| Transcript
for the Math
Jam "2007 AMC 10/12 B Math Jam"
on Feb 23. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk (19:03:16)
Welcome to the 2007 AMC 10B/12B Math Jam!
rrusczyk (19:03:23)
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
rrusczyk (19:03:28)
The classroom is moderated, meaning that students can type into the classroom, but only the moderator can choose a comment to drop into the classroom. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
rrusczyk (19:03:37)
When I post a problem, I will also post a link to the problem, like so:
rrusczyk (19:03:58)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-79145746.gif
rrusczyk (19:04:03)
This allows you to click on the link and view the problem in a separate window as we work through the solution. Depending on your computer's configuration, you may have to hold down the Ctrl key while clicking on the link. If it still doesn't work, try disabling your pop-up browser. If it still doesn't work, sorry, I'm afraid you're out of luck.
rrusczyk (19:04:09)
Also there will be some images in this session. The images should appear directly in the window, but I will also provide a link to images, like so:
rrusczyk (19:04:17)
rrusczyk (19:04:19)
http://www.artofproblemsolving.com/Classes/IntroNumbers/L1/logo.gif
rrusczyk (19:04:29)
There are a lot of students here! As I said, only well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted.
rrusczyk (19:04:39)
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. ""Working through the solutions"" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the answer. That's not why we're here. We're going to work through the problems step-by-step, and people who post comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, are going to be ignored.
rrusczyk (19:04:45)
Before we continue, I'd like to remind everyone about some upcoming Art of Problem Solving classes.
rrusczyk (19:04:50)
First, for those of you who passed the AMC tests and will be taking the AIME, we will be hosting a weekend seminar for AIME preparation. This Special AIME Problem Seminar is on Saturday, March 3, and Sunday, March 4 from 3:30 - 6:30 PM ET (12:30 - 3:30 PM PT) each day. During this class we will discuss both general test-taking strategies and specific mathematical tactics for the AIME. We will discuss several problems both from past AIMEs and from other challenging competitions that exemplify some of the most common problem-solving tactics on the AIME.
rrusczyk (19:04:57)
I will be teaching the Special AIME Problem Seminar. I earned the only perfect score on the AIME in 1989 (yeah, that long ago), and I'll be sharing some of the general approaches I used to do so.
rrusczyk (19:05:02)
Next, the Intermediate Number Theory Seminar begins on Wednesday, February 28. Classes are from 7:30 - 9:00 PM ET (4:30 - 6:00 PM PT) on Wednesdays from Feb. 28 through Apr. 18. This class covers advanced number theory problem solving tactics such as those needed for success on the AIME and USAMO. The course is taught by 2-time IMO medalist Naoki Sato.
rrusczyk (19:05:08)
Finally, the in Intermediate Algebra class begins on Tuesday, March 6. Classes are from 7:30 - 9:00 PM ET (4:30 - 6:00 PM PT) on Tuesdays from Mar. 6 through May 22. This class will cover algebraic problem solving tactics that are useful for advanced AMC problems, many AIME problems, and even beginning Olympiad problems. The course is taught by 2-time IMO participant Valentin Vornicu.
rrusczyk (19:05:18)
We will be hosting a Math Jam on Monday to discuss the AIME Seminar and the Intermediate Algebra classes. You can also ask questions about the courses at the end of this Math Jam.
not_trig (19:05:13)
yay i signed up for AIME class
rrusczyk (19:05:26)
I'll see you there!
rrusczyk (19:05:32)
And now, back to the Math Jam.
rrusczyk (19:05:54)
The Math Jam will proceed as follows:
rrusczyk (19:06:00)
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B. (#24 on the 12B is also #25 on the 10B, so that'll be a total of 9 problems.) After that, time permitting, I will take requests for some other problems for discussion.
rrusczyk (19:06:09)
rrusczyk (19:06:26)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-256025106.gif
rrusczyk (19:06:32)
rrusczyk (19:06:33)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/28d85612d853145b94a871ecf92fd9a4.png
rrusczyk (19:06:40)
Where do we start?
DiscreetFourierTransform (19:06:54)
assign side names
calc rulz (19:07:03)
Label lengths first (x for the square side length)
rrusczyk (19:07:11)
We can start by labeling our diagram. We want the side of the square, which we'll call s.
rrusczyk (19:07:17)
rrusczyk (19:07:21)
What else can we label in terms of s?
mario_jda128 (19:06:46)
this has a LOT of 3:4:5 ratios
not_trig (19:06:50)
similar triangles?
whynot (19:06:59)
b is a right angle
tjhance (19:07:00)
similar triangles
DiscreetFourierTransform (19:07:08)
and use similarities between triangles ABC and BWZ
rrusczyk (19:07:57)
Right angles and parallel lines often give us similar triangles. We have WZ || AC, so triangle BWZ ~ BAC. So, what do we know about BW and BZ?
mario_jda128 (19:08:12)
using the 3:4:5 ratio, BW is 3s/5 and BZ is 4s/5
calc rulz (19:08:19)
BW=3s/5, BZ=4s/5
jnApp66 (19:08:19)
BW/BZ = 3/4
rrusczyk (19:08:34)
Both BWZ and BAC are 3-4-5 triangles, so we have BW = 3s/5 and BZ = 4s/5. So, what other lengths can we find in terms of s?
rrusczyk (19:09:20)
There certainly are a lot of ways to do this problem.
rrusczyk (19:09:24)
Here are a couple:
calc rulz (19:08:51)
WA=5s/4
tjhance (19:09:02)
ZC=5/3*s
rrusczyk (19:10:07)
These two are using the fact that the right and left triangles are 3-4-5 triangles.
rrusczyk (19:10:10)
We can also do this:
DiscreetFourierTransform (19:09:47)
WBZ is similar to ABC, so CZ = 4 - 4s/5
rrusczyk (19:10:18)
We can also find AW and CZ in terms of s, as shown below:
rrusczyk (19:10:25)
rrusczyk (19:10:30)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/74a2e5e16640cd5d465e9187ed2c8f90.png
rrusczyk (19:10:38)
There are many ways to finish from here. Show me some.
DiscreetFourierTransform (19:11:21)
and use that so that S/(4 - (4S/5)) = 3/5
13375P34K43V312 (19:11:26)
similar triangles
rrusczyk (19:11:45)
We have ZCY ~ ACB, so we have ZY/ZC = AB/AC.
rrusczyk (19:11:49)
rrusczyk (19:11:52)
What's s?
DiscreetFourierTransform (19:12:05)
Which you can solve easily, giving S = 60/37
mario_jda128 (19:12:10)
60/37
13375P34K43V312 (19:12:27)
60/37
rrusczyk (19:12:34)
Solving this equation gives us s = 60/37.
now a ranger (19:12:47)
how else can you solve it?
rrusczyk (19:13:10)
See if you can find another solution by dropping an altitude from B to AC.
13375P34K43V312 (19:13:16)
you could use areas i think
rrusczyk (19:13:23)
Exactly.
rrusczyk (19:13:33)
Next problem:
rrusczyk (19:13:34)
rrusczyk (19:13:42)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-170191780.gif
rrusczyk (19:13:47)
How do we find expected value?
DiscreetFourierTransform (19:14:02)
It's the values of the outcomes times the probabilities of the outcomes
not_trig (19:14:10)
sum of (chance of case x amount you get per case)
rrusczyk (19:14:25)
We find expected value by finding product of the probability of each outcome times the value of each outcome, then adding all the resulting products.
rrusczyk (19:14:28)
So, what is the probability we win $2?
13375P34K43V312 (19:14:33)
1/16
mario_jda128 (19:14:36)
1/16
DiscreetFourierTransform (19:14:40)
1/4 * 1/4
rrusczyk (19:14:47)
To win, our number has to be on the bottom of both dice. We have a 1/4 probability for our number to be on the bottom of one die, so the probability it's on the bottom of both is (1/4)^2 = 1/16.
rrusczyk (19:14:53)
Should we do the win $1 or lose $1 case next?
