| Transcript
for the Math
Jam "2009 AMC 10/12 A Math Jam"
on Feb 11. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:00:29
Welcome to the 2009 AMC 10A/12A Math Jam!
rrusczyk19:00:35
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.
rrusczyk19:00:44
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.
rrusczyk19:01:45
There are a lot of students here! As I said, only well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
rrusczyk19:01:59
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go (and certainly not to 200+ students -- this Math Jam won't be as orderly or as thorough as our typical online classes.)!
rrusczyk19:02:20
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.
rrusczyk19:02:40
The Math Jam will proceed as follows:
rrusczyk19:02:46
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. After that, time permitting, I will take requests for some other problems for discussion.
rrusczyk19:03:07
Let's do some problems.
rrusczyk19:03:10
rrusczyk19:03:22
rrusczyk19:03:43
Where do we start?
nikeballa9619:03:57
CONNECT THE CENTERS!
aleph019:03:57
draw square between circle centers
jhangil19:03:57
draw a square with the radius of circles
brightzhu19:03:57
Draw square with vertices in the centers of the small circles
rrusczyk19:04:13
Connecting the centers of the circles forms a square.
rrusczyk19:04:19
How will we work with that square?
imakittykat2319:04:22
we can assume the radious of the smaller circles is 1
rrusczyk19:04:30
That does indeed make life a little easier.
rrusczyk19:04:37
Let's let 1 be the radius of the small circles.
imakittykat2319:04:57
the side of the square is 2
eragon9719:04:58
the square has side length 2 and diagonal 2 rt 2
GeniusAmI19:04:58
square has side length of 2
lisaz9419:04:58
so a square with side length 2
alligator11219:04:58
then the diagonal of the square is 2 sqrt(2)
gh62519:05:01
The diagonal of the square is then 2\sqrt{2}.
rrusczyk19:05:07
Then the centers of the 4 small circles form a square of side length 2.
rrusczyk19:05:12
alligator11219:05:34
and the diameter of the large circle will be 2sqrt(2) + 2
kyuubi919:05:35
the the big circle's radius should then be 1+ sqrt(2)
matheoteque19:05:35
so the diameter of the big circle is 2sqrt(2) + 2
lisaz9419:05:35
then the radius of the larger circle would be sqrt(2) + 1
rrusczyk19:06:14
imakittykat2319:06:28
how do we know the lenth of the diagonal??
rrusczyk19:06:41
thepianistalex19:06:43
45-45-90 triangle
grn_trtle19:06:43
pythagorean theorem
rrusczyk19:06:46
That's why.
rrusczyk19:06:57
So, we have the diameter of the big circle. What's its area?
MiddleSchooler19:07:19
and the big circle's area is 3+2sqrt{2}
TheWorstPlayer19:07:19
(3+ 2sqrt{2})pi
blackbelt1425319:07:19
pi(3+2rt2)
Geneva19:07:19
So the area of the big circle would be 2+2sqrt(2)+1
theprodigy19:07:19
pi*(3+2 sqrt2)
Guest19:07:22
how did we get the diameter of the big circle?
rrusczyk19:07:41
It equals a diameter of the square, plus radii of two of the small circles.
suma_milli19:07:45
We added the diagonal of the square to the radius of two of the smaller circles
prophet88619:07:47
we added the square diagonal and the radii of 2 smaller circles
rrusczyk19:07:50
Exactly.
rrusczyk19:07:55
rrusczyk19:07:59
rrusczyk19:08:07
And the areas of the small circles?
nikeballa9619:08:25
4pi
dominator1119:08:25
4pi
modx0719:08:25
4pi
thepianistalex19:08:25
4pi
TheWorstPlayer19:08:25
4pi
rrusczyk19:08:36
daman7169219:08:45
you can cancel out the pi's for the purpose of a ratio
rrusczyk19:08:48
rrusczyk19:08:59
How do we simplify?
nikeballa9619:09:23
multiply the denominator and numerator by (3-2sqrt2)
FantasyLover19:09:23
multiply by conjugate
12345678919:09:23
Then we can rationalize.
darksigma19:09:23
Now rationalize
daultimate119:09:23
Rationalize the denominator.
blackbelt1425319:09:23
rationalize the denom
charles.du19:09:23
rationalize
AIME_is_hard19:09:29
Multiply by (3-2sqrt2)/(3-2sqrt2)
rrusczyk19:09:32
And what do we get?
