)
What is our final answer?
zabelman (19:51:02)
2*24 gives 13 and 11, and thats the only one
isaacchao (19:51:22)
11,13
mathenthusiast8 (19:51:25)
(b), 1
RC-7th_? (19:51:30)
1
mathenthusiast8 (19:51:30)
only (11, 13) works
RC-7th_? (19:51:33)
1 (B)
Ars (19:51:47)
1
DPatrick (19:52:19)
DPatrick (19:52:59)
A lot of you jumped to the conclusion that since we have 3 solutions for p and q, that the answer was 3. Be careful!
mna851 (19:52:33)
That was a nice problem
DPatrick (19:53:36)
Finally, #25:
DPatrick (19:53:52)
DPatrick (19:54:12)
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DPatrick (19:54:20)
How can we begin here?
RC-7th_? (19:54:14)
Draw a picture
mna851 (19:54:17)
We need a picture
DPatrick (19:54:29)
It makes sense to draw a diagram and get a good idea of what is going on visually. Is there any particular way we might think to draw the diagram?
Elemennop (19:54:48)
An acute triangle, since 25^2 + 39^2 > 42^2
probability1.01 (19:55:09)
let us have the base as AB
DPatrick (19:55:22)
I like the idea of making either AC or AB parallel to the "x-axis" of my drawing. This is because those are the sides from which the smaller triangle is cut from the larger. As we have seen, being able to compare bases sometimes helps us with area comparison problems.
Monkey (19:55:19)
Think of height and base ratios
DPatrick (19:55:38)
Yes. As it happens, I drew is with AC as the base.
DPatrick (19:55:41)
DPatrick (19:55:43)
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DPatrick (19:55:49)
Now, how can we compare the area of triangle ADE to quadrilateral BCED?
sam3.14 (19:56:19)
[DBCE]=[ABC]-[ADE]
pika (19:56:29)
compare ADE to ABC?
probability1.01 (19:56:30)
sort of--but you must compare it indirectly via comparison to the area of triangle ABC
mathenthusiast8 (19:56:31)
small triangle/(large triangle area - small triangle)
DPatrick (19:56:49)
Since the two parts are comparing together make up the area of triangle ABC, it makes sense to try and compare one of the smaller figures to triangle ABC. Triangle ADE seems like the most reasonable candidate as they two triangles have two sides each on the same lines and a common angle.
DPatrick (19:56:53)
How can we compare the area of ADE to ABC?
mathenthusiast8 (19:57:33)
heights seem helpful, since we know the bases, but they seem to be difficult to attain
Scrambled (19:57:42)
ratio of heights
pika (19:58:02)
Height and base?
DPatrick (19:58:19)
Since the area of a triangle can be found by the formula bh/2, we can compare the areas using the ratios of the bases and the heights of the triangles.
DPatrick (19:58:28)
What is the ratio of the bases of ADE and ABC?
Scrambled (19:58:36)
1:3
RC-7th_? (19:58:36)
1:3
gleb (19:58:37)
1/3
Monkey (19:58:39)
The ratio of bases is 1/3
Elemennop (19:58:41)
1/3
DPatrick (19:58:53)
We have the ratio between the bases of ADE and ABC. AE is AC/3.
DPatrick (19:58:58)
How can we find the ratio of the heights?
pika (19:58:26)
drop a perpendiclar for the height of 1 and then the other
probability1.01 (19:59:12)
The heights can be obtained by drawing perpendiculars to AC from B and D, then comparing similar triangles
Scrambled (19:59:12)
similar triangles
vfiroiu_2 (19:59:22)
drop perpendiculars from D and B to AC
DPatrick (19:59:33)
DPatrick (19:59:38)
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RC-7th_? (19:59:26)
19/25
Hexahedron (19:59:31)
it is just 19:25
d343seven (19:59:37)
19/25?
mandy_pal (19:59:38)
19/25
DPatrick (19:59:53)
All we need is the ratio of the heights that we drop down to the bases of the triangles. We find this ratio by noting that the perpendiculars give us two right triangles that must be similar because they share an acute angle. We can now say that the ratio of the heights is the same as the ratio of the hypotenuses which is 19/25.
DPatrick (20:00:03)
What is the ratio of the area of the triangles?
RC-7th_? (19:59:59)
The ratio of [ADE] to [ABC] is 19/75
mandy_pal (20:00:11)
19/75
fedwinri (20:00:16)
19/75
Monkey (20:00:19)
19/75
DPatrick (20:00:29)
DPatrick (20:00:40)
How does the ratio of the triangles give us the ratio of ADE to BCED?
RC-7th_? (20:00:32)
The answer is 19/56 (D)
gleb (20:00:59)
19/(75-19)
isaacchao (20:01:00)
19/56
Wumbate (20:01:04)
It is simply 19/(75-19)=19/56
mathenthusiast8 (20:01:06)
so 19/56, or (d) because 19/(75-19)
mna851 (20:01:13)
Area of BCED is equal to ABC is ADE
pilot (20:01:15)
19/(75-19)
DPatrick (20:01:32)
Since the quadrilateral is all of the region in the big triangle but outside the smaller triangle, it's area is 1 - 19/75 = 56/75 as large (in area) as the big triangle.
DPatrick (20:01:44)
DPatrick (20:01:58)
In order to solve this problem we once again used ratios of bases and heights to compare areas. We also used convenient similar triangles which we can often construct by dropping altitudes within a diagram.
DPatrick (20:02:41)
As a closing note: those of you that know some trigonometry -- in particular, the Law of Sines -- can solve this problem using that method. But since the AMC 10 doesn't require trig, I won't go into that solution here.
DPatrick (20:02:59)
Now I'll turn things back over to Mathew, for the AMC 12.
riposte (20:03:39)
How much time do we have in AMC 10 and 12 for 5 problems?
MCrawford (20:04:13)
We'll take as much time as we need.
MCrawford (20:04:19)
MCrawford (20:04:22)
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MCrawford (20:04:31)
This is probelm 21 from the AMC 12 A
MCrawford (20:04:37)
How might we begin with this problem?
countersixte_2 (20:04:52)
Express in exponential form?
mo (20:04:56)
write a^c^2005=b
nonquotidian (20:04:56)
exponent out the log: a^c^2005=b
eastsoldier (20:04:58)
set it in exponetial form
pilot (20:04:52)
try some numbers
dts (20:04:53)
c>=2 is ridiculuous; b would have to be enormous.
Elemennop (20:04:58)
c must be either 1 or 0, otherwise b would be too great to fit a + b + c = 2005
MCrawford (20:05:21)
We have two equations to work with, but the second looks more like an equation meant to give boundaries on the variables, so we'll work with the first.
MCrawford (20:05:24)
ChrisChang (20:05:41)
so we look at values of a and b that give c = 0 or 1
Hexahedron (20:05:11)
just looking at this, we can tell that c<2 because if c is greater then or equal to 2 then the other numbers would have to be huge
towersfreak2006 (20:05:54)
since a is an integer, and b<2006, c<2
MCrawford (20:06:05)
The second equation helps us most when we use the fact that a is at least 2. The exponent in our exponential equation can be at most 10 in order for b not to exceed 2005, the sum of a, b, and c.
MCrawford (20:06:11)
MCrawford (20:06:19)
The only possible values for c are 0 and 1. What do we do now?
blahblahblah (20:06:26)
cases
towersfreak2006 (20:06:27)
plug in the values and solve!
eastsoldier (20:06:30)
plug it in
ChrisChang (20:06:14)
when c = 0, b = 1 and a = 2004
ChrisChang (20:06:44)
when c = 1, a = b so a = 1002, b = 1002, c = 1
blahblahblah (20:06:51)
mo (20:06:51)
well we know that when c=1, then a=b. So a+b+c=2005, a=1002, b=1002.
Hexahedron (20:06:54)
if C=1 then the pair (1002,1002,1) works and if C=0 then we get the pair (2004,1,0)
MCrawford (20:07:13)
MCrawford (20:07:28)
From case (1) we see that b = 1 and c = 0. This means a = 2004.
From case (2) we see that a = b and since c = 1, a = b = 1002.
RC-7th_? (20:07:14)
So C
MCrawford (20:07:38)
There are 2 solutions and the answer is (C).
This problem teaches several important lessons. First, it makes sense not to focus too much on a "bounding equation" until it's useful. Next, and possibly most important, it's often easier to examine the exponential form of logarithmic equations. Finally, restricting solutions to integers makes it easier to hone in on our final answer.
MCrawford (20:07:49)
Now, let's take a look at problem 22.
MCrawford (20:07:57)
MCrawford (20:08:02)
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MCrawford (20:08:05)
What's the first thing we should do with a problem like this?
eastsoldier (20:08:15)
set up equations
towersfreak2006 (20:08:20)
gather our information, and convert them to algebraic expressions
probability1.01 (20:08:21)
Assign dimensions to the box: x, y, z
dts (20:08:21)
Let x, y, z be the side lengths.