13375P34K43V312 (19:15:04)
lose $1, it's easier to do directly
mario_jda128 (19:15:05)
lose $1
rrusczyk (19:15:12)
The lose $1 case is a little easier - what do we find?
mario_jda128 (19:15:15)
9/16
13375P34K43V312 (19:15:18)
9/16
DiscreetFourierTransform (19:15:25)
You have two dice, so you have a 3/4 chance both times of being worng
not_trig (19:15:26)
3/4x3/4
rrusczyk (19:15:33)
We lose $1 if both dice fail to have our number on the bottom. This happens with probability (1-1/4)^2 = 9/16.
rrusczyk (19:15:36)
So, what is the probability we win $1?
mario_jda128 (19:15:31)
this leaves 6/16 for winning $1
13375P34K43V312 (19:15:39)
then win $1 is the rest, 3/8
not_trig (19:15:49)
win 1: 1/4x3/4 + 3/4x1/4
BOGTRO (19:15:54)
3/8
led14pa2 (19:15:56)
6/16 or 3/8
jnApp66 (19:15:57)
2(3/4)(1/4)
rrusczyk (19:16:10)
We subtract our other two from 1 to get 1 - 1/16 - 9/16 = 3/8.
rrusczyk (19:16:11)
What is our expected value?
mario_jda128 (19:16:18)
6/16 + (1/16)*2 - 9/16 = -1/16 or B
DiscreetFourierTransform (19:16:43)
-1(9/16) + 1(6/16) + 2(1/16)
jnApp66 (19:16:54)
2(1/16)-1(9/16)+1(3/8)
tjhance (19:16:54)
(3/8)(1)+(9/16)(-1)+(1/16)2
calc rulz (19:17:01)
rrusczyk (19:17:10)
rrusczyk (19:17:21)
On to #23:
rrusczyk (19:17:30)
rrusczyk (19:17:45)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-220031396.gif
rrusczyk (19:17:59)
Those of you in the Intro Geometry class should have gotten this one - we did a problem almost exactly like this two weeks ago.
rrusczyk (19:18:06)
What do we know about the smaller pyramid and the larger pyramid that will help us?
13375P34K43V312 (19:18:13)
they are similar
tjhance (19:18:14)
they are similar
mario_jda128 (19:18:15)
similar
rrusczyk (19:18:26)
We know that the two pyramids are similar. How does this help?
LLaMaLuv40 (19:18:45)
We can create a ratio
ark_xm (19:18:45)
area to altitude ratio
rrusczyk (19:18:55)
Explain.
DiscreetFourierTransform (19:18:46)
We can use the fact that if the edgelengths are increased by a factor of A, the area is increased by a^2
not_trig (19:18:50)
ratio of sides squared = ratio of surface areas
rrusczyk (19:19:34)
Because they are similar, the ratio of their surface areas equals the square of the ratio of corresponding lengths in the figure. What such lengths should we focus on?
tjhance (19:19:45)
altitudes
DiscreetFourierTransform (19:19:47)
the original altitude, and altitude - 2
13375P34K43V312 (19:19:49)
the altitudes
Bictor717 (19:19:50)
the altitudes
rrusczyk (19:20:01)
Let's focus on the height, since that's what we care about. We let the original pyramid have height h. Then what?
humDdum (19:20:13)
THe height of the smaller one is h-2
13375P34K43V312 (19:20:19)
the other is h-2
rrusczyk (19:20:37)
We know the small pyramid has height h - 2 because the plane cuts the large pyramid 2 units from the base of the large pyramid. So, what equation do we have?
calc rulz (19:20:24)
tjhance (19:21:03)
h/(h-2)=sqrt(2)
not_trig (19:21:10)
sqrt(2)(h-2)=h
rrusczyk (19:21:20)
We know that the ratio of the height of the large pyramid to the height of the smaller pyramid is the square root of the ratio of their surface areas, so we have
rrusczyk (19:21:25)
rrusczyk (19:21:55)
Solving this equation for h gives us what?
DiscreetFourierTransform (19:22:08)
13375P34K43V312 (19:22:13)
4+2rt2
not_trig (19:22:16)
(sqrt(2)-1)h=2sqrt(2) => h=2sqrt(2)/(sqrt(2)-1)
rrusczyk (19:22:28)
We multiply both sides by h - 2 to find
rrusczyk (19:22:34)
DiscreetFourierTransform (19:22:30)
There's been a lot of similarity on this AMC10
rrusczyk (19:22:50)
Similarity is a very powerful problem solving tool!
rrusczyk (19:23:25)
rrusczyk (19:23:28)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-227013940.gif
rrusczyk (19:23:33)
What does the fact that the number is divisible by 4 tell us?
mario_jda128 (19:23:37)
(B) is berserk because 4494 isn't divisible by 4
Secret Asian (19:23:43)
the last 2 digits are divisible by 4?
DiscreetFourierTransform (19:23:44)
Clearly, it must end in ...44
whynot (19:23:44)
b is not a choice
ark_xm (19:23:49)
last 2 digits divisible by 4
rrusczyk (19:24:01)
A number is divisible by 4 if and only if the number formed by its final two digits is divisible by 4. Therefore, our number must end in 44. No number that ends in 94, 99, or 49 is divisible by 4.
rrusczyk (19:24:06)
What does the fact that the number is divisible by 9 tell us?
tjhance (19:24:18)
sum of digits is divisible by 9
not_trig (19:24:23)
sum of digits is divisible by 9
LLaMaLuv40 (19:24:26)
It's digits add up to a multiple of 9
BOGTRO (19:24:26)
THe digits add up to be a multiple of 9
humDdum (19:24:27)
They must add up to a number divisible by 9
rrusczyk (19:24:35)
A number is divisible by 9 if and only if the sum of its digits is divisible by 9. How does this help?
calc rulz (19:24:19)
There are 9 4s
tjhance (19:24:57)
the number of 4's is divisible by 9
rrusczyk (19:25:26)
All 9's are divisible by 9, so we disregard them. We need nine 4's to get a sum of 4's that is divisible by 9.
rrusczyk (19:25:33)
So, what do we know about our number?
DiscreetFourierTransform (19:25:17)
and we have at least one nine
DiscreetFourierTransform (19:25:22)
so the number is a ten digit number
mario_jda128 (19:25:40)
since there must be a 9, though, smallest is 4444444944
DiscreetFourierTransform (19:25:55)
the smallest such number is 4,444,444,944
calc rulz (19:26:01)
It has at least one 9, at least one 4, it has 9 4's in fact, and its last two digits are 44...and it's the smallest one
rrusczyk (19:26:25)
Our number has at least nine 4's, one 9 (we must have one of each), and must end in 44.
rrusczyk (19:26:45)
The smallest such number is 4444444944, so our answer is (C), 4944.
rrusczyk (19:27:02)
And now for #25:
rrusczyk (19:27:12)
rrusczyk (19:27:23)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-61460297.gif
rrusczyk (19:27:29)
This looks pretty tough. Where do we start?
DiscreetFourierTransform (19:27:39)
Make a variable substitution?
not_trig (19:27:49)
combine terms
13375P34K43V312 (19:27:53)
set equal to some integer k?
DiscreetFourierTransform (19:28:22)
rrusczyk (19:28:31)
All three of these approaches will work. (The second and third will end up being essentially the same.)
rrusczyk (19:28:44)
We'll set our expression equal to n. This lets us use some algebraic manipulation. We now wish to find all pairs of positive integers (a,b) such that there is an integer n for which
rrusczyk (19:28:47)
rrusczyk (19:28:55)
Now what?