mz9419:10:08
4(3-2sqrt2)/1=4(3-2sqrt2)=C
alligator11219:10:08
12-8sqrt(2)
eragon9719:10:08
4(3-2 sqrt 2)
jhangil19:10:08
4(3-2sqrt2) C
themehlman19:10:08
4(3-2sqrt(2))
rrusczyk19:10:13
rrusczyk19:10:20
Answer (C)
rrusczyk19:10:31
Here's an interesting point:
alexhhmun19:10:33
We don't need to compute the areas individually. Compare radius 1 to radius 1+sqrt(2) and square the ratio to get the ration of one of the smaller circle's area to the larger, then multiply by 4.
rrusczyk19:10:55
Here's another way we can approach the problem:
mathemagician172919:11:00
you can draw a right triangle from the center of one of the small circles
rrusczyk19:11:32
If we do this, and let the big circle have radius 1, and the small circles have radius r, then you get something like this:
candyhyperalert19:11:34
what would the hypotnuse be? the diagonal of the square?
rrusczyk19:11:36
mathguy99919:11:40
How do we know it is a right triangle?
rrusczyk19:11:55
If we do this for each pair of tangent circles, you'll have 4 angles around a point.
Guest_8219:12:09
what does r represent? the radius?
rrusczyk19:12:16
Of each little circle, yes.
rrusczyk19:12:30
Now, how can you use this to find r?
Yongyi78119:12:34
And we're assuming the radius of the large circle is 1
panjia12319:12:34
1 is the radius of the large circle?
rrusczyk19:12:36
Correct.
ritwik_anand19:12:46
pythagoreas theorem
blackbelt1425319:12:46
pythagorean theorem
addicted2sharpiz19:12:46
Pythagorean Theorem
rrusczyk19:12:49
Or?
EcstaticPotter19:12:54
1-r times \sqrt 2 = 2r
mathguy99919:12:58
45 45 90
modx0719:12:58
45-45-90
rrusczyk19:13:01
Exactly.
rrusczyk19:13:11
rrusczyk19:13:20
And now, we can crank out what r is.
Guest_8219:13:28
how did you get that?
rrusczyk19:13:51
nikeballa9619:13:59
right triangles, in a right triangle with two angles equaling 45, the ratio of the sides is x-x-xsqrt2
snowboarder32119:13:59
45-45-90 the legs times sqrt(2) is the hypotenuse
rrusczyk19:14:02
Exactly.
dominator1119:14:05
How are the legs each 1-r?
matheoteque19:14:33
big radius minus little radius, and we assumed the big radius to be 1
nikeballa9619:14:33
its the large circle-which has radius 1 minus the radius of the small circle, which is r
rrusczyk19:14:35
The radius of the large circle is 1 (we can choose whatever we like for that), and the radius of the little circle is r.
suma_milli19:14:57
How did you know the triangle is a 45-45-90 triangle?
rrusczyk19:15:03
Legs have same length.
rrusczyk19:15:19
rrusczyk19:15:38
And from here, we just have arithmetic that is much like our first solution.
rrusczyk19:16:08
In the interest of not being here all night, I'll let you work that out on your own!
dominator1119:16:11
So the answer is C then?
rrusczyk19:16:26
Correct. And now, we'll move on to the next problem.
rrusczyk19:16:39
rrusczyk19:17:14
What's the key to knocking this problem off quickly?
rrusczyk19:18:43
I see a lot of you saying that it doesn't matter if we take the numbers off the dice at all -- that the probability is the same as just rolling two fair dice.
rrusczyk19:18:55
That's a pretty bold claim (that I think you'll have trouble justifying!!)
rrusczyk19:19:20
How can we think about this problem to get to the answer quickly (and accurately!)
mz9419:19:24
i think you can do this by just thinking of the problem differently: we first roll 2 empty-faced dice, then we randomly reach in the bag and pick 2 #s
samk19:19:24
change the scenario to that of pulling out two numbers out of a bag, where the choices are [1,1,2,2...6,6]
rrusczyk19:19:41
Exactly!
rrusczyk19:19:46
Imagine doing the two operations in reverse. That is, roll the blank dice first, then attach all the numbers.
rrusczyk19:19:53
Think about it -- this doesn't in any way affect the outcome of the experiment!