ChrisChang (20:08:10)
draw a picture
boblard (20:08:13)
draw a figuires
blahblahblah (20:08:18)
draw a diagram
MCrawford (20:08:55)
Okay, so we have the first two things...
MCrawford (20:09:01)
Here we have a word problem, so the first thing we must do is translate the words into mathematics. Making the correct analogies between the information given and the mathematical statements we are going to work with may be the most important challenge in this problem.
Elemennop (20:08:43)
edges a, b, c, then 2ab + 2bc + 2ac = 384 and 4a + 4b + 4c = 112 or a + b + c = 28
pika (20:08:53)
and x+y+z=28
pika (20:08:25)
192=xy+yz+xz
MCrawford (20:09:43)
RC-7th_? (20:09:38)
we want sqrt(x^2+y^2+z^2), so we square 28 and subtract 384
RC-7th_? (20:08:27)
2r is the length of the interior diagonal
MCrawford (20:10:23)
Before we begin diving into the equations, it makes sense to analyze what exactly we are looking for. It's important to keep an eye on the ball. This allows us to work more efficiently and with fewer errors.
MCrawford (20:10:28)
Our goal is to find the value of r. Really this means finding the value of half of d, the diameter of the sphere which is also the spatial diagonal of the box:
MCrawford (20:10:35)
vfiroiu_2 (20:09:29)
blahblahblah (20:10:04)
Elemennop (20:10:21)
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 784, Then subtract the first equation from this new one and you get a^2 + b^2 + c^2 = 400
JHu (20:10:25)
RC-7th_? (20:10:35)
Then we take the square root of 784-384=400 to get 20/2=10 as the answer
Bictor717 (20:10:47)
ChrisChang (20:11:00)
sqrt(x^2 + y^2 + z^2) = diameter
pika (20:10:44)
10 is the answer?
gleb (20:11:08)
In order to find the radius, first find the diameter in terms of our variables
MCrawford (20:11:33)
MCrawford (20:11:42)
Elemennop (20:11:25)
sqrt(a^2+b^2+c^2)/2 = r = 10
MCrawford (20:12:01)
We are looking for r which is d/2, so our answer is 10. (B).
MCrawford (20:12:18)
There were four tasks in this problem. First, we have to convert the words into math. We did this by assigning variables to the side lengths -- quantities that we did not know, but that we knew how work with. Next we recognized that the diameter of the sphere was the spatial diagonal of the box. This is apparent due to the symmetry of the box and sphere. Third, we used two applications of the Pythagorean theorem to find the length of the diagonal in terms of the lengths of the sides of the box. Finally, we used the system of equations we setup at the beginning to find the value of the sum of the squares of the sides. None of these tasks was particularly difficult on its own, but problem solving often means putting simpler pieces together one by one.
zabelman (20:08:56)
a very good idea (not my own): Assume the box is flat, i.e. one of the dimensions is 0
MCrawford (20:12:47)
It is often helpful to make special cases -- even extreme ones.
MCrawford (20:13:00)
Let's take a look at problem 23...
MCrawford (20:13:08)
MCrawford (20:13:14)
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MCrawford (20:13:22)
How can we begin with a problem such as this one?
eastsoldier (20:13:36)
find the all possible number of set
MCrawford (20:14:05)
First, we should recognize this as a counting probability problem. Our task is to first count the total number of ways in which we can select a and b. This will be our denominator. Then we want to count the number of ways in which the logarithm given will be an integer. The total number of these "successful outcomes" will comprise our numerator.
MCrawford (20:14:09)
Now, in how many ways can we select a and b?
mathenthusiast8 (20:14:02)
which is 25*24, since they must be distinct = 600
eastsoldier (20:14:08)
then the number of set that satisfies the thing
MCrawford (20:14:26)
MCrawford (20:14:41)
We have the denominator in our probability. We must now count the number of successful outcomes to determine the numerator.
blahblahblah (20:13:46)
SpaceTerr808 (20:13:41)
simply the expression log basea of b
pika (20:13:44)
loga(b)=log2(b)/log2(a)?
Bictor717 (20:14:59)
MCrawford (20:15:32)
mathenthusiast8 (20:15:12)
or the power on b should be divisible by a. the other statement was too broad
probability1.01 (20:15:14)
then, if we choose 2^n and 2^k, for a and b respectively, we want it so that n | k
dts (20:13:44)
Consider the easier, equivalent problem of the set (1, 2, 3, ..., 25), where the second element chosen must be divisible by the first.
riposte (20:14:18)
a = 2^i and b = 2^j. we need j to be a multple of i
gleb (20:14:39)
we know that b must be a multiple of a
mathenthusiast8 (20:14:54)
to have the log be an integer, b should be divisible by a since they share bases
Jimmy (20:14:55)
a<b
pika (20:15:26)
just switch it into 1, 2, 3..., 25
gleb (20:16:20)
so find number of multiples for each 1-25
MCrawford (20:16:45)
MCrawford (20:16:52)
We can simplify the problem at hand by noting that our job is to count the number of ways in which we can select m and n from {1, 2, 3,..., 25} such that m/n is an integer.
mathenthusiast8 (20:16:49)
in this problem, there are 24+11+7+5+4+3+2+2+1+1+1+1=62 good pairs.
probability1.01 (20:16:59)
I just systematically counted at this point, but I think there might be a better way....
vfiroiu_2 (20:16:17)
count on fingers
MCrawford (20:17:30)
Unless you're more dextrous with your toes.
RC-7th_? (20:17:15)
There are 62 ways, so B is the answer
MCrawford (20:17:41)
The easiest way to count seems to be to start at 1 and count the number of other integers in the set that are multiples of 1.
MCrawford (20:17:44)
There are 24 multiples of 1 in the set besides 1.
There are 11 multiples of 2 in the set besides 2.
There are 7 multiples of 3 in the set besides 3.
There are 5 multiples of 4 in the set besides 4.
There are 4 multiples of 5 in the set besides 5.
There are 3 multiples of 6 in the set besides 6.
There are 2 multiples of 7 in the set besides 7.
There are 2 multiples of 8 in the set besides 8.
There is 1 multiple of 9 in the set besides 9.
There is 1 multiple of 10 in the set besides 10.
There is 1 multiple of 11 in the set besides 11.
There is 1 multiple of 12 in the set besides 12.
mathenthusiast8 (20:17:32)
62/600=31/300, or (B)
blahblahblah (20:17:39)
62/600=31/300
MCrawford (20:17:54)
MCrawford (20:18:44)
Now let's conquer 24 -- one of the two problems that gave top students problems.
MCrawford (20:18:54)
MCrawford (20:18:57)
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MCrawford (20:19:04)
How can we begin with this problem?
blahblahblah (20:19:10)
I really liked 24). The first thing that you absolutely need to notice is that deg(Q(x))=2
MCrawford (20:19:43)
How do we know the degree of Q?
probability1.01 (20:19:42)
(Q(x) - 1)(Q(x) - 2)(Q(x) - 3) = (x - 1)(x - 2)(x - 3)R(x)
Elemennop (20:19:42)
P((Q(X)) must have degree 6, so Q(X) has degree 2
MCrawford (20:20:08)
Demon (20:19:58)
∀x∈{1,2,3} Q(x)∫{1,2,3}
MCrawford (20:20:30)
Since R(x) is a third degree polynomial, each side is a sixth degree polynomial. Furthermore, the sixth degree polynomial has factors of (x - 1), (x - 2), and (x - 3). That gives 1, 2, and 3 as roots of the sixth degree polynomial.
MCrawford (20:20:45)
We now know that one of Q(x) - 1, Q(x) - 2, and Q(x) - 3 for each x = 1, 2, and 3.
Now, how can we count the total number of possible Q(x)?
MCrawford (20:21:27)
We now know that one of Q(x) - 1, Q(x) - 2, and Q(x) - 3 is 0 for each x = 1, 2, and 3.
Now, how can we count the total number of possible Q(x)?
Scrambled (20:20:56)
both polynomials must have roots of 1,2,and 3
kostya (20:21:15)
blahblahblah (20:21:21)
this gives 27 - but there's a catch
dts (20:21:18)
3*3*3=27
MCrawford (20:21:53)
Monkey (20:21:44)
Let Q(x) equal 1,2, or 3
Bictor717 (20:21:54)
1, 2, 3 are roots, Q(1), Q(2), Q(3) must all be 1, 2, or 3
Scrambled (20:21:19)
combinations of ways this can be done possibly?
dts (20:22:45)
Some possibilities can't have degree 2.
MCrawford (20:22:57)
Q(x) must be a quadratic polynomial. Otherwise the two halves of our equation could not be sixth degree polynomials. This means Q(x) cannot have three roots. This eliminates 3 possible ways to arrange 1, 2, and 3 as roots of Q(x) - 1, Q(x) - 2, and Q(x) - 3.