13375P34K43V312 (19:28:54)
multiply through by 9ab
tjhance_2 (19:29:00)
multiply by 9ab
rrusczyk (19:29:12)
We're not too happy about fractions, so we can multiply through by 9ab to get rid of them.
rrusczyk (19:29:15)
rrusczyk (19:29:18)
How does this help?
rrusczyk (19:29:34)
Remember, we must have a, b, and n be positive integers.
not_trig (19:29:27)
14b^2 must be divisible by 9
DiscreetFourierTransform (19:30:08)
and 9a^2 is a multiple of b
DiscreetFourierTransform (19:30:27)
so 9 is divisible by b
rrusczyk (19:31:32)
Putting these together gives us what for options for b?
DiscreetFourierTransform (19:31:10)
which means B can only be 1, 3, or 9
13375P34K43V312 (19:31:36)
3, 9
whynot (19:31:44)
b = 1, 3, 9
calc rulz (19:31:53)
1,3,9
rrusczyk (19:32:08)
Can we eliminate b=1, as 133. . . suggests?
not_trig (19:32:24)
yes, since 3|b...
calc rulz (19:32:39)
Well yes if b^2 is a multiple of 9, b is a multiple of 3.
rrusczyk (19:32:49)
rrusczyk (19:33:11)
We also know from our equation that 9 divides 14b^2, so 9 divides b^2. Therefore, b can't be 1.
rrusczyk (19:33:17)
What about a?
13375P34K43V312 (19:33:28)
must be a factor of 14
DiscreetFourierTransform (19:33:31)
a must be a factor of 14
calc rulz (19:33:38)
a can be a divisor of 14
rrusczyk (19:34:01)
rrusczyk (19:34:47)
Now what?
13375P34K43V312 (19:34:36)
(1,3),(2,3),(7,3),(14,3) all work
DiscreetFourierTransform (19:34:41)
Now we have 2 times 4 cases to check
calc rulz (19:35:02)
Try 3 and 9 for b and see if they work
DiscreetFourierTransform (19:35:04)
Now we check each of the eight cases available: Either B = 3, or B = 9
rrusczyk (19:35:20)
We only have 8 possibilities, and we can run through them pretty quickly.
DiscreetFourierTransform (19:35:18)
B = 9 doesn't work
rrusczyk (19:35:47)
We do have to check each case - we've only limited ourselves to possibilities.
rrusczyk (19:35:55)
We have to check each possibility.
rrusczyk (19:36:24)
When we check them all (where b = 3 or 9 and a = 1, 2, 7, or 14), we find that all the ones with b=9 fail.
rrusczyk (19:36:27)
The only ones that work are (1,3), (2,3), (7,3), and (14,3), so there are 4 pairs that satisfy the problem.
13375P34K43V312 (19:36:15)
(1,9), (2,9), (7, 9), (14,9) don't work, there are four solutions
rrusczyk (19:36:38)
Exactly.
rrusczyk (19:36:46)
See if you can find another solution starting with the substitution u = a/b. This substitution has the advantage of reducing the number of variables we have to deal with by 1.
rrusczyk (19:37:16)
Now we'll look at the last 5 problems of the AMC 12 B.
rrusczyk (19:37:27)
One of these is the same as the problem we just did, so we'll skip that one.
rrusczyk (19:37:33)
rrusczyk (19:37:47)
Where should we start?
tjhance_2 (19:37:56)
find the least power of 3 above 2007
rrusczyk (19:38:10)
Which is?
DiscreetFourierTransform (19:38:17)
3^7
DiscreetFourierTransform (19:38:20)
2187
tjhance_2 (19:38:24)
3^7
rrusczyk (19:38:37)
We first observe that 3^6 < 2007 < 3^7, so we only have to deal with numbers that have 7 or fewer digits in base 3.
rrusczyk (19:38:40)
How can we count these?
calc rulz (19:39:14)
Divide into cases based on number of digits
rrusczyk (19:39:27)
We can tackle this with a little organized casework. First, the 1-digit palindromes. How many are there?
theone853 (19:39:31)
2
DiscreetFourierTransform (19:39:34)
1 and 2
tjhance_2 (19:39:39)
2
humDdum (19:39:42)
2
13375P34K43V312 (19:39:43)
2 (trivial)
rrusczyk (19:39:49)
There are 2: 1, 2.
rrusczyk (19:39:54)
How about 2-digit palindromes?
DiscreetFourierTransform (19:39:43)
then 11 and 22
theone853 (19:40:00)
2: 11, 22
calc rulz (19:40:02)
11 22
13375P34K43V312 (19:40:03)
there are 2 two digit palindromes as well
rrusczyk (19:40:10)
There are 2: 11, 22.
rrusczyk (19:40:13)
3-digit?
13375P34K43V312 (19:40:19)
2*3=6
tjhance_2 (19:40:29)
6: 2 possiblities for first and last digit, 3 possibilities for middle
humDdum (19:40:39)
6
rrusczyk (19:41:01)
We have 2 choices for the first digit, 3 for the middle digit, then the last digit must be the same as the first. That gives us 6 three-digit palindromes.
rrusczyk (19:41:05)
4-digit?
13375P34K43V312 (19:41:14)
same, 2*3
tjhance_2 (19:41:14)
6 again; this time there are two middle digits
humDdum (19:41:25)
6 again
ark_xm (19:41:28)
2*3 = 6 again
rrusczyk (19:41:33)
We could take the same approach as with the three-digit numbers: 2 choices for the first digit, 3 for the second, then the third digit matches the second and the fourth matches the first. So, there are 2*3 = 6 four-digit palindromes. Seeing that there are the same number of three-digit and four-digit palindromes, we might look for a reason they're the same. Why are they the same?
theone853 (19:41:31)
because 4-digit depends only on the 1st 2 digits, it's the same as 3-digits, 6
tjhance_2 (19:42:11)
the only difference is that there is an extra middle digit; the two middle digits must be the same
rnwang2 (19:42:15)
the palindromes are determined by the first two digits
rrusczyk (19:42:23)
The middle two digits of any four-digit palindrome are the same. If we replace these with just a single copy of that digit, we get a three-digit palindrome. For example, 1221 becomes 121 and 2002 becomes 202. This process runs in reverse, too. We can create a four-digit palindrome from any three-digit palindrome by reversing this process. So, we say that the three-digit palindromes and the four-digit palindromes are in 1-1 correspondence, and we know that there is the same number of each.
rrusczyk (19:42:34)
How many five-digit palindromes?
13375P34K43V312 (19:42:42)
2*3*3=18
calc rulz (19:42:45)
2*3*3=18
whynot (19:42:54)
2 x 3 x 3 = 18
ark_xm (19:42:56)
2 * 3 * 3 = 18
rrusczyk (19:43:01)
2 choices for the first digit, 3 for the second, 3 for the third. The fourth and fifth digits are determined by our earlier choices, so we have 2*3*3 = 18 five-digit palindromes.
rrusczyk (19:43:04)
Six-digit?
13375P34K43V312 (19:43:07)
same number of 6 digit
ark_xm (19:43:08)
same, 18
hadasah (19:43:12)
18 again
calc rulz (19:43:14)
same=18
rrusczyk (19:43:22)
Just as there is the same number of three-digit palindromes as four-digit palindromes, there are the same number of five- and six-digit palindromes. (We can just repeat the middle digit of any 5-digit palindrome to make a 6-digit palindrome.) So, there are 18 six-digit palindromes.
rrusczyk (19:43:25)
Seven-digits?