rrusczyk19:20:02
So all the question is really asking is the probability that two randomly drawn numbers from our bag sum to 7.
rrusczyk19:20:06
What is this probability?
darksigma19:20:41
Therefore, since 7 can be summed by any of the six numbers, we have no restricions on the first pick. However on the second, we have two choices left out of the 11 that can make the sum to 11. THerefore, the probabliiity is 2/11
nikeballa9619:20:42
good point. then, we have 2 ways to get 7 after the frist rolol, no matter what
brightzhu19:20:42
obly the second dice matters
alpha12319:20:42
the first one that we pick doesnt matter, there are 2 out of 11 choices for the second, so 2/11 D)
rrusczyk19:21:08
No matter what number we draw first, there are 2 numbers we can draw second to get a sum of 7, out of the 11 numbers left in the bag.
rrusczyk19:21:17
Therefore, the desired probability is 2/11.
rrusczyk19:21:45
Those of you who are here for the Olympiad Geometry class should leave the classroom and reload the Classroom page on the website. You should be able to access the Olympiad Geometry class now.
candyhyperalert19:21:52
That is genius.
rrusczyk19:21:59
Yeah, that's a pretty nice problem!
SemonR19:22:08
why are there two numbers?
Poincare19:22:10
i don't understand why there are two choices?
grn_trtle19:22:43
If you were to pick 1 first, you'd need 6 to make 7, and there are 2 sixes
27r19:22:46
For whatever number you drew first, there are 2 tiles of the corresponding number that will sum to seven.
rrusczyk19:22:48
There are 2 of each number, 1 through 6. If you draw k on the first draw, you have to draw 7-k on the second draw. There are two slips with 7-k on it.
KLKLKL19:22:55
I am confused why there are 11 choices
rrusczyk19:23:08
Once you draw out the first slip, there are 11 slips left for the second draw.
rrusczyk19:23:21
Let's move on to Problem 23.
rrusczyk19:23:41
leafiness019:24:06
it's a trapezoid?
jhangil19:24:07
Can we assume that it's a trapezoid?
rrusczyk19:24:15
Can anyone *prove* it's a trapezoid?
jhangil19:24:42
It has equal areas for AED and BEC
mathguy99919:24:42
Because there are equal areas.
candyhyperalert19:24:42
The bases must be the same and so must be the height => paralell sides
KLKLKL19:24:47
different sized bases and equal diagonals makes it isosceles?
billybob4219:24:54
We can add [dec] to the area of the original triangles
blackbelt1425319:24:59
since ABD and ABC have equal bases and equal areas, they have equal heights, so AB\\CD
rrusczyk19:25:08
Exactly. Here's a diagram:
rrusczyk19:25:08
rrusczyk19:25:45
Triangles ACD and BCD have the same area since both equal [EDC] plus a grey triangle, and the grey triangles have the same area.
rrusczyk19:25:49
So the height from A to CD equals the height from B to CD.
rrusczyk19:25:53
Therefore AB is parallel to CD, and thus ABCD is a trapezoid.
rrusczyk19:25:54
Now what?
alpha12319:26:28
AEB is similar to CED
vallon2219:26:28
we have similar trianges ABE and CDE
mround19:26:28
similar triangles...
addicted2sharpiz19:26:32
Note how ABE and DCE are similar triangles
rrusczyk19:26:38
Trapezoids and diagonals lead to similar triangles.
rrusczyk19:26:41
AEB and CED are similar.
rrusczyk19:26:46
How does this help?
james4l19:27:11
then AE/EC=AB/CD=3/4, so AE=3/(3+4)*14=6.
alligator11219:27:11
AB/CD=AE/CE=9/12=3/4
leafiness019:27:11
similarity ratio
EcstaticPotter19:27:11
AC is divided into AE and CE into the ratio 3:4
imakittykat2319:27:11
their sides are proportinal
rrusczyk19:27:21
Our similar triangles gives is:
rrusczyk19:27:25
*give us:
rrusczyk19:27:28
Yongyi78119:27:55
Therefore, AE = 3/7 AC = 3/7 * 14 = 6
jinglesmells99919:28:23
Total length of 14 divided by 7, gives 2, so multiply 2*3 to get 6
rrusczyk19:28:27
AIME1519:28:32
When just beginning to approach this problem how would we think of trapezoids?
rrusczyk19:28:46
I'd start from trying to draw conclusions from the equal areas.