MCrawford (20:23:06)
How can we check the rest of the possibilities?
blahblahblah (20:23:00)
The problem is that if Q(x)=x for 1,2,3
blahblahblah (20:23:11)
then Q(x) is linear
dts (20:23:29)
It also can't be linear.
Bictor717 (20:23:28)
graph x vs Q(x)
mo (20:23:36)
when (1,1,1), (2,2,2,), (3,3,3,), (1,2,3), and (3,2,1) must be eliminated b/c when chosen, their degree is incorrect.
blahblahblah (20:19:38)
and that three points determine a quadratic
blahblahblah (20:23:43)
Q(x)=x, Q(x)=-x, Q(x)=1,Q(x)=2,Q(x)=3 --> subtract 5 possibilities
MCrawford (20:24:31)
MCrawford (20:24:40)
Of the 6 remaining arrangements, we need only be sure that none of them are lines.
MCrawford (20:24:44)
Two of the 6 remaining arrangements form lines. They are
Q(1) = 1, Q(2) = 2, Q(3) = 3, and
Q(1) = 3, Q(2) = 2, Q(3) = 1.
gleb (20:24:23)
so 27-5=22 (B)
MCrawford (20:25:05)
We can now see that there are 18 + 4 = 22 polynomials Q(x) for which the proposed functional equation holds as defined. (B).
MCrawford (20:25:29)
This problem shows us how important it is to recognize whether or not steps that we are taking are rigorous within the definitions of the problem.
MCrawford (20:25:43)
If we pay attention to detail, this problem is not so rough.
MCrawford (20:26:20)
I thougt number 25 was the most challenging problem on the A exams...
MCrawford (20:26:25)
MCrawford (20:26:36)
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MCrawford (20:26:39)
Any ideas that could give us a start?
Bictor717 (20:26:45)
Consider a unit cube first
eastsoldier (20:26:48)
My friend set up 8 cubes
Mr.Ocax (20:26:57)
find all the points first?
mathenthusiast8 (20:27:00)
recognizing what shape the points form
MCrawford (20:27:24)
I personally prefer to avoid these starting points...
MCrawford (20:27:44)
I would like to explore this problem a bit before jumping straight to the count.
MCrawford (20:27:58)
It's possible I'll discover things that will help me count more clearly -- without error.
soulzmischief (20:28:02)
different possible side lengths of the triangles?
MCrawford (20:28:25)
rcv (20:28:29)
Color the vertices black or white like a checkerboard. Parity eliminates a lot of possibilities.
MCrawford (20:28:42)
What might help us in regards to finding situations in which these distances will all be the same?
probability1.01 (20:29:12)
the squares of the differences in coordinates must be 0, 1, or 4
MCrawford (20:30:03)
MCrawford (20:30:12)
Is there some way we can eliminate some of these from contention?
soulzmischief (20:29:13)
the distances must have the same parity
MCrawford (20:30:22)
What do you mean?
soulzmischief (20:31:02)
the different side lengths of the equaliteral triangles must at least ahve the same parity to be equal, so that would eliminate some possibilities
MCrawford (20:31:26)
Often times a little investigation can help us get a better grip on a problem whether or not we think we need it. Is there some trivial example of an equilateral triangle we could take a loot at?
probability1.01 (20:31:52)
(1,0,0)(0,1,0)(0,0,1)
MCrawford (20:32:08)
Suppose we start at the point (1, 0, 0), then move to (0, 1, 0), then move to (0, 0, 1) and then back to (1, 0, 0). In this way we trace out an equilateral triangle by adding 1 to one of the coordinates and subtracting 1 from another. What have we noticed from this most trivial of examples?
eastsoldier (20:32:35)
it is done in a unit cube
MCrawford (20:32:51)
I'm looking for something more general.
MCrawford (20:33:01)
That will apply to all the possible triangles.
probability1.01 (20:33:21)
We can shift stuff by adding the same amount to a certain coordinate of each point
RC-7th_? (20:33:26)
We use the same transformation to move from one vertex to another
MCrawford (20:34:11)
What do happens after three shifts?
pika (20:34:25)
we go back to start
origamimasterjared (20:34:27)
It's back where we started
MCrawford (20:35:00)
What does that tell us about the nature of the shifts -- 3 shifts leave everything cancelled?
MCrawford (20:36:13)
We showed a shift in a (+1, -1, 0) permuted pattern...
MCrawford (20:36:27)
Could we have a 2/1/0 pattern (with various signs at the right times?
MCrawford (20:37:18)
Could three such shifts bring us back to the same point?
Bictor717 (20:37:21)
No, they can't add to 0
MCrawford (20:37:34)
Why not?
origamimasterjared (20:38:20)
2?1?0
MCrawford (20:38:44)
That's for one shift -- we get three.
Hexahedron (20:39:16)
2+-1+-1+-1 still can't equal 0
MCrawford (20:39:25)
In order to move three times and get back to the starting point, the sum of the directional distances at each move must be even. If it were odd, three movements would involve an odd number of unit shifts, so the result of all the shifts could not cancel out to 0.
MCrawford (20:39:43)
This kind of parity argument is often useful when dealing with circuitous paths and symmetries (and here we're dealing with both).
MCrawford (20:39:48)
MCrawford (20:39:56)
Can we eliminate any more?
mathenthusiast8 (20:31:07)
we can eliminate the space diagonal...it will not have any cohorts
MCrawford (20:40:42)
Any others?
eastsoldier (20:41:02)
second is eliminated b/c there are two non-moving
MCrawford (20:41:19)
Why is that bad?
eastsoldier (20:41:38)
it cannot go back to where it started unless its right triangle
probability1.01 (20:41:55)
that means we are only using the "grid," which is all right angles... if you know waht I mean
countersixte_2 (20:42:00)
We can only change one dimension at a time
Bictor717 (20:42:11)
two nonmoving means that we are adding 3 odd-shifts
MCrawford (20:42:20)
First, we notice that we cannot simply shift in one axis at a time. A similar parity argument as before prevents such a shift from returning to the original vertex.
Next, we notice that a shift of (2, 2, 2) could only happen between diametrically opposite vertices of the cube. Since these only come in pairs, we can't make a triangle using these shifts.
MCrawford (20:42:25)
MCrawford (20:42:35)
Can we construct at least one example for each of these three possibilities?
eastsoldier (20:43:01)
(0,0,1)(1,0,0)(0,1,0) as we did up there
mathenthusiast8 (20:43:05)
sqrt(2) is obviously the one already exemplified
probability1.01 (20:43:22)
d = sqrt(2) we already did above, and d = 2sqrt(2) would be a similar one with (2, 0, 0), (0, 2, 0), (0, 0, 2)
mathenthusiast8 (20:43:27)
2sqrt(2) connects the 2x2x2 cube vertices
MCrawford (20:44:44)
vfiroiu_2 (20:44:22)
(0,1,2)(1,2,0)(2,0,1)
Mr.Ocax (20:44:24)
(2,1,0), (1,0,2), (0,2,1)
MCrawford (20:44:55)
MCrawford (20:45:06)
Now, we can begin our the counting. How can we most easily count the number of the smallest triangles?
Elemennop (20:34:19)
It also has 8 orientations in a single unit cube
mathenthusiast8 (20:45:24)
how many per unit cube...n*8
Monkey (20:45:35)
There are 8 per small cube,and there are 8 small cubes, with a total of 64.
MPS-vras (20:45:54)
for each vertex there is one in a cube
MCrawford (20:46:07)
First, we note that there are 8 smaller cubes that contain any version of the smallest triangle:
MCrawford (20:46:11)
mathenthusiast8 (20:45:53)
combining the two statements, 8*8
MCrawford (20:46:25)
MCrawford (20:46:32)
Are we worried that some of these reflections could produce the same triangle? Why or why not?
Bictor717 (20:46:54)
The triangles isolate a vertex. Each vertex gives a different triangle.
Elemennop (20:47:07)
No, because triangles within one unit cube won't coincide with ones in another, and one can manually check the 8 orientations do not coincide within the same unit cube
avolfson (20:47:32)
no because the part that the triangle cuts off and the part left over aren't the same
MCrawford (20:47:45)
We should consider whether or not any triangle is reproduced by some combination of reflections. However, none of the sides of the triangle move parallel to any of the coordinate axes meaning that none of the axes of symmetry of the triangles are parallel with the coordinate axes. This means each of the 8 orientations is necessarily distinct.
MCrawford (20:47:55)
MCrawford (20:48:01)
How many of each of the larger triangles?
Hexahedron (20:48:16)
8
avolfson (20:48:18)
8 ?
MGriffin487 (20:48:21)
8 and 8
Monkey (20:48:21)
8
MCrawford (20:48:44)
Why just 8 for each?