DiscreetFourierTransform (19:43:38)
2 * 3 * 3 * 3
BOGTRO (19:43:40)
2*3*3*3=54
ark_xm (19:43:40)
2 * 3 * 3 * 3 = 54
whynot (19:43:44)
2 x 3 x 3 x 3 = 54
tjhance_2 (19:43:45)
18*3=54
calc rulz (19:43:45)
2*3*3*3=54
rrusczyk (19:43:54)
2 choices for the first digit, 3 choices for each of the second, third, and fourth digits. The fifth, sixth, and seventh are determined by the third, second, and first, respectively. So, we have 2*3*3*3 = 54 seven-digit palindromes.
rrusczyk (19:44:00)
Do we keep going to 8 digits?
DiscreetFourierTransform (19:44:03)
No
BOGTRO (19:44:04)
No
humDdum (19:44:09)
no, its too big
ark_xm (19:44:10)
no because that would be over 2007
rrusczyk (19:44:24)
Are all our 7-digit palindromes ok?
DiscreetFourierTransform (19:44:29)
NO!
calc rulz (19:44:30)
But what if some of the palindromes are over 2007?
DiscreetFourierTransform (19:44:33)
Some of them are larger tha 2007
rrusczyk (19:44:40)
Not all of our seven-digit palindromes are among the first 2007 positive integers. How do find out how many are?
DiscreetFourierTransform (19:44:45)
2007 in base three is 2202100
calc rulz (19:44:49)
Find 2007 in base 3
ark_xm (19:44:56)
find what 2007 is in base 3
rrusczyk (19:45:02)
We convert 2007 to base 3, which gives us 2202100. How do we quickly count our seven-digit palindromes then?
13375P34K43V312 (19:45:30)
2210, 2211, 2212, 2220, 2221, 2222 don't work so subtact 6
DiscreetFourierTransform (19:46:20)
I'm getting six extraneous solutions...
DiscreetFourierTransform (19:46:27)
I'm not sure if that's the fastest way though...
rrusczyk (19:46:54)
We can easily list the ones that are greater than 2202100: 2222222, 2221222, 2220222, 2212122, 2211122, 2210122. The next lowest palindrome is 2202022, which is smaller than 2202100. So, there are 6 seven-digit palindromes greater than 2007. This leaves us 54 - 6 = 48 that are less than 2007.
rrusczyk (19:47:10)
(Counting the bad ones is a lot easier than counting the good ones)
rrusczyk (19:47:14)
How many palindromes total do we have?
DiscreetFourierTransform (19:47:32)
BOGTRO (19:47:37)
100
ark_xm (19:47:24)
100
13375P34K43V312 (19:47:56)
2+2+6+6+18+18+54-6=100
rrusczyk (19:48:05)
We have 2*2 + 2*6 + 2*18 + 48 = 100, which is choice (A).
rrusczyk (19:48:12)
Next up:
rrusczyk (19:48:16)
rrusczyk (19:48:30)
Where can we start to get a feel for the problem?
13375P34K43V312 (19:48:21)
diagram please
rrusczyk (19:48:41)
We certainly want to draw a diagram.
rrusczyk (19:48:45)
rrusczyk (19:48:47)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/8f3ad093b446474400fe7833db4a9be6.png
rrusczyk (19:48:51)
There's our starting set-up, with the dot representing the first midpoint on the boundary of R. What might we do now?
not_trig (19:48:32)
find the midpoints of the three medians first
DiscreetFourierTransform (19:49:02)
Draw some more lines?
humDdum (19:49:04)
try some other easy points
rrusczyk (19:49:21)
We can find a few more points on the boundary of R. When the first particle gets to B, the second will be at the midpoint of AC. When the first particle gets to C, the second particle is at the midpoint of AB. This gives us a couple more points on the boundary of R:
rrusczyk (19:49:26)
rrusczyk (19:49:27)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/7757fc38cdb66a92cea7c3ef24f0b538.png
13375P34K43V312 (19:49:17)
find some more points, looks like an equilateral triangle
DiscreetFourierTransform (19:49:35)
Aha!
rrusczyk (19:49:45)
Now we have a really good guess at what our region is. We're particularly helped by the fact that all of our guesses are rational. Had they given a couple choices that included pi, we'd have to worry about the boundary of our region being curved. But here, we can be pretty sure that our region R is triangle DEF.
13375P34K43V312 (19:49:46)
we need to prove that it's an equilateral triangle
rrusczyk (19:50:06)
Not on the AMC - but that's a good instinct.
rrusczyk (19:50:07)
Can anyone give a quick explanation why the midpoint of the segment that connects the two particles goes from D to E as the first particle goes from A to B and the second goes from M to C to N?
rrusczyk (19:50:21)
(We'll sketch a quick proof for fun)
calc rulz (19:49:02)
Set up coordinates, and parametric ones for their paths
rrusczyk (19:50:42)
Could do it this way - does anyone have a geometric solution?
DiscreetFourierTransform (19:50:30)
It's homothetic about O (center of ABC)
rrusczyk (19:51:30)
The two particles aren't, but the two triangles are.
rrusczyk (19:51:41)
rrusczyk (19:52:03)
What happens to the second particle as the first goes from A to P?
not_trig (19:52:12)
M -> C
humDdum (19:52:21)
second particle M to C
BOGTRO (19:52:21)
IT Goes to C
rrusczyk (19:52:32)
Let's look at what happens as the first particle goes from A to P and the second goes from M to C.
rrusczyk (19:52:40)
What do we know about segments MP and AC?
13375P34K43V312 (19:52:52)
2MP=AC
rrusczyk (19:53:02)
What else?
13375P34K43V312 (19:53:08)
parallel
rnwang2 (19:53:09)
they're parallel
tjhance_2 (19:53:14)
parallel
rrusczyk (19:53:20)
Lines MP and AC are parallel by the Midline Theorem. (We can prove this by noting that BP/BA = BM/BC = 1/2, so BPM ~ BAC by SAS Similarity. Therefore, <BPM = <BAC, so PM || AC.)
rrusczyk (19:53:27)
rrusczyk (19:53:36)
Since D is the midpoint of AM, it is on the line between MP and AC that is parallel to both. Since F is the midpoint of CP, it also is on the line between MP and AC that is parallel to both. As the first particle goes from A to P, it moves from the line AC to MP at the same rate as the second particle goes from line MP to line AC. Therefore, the midpoint of the segment connecting them will always be on the line parallel to AC and MP and midway between them.
rrusczyk (19:53:44)
(You could make this more formal with either Euclidean geometry or with coordinates, but this intuitive explanation should be good enough to convince us that our region R is triangle DEF.)
rrusczyk (19:53:50)
rrusczyk (19:54:07)
How can we find the ratio of the area of DEF to the area of ABC?
rrusczyk (19:54:51)
What do these triangles have in common?
tjhance_2 (19:55:00)
center
Bictor717 (19:55:04)
same center
rrusczyk (19:55:11)
Both triangles have the same center, which we'll call O:
rrusczyk (19:55:16)
rrusczyk (19:55:20)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/647c41b6c8769e90b71da4f69a9ddd15.png
rrusczyk (19:55:21)
How does this help?