Geniuscide19:28:53
same area triangles
rrusczyk19:29:06
The equal areas give us more equal area triangles, and then the trapezoid pops out.
Lulze19:29:08
The equal areas and the fact that we have nothing else to go on. =)
rrusczyk19:29:24
There's also a little bit of wishful thinking that might inspire you to look for it.
all4math19:29:26
How would we know this without drawing anything?
rrusczyk19:29:39
You might not, which is why you draw a diagram!
Guest_8219:29:43
i still do not understand the last step
venkataraman19:29:43
Why is it 3/7?
rrusczyk19:29:49
We have AE/EC = 3/4.
rrusczyk19:30:12
So, AE is 3 parts to EC's 4 parts, which means the whole length AC is 7 parts, so AE is 3/7 of AC.
SemonR19:30:15
waaait, why are the triangles similar?
rrusczyk19:30:26
AB and CD are parallel because ABCD is a trapezoid.
TheWorstPlayer19:30:34
I don't understand how we can just conlude that they have equal heights and bases from the fact that they are equal triangles.
rrusczyk19:30:49
We know that the bases are the same: ACD and BCD have the same base in base CD.
rrusczyk19:30:57
They have the same area, so they must have the same height.
rrusczyk19:31:28
A few of you noted that you found the solution by just assuming the trapezoid is isosceles.
rrusczyk19:31:54
You get lucky here in that it works! But you have to be careful with this tactic -- it can be very dangerous.
rrusczyk19:32:04
(It's discussed in more detail in the Intro Geometry book and class.)
Guest_8219:32:08
then how would you know that it is a trapezoid?
rrusczyk19:32:18
We showed that A and B are the same distance from CD.
MiddleSchooler19:32:22
is that a legitimate approach? because i used that...
rrusczyk19:32:54
For this problem, yes, but it doesn't constitute a proof that AE must always be 6. Basically, the statement of the problem implies that AE is always some constant value.
rrusczyk19:33:12
Let's move on to #24.
rrusczyk19:33:36
Sorry I can't address all everyone's questions -- there are over 200 people in class now, which is way larger (way, way larger) than our typical classes!
rrusczyk19:33:43
Problem 24!
rrusczyk19:33:44
rrusczyk19:33:56
How do we approach this?
Yongyi78119:34:06
The ones which do NOT contain the interior of the cube are the ones that coincide with a face.
nikeballa9619:34:06
okay, we see that the only ways for a point to NOT be contained within the plane is if all three chosen points are on the one face.
darksigma19:34:06
I used compimentary counting, recognizing the fact that the plane doesn't contain points if it has three points all on the same face
rrusczyk19:34:18
Exactly! We count what we don't want!
rrusczyk19:34:24
We only fail to get points inside the cube if the plane that we get contains one of the faces of the cube.
rrusczyk19:34:27
When does this occur?
veezbo19:34:46
what I did to this problem was all three points have to be on the same face
theprodigy19:34:46
when all 3 points are on 1 face
remy114019:34:46
The three points are on the same face.
rrusczyk19:34:50
When our three points are all vertices of the same face.
rrusczyk19:34:53
How many sets of 3 points are vertices of the same face?
nikeballa9619:35:18
when they are all on one face. this can happen four ways on each face...times six faces for a total of 24 ways.
FantasyLover19:35:18
4*6=24
brightzhu19:35:18
4 triangles on each face so 24 total
blackbelt1425319:35:18
24
rrusczyk19:35:23
There are 6 faces, and for each face, there are 4 sets of three corners of that face.
rrusczyk19:35:27
So 24 sets of 3 vertices will not contain points inside the cube
rrusczyk19:35:30
How many overall sets of 3 vertices are there?
prophet88619:35:48
8C3
blackbelt1425319:35:48
8C3=56
daveshail19:35:48
8 combination 3
FantasyLover19:35:48
8C3=56
MiddleSchooler19:35:48
8C3 = 56
mathguy99919:35:48
8C3
rrusczyk19:36:01
rrusczyk19:36:18
(If you don't know what that symbol is, check out our Intro Counting class or book!)
rrusczyk19:36:24
How do we finish?