MCrawford (20:49:19)
We were able to shift the smaller triangles -- what keeps us from shifting over dimensions?
Hexahedron (20:49:14)
because they can only be put inside cubes of side length 2, and there is only 1 cube with side length 2
MGriffin487 (20:49:34)
because unlike the triangle w/ the side lengths of sqrt, the side lengths take up the whole 2^3 cube
MCrawford (20:49:46)
Each of the larger triangles is bound by the larger (2 x 2 x 2) cube with no wiggle room. We can't reproduce these triangles by shifting them over as we could the smaller triangles. This means that we can only product distinct versions of each of the larger triangles by reorienting them. Again, we have 8 possible orientations for each.
ChrisChang (20:46:01)
so its 80?
Hexahedron (20:48:35)
so 64+8+8=80 which is answer choice C
MCrawford (20:50:04)
MCrawford (20:50:11)
This problem displays a lot of interesting problem solving concepts.
MCrawford (20:50:16)
First and foremost -- organize your casework. A lot of people missed this problem answering (A) because they never found the triangles with side length sqrt(6). This won't happen if you hunt for cases in a rigorous way.
MCrawford (20:50:19)
Find a way to model the problem so that you aren't guessing. This relates directly to the first point about organizing casework. In this case, we modeled the problem by evaluating trios of "coordinate shifts" which together gave possible distances and created a triangular path.
MCrawford (20:50:30)
Consider symmetries in geometric counting problems. If you give consideration to the sources of these symmetries (in this case, the axes), then calculating the way they affect your count should not be a nightmarish task.
MCrawford (20:50:37)
I would like to follow this problem up with a fun problem with an interesting problem a friend of mine gave me a few months ago. Suppose that there is a robber on an n-dimension unit cube running from n cops. The cops and robber can only move between adjacent vertices of the cube. However, the cops take turns with the robber moving. The robber moves first, then the cops, then the robber, etc. If everyone of the cube always knows where everyone else is, can the cops form a strategy to catch the robber? If not, why not? If so, what's the most efficient strategy?
MCrawford (20:51:15)
Once again, we have a seminar class at the end of this month that some of you might be interested in taking. The special AIME seminar involves two three hours classes, including a half hour break each day, plus a mock AIME designed to help students prepare for the AIME. Time will be spent on AIME problems covering algebra, geometry, number theory, counting, and probability. The cost of this seminar is $40 and you can find details here:
MCrawford (20:51:23)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class4
pika (20:51:10)
does it depend on the dimension?
MCrawford (20:51:42)
The cops and robber problem has a general algorithm that spans all dimensions.
Hexahedron (20:51:57)
even for hypercubes?
MCrawford (20:52:18)
Yes, the point of the problem is to go out to many dimensions -- it's easy in 2 or 3.
probability1.01 (20:51:59)
there are 2^n possible points, with each point connecting to n other points.
mna851 (20:52:37)
I dont see how any student could get that question in the alotted time even if it was the only question on the test.
MCrawford (20:53:56)
I was trying to provoke a lot of ideas with that solution, it doesn't need to take nearly that long.
MCrawford (19:28:20)
Before we begin, we have a seminar class at the end of this month that some of you might be interested in taking. The special AIME seminar involves two three hours classes, including a half hour break each day, plus a mock AIME designed to help students prepare for the AIME. Time will be spent on AIME problems covering algebra, geometry, number theory, counting, and probability. The cost of this seminar is $40 and you can find details here:
MCrawford (19:28:26)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class4
MCrawford (19:28:33)
At this time Dave Patrick will walk through the last 5 problems from the AMC 10 A.
DPatrick (19:28:50)
21.
DPatrick (19:28:52)
DPatrick (19:29:00)
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DPatrick (19:29:15)
(If you click on the link, you'll get a copy of the problem in a separate window, which is easier to follow along with)
DPatrick (19:29:25)
What seems like a reasonable first step with this problem?
NightFlarer (19:29:13)
write it as (n(n+1))/2
pika (19:29:18)
n(n+1)/2
everyday847 (19:29:19)
Well, 1+2+...+n = n(n+1)/2
mna851 (19:29:29)
n(n+1)/2
DPatrick (19:29:49)
DPatrick (19:30:02)
If this is not a sum that you are familiar with, it would be a good idea to discuss this in
DPatrick (19:30:10)
the Basics or Getting Started forums.
DPatrick (19:30:21)
Now, we have rewritten the given problem in a simpler form -- this is already a great leap in the challenge of problem solving. We like easier problems. We are now asking for how many positive integers n(n+1)/2 evenly divides 6n. Where can we go from here?
pika (19:29:51)
When is 6n/(n(n+1)/2) an integer?
RC-7th_? (19:30:00)
Then 12/(n+1) must be an integer
NightFlarer (19:30:11)
so (n(n+1)/2 has to divide 6n, cancel out the n and multiply to get 12/(n+1) = some integer k
mna851 (19:30:36)
n+1 must divide 12
everyday847 (19:30:47)
Thus 6n / (n(n+1)/2) is an integer.
DPatrick (19:30:58)
DPatrick (19:31:10)
How can we count the number of positive integers n for which this is true?
mathenthusiast8 (19:31:21)
factors of 12
zabelman (19:31:22)
look at divisors of 12
everyday847 (19:31:23)
Factors of 12 are 1,2,3,4,6,12
aznarchmage (19:31:33)
find the factors of 12
DPatrick (19:31:48)
everyday847 (19:31:47)
And n-values can be 0 - not valid, because it's not positive, 1,2,3,5,11
isaacchao (19:31:49)
omit 1, 0 is not possitive integer
Scrambled (19:32:00)
0 cannot be n
sam3.14 (19:32:03)
n+1 cannot equal 1
DPatrick (19:32:17)
We know that n + 1 is a positive divisor of 12, but n must be a positive integer. This means that we count all of the positive divisors of 12 except 1, so our answer is 5. (B).
Scrambled (19:32:08)
wait a minute; does 1+2+...+n mean that n cannot be 1 or 2? (because there was an answer choice for people who assumed that n cant be 1 or 2and for those who assumed it could be)
DPatrick (19:33:18)
No: 1 + 2 + ... + n is a notational convenience for the sum of the positive integers from 1 through n.
DPatrick (19:33:41)
So even though it may look a little strange, n=1 or n=2 is valid.
DPatrick (19:34:00)
22.
DPatrick (19:34:13)
DPatrick (19:34:15)
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DPatrick (19:34:25)
What can we first do with this problem?
everyday847 (19:34:44)
To be common, they'd have to be divisible by 12, the LCM.
RC-7th_? (19:34:48)
Look at the LCM of 4 and 6
RC-7th_? (19:35:40)
The common elements are divisible by 4 and 6, so 12
mathenthusiast8 (19:35:41)
so let me rephrase: find the LCM of 4 and 6
kostya (19:35:45)
#'s that divide by 12 are in both sets
everyday847 (19:35:31)
I'd say that thus, every other member of T would be in S, and every third member of S would be in T.
DPatrick (19:36:45)
In light of this last comment -- the first thing that we can do is rephrase the problem such that we are asking a more simple question. Of the two sets S and T, S has a smaller range. This means we could just search for the number of multiples of 6 in set S.
DPatrick (19:37:04)
How can we easily count the number of multiples of 6 in set S?
gleb (19:36:43)
Rewrite the sets as {2^2, 2^2n,...2^2n^2004} and {2*3, 2*3n,...2*3n^2004}
pika (19:36:56)
int(2005/3)=668
mna851 (19:37:17)
They follow a pattern of every third number.
Ars (19:37:25)
divide by 3
mo (19:37:36)
They occur for once for every 3 multiples of 4.
sam3.14 (19:37:37)
it's every three
everyday847 (19:37:39)
Every third multiple of 4 is a multiple of 6.
zabelman (19:37:46)
every third element is a multiple of 6
DPatrick (19:37:53)
Now, in order for an integer to be a multiple of 6, it must be both a multiple of 2 and 3. Clearly ever member of S is a multiple of 2, so we are left counting the members of S that are multiples of 3.
DPatrick (19:38:02)
pokey (19:38:18)
Why is there no E choice?
DPatrick (19:38:45)
Sorry, it got chopped off. (E) was 1001.
DPatrick (19:39:06)
Let's move on to #23
DPatrick (19:39:11)
DPatrick (19:39:14)
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DPatrick (19:39:19)
DPatrick (19:39:22)
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DPatrick (19:39:42)
Let's start by labeling the points that are left unlabeled. Are there any points that we might give a particular label (that might be helpful)?
zabelman (19:39:50)
O
Scrambled (19:39:56)
the point opposite to d
mathenthusiast8 (19:39:58)
F, as in chord DF
joml88 (19:39:59)
center...O
pika (19:39:59)
Intersection of DC and the circdle
DPatrick (19:40:34)
DPatrick (19:40:39)
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pilot (19:40:02)
I would set r=3 AC=2 and BC=4 since we're dealing with a ratio
kostya (19:40:17)
AC=2
DPatrick (19:40:48)
I agree. We can go about hunting down relative lengths, though I would personally start by assigning a length to some segment and working from there. I personally find it makes my work neater whether or not it actually comes in handy in the problem. In this case I'll let AC = 2. This means BC = 4.