DiscreetFourierTransform (19:56:29)
Segment OC = OF + FC
rrusczyk (19:57:02)
I'm going to use OA: OA = OD + AD. How does this help? What do we know about OA or AD?
tjhance_2 (19:57:21)
AD=AM/2
rrusczyk (19:57:41)
D is the midpoint of AM, so AD = AM/2.
calc rulz (19:57:49)
OA=2AM/3
rnwang2 (19:58:01)
AO = (2/3)(AM)
rrusczyk (19:58:21)
O is the centroid of both triangles. So, we know that AO = (2/3)(AM). We also know that AD = (1/2)(AM), so?
rrusczyk (19:58:27)
rrusczyk (19:59:14)
We're looking for a ratio of areas, so if we find a ratio of corresponding sides, we'll be happy. We have AO = (2/3)(AM), and now we find:
humDdum (19:58:40)
OD is 1/6 of AM
tjhance_2 (19:59:05)
(2/3)AM=(1/2)AM+OD
rrusczyk (19:59:20)
DO = AO - AD = (1/6)(AM).
rrusczyk (19:59:26)
So, how does this help?
humDdum (19:59:38)
So OD/OA is 1/4
rrusczyk (19:59:54)
We now have DO/AO = (1/6)/(2/3) = 1/4. How does this help?
humDdum (19:59:51)
and we square that ratio to get the ratio of areas
humDdum (20:00:02)
so its 1/16!
tjhance_2 (20:00:06)
ratio of areas is square of that
calc rulz (20:00:07)
This is the ratio of the circumradii of the triangles, so the area ratio is 1/16
ark_xm (20:00:11)
(1/4)^2 = Area of DEF / Area of ABC
not_trig (20:00:11)
the areas are in the ratios of the sides squared, so their ratio is 1/16
rrusczyk (20:00:20)
Triangles DFO and ACO are similar, so we have DF/AC = DO/AO = 1/4, so our desired ratio of areas is (1/4)^2 = 1/16.
rrusczyk (20:00:31)
Next up:
rrusczyk (20:00:36)
rrusczyk (20:00:49)
I forgot to include the choices. A little AIME practice ;)
rrusczyk (20:00:53)
How do we set this up?
13375P34K43V312 (20:00:54)
let the sides be x, y
jiang88 (20:01:07)
put the algebra form
rrusczyk (20:01:23)
We let the legs have lengths x and y and we write an equation. The area is simply xy/2. The perimeter is x + y + sqrt(x^2 + y^2), so we have
rrusczyk (20:01:26)
rrusczyk (20:01:31)
What's ugly here?
DiscreetFourierTransform (20:01:34)
radical@
not_trig (20:01:36)
SQUARE ROOT
humDdum (20:01:37)
the sqrt
now a ranger (20:01:38)
the radical
rrusczyk (20:01:48)
What do we do about that?
jiang88 (20:01:47)
collect the radical on one side
13375P34K43V312 (20:01:48)
the square root, so isolate it on one side
DiscreetFourierTransform (20:01:49)
isolate that puppy
DiscreetFourierTransform (20:01:52)
and square away
now a ranger (20:01:53)
square
rrusczyk (20:02:00)
The square root is ugly. The fraction is no fun, either. We get rid of the fraction, isolate the radical, then square to get rid of the radical.
rrusczyk (20:02:03)
rrusczyk (20:02:06)
rrusczyk (20:02:16)
rrusczyk (20:02:23)
Now what?
DiscreetFourierTransform (20:02:26)
Which factors nicely!
not_trig (20:02:27)
divide by xy
13375P34K43V312 (20:02:35)
divide both sides by xy
not_trig (20:02:38)
since xy>0
jiang88 (20:02:40)
divide xy
rrusczyk (20:02:45)
rrusczyk (20:02:54)
rrusczyk (20:02:59)
What do we do?
13375P34K43V312 (20:02:48)
then be cool and use simon's favorite factoring trick
DiscreetFourierTransform (20:02:56)
Use simons favorite factori trick
calc rulz (20:03:04)
SFFT
tjhance_2 (20:03:13)
simons favorite factoring trick
rrusczyk (20:03:26)
It's time for Simon's Favorite Factoring Trick!
rrusczyk (20:03:39)
How?
DiscreetFourierTransform (20:03:53)
Add 72 to both sides
13375P34K43V312 (20:03:54)
(x-12)(y-12)=72
tjhance_2 (20:03:49)
add 72 to both sides
calc rulz (20:03:51)
add 72
DiscreetFourierTransform (20:04:10)
and because we'll get xy - 12x - 12y + 144
rrusczyk (20:04:20)
We hope to factor this as (x+a)(y+b) for some numbers a and b. The coefficients of x and y in xy - 12x - 12y + 72 tell us that a = b = -12, so we think we can factor the left side as (x-12)(y-12). Expanding this gives xy - 12x - 12y + 144, so we see that we have to add 72 to both sides to give
rrusczyk (20:04:25)
DiscreetFourierTransform (20:04:39)
which becomes (x - 12)(y - 12) = 72
rrusczyk (20:05:06)
Now we can factor, and we have (x-12)(y-12) = 72.
jiang88 (20:05:19)
now solve for integral solution
rrusczyk (20:05:40)
How?
not_trig (20:05:28)
divide, and y-12 is a factor of 72
calc rulz (20:05:54)
factor 72, then x and y are 12 plus each of the factors
whynot (20:05:56)
prime factorize 72?
rrusczyk (20:06:13)
We want positive integer solutions in x and y, so we can use the possible ways to factor 72 as the product of two integers to find x and y:
rrusczyk (20:06:16)
DiscreetFourierTransform (20:06:20)
The factors of 72 are (1,72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9)
humDdum (20:06:42)
so there are 6 ways
tjhance_2 (20:06:58)
that's 6 possible solutions, one for each factorization
rrusczyk (20:07:12)
Each one of these gives us a different right triangle. For example, x - 12 = 1, y - 12 = 72 gives us x = 13, y = 84.
rrusczyk (20:07:20)
We also have to worry about factoring 72 as the product of two negative numbers, such as 72 = (-8)(-9). However, the only such product for which x and y are still positive is (-8)(-9). Solving x - 12 = -8, y - 12 = -9 gives x = 4, y = 3, which is an extraneous solution to the original equation we wrote. This scares us a little. How can we make sure none of the other solutions is extraneous?
tjhance_2 (20:07:34)
quickly check them; there are only 6
Secret Asian (20:07:47)
doesn't this have to be a triangle, so there can't be negative solutions?
rrusczyk (20:08:14)
We could check them all, but we also could note we don't want any negative solutions.
rrusczyk (20:08:18)
Or, we could look back at:
rrusczyk (20:08:24)
rrusczyk (20:08:34)
This equation shows us quickly that x = 4, y = 3 is extraneous (the right is negative, the left is positive). All our other solutions have x and y greater than 12. So, the left will be positive for each of these solutions, and these solutions are not extraneous.
rrusczyk (20:08:42)
So, we have 6 non-congruent right triangles with positive integer leg lengths such that their areas are numerically equal to 3 times their perimeters.
rrusczyk (20:09:36)
Problem 24 on the AMC 12 B is the same as Problem 25 on the AMC 10 B, which we tackled earlier.
rrusczyk (20:09:44)
So, on to Problem 25.
rrusczyk (20:09:52)
rrusczyk (20:10:02)
Oh no! 3D geometry! It must be hard.
rrusczyk (20:10:14)
Or maybe not . . .
13375P34K43V312 (20:10:08)
somehow draw a diagram
humDdum (20:10:10)
and icky diagram
rrusczyk (20:10:32)
Let's figure out how.
rrusczyk (20:10:34)
We have to find a way to think about the problem. Rather than thinking about all 5 points or all 5 segments at once, let's focus on one segment first.
rrusczyk (20:10:37)
Which segment should we focus on, and why?
13375P34K43V312 (20:10:38)
i kinda died at the drawing step
rrusczyk (20:10:59)
Don't feel bad - that's the whole problem right there, and it's not easy!!
DiscreetFourierTransform (20:10:48)
DE
DiscreetFourierTransform (20:10:54)
because it has something paralell to it?
calc rulz (20:10:59)
The fact htat ABC are in a plane and DE is parallel to them
tjhance_2 (20:11:06)
focus on DE, because it is specifically mentioned
MPS-vras (20:11:18)
focus on DE as it is parallel to a face of the figure
rrusczyk (20:11:30)
We zero in on segment DE because it has right angles at both ends (<CDE = <DEA = 90) and because it is parallel to a face of the figure.
rrusczyk (20:11:35)
So, put your 3D thinking caps on. Let's look at <CDE = 90 first. What does this tell us?
not_trig (20:11:53)
CD is perpendicular to DE? :)
guyinPA (20:11:57)
it is a right angle
rrusczyk (20:12:19)
Yes, but what does this tell us about point C?