brightzhu19:37:15
1-24/56=32/56=4/7
mcsquared19:37:15
(56-24) /56 = 4/7, so the answer is C
snowboarder32119:37:15
1-(24/56)=4/7
prophet88619:37:15
24/56=3/7, 1-(3/7)=4/7 is our probability of getting interior pts when choosing 3 vertices.
epic200619:37:19
1 - (24/56) = 4/7
rrusczyk19:37:23
rrusczyk19:37:27
Thus the probability of containing points inside the cube is 4/7. Answer (C).
rrusczyk19:37:40
Counting what you don't want -- that's one of my favorite counting strategies!
rrusczyk19:37:44
Next problem!
rrusczyk19:37:56
rrusczyk19:38:19
How do we deal with this?
jvenezuela71619:38:27
The notation 10...064 isn't very nice, so rewrite as 10^{k+2}+64
peregrinefalcon8819:38:27
we can write this as 10^(k+2)+2^6
Ignite16819:38:27
Rewrite it as I = 10^(k+2) + 64, to make it easier to think about
rrusczyk19:38:43
rrusczyk19:38:54
How does this help?
Guest_3719:39:10
how do we get that in the first place?
rrusczyk19:39:58
1000 ... 064 is a 1 with k zeros, and then 2 more digits. In other words, it is 64 more than 1 with k+2 0's after it, or 10^{k+2} + 64
rrusczyk19:40:02
Now what do we do?
jvenezuela71619:40:07
KaneHsu19:40:14
This expression can be rewritten as 2^(k+2) * 5^(k+2) +64
rrusczyk19:40:35
We care most about factors of 2, so we write 10 as 2 times 5, and we have:
rrusczyk19:40:51
rrusczyk19:40:56
Um, now what?
james4l19:41:12
Factor out 2^6: 2^6[2^(k-4)*5^(k+2)+1]
Ignite16819:41:12
Factor 2^6 out, to get (2^{k-4}*5^{k+2} + 1}
mz9419:41:12
factor out a 2^6?
charles.du19:41:12
factor out 2^6
rrusczyk19:41:20
OK, that gives us:
rrusczyk19:41:22
rrusczyk19:41:25
Now what?
panjia12319:42:18
see if the 2^(k-4)5^(k+2)+1 can be divisible by 2
rrusczyk19:42:23
Good idea. Can it?
Geniuscide19:42:29
if the expression inside parenthesis is even we have another factor of 2
alligator11219:42:29
If (2^(k-4) * 5^(k+2) +1) is even, then we get another factor of 2
math_galois19:42:35
k-4=0
aznlord133719:42:35
let k = 4, only way for inside to be even
rrusczyk19:42:47
Only if k = 4 do we have another factor of 2
rrusczyk19:43:18
If k > 4, then the factor in the parentheses is odd (if k < 4, then it's a fraction, and I_k has less than 6 twos).
mz9419:43:22
so that makes our final answer 7, or B
epic200619:43:24
Answer is 7: B
lowtopology19:43:24
So the answer is 7
rrusczyk19:43:27
Why not 8?
rrusczyk19:43:44
We've shown that it is possible to have 7 factors of 2. Why are there not 8?
doonie19:43:53
5^6+1 not divisible by 4
Kyung-Hee Cha19:43:53
you can only factor out one more two
MiddleSchooler19:43:53
because the inside of the parentheses is 15626
ckck19:44:18
5^6 ends with something 25+1
rrusczyk19:44:33
Exactly. When k =4, that second factor is 5^6 + 1, which ends in 26, which means that second factor cannot be divisible by 4.
VDLmath19:44:38
This was on the amc 12, right?
rrusczyk19:44:41
Yes.
jvenezuela71619:44:47
#18
107740019:44:48
#18
jhangil19:44:48
#18
rrusczyk19:44:53
#18, I'll bet :)
rrusczyk19:45:06
(And #25 on the AMC 10)
Guest_8219:45:11
are we done with amc 10?
rrusczyk19:45:45
Here's an interesting thought:
nikeballa9619:45:49
i have an alternate way to do it. we notice that $10^n$ is always divisible by $2^n$ but NOT $2^{n+1}$. therefore, $10^6$, but not $10^7$. by adding that lst 64, we see that it becomes divisibly by $2^7$, so our answer is 7.
rrusczyk19:45:56
I'll let you chew on that on your own.
rrusczyk19:46:13
We'll now do the AMC 12, and then take questions about other problems.