DPatrick (19:41:00)
Now, keeping in mind that trig isn't required for the AMC10, what can we do to compare the areas of DCE and ABD?
MGriffin487 (19:41:30)
compare the parts that you multiply to find the area
MGriffin487 (19:41:34)
aka base and height.
fedwinri (19:41:36)
Compare the lengths of the bases (since the hights are the same).
DPatrick (19:42:02)
We see several triangles that share bases, parts of bases, or heights.
pika (19:41:14)
Altitudes and bases
zabelman (19:41:15)
calculate them :)
gleb (19:41:57)
find the ratio of their altitudes
DPatrick (19:42:43)
Let's compare DCO and ACB. They have the same height. What is the ratio of their bases?
DPatrick (19:42:54)
Sorry...I meant DCO and ADB.
pilot (19:43:05)
1:6
RC-7th_? (19:43:07)
1:6
SuperNova (19:43:09)
1:6
MaThWhIz2004 (19:43:15)
1/6
zabelman (19:43:16)
1/6
DPatrick (19:43:37)
Right: CO = 1 and AB = 6, so the bases have ratio 1:6. This means the area of DCO is 1/6 of the area of ABD.
DPatrick (19:43:49)
What other triangles can we compare?
DPatrick (19:44:04)
Remember that we want DCE in the end...
zabelman (19:44:01)
DCO and COE
Bictor717 (19:44:05)
DCO and OCE
fedwinri (19:44:11)
And ECO = DCO
RC-7th_? (19:44:12)
[DOE]=[DCO]
riposte (19:44:13)
DCO and COE
pilot (19:44:15)
DCO=COE
DPatrick (19:44:26)
By symmetry, if we drop a perpendicular from E to AB, it would be the same length as DC. This means triangles DCO and COE share a base and have equal heights. The area of DCE is twice the area of either of these triangles.
DPatrick (19:44:36)
What is our final answer?
RC-7th_? (19:44:23)
So the answer is 1/3
zabelman (19:44:43)
1/3
d343seven (19:44:44)
1/3
fedwinri (19:44:46)
1/3
Monkey (19:44:47)
1/3
pokey (19:44:48)
1/3
pilot (19:44:49)
1/3
joyofpi (19:44:49)
1/3
riposte (19:44:49)
1/3
DPatrick (19:45:11)
Since DCE has twice the area of DCO, the ratio of the area of DCE to the area of ABD is 1/3, (C).
DPatrick (19:45:14)
Here we used our understanding of symmetry and the formula bh/2 for calculating the area of a triangle. Remember that this formula allows us to compare areas by comparing lengths of sides of triangle. That often comes in handy.
DPatrick (19:45:22)
Now for #24:
DPatrick (19:45:30)
DPatrick (19:45:33)
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DPatrick (19:46:01)
This problem is going to require a bit of exploration. It's important to begin mathematical exploration on solid footing, so let's collect some facts. What facts can we derive from the problem statement?
NightFlarer (19:45:52)
n must be the square of a prime number
RC-7th_? (19:46:13)
n and n+48 are both squares
solafidefarms_2 (19:46:15)
n is a square
pika (19:46:16)
and n+48 is the square of a prime
DPatrick (19:46:41)
We know that n must be a perfect square in order for the square root of n to be a prime (integer).
RC-7th_? (19:46:32)
n+48 is also the square of a prime number
aznarchmage (19:46:33)
n+48 must also be the square of a prime number
DPatrick (19:46:54)
The functional equations we are given tell us that n and n + 48 are each square of not just integers, but of prime integers. How can we use this fact?
isaacchao (19:47:20)
p^2 + 48 = q^2
RC-7th_? (19:47:22)
(x+y)(x-y)=48, where x and y are primes?
NightFlarer (19:47:30)
call n p^2 and let n+ 48 = q^2
DPatrick (19:47:37)
Many number theory problems require us to step into the realm of algebra. This is one of them. We can help ourselves by labeling the primes that we intend to discuss.
DPatrick (19:47:44)
DPatrick (19:47:55)
mna851 (19:48:05)
Difference of squares
everyday847 (19:48:05)
Well, q^2 - p^2 = 48
zabelman (19:48:14)
(q-p)(q+p)=48, and look at divisors
DPatrick (19:48:23)
DPatrick (19:48:40)
DPatrick (19:48:56)
What does this equation tell us?
mathenthusiast8 (19:48:15)
their sum and difference are factors of 48
mandy_pal (19:49:03)
find factors of 48
Scrambled (19:49:07)
we can find factors of 48
pika (19:49:10)
factors of 48
DPatrick (19:49:21)
The equation now tells us that both (q + p) and (q - p) are divisors of 48 that pair up so that their product is 48. How can we work with that information?
Jimmy (19:49:36)
decompose 48
zabelman (19:49:38)
list factor pairs of 48 and solve for p and q in each case
ekrem (19:49:50)
try the factors
DPatrick (19:50:09)
DPatrick (19:50:12)
For which of these pairs of divisors are there prime numbers p and q such that the sum and difference of the primes is equal to each divisor in the pair?
mathenthusiast8 (19:49:34)
the factors must be of the same parity
mna851 (19:49:50)
q and p must also have the same parity
DPatrick (19:50:52)
Experienced students of number theory might quicken their search knowing that q + p and q - p must both be even or both be odd. This is because their difference is 2p -- an even number. Since the divisors can't both be odd, we are left with only three pairs to examine.
DPatrick (19:51:11)
DPatrick (19:51:18)
What is our final answer?
zabelman (19:51:02)
2*24 gives 13 and 11, and thats the only one
isaacchao (19:51:22)
11,13
mathenthusiast8 (19:51:25)
(b), 1
RC-7th_? (19:51:30)
1
mathenthusiast8 (19:51:30)
only (11, 13) works
RC-7th_? (19:51:33)
1 (B)
Ars (19:51:47)
1
DPatrick (19:52:19)
DPatrick (19:52:59)
A lot of you jumped to the conclusion that since we have 3 solutions for p and q, that the answer was 3. Be careful!
mna851 (19:52:33)
That was a nice problem
DPatrick (19:53:36)
Finally, #25:
DPatrick (19:53:52)
DPatrick (19:54:12)
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DPatrick (19:54:20)
How can we begin here?
RC-7th_? (19:54:14)
Draw a picture
mna851 (19:54:17)
We need a picture
DPatrick (19:54:29)
It makes sense to draw a diagram and get a good idea of what is going on visually. Is there any particular way we might think to draw the diagram?
Elemennop (19:54:48)
An acute triangle, since 25^2 + 39^2 > 42^2
probability1.01 (19:55:09)
let us have the base as AB
DPatrick (19:55:22)
I like the idea of making either AC or AB parallel to the "x-axis" of my drawing. This is because those are the sides from which the smaller triangle is cut from the larger. As we have seen, being able to compare bases sometimes helps us with area comparison problems.
Monkey (19:55:19)
Think of height and base ratios
DPatrick (19:55:38)
Yes. As it happens, I drew is with AC as the base.
DPatrick (19:55:41)
DPatrick (19:55:43)
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DPatrick (19:55:49)
Now, how can we compare the area of triangle ADE to quadrilateral BCED?
sam3.14 (19:56:19)
[DBCE]=[ABC]-[ADE]
pika (19:56:29)
compare ADE to ABC?
probability1.01 (19:56:30)
sort of--but you must compare it indirectly via comparison to the area of triangle ABC
mathenthusiast8 (19:56:31)
small triangle/(large triangle area - small triangle)
DPatrick (19:56:49)
Since the two parts are comparing together make up the area of triangle ABC, it makes sense to try and compare one of the smaller figures to triangle ABC. Triangle ADE seems like the most reasonable candidate as they two triangles have two sides each on the same lines and a common angle.
DPatrick (19:56:53)
How can we compare the area of ADE to ABC?
mathenthusiast8 (19:57:33)
heights seem helpful, since we know the bases, but they seem to be difficult to attain
Scrambled (19:57:42)
ratio of heights
pika (19:58:02)
Height and base?
DPatrick (19:58:19)
Since the area of a triangle can be found by the formula bh/2, we can compare the areas using the ratios of the bases and the heights of the triangles.
DPatrick (19:58:28)
What is the ratio of the bases of ADE and ABC?
Scrambled (19:58:36)
1:3
RC-7th_? (19:58:36)
1:3
gleb (19:58:37)
1/3
Monkey (19:58:39)
The ratio of bases is 1/3
Elemennop (19:58:41)
1/3
DPatrick (19:58:53)
We have the ratio between the bases of ADE and ABC. AE is AC/3.