DiscreetFourierTransform (20:12:33)
It lies in a plane with D?
rrusczyk (20:12:56)
And what do we know about that plane?
rrusczyk (20:13:26)
We know DE = 2. We know CD = 2. We know <CDE = 90. What does this tell us about the location of point C?
rrusczyk (20:14:01)
Picture in your mind: a segment DE with length 2.
rrusczyk (20:14:20)
We also know that CD = 2 and, even better, that <CDE = 90.
rrusczyk (20:14:29)
What set of points describes where C could be?
not_trig (20:14:41)
circle
DiscreetFourierTransform (20:14:43)
The locus is a circle
tjhance_2 (20:14:44)
a circle
rrusczyk (20:14:55)
And where is this circle?
tjhance_2 (20:15:02)
center of D
calc rulz (20:15:05)
perpendicular to DE
not_trig (20:15:12)
2 away from D
rrusczyk (20:15:24)
Because <CDE = 90, C is in the plane through D perpendicular to DE.
rrusczyk (20:15:35)
Because CD = 2, it is on the circle in this plane with radius 2 and center D.
rrusczyk (20:16:05)
(In your mind now should be a segment, DE, and a circle centered at D that C lives on)
rrusczyk (20:16:21)
Hmm. . . How about <DEA = 90; what does this tell us?
humDdum (20:16:36)
Another circle!
tjhance_2 (20:16:40)
A is on a circle centered at E parallel to the other circle
guyinPA (20:16:51)
circle with center e rad. 2
calc rulz (20:16:57)
A is in the plane perpendicular to DE through E
rrusczyk (20:17:12)
Because <DEA = 90, point A is in the plane through E perpendicular to D. As before, we combine this with EA = 2 to see that A is on the circle in this plane with radius 2 and center E.
rrusczyk (20:17:16)
What geometric figure should we be thinking about now?
humDdum (20:17:27)
a cylinder?
tjhance_2 (20:17:28)
cylinder
calc rulz (20:17:35)
cylinder
rrusczyk (20:17:44)
We have segment DE, two planes through D and E perpendicular to DE, and circles in each plane centered at D and E, respectively. We're talking about a cylinder!
rrusczyk (20:18:01)
rrusczyk (20:18:04)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/cfc1672c1a1368addf4a9f02e38781d1.png
rrusczyk (20:18:30)
Point C is on the circumference of the top base and point A is on circumference of the bottom base!
rrusczyk (20:18:42)
Now, what do we know about A and C that might help us decide how these are situated on the two bases?
not_trig (20:19:03)
angle ABC=90
tjhance_2 (20:19:09)
angle ABC=90
13375P34K43V312 (20:19:17)
AC=2rt2
hadasah (20:19:19)
ABC is a right angle?
rrusczyk (20:19:35)
We know that <ABC = 90 and AB = BC = 2, so AC = 2*sqrt(2). How does this help?
rrusczyk (20:19:41)
calc rulz (20:19:50)
AC is in a plane perpendicular to DE, so AC and DE are skew
rrusczyk (20:20:24)
True. But where are they? Suppose we pick a point on the top base to be C. How do we find A?
not_trig (20:20:25)
... not perpendicular... parallel
rrusczyk (20:20:44)
Right, parallel
not_trig (20:22:08)
there are not many A, C on the circles such that AC = 2sqrt(2)
rrusczyk (20:22:40)
Once we pick a point on the top base to be C, there are only two points on the bottom base that are 2*sqrt(2) away from C. (To see why, let X be the point on the bottom base that is directly below C. Since AC = 2*sqrt(2) and CX = 2, we have AX = 2, so point A must be on the circle with center X and radius 2. This circle meets the circumference of the bottom base at two points. These are the two points that could be A.)
rrusczyk (20:22:48)
Now we're close. We'll just focus on one of these two points that could be A (the other is the mirror image of A in plane CDE)
rrusczyk (20:22:55)
rrusczyk (20:22:57)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/69202cc883c186d38687d3223a49cd8c.png
rrusczyk (20:23:20)
Now what?
DiscreetFourierTransform (20:22:17)
So we need to find point B to establish the triangle BDE
rrusczyk (20:23:38)
Indeed. What do we know about B?
not_trig (20:19:57)
and ABC is parallel to DE
rrusczyk (20:24:02)
What does this tell us about where point B is?
rrusczyk (20:24:46)
Plane ABC is parallel to line DE. Can we add this plane to our diagram?
whynot (20:25:48)
yes, cutting the cylinder vertically?
rrusczyk (20:25:59)
Because plane ABC is parallel to DE, we know that B is in the plane that is through A and C, and parallel to the axis of the cylinder:
rrusczyk (20:26:04)
rrusczyk (20:26:08)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/67bc284072eb4f848e3f4889c8b2d059.png
rrusczyk (20:26:20)
B is on that plane in the back of the cylinder.
rrusczyk (20:26:22)
But where?
rrusczyk (20:26:29)
What else do we know about B?
tjhance_2 (20:26:49)
ABC is right angle, and the hypotnuse of ABC is sqrt(2)
MPS-vras (20:26:56)
vertex of a right angle
DiscreetFourierTransform (20:26:58)
BC = 2 = AB
rrusczyk (20:27:24)
What does the fact that <ABC = 90 tell us about where B is?
Bictor717 (20:28:03)
On the circle with diameter AC
rrusczyk (20:28:17)
Because <ABC = 90, we know that B is on the sphere with diameter AC. The intersection of this sphere with the plane that we know is B must be in is a circle.
rrusczyk (20:28:32)
rrusczyk (20:28:34)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/46a932d437c7e8c18240a7dc349c6874.png
rrusczyk (20:28:37)
Almost there!
rrusczyk (20:28:41)
Where is B?
whynot (20:28:51)
one of the corners of the square
tjhance_2 (20:29:07)
one of the right angles, either right above A or below C
rrusczyk (20:29:23)
Finally, we know that AB = BC = 2, which tells us that B is one of the endpoints of the diameter of this circle that is perpendicular to AC. In other words, B is one of the 2 points besides A and C where this circle hits the cylinder. Both are the same distance from DE. Here's the diagram with one of them labeled:
rrusczyk (20:29:30)
rrusczyk (20:29:31)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/c69b6cab6b7958ef2d7ec89fc7560bde.png
rrusczyk (20:29:38)
Now we're home. What is the area of BDE?
rrusczyk (20:29:43)
The radius and the height of the cylinder both are 2, so BDE is an isosceles right triangle with area (2)(2)/2 = 2.
humDdum (20:29:48)
2
13375P34K43V312 (20:30:12)
oh wow
DiscreetFourierTransform (20:30:20)
Drawing the diagram was ~90% of the work on the problem.
rrusczyk (20:30:41)
Yep. It's a 'put down your pencil and think' problem.