rrusczyk19:46:37
2009 AMC 12 A
rrusczyk19:46:53
A warning before we start:
rrusczyk19:47:06
Some of the math we need for the AMC 12 is much more advanced than that for the AMC 10.
rrusczyk19:47:17
I will not be able to teach all the math needed for these problems as we go.
rrusczyk19:47:29
So, those of you here for the AMC 10 may be a good bit lost on some of these problems.
rrusczyk19:47:55
I won't be able to bring you up to speed on these areas of math in one night (but give me a couple years and you'll be fine :) )
mz9419:48:02
bring it on ;]
rrusczyk19:48:06
Will do:
rrusczyk19:48:08
rrusczyk19:48:24
Yongyi78119:48:35
First, we see x^12 + x^8 + x^4 + 1 is just p(x^4)
matheoteque19:48:35
second equation = p(x^4)
Geniuscide19:48:35
well, this is just p(x^4)
rrusczyk19:48:52
How does this help?
Wikipedian133719:49:01
jvenezuela71619:49:01
So the roots of g are just the fourth roots of p, which are given.
matheoteque19:49:01
so all the roots of g(x) are fourth roots of the roots of p(x)
rrusczyk19:49:07
Naonao19:49:16
they gave us the roots of p(x) as 2009 9002 and 2009+9002i
rrusczyk19:49:19
darksigma19:49:23
Therefore, x^4 = all of the zeros
Yongyi78119:49:23
So g(x) = (x^4 - 2009 - 9002 pi i)(x^4 - 2009)(x^4 - 9002)
rrusczyk19:49:26
Any root of g must be a fourth root of one of these three roots. How does this finish the problem?
Yongyi78119:50:29
We take each of the three factors separately
rrusczyk19:50:35
OK, and what do we find?
Wikipedian133719:50:52
Ignite16819:50:52
There are a positive and negative real root of 2009 and 9002 that can work, giving us 4 real roots
Yongyi78119:50:52
For the first one, all four roots are complex
Yongyi78119:50:52
For the second and third, there are two complex roots for each.
james4l19:50:52
x^4-2009 has 2 nonreal zeros, x^4-9002 has 2 nonreal zeros, and that other equation has 4 nonreal zeros: add up to get 8.
lifeisacircle19:50:52
2009 and 9002 both have 2 imaginary roots, and 2009 + 9002pi i has 4 imaginary roots, so we have 8 imaginary roots total.
rrusczyk19:51:02
rrusczyk19:51:10
THECHAMP19:51:16
it's c
mz9419:51:17
(x^4-2009-9002pi*i) yields 4 nonreal roots, and each of the other 2 have 2 nonreal roots , so our answer is 8
Complex_Ninja19:51:17
8
rrusczyk19:51:22
That gives us a total of 2 + 2 + 4 = 8 nonreal roots. The answer is (C).
all4math19:51:46
What does g(x) means? I have sean the f(x) but not g(x).
rrusczyk19:51:54
I defined the second polynomial to be g(x)
darksigma19:51:57
Why aren't there 3 nonreal roots for the real #'s?
rrusczyk19:52:05
Positive and negative fourth roots.
rrusczyk19:52:15
And now, on to number 22.
lowtopology19:52:21
Wow...that problem really wasn't that bad, just scary at first
rrusczyk19:52:24
Indeed.
rrusczyk19:52:26
rrusczyk19:52:40
First up, what is the polygon?
undefined11719:52:50
The cross-section is a hexagon
jvenezuela71619:52:50
Hexagon.
bpgbcg19:52:50
A regular hexagon
james4l19:52:50
notice that the crossection is a regular hexagon
rrusczyk19:52:54
How can we quickly see that the polygon is a regular hexagon *without even drawing a diagram*?