DPatrick (19:58:58)
How can we find the ratio of the heights?
pika (19:58:26)
drop a perpendiclar for the height of 1 and then the other
probability1.01 (19:59:12)
The heights can be obtained by drawing perpendiculars to AC from B and D, then comparing similar triangles
Scrambled (19:59:12)
similar triangles
vfiroiu_2 (19:59:22)
drop perpendiculars from D and B to AC
DPatrick (19:59:33)
DPatrick (19:59:38)
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RC-7th_? (19:59:26)
19/25
Hexahedron (19:59:31)
it is just 19:25
d343seven (19:59:37)
19/25?
mandy_pal (19:59:38)
19/25
DPatrick (19:59:53)
All we need is the ratio of the heights that we drop down to the bases of the triangles. We find this ratio by noting that the perpendiculars give us two right triangles that must be similar because they share an acute angle. We can now say that the ratio of the heights is the same as the ratio of the hypotenuses which is 19/25.
DPatrick (20:00:03)
What is the ratio of the area of the triangles?
RC-7th_? (19:59:59)
The ratio of [ADE] to [ABC] is 19/75
mandy_pal (20:00:11)
19/75
fedwinri (20:00:16)
19/75
Monkey (20:00:19)
19/75
DPatrick (20:00:29)
DPatrick (20:00:40)
How does the ratio of the triangles give us the ratio of ADE to BCED?
RC-7th_? (20:00:32)
The answer is 19/56 (D)
gleb (20:00:59)
19/(75-19)
isaacchao (20:01:00)
19/56
Wumbate (20:01:04)
It is simply 19/(75-19)=19/56
mathenthusiast8 (20:01:06)
so 19/56, or (d) because 19/(75-19)
mna851 (20:01:13)
Area of BCED is equal to ABC is ADE
pilot (20:01:15)
19/(75-19)
DPatrick (20:01:32)
Since the quadrilateral is all of the region in the big triangle but outside the smaller triangle, it's area is 1 - 19/75 = 56/75 as large (in area) as the big triangle.
DPatrick (20:01:44)
DPatrick (20:01:58)
In order to solve this problem we once again used ratios of bases and heights to compare areas. We also used convenient similar triangles which we can often construct by dropping altitudes within a diagram.
DPatrick (20:02:41)
As a closing note: those of you that know some trigonometry -- in particular, the Law of Sines -- can solve this problem using that method. But since the AMC 10 doesn't require trig, I won't go into that solution here.
DPatrick (20:02:59)
Now I'll turn things back over to Mathew, for the AMC 12.
riposte (20:03:39)
How much time do we have in AMC 10 and 12 for 5 problems?
MCrawford (20:04:13)
We'll take as much time as we need.
MCrawford (20:04:19)
MCrawford (20:04:22)
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MCrawford (20:04:31)
This is probelm 21 from the AMC 12 A
MCrawford (20:04:37)
How might we begin with this problem?
countersixte_2 (20:04:52)
Express in exponential form?
mo (20:04:56)
write a^c^2005=b
nonquotidian (20:04:56)
exponent out the log: a^c^2005=b
eastsoldier (20:04:58)
set it in exponetial form
pilot (20:04:52)
try some numbers
dts (20:04:53)
c>=2 is ridiculuous; b would have to be enormous.
Elemennop (20:04:58)
c must be either 1 or 0, otherwise b would be too great to fit a + b + c = 2005
MCrawford (20:05:21)
We have two equations to work with, but the second looks more like an equation meant to give boundaries on the variables, so we'll work with the first.
MCrawford (20:05:24)
ChrisChang (20:05:41)
so we look at values of a and b that give c = 0 or 1
Hexahedron (20:05:11)
just looking at this, we can tell that c<2 because if c is greater then or equal to 2 then the other numbers would have to be huge
towersfreak2006 (20:05:54)
since a is an integer, and b<2006, c<2
MCrawford (20:06:05)
The second equation helps us most when we use the fact that a is at least 2. The exponent in our exponential equation can be at most 10 in order for b not to exceed 2005, the sum of a, b, and c.
MCrawford (20:06:11)
MCrawford (20:06:19)
The only possible values for c are 0 and 1. What do we do now?
blahblahblah (20:06:26)
cases
towersfreak2006 (20:06:27)
plug in the values and solve!
eastsoldier (20:06:30)
plug it in
ChrisChang (20:06:14)
when c = 0, b = 1 and a = 2004
ChrisChang (20:06:44)
when c = 1, a = b so a = 1002, b = 1002, c = 1
blahblahblah (20:06:51)
mo (20:06:51)
well we know that when c=1, then a=b. So a+b+c=2005, a=1002, b=1002.
Hexahedron (20:06:54)
if C=1 then the pair (1002,1002,1) works and if C=0 then we get the pair (2004,1,0)
MCrawford (20:07:13)
MCrawford (20:07:28)
From case (1) we see that b = 1 and c = 0. This means a = 2004.
From case (2) we see that a = b and since c = 1, a = b = 1002.
RC-7th_? (20:07:14)
So C
MCrawford (20:07:38)
There are 2 solutions and the answer is (C).
This problem teaches several important lessons. First, it makes sense not to focus too much on a "bounding equation" until it's useful. Next, and possibly most important, it's often easier to examine the exponential form of logarithmic equations. Finally, restricting solutions to integers makes it easier to hone in on our final answer.
MCrawford (20:07:49)
Now, let's take a look at problem 22.
MCrawford (20:07:57)
MCrawford (20:08:02)
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MCrawford (20:08:05)
What's the first thing we should do with a problem like this?
eastsoldier (20:08:15)
set up equations
towersfreak2006 (20:08:20)
gather our information, and convert them to algebraic expressions
probability1.01 (20:08:21)
Assign dimensions to the box: x, y, z
dts (20:08:21)
Let x, y, z be the side lengths.
ChrisChang (20:08:10)
draw a picture
boblard (20:08:13)
draw a figuires
blahblahblah (20:08:18)
draw a diagram
MCrawford (20:08:55)
Okay, so we have the first two things...
MCrawford (20:09:01)
Here we have a word problem, so the first thing we must do is translate the words into mathematics. Making the correct analogies between the information given and the mathematical statements we are going to work with may be the most important challenge in this problem.
Elemennop (20:08:43)
edges a, b, c, then 2ab + 2bc + 2ac = 384 and 4a + 4b + 4c = 112 or a + b + c = 28
pika (20:08:53)
and x+y+z=28
pika (20:08:25)
192=xy+yz+xz
MCrawford (20:09:43)
RC-7th_? (20:09:38)
we want sqrt(x^2+y^2+z^2), so we square 28 and subtract 384
RC-7th_? (20:08:27)
2r is the length of the interior diagonal
MCrawford (20:10:23)
Before we begin diving into the equations, it makes sense to analyze what exactly we are looking for. It's important to keep an eye on the ball. This allows us to work more efficiently and with fewer errors.
MCrawford (20:10:28)
Our goal is to find the value of r. Really this means finding the value of half of d, the diameter of the sphere which is also the spatial diagonal of the box:
MCrawford (20:10:35)
vfiroiu_2 (20:09:29)
blahblahblah (20:10:04)
Elemennop (20:10:21)
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 784, Then subtract the first equation from this new one and you get a^2 + b^2 + c^2 = 400
JHu (20:10:25)
RC-7th_? (20:10:35)
Then we take the square root of 784-384=400 to get 20/2=10 as the answer
Bictor717 (20:10:47)
ChrisChang (20:11:00)
sqrt(x^2 + y^2 + z^2) = diameter
pika (20:10:44)
10 is the answer?
gleb (20:11:08)
In order to find the radius, first find the diameter in terms of our variables
MCrawford (20:11:33)
MCrawford (20:11:42)
Elemennop (20:11:25)
sqrt(a^2+b^2+c^2)/2 = r = 10
MCrawford (20:12:01)
We are looking for r which is d/2, so our answer is 10. (B).
MCrawford (20:12:18)
There were four tasks in this problem. First, we have to convert the words into math. We did this by assigning variables to the side lengths -- quantities that we did not know, but that we knew how work with. Next we recognized that the diameter of the sphere was the spatial diagonal of the box. This is apparent due to the symmetry of the box and sphere. Third, we used two applications of the Pythagorean theorem to find the length of the diagonal in terms of the lengths of the sides of the box. Finally, we used the system of equations we setup at the beginning to find the value of the sum of the squares of the sides. None of these tasks was particularly difficult on its own, but problem solving often means putting simpler pieces together one by one.
zabelman (20:08:56)
a very good idea (not my own): Assume the box is flat, i.e. one of the dimensions is 0
MCrawford (20:12:47)
It is often helpful to make special cases -- even extreme ones.