King Paul (20:30:33)
what a solution!
rrusczyk (20:31:03)
:) The official solution uses analytic geometry and is far less pretty, in my opinion.
rrusczyk (20:31:32)
Before taking requests for more problems, I'd like to remind everyone about our upcoming classes.
rnwang2 (20:31:30)
wait, so you came up with this solution?
rrusczyk (20:31:42)
Yep :)
rrusczyk (20:31:44)
First, for those of you who passed the AMC tests and will be taking the AIME, we will be hosting a weekend seminar for AIME preparation. This Special AIME Problem Seminar is on Saturday, March 3, and Sunday, March 4 from 3:30 - 6:30 PM ET (12:30 - 3:30 PM PT) each day. During this class we will discuss both general test-taking strategies and specific mathematical tactics for the AIME. We will discuss several problems both from past AIMEs and from other challenging competitions that exemplify some of the most common problem-solving tactics on the AIME.
rrusczyk (20:31:48)
I will be teaching the Special AIME Problem Seminar. I earned the only perfect score on the AIME in 1989 (yeah, that long ago), and I'll be sharing some of the general approaches I used to do so.
rrusczyk (20:31:52)
Next, the Intermediate Number Theory Seminar begins on Wednesday, February 28. Classes are from 7:30 - 9:00 PM ET (4:30 - 6:00 PM PT) on Wednesdays from Feb. 28 through Apr. 18. This class covers advanced number theory problem solving tactics such as those needed for success on the AIME and USAMO. The course is taught by 2-time IMO medalist Naoki Sato.
rrusczyk (20:31:56)
Finally, the in Intermediate Algebra class begins on Tuesday, March 6. Classes are from 7:30 - 9:00 PM ET (4:30 - 6:00 PM PT) on Tuesdays from Mar. 6 through May 22. This class will cover algebraic problem solving tactics that are useful for advanced AMC problems, many AIME problems, and even beginning Olympiad problems. The course is taught by 2-time IMO participant Valentin Vornicu.
rrusczyk (20:32:03)
We will be hosting a Math Jam on Monday to discuss the AIME Seminar and the Intermediate Algebra classes. You can also ask questions about the courses at the end of this Math Jam.
rrusczyk (20:32:26)
And now, back to the Math Jam.
King Paul (20:32:29)
wow 1989 is very long ago.
rrusczyk (20:32:37)
:(
rrusczyk (20:32:53)
Let's take some requests.
DiscreetFourierTransform (20:30:57)
Can I request we go over problem 20 on the 12B?
PenguinIntegral (20:32:57)
The AIME classes are great. They teach a lot about how to think in an organized fashion about problems, and I think that approach really helped me in general.
rrusczyk (20:33:09)
Thanks :)
rrusczyk (20:33:28)
rrusczyk (20:34:11)
What's your first reaction on reading this problem?
humDdum (20:34:25)
yuck
rrusczyk (20:34:33)
Mine too - this definitely isn't my favorite problem on the test.
rrusczyk (20:34:41)
Let's get over that and see what we can do. What can we determine about the first parallelogram?
DiscreetFourierTransform (20:34:46)
Find Y intercepts?
rrusczyk (20:34:59)
We see that two equations have the y-intercept (0,c) and two have the y-intercept (0,d). We have two vertices of our parallelogram. Can we determine anything easily about the other two?
not_trig (20:35:26)
it's ugly, but we can solve for x
DiscreetFourierTransform (20:35:27)
The parallelogram has two vertices that are vertically spaced
rrusczyk (20:36:02)
What is x for each of the other two points?
not_trig (20:36:39)
plus or minus (d-c)/(b-a)
DiscreetFourierTransform (20:36:39)
rrusczyk (20:37:07)
From y = ax + d and y = bx + c, we have ax + d = bx + c, so x = (c-d)/(b-a) for one of our points.
rrusczyk (20:37:12)
From y = ax + c and y = bx + d, we have ax + c = bx + d, so x = -(c-d)/(b-a) for the last vertex.
rrusczyk (20:37:13)
Do we need to find the y-coordinates of these other two vertices?
DiscreetFourierTransform (20:37:54)
No
rrusczyk (20:38:03)
No! These vertices are on opposite sides of the y-axis because their x-coordinates are negatives. The first two vertices we found are on the y-axis, so we can view our parallelogram as two triangles, both with the segment from (0,c) to (0,d) as a base.
DiscreetFourierTransform (20:38:06)
We have two triangles formed by the x coords and the y intercepts
rrusczyk (20:38:14)
The height of each of these triangles to this base is |(c-d)/(b-a)|, which we get from the x-coordinates of the last two vertices of the parallelogram we found.
rrusczyk (20:38:17)
So, in terms of a, b, c, and d, what is the area of the parallelogram?
not_trig (20:38:35)
(d-c)^2/(b-a)
not_trig (20:38:42)
assuming b>a
DiscreetFourierTransform (20:38:59)
DiscreetFourierTransform (20:39:21)
so multiply that by two
rrusczyk (20:39:30)
rrusczyk (20:39:33)
This equals 18, so we have
rrusczyk (20:39:38)
rrusczyk (20:39:42)
What does doing this same analysis on the other parallelogram give?
DiscreetFourierTransform (20:40:30)
rrusczyk (20:40:43)
The only difference between the parallelograms is the area and we have -d in the second set of equations where we have +d in the first set of equations. So, we replace 18 with 72 and +d with -d in our final equation above to get a second equation for the second parallelogram:
rrusczyk (20:40:48)
rrusczyk (20:40:55)
What do we do with our two equations?
PenguinIntegral (20:41:16)
Subtract for difference of squares?
tjhance_2 (20:41:23)
subtract equations
rrusczyk (20:41:31)
Subtracting the first equation from the second knocks out the c^2 and d^2 and leaves 54|b -a| = 4cd, so 27|b-a| = 2cd.
rrusczyk (20:41:41)
We want to minimize a+b+c+d, remember.
rrusczyk (20:41:52)
What does our equation tell us about a, b, c, and d?
DiscreetFourierTransform (20:42:03)
|b-a| is even
not_trig (20:42:04)
and 27|cd
rrusczyk (20:42:21)
From 27|b-a| = 2cd, we know that b-a is even and cd is a multiple of 27
rrusczyk (20:42:30)
So, how do we minimize a+b+c+d?
not_trig (20:42:45)
let cd = 27, c=3, d=9; let b=3, a=1
rrusczyk (20:43:09)
Therefore, we minimize a+b by letting a and b be 1 and 3 and we minimize c+d by letting c and d be 3 and 9. So, our minimum is 1 + 3 + 3 + 9 =16. (Our problem is satisfied when we let (a,b,c,d) = (1,3,3,9).)
rrusczyk (20:43:32)
Any more requests?
King Paul (20:43:36)
19 on 10B
rrusczyk (20:44:15)
rrusczyk (20:44:32)
DiscreetFourierTransform (20:44:56)
We need to look at remainders
rrusczyk (20:45:59)
What cases do the shaded squares correspond to?
PenguinIntegral (20:46:15)
same parity?
tjhance_2 (20:46:26)
the remainders are of the same parity
Bictor717 (20:46:46)
When the remainders are of the same parity, the square is shaded. Otherwise, the square isn't.
rrusczyk (20:46:54)
The shaded squares denote the cases where both remainders are odd or both are even (i.e., same parity)
rrusczyk (20:47:14)
So, what is the probability the first remainder is even?
tjhance_2 (20:47:25)
1/3
DiscreetFourierTransform (20:47:31)
1/3
hadasah (20:47:54)
1/3
rrusczyk (20:48:02)
The first remainder is even with probability 2/6 = 1/3 and odd with probability 2/3.
rrusczyk (20:48:08)
And the second remainder?
rrusczyk (20:48:26)
The second remainder is even with probability 3/6 = 1/2 and odd with probability 1/2. So, what's our answer?
DiscreetFourierTransform (20:48:23)
1/2 odd, 1/2 even
tjhance_2 (20:48:25)
1/2
King Paul (20:48:32)
1/2
DiscreetFourierTransform (20:48:50)
1/2 * 1/3 + 1/2 * 2/3
rrusczyk (20:49:02)
We have (1/3)(1/2) + (2/3)(1/2) = 1/2 for our total probability.
rrusczyk (20:49:25)
Next request.
whynot (20:44:03)
12 B, problem 19
rrusczyk (20:50:07)
rrusczyk (20:50:11)
What do we know about our cylinder?
rrusczyk (20:51:41)
What part of the rhombus will become what part of the cylinder?
tjhance_2 (20:50:58)
it has height of 6
tjhance_2 (20:52:16)
BC and AD are circumfrances of base
PenguinIntegral (20:52:10)
one side will be the bottom, and one will be the top
rrusczyk (20:52:56)
Two sides of the rhombus will be circumferences of the bases.
rrusczyk (20:53:07)
We know that the circumference of the base is 6. How does this help? What does
this tell us about the cylinder?
tjhance_2 (20:53:55)
we can find the radius from that and then thus find the volume in terms of the height
rrusczyk (20:54:24)
rrusczyk (20:54:28)
In terms of the rhombus, what is the height of the cylinder?
DiscreetFourierTransform (20:54:48)
and the altitude is 6 sin x
DiscreetFourierTransform (20:54:57)
where X = <ABC
rrusczyk (20:55:04)
Let's see why.
rrusczyk (20:55:09)
The height of the cylinder is just the distance between opposite sides of the rhombus, since BC becomes the circumference of the top and AD becomes the circumference of the bottom.
rrusczyk (20:55:18)
rrusczyk (20:55:21)
And how can we express this distance between opposite sides?
rrusczyk (20:55:51)
What should we add to the diagram?
rrusczyk (20:56:26)
We want to find the distance between opposite sides of the rhombus.
rrusczyk (20:56:37)
Nowhere in this diagram do we have that distance.
rrusczyk (20:56:41)
Where should we add it?
nasafan (20:56:48)
sqrt(AD^2-AX^2) where is a point added to the diagram forming a right angle between AX and DX
tjhance_2 (20:56:54)
extend AB, drop perpendicular to it from D to E on AB, and then you know that DAE=ABC
rrusczyk (20:57:12)
We add an altitude
rrusczyk (20:57:15)
We draw the altitude from C to AB, inspired by the fact that we must find sin <ABC:
rrusczyk (20:57:20)
rrusczyk (20:57:22)
How can we express CE?
Bictor717 (20:56:37)
6sinABC
tjhance_2 (20:58:01)
6sin(<ABC)
rrusczyk (20:58:11)
rrusczyk (20:58:16)
Almost there. How do we finish?
DiscreetFourierTransform (20:58:32)
The volume formula for a cylinder
rrusczyk (20:58:52)
rrusczyk (20:58:55)
So?
rrusczyk (20:59:37)
What do we do with this?
DiscreetFourierTransform (20:59:39)
set that equal to 6
Bictor717 (20:59:43)
Set it equal to 6
tjhance_2 (20:59:57)
and we know the volume is 6! so set that equal to that and solve
rrusczyk (21:00:20)
Setting this equal to the given volume, 6, we have sin(<ABC) = pi/9.
rrusczyk (21:00:32)
We also had a few requests for:
rrusczyk (21:00:35)
rrusczyk (21:00:46)
(This is from the B test)
rrusczyk (21:01:05)
How many ways can we choose the first block?
tjhance_2 (21:01:09)
25
PenguinIntegral (21:01:09)
25
rrusczyk (21:01:17)
Then what?
PenguinIntegral (21:01:12)
then 16
Bictor717 (21:01:29)
16 blocks left
DiscreetFourierTransform (21:01:31)
then there are 16 not in it's row or column
whynot (21:01:35)
4 x 4 for the second block
rrusczyk (21:01:42)
After choosing the first, we eliminate that row and column, leaving a 4x4 grid of blocks.
rrusczyk (21:01:50)
So, we have 16 choices.
rrusczyk (21:01:53)
Then what?
DiscreetFourierTransform (21:01:54)
and then there are 9 not in the row or columns already taken up
PenguinIntegral (21:02:01)
9
Bictor717 (21:02:02)
9 choices
rrusczyk (21:02:12)
In the same way, we now have a 3x3 grid left, so we have 9 choices.
rrusczyk (21:02:21)
So, is our answer 25 x 16 x 9?
DiscreetFourierTransform (21:02:23)
however, we can pick them in any order
DiscreetFourierTransform (21:02:30)
we need to divide by 3!
tjhance_2 (21:02:33)
no, we need to divide by the number of permutations
PenguinIntegral (21:02:39)
Because order doesn't matter
rrusczyk (21:03:26)
Exactly, we must divide by 3!, because our approach above will produce each possible group of 3 blocks 3! times, once for each order in which we can choose the 3 blocks.
rrusczyk (21:03:31)
Any more requests?
nasafan (21:03:46)
# 16 12 B?
rrusczyk (21:04:22)
This'll be the last one - it's getting late on Friday night and I have to get up early for Chapter MATHCOUNTS tomorrow. . .
rrusczyk (21:04:25)
rrusczyk (21:05:19)
What cases can we split this problem into?
Bictor717 (21:05:30)
How many different colors are used
rrusczyk (21:05:55)
We can divide the colorings into cases based on how many colors are used.
rrusczyk (21:06:15)
When you use casework, always find a way to be organized about it!
rrusczyk (21:06:20)
First, just 1 color. Obviously, we can do this in 3 ways because we have 3 colors.
rrusczyk (21:06:43)
Next, 2 colors. How many ways can we choose 2 colors to use?
tjhance_2 (21:07:05)
3 choose 2 = 3
Bictor717 (21:07:10)
3C2=3 ways
rrusczyk (21:07:17)
We have 3 ways to omit a color, so we have 3 ways to choose 2 colors.
rrusczyk (21:07:24)
What possibilities do we have to consider once we've chosen our two colors?
tjhance_2 (21:07:45)
2 sides for both colors, or one side for one color and 3 sides for the other
rrusczyk (21:07:55)
We could paint one face with one color and the other three with the other. How many ways can we do this?
tjhance_2 (21:08:05)
2
rrusczyk (21:08:14)
We have 2 ways to pick the color we use once, then only one indistinguishable way to paint the tetrahedron after that.
rrusczyk (21:08:17)
We could paint two faces with one color and the other two with the other color. How many ways can we do that?
tjhance_2 (21:08:24)
just one
Bictor717 (21:08:35)
1
rrusczyk (21:08:42)
There's only one way to do this - the two faces with color A share and edge and the two with color B share the opposite edge. Any tetrahedron with two faces with color A and two with color B fits this description, so they can all be rotated to appear identical.
rrusczyk (21:08:46)
So, how many colorings with 2 colors do we have?
Bictor717 (21:09:10)
3*(2+1)=9
rrusczyk (21:09:17)
We have three ways to pick the colors and three ways to do the coloring once the colors are chosen. So, there are 3(1+2) = 9 ways to color with 2 colors.
rrusczyk (21:09:20)
All that's left is using all three colors. How many ways can we color if we use all three colors?
tjhance_2 (21:09:38)
one color is used twice, there are 3 choices for that color
Bictor717 (21:09:45)
2 faces of 1 color, 1 each of the other colors
rrusczyk (21:10:08)
And how many ways to color the tetrahedron once we've chosen which color to duplicate?
Bictor717 (21:10:21)
1
tjhance_2 (21:10:27)
just one
rrusczyk (21:10:33)
We must use one color twice. We can choose the duplicate color in 3 ways. Once we have chosen which color to duplicate, we only have 1 way to paint the tetrahedron. For example, if we use 2 blues, we can always rotate the tetrahedron so that red is on bottom and white is facing towards us.
rrusczyk (21:10:36)
So, how many total colorings do we have?
Bictor717 (21:11:32)
3 for 3 colors, so 3+9+3=15 in total
nasafan (21:11:33)
15
rrusczyk (21:11:40)
We have 3 + 9 + 3 = 15 colorings.
rrusczyk (21:11:58)
That's it for the Math Jam tonight! Good luck to those of you who qualified for the AIME, and we'll see you in the AIME Math Jams!