MCrawford (20:13:00)
Let's take a look at problem 23...
MCrawford (20:13:08)
MCrawford (20:13:14)
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MCrawford (20:13:22)
How can we begin with a problem such as this one?
eastsoldier (20:13:36)
find the all possible number of set
MCrawford (20:14:05)
First, we should recognize this as a counting probability problem. Our task is to first count the total number of ways in which we can select a and b. This will be our denominator. Then we want to count the number of ways in which the logarithm given will be an integer. The total number of these "successful outcomes" will comprise our numerator.
MCrawford (20:14:09)
Now, in how many ways can we select a and b?
mathenthusiast8 (20:14:02)
which is 25*24, since they must be distinct = 600
eastsoldier (20:14:08)
then the number of set that satisfies the thing
MCrawford (20:14:26)
MCrawford (20:14:41)
We have the denominator in our probability. We must now count the number of successful outcomes to determine the numerator.
blahblahblah (20:13:46)
SpaceTerr808 (20:13:41)
simply the expression log basea of b
pika (20:13:44)
loga(b)=log2(b)/log2(a)?
Bictor717 (20:14:59)
MCrawford (20:15:32)
mathenthusiast8 (20:15:12)
or the power on b should be divisible by a. the other statement was too broad
probability1.01 (20:15:14)
then, if we choose 2^n and 2^k, for a and b respectively, we want it so that n | k
dts (20:13:44)
Consider the easier, equivalent problem of the set (1, 2, 3, ..., 25), where the second element chosen must be divisible by the first.
riposte (20:14:18)
a = 2^i and b = 2^j. we need j to be a multple of i
gleb (20:14:39)
we know that b must be a multiple of a
mathenthusiast8 (20:14:54)
to have the log be an integer, b should be divisible by a since they share bases
Jimmy (20:14:55)
a<b
pika (20:15:26)
just switch it into 1, 2, 3..., 25
gleb (20:16:20)
so find number of multiples for each 1-25
MCrawford (20:16:45)
MCrawford (20:16:52)
We can simplify the problem at hand by noting that our job is to count the number of ways in which we can select m and n from {1, 2, 3,..., 25} such that m/n is an integer.
mathenthusiast8 (20:16:49)
in this problem, there are 24+11+7+5+4+3+2+2+1+1+1+1=62 good pairs.
probability1.01 (20:16:59)
I just systematically counted at this point, but I think there might be a better way....
vfiroiu_2 (20:16:17)
count on fingers
MCrawford (20:17:30)
Unless you're more dextrous with your toes.
RC-7th_? (20:17:15)
There are 62 ways, so B is the answer
MCrawford (20:17:41)
The easiest way to count seems to be to start at 1 and count the number of other integers in the set that are multiples of 1.
MCrawford (20:17:44)
There are 24 multiples of 1 in the set besides 1.
There are 11 multiples of 2 in the set besides 2.
There are 7 multiples of 3 in the set besides 3.
There are 5 multiples of 4 in the set besides 4.
There are 4 multiples of 5 in the set besides 5.
There are 3 multiples of 6 in the set besides 6.
There are 2 multiples of 7 in the set besides 7.
There are 2 multiples of 8 in the set besides 8.
There is 1 multiple of 9 in the set besides 9.
There is 1 multiple of 10 in the set besides 10.
There is 1 multiple of 11 in the set besides 11.
There is 1 multiple of 12 in the set besides 12.
mathenthusiast8 (20:17:32)
62/600=31/300, or (B)
blahblahblah (20:17:39)
62/600=31/300
MCrawford (20:17:54)
MCrawford (20:18:44)
Now let's conquer 24 -- one of the two problems that gave top students problems.
MCrawford (20:18:54)
MCrawford (20:18:57)
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MCrawford (20:19:04)
How can we begin with this problem?
blahblahblah (20:19:10)
I really liked 24). The first thing that you absolutely need to notice is that deg(Q(x))=2
MCrawford (20:19:43)
How do we know the degree of Q?
probability1.01 (20:19:42)
(Q(x) - 1)(Q(x) - 2)(Q(x) - 3) = (x - 1)(x - 2)(x - 3)R(x)
Elemennop (20:19:42)
P((Q(X)) must have degree 6, so Q(X) has degree 2
MCrawford (20:20:08)
Demon (20:19:58)
∀x∈{1,2,3} Q(x)∫{1,2,3}
MCrawford (20:20:30)
Since R(x) is a third degree polynomial, each side is a sixth degree polynomial. Furthermore, the sixth degree polynomial has factors of (x - 1), (x - 2), and (x - 3). That gives 1, 2, and 3 as roots of the sixth degree polynomial.
MCrawford (20:20:45)
We now know that one of Q(x) - 1, Q(x) - 2, and Q(x) - 3 for each x = 1, 2, and 3.
Now, how can we count the total number of possible Q(x)?
MCrawford (20:21:27)
We now know that one of Q(x) - 1, Q(x) - 2, and Q(x) - 3 is 0 for each x = 1, 2, and 3.
Now, how can we count the total number of possible Q(x)?
Scrambled (20:20:56)
both polynomials must have roots of 1,2,and 3
kostya (20:21:15)
blahblahblah (20:21:21)
this gives 27 - but there's a catch
dts (20:21:18)
3*3*3=27
MCrawford (20:21:53)
Monkey (20:21:44)
Let Q(x) equal 1,2, or 3
Bictor717 (20:21:54)
1, 2, 3 are roots, Q(1), Q(2), Q(3) must all be 1, 2, or 3
Scrambled (20:21:19)
combinations of ways this can be done possibly?
dts (20:22:45)
Some possibilities can't have degree 2.
MCrawford (20:22:57)
Q(x) must be a quadratic polynomial. Otherwise the two halves of our equation could not be sixth degree polynomials. This means Q(x) cannot have three roots. This eliminates 3 possible ways to arrange 1, 2, and 3 as roots of Q(x) - 1, Q(x) - 2, and Q(x) - 3.
MCrawford (20:23:06)
How can we check the rest of the possibilities?
blahblahblah (20:23:00)
The problem is that if Q(x)=x for 1,2,3
blahblahblah (20:23:11)
then Q(x) is linear
dts (20:23:29)
It also can't be linear.
Bictor717 (20:23:28)
graph x vs Q(x)
mo (20:23:36)
when (1,1,1), (2,2,2,), (3,3,3,), (1,2,3), and (3,2,1) must be eliminated b/c when chosen, their degree is incorrect.
blahblahblah (20:19:38)
and that three points determine a quadratic
blahblahblah (20:23:43)
Q(x)=x, Q(x)=-x, Q(x)=1,Q(x)=2,Q(x)=3 --> subtract 5 possibilities
MCrawford (20:24:31)
MCrawford (20:24:40)
Of the 6 remaining arrangements, we need only be sure that none of them are lines.
MCrawford (20:24:44)
Two of the 6 remaining arrangements form lines. They are
Q(1) = 1, Q(2) = 2, Q(3) = 3, and
Q(1) = 3, Q(2) = 2, Q(3) = 1.
gleb (20:24:23)
so 27-5=22 (B)
MCrawford (20:25:05)
We can now see that there are 18 + 4 = 22 polynomials Q(x) for which the proposed functional equation holds as defined. (B).
MCrawford (20:25:29)
This problem shows us how important it is to recognize whether or not steps that we are taking are rigorous within the definitions of the problem.
MCrawford (20:25:43)
If we pay attention to detail, this problem is not so rough.
MCrawford (20:26:20)
I thougt number 25 was the most challenging problem on the A exams...
MCrawford (20:26:25)
MCrawford (20:26:36)
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MCrawford (20:26:39)
Any ideas that could give us a start?
Bictor717 (20:26:45)
Consider a unit cube first
eastsoldier (20:26:48)
My friend set up 8 cubes
Mr.Ocax (20:26:57)
find all the points first?
mathenthusiast8 (20:27:00)
recognizing what shape the points form
MCrawford (20:27:24)
I personally prefer to avoid these starting points...
MCrawford (20:27:44)
I would like to explore this problem a bit before jumping straight to the count.
MCrawford (20:27:58)
It's possible I'll discover things that will help me count more clearly -- without error.
soulzmischief (20:28:02)
different possible side lengths of the triangles?
MCrawford (20:28:25)
rcv (20:28:29)
Color the vertices black or white like a checkerboard. Parity eliminates a lot of possibilities.
MCrawford (20:28:42)
What might help us in regards to finding situations in which these distances will all be the same?
probability1.01 (20:29:12)
the squares of the differences in coordinates must be 0, 1, or 4
MCrawford (20:30:03)
MCrawford (20:30:12)
Is there some way we can eliminate some of these from contention?
soulzmischief (20:29:13)
the distances must have the same parity
MCrawford (20:30:22)
What do you mean?
soulzmischief (20:31:02)
the different side lengths of the equaliteral triangles must at least ahve the same parity to be equal, so that would eliminate some possibilities
MCrawford (20:31:26)
Often times a little investigation can help us get a better grip on a problem whether or not we think we need it. Is there some trivial example of an equilateral triangle we could take a loot at?
probability1.01 (20:31:52)
(1,0,0)(0,1,0)(0,0,1)
MCrawford (20:32:08)
Suppose we start at the point (1, 0, 0), then move to (0, 1, 0), then move to (0, 0, 1) and then back to (1, 0, 0). In this way we trace out an equilateral triangle by adding 1 to one of the coordinates and subtracting 1 from another. What have we noticed from this most trivial of examples?
eastsoldier (20:32:35)
it is done in a unit cube
MCrawford (20:32:51)
I'm looking for something more general.
MCrawford (20:33:01)
That will apply to all the possible triangles.
probability1.01 (20:33:21)
We can shift stuff by adding the same amount to a certain coordinate of each point
RC-7th_? (20:33:26)
We use the same transformation to move from one vertex to another
MCrawford (20:34:11)
What do happens after three shifts?
pika (20:34:25)
we go back to start
origamimasterjared (20:34:27)
It's back where we started
MCrawford (20:35:00)
What does that tell us about the nature of the shifts -- 3 shifts leave everything cancelled?
MCrawford (20:36:13)
We showed a shift in a (+1, -1, 0) permuted pattern...
MCrawford (20:36:27)
Could we have a 2/1/0 pattern (with various signs at the right times?
MCrawford (20:37:18)
Could three such shifts bring us back to the same point?
Bictor717 (20:37:21)
No, they can't add to 0
MCrawford (20:37:34)
Why not?
origamimasterjared (20:38:20)
2?1?0
MCrawford (20:38:44)
That's for one shift -- we get three.
Hexahedron (20:39:16)
2+-1+-1+-1 still can't equal 0
MCrawford (20:39:25)
In order to move three times and get back to the starting point, the sum of the directional distances at each move must be even. If it were odd, three movements would involve an odd number of unit shifts, so the result of all the shifts could not cancel out to 0.
MCrawford (20:39:43)
This kind of parity argument is often useful when dealing with circuitous paths and symmetries (and here we're dealing with both).
MCrawford (20:39:48)
MCrawford (20:39:56)
Can we eliminate any more?
mathenthusiast8 (20:31:07)
we can eliminate the space diagonal...it will not have any cohorts
MCrawford (20:40:42)
Any others?
eastsoldier (20:41:02)
second is eliminated b/c there are two non-moving
MCrawford (20:41:19)
Why is that bad?
eastsoldier (20:41:38)
it cannot go back to where it started unless its right triangle
probability1.01 (20:41:55)
that means we are only using the "grid," which is all right angles... if you know waht I mean
countersixte_2 (20:42:00)
We can only change one dimension at a time
Bictor717 (20:42:11)
two nonmoving means that we are adding 3 odd-shifts
MCrawford (20:42:20)
First, we notice that we cannot simply shift in one axis at a time. A similar parity argument as before prevents such a shift from returning to the original vertex.
Next, we notice that a shift of (2, 2, 2) could only happen between diametrically opposite vertices of the cube. Since these only come in pairs, we can't make a triangle using these shifts.
MCrawford (20:42:25)
MCrawford (20:42:35)
Can we construct at least one example for each of these three possibilities?
eastsoldier (20:43:01)
(0,0,1)(1,0,0)(0,1,0) as we did up there
mathenthusiast8 (20:43:05)
sqrt(2) is obviously the one already exemplified
probability1.01 (20:43:22)
d = sqrt(2) we already did above, and d = 2sqrt(2) would be a similar one with (2, 0, 0), (0, 2, 0), (0, 0, 2)
mathenthusiast8 (20:43:27)
2sqrt(2) connects the 2x2x2 cube vertices
MCrawford (20:44:44)
vfiroiu_2 (20:44:22)
(0,1,2)(1,2,0)(2,0,1)
Mr.Ocax (20:44:24)
(2,1,0), (1,0,2), (0,2,1)
MCrawford (20:44:55)
MCrawford (20:45:06)
Now, we can begin our the counting. How can we most easily count the number of the smallest triangles?
Elemennop (20:34:19)
It also has 8 orientations in a single unit cube
mathenthusiast8 (20:45:24)
how many per unit cube...n*8
Monkey (20:45:35)
There are 8 per small cube,and there are 8 small cubes, with a total of 64.
MPS-vras (20:45:54)
for each vertex there is one in a cube
MCrawford (20:46:07)
First, we note that there are 8 smaller cubes that contain any version of the smallest triangle:
MCrawford (20:46:11)
mathenthusiast8 (20:45:53)
combining the two statements, 8*8
MCrawford (20:46:25)
MCrawford (20:46:32)
Are we worried that some of these reflections could produce the same triangle? Why or why not?
Bictor717 (20:46:54)
The triangles isolate a vertex. Each vertex gives a different triangle.
Elemennop (20:47:07)
No, because triangles within one unit cube won't coincide with ones in another, and one can manually check the 8 orientations do not coincide within the same unit cube
avolfson (20:47:32)
no because the part that the triangle cuts off and the part left over aren't the same
MCrawford (20:47:45)
We should consider whether or not any triangle is reproduced by some combination of reflections. However, none of the sides of the triangle move parallel to any of the coordinate axes meaning that none of the axes of symmetry of the triangles are parallel with the coordinate axes. This means each of the 8 orientations is necessarily distinct.
MCrawford (20:47:55)
MCrawford (20:48:01)
How many of each of the larger triangles?
Hexahedron (20:48:16)
8
avolfson (20:48:18)
8 ?
MGriffin487 (20:48:21)
8 and 8
Monkey (20:48:21)
8
MCrawford (20:48:44)
Why just 8 for each?
MCrawford (20:49:19)
We were able to shift the smaller triangles -- what keeps us from shifting over dimensions?
Hexahedron (20:49:14)
because they can only be put inside cubes of side length 2, and there is only 1 cube with side length 2
MGriffin487 (20:49:34)
because unlike the triangle w/ the side lengths of sqrt, the side lengths take up the whole 2^3 cube
MCrawford (20:49:46)
Each of the larger triangles is bound by the larger (2 x 2 x 2) cube with no wiggle room. We can't reproduce these triangles by shifting them over as we could the smaller triangles. This means that we can only product distinct versions of each of the larger triangles by reorienting them. Again, we have 8 possible orientations for each.
ChrisChang (20:46:01)
so its 80?
Hexahedron (20:48:35)
so 64+8+8=80 which is answer choice C
MCrawford (20:50:04)
MCrawford (20:50:11)
This problem displays a lot of interesting problem solving concepts.
MCrawford (20:50:16)
First and foremost -- organize your casework. A lot of people missed this problem answering (A) because they never found the triangles with side length sqrt(6). This won't happen if you hunt for cases in a rigorous way.
MCrawford (20:50:19)
Find a way to model the problem so that you aren't guessing. This relates directly to the first point about organizing casework. In this case, we modeled the problem by evaluating trios of "coordinate shifts" which together gave possible distances and created a triangular path.
MCrawford (20:50:30)
Consider symmetries in geometric counting problems. If you give consideration to the sources of these symmetries (in this case, the axes), then calculating the way they affect your count should not be a nightmarish task.
MCrawford (20:50:37)
I would like to follow this problem up with a fun problem with an interesting problem a friend of mine gave me a few months ago. Suppose that there is a robber on an n-dimension unit cube running from n cops. The cops and robber can only move between adjacent vertices of the cube. However, the cops take turns with the robber moving. The robber moves first, then the cops, then the robber, etc. If everyone of the cube always knows where everyone else is, can the cops form a strategy to catch the robber? If not, why not? If so, what's the most efficient strategy?
MCrawford (20:51:15)
Once again, we have a seminar class at the end of this month that some of you might be interested in taking. The special AIME seminar involves two three hours classes, including a half hour break each day, plus a mock AIME designed to help students prepare for the AIME. Time will be spent on AIME problems covering algebra, geometry, number theory, counting, and probability. The cost of this seminar is $40 and you can find details here:
MCrawford (20:51:23)
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pika (20:51:10)
does it depend on the dimension?
MCrawford (20:51:42)
The cops and robber problem has a general algorithm that spans all dimensions.
Hexahedron (20:51:57)
even for hypercubes?
MCrawford (20:52:18)
Yes, the point of the problem is to go out to many dimensions -- it's easy in 2 or 3.
probability1.01 (20:51:59)
there are 2^n possible points, with each point connecting to n other points.
mna851 (20:52:37)
I dont see how any student could get that question in the alotted time even if it was the only question on the test.
MCrawford (20:53:56)
I was trying to provoke a lot of ideas with that solution, it doesn't need to take nearly that long.
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