Lulze19:53:29
Hexagon, as it intercepts all of the faces except for 2, so it has 6 sides.
undefined11719:53:29
the plane passes through six sides
Yongyi78119:53:29
There are 8 faces, but it's parallel to 2, so it must cut through 6 faces
matheoteque19:53:29
octahedron on a side: 6 vertically going edges
rrusczyk19:54:18
An octahedron has 8 faces. The plane doesn't intersect 2 of the faces, but intersects the other 6 faces. By symmetry, the resulting hexgon is regular.
rrusczyk19:54:31
Here's a picture that's worth those couple dozen words:
rrusczyk19:54:35
Yongyi78119:54:43
Now we only need to find the side length.
rrusczyk19:54:47
And how do we do that?
charles.du19:55:05
saide length is one half of base of triangle
james4l19:55:05
its sidelength is 1/2 (midpoints of BE, BD)
aznlord133719:55:05
midline
Geniuscide19:55:05
half of triangle length
worthawholebean19:55:10
1/2 the side length of the octahedron
bpgbcg19:55:10
It is 1/2 way from D to BC, so 1/2
rrusczyk19:55:14
Each side is formed by connecting the midpoints of the sides of an equilateral triangle with side length 1. Therefore, each side has length 1/2. (Let M and N be the midpoints of BD and CD, respectively. MDN is similar to BDC and MD = BD/2, so MN = BC/2 = 1/2.)
rrusczyk19:55:23
How can we quickly find the area of a regular hexagon with side length 1/2?
ak90909019:55:51
split into six equilateral triangles
bpgbcg19:55:52
Find the triangle and multiply by 6
Yongyi78119:55:52
So it's 6 * s^2\sqrt{3} / 4
Geniuscide19:55:52
6 equilateral triangles of length 1/2
rrusczyk19:55:58
Connecting each vertex of a regular hexagon to the center of the hexagon produces 6 equilateral triangles with side length equal to that of the hexagon.
rrusczyk19:56:04
And then we just bash.
worthawholebean19:56:15
lowtopology19:56:16
Find area of equilateral triangle and multiply by 6
MiddleSchooler19:56:16
6*sqrt{3}/4*(1/2)^2
Ignite16819:56:16
You get 3rt(3)/8, 3+3+8 = 14, E
rrusczyk19:56:25
rrusczyk19:56:28
The answer is 3+3+8=14, which is choice (E).
darksigma19:56:45
How are you supposed to picutre this?
rrusczyk19:56:50
Just as shown above.
nikeballa96_919:56:53
another intimidating but not too difficult one.
rrusczyk19:57:04
OK - now on to the intimidating *and* difficult ones :)
rrusczyk19:57:26
(Also, a reminder - I won't be taking general AMC questions until after we finish all the math)
rrusczyk19:57:32
rrusczyk19:57:40
Yikes. Where do we start?
ak90909019:58:02
reading it a few more times....
rrusczyk19:58:05
Good idea.
alpha12319:58:14
go for the 1.5 points!!!
rrusczyk19:58:22
That might be a good idea, too :)
Periodic19:58:29
draw a sample graph out
james4l19:58:29
graph them
HydraPyros19:58:29
graphj it on paper.
rrusczyk19:58:51
With a lot of cleverness, we can draw some deductions by considering the geometric transformations of the graph of y = f(x) that produce the graph of y = g(x). Unfortunately, that turns out to be the long way to do the problem (I know from experience -- if any of you have an elegant solution with this approach, please let me know!).
rrusczyk19:59:01
What clue in the problem suggests that there might not be a simple transformational solution?
worthawholebean19:59:17
the answer format
Wikipedian133719:59:17
m + n \root p
rrusczyk19:59:23
rrusczyk19:59:35
I will offer this interesting observation for you to think about later:
Complex_Ninja19:59:37
They are 180 degree rotations of eachother about (50,0)
rrusczyk20:00:06
This is how I did the problem the first time around -- but I'm not convinced it's the best way -- if you see a quick way to finish from here, let me know.
rrusczyk20:00:25
We don't have time tonight to explore why this statement is true.
rrusczyk20:00:57
If you have the Intro or Intermediate Algebra books (or if you can travel to the future and get our Precalculus book), you can read about functional transformations and learn why it is true.
nikeballa96_920:00:59
what statement?
rrusczyk20:01:08
I refer to Complex_Ninja's statement.
rrusczyk20:01:14
Now, back to our story.
rrusczyk20:01:27
We're probably not going to get away with a slick transformational solution.
rrusczyk20:01:34
So, we're going to dive into the algebra.
rrusczyk20:01:39
How do we set the problem up?
Geniuscide20:02:08
f(x)=a(x-h)^2+k
gh62520:02:09
let f(x)=ax^2+bx+c
mz9420:02:09
assume f(x)=ax^2+bx+c, brute force, and hope we have enough time?
gomath88820:02:09
g(x)=ax^2+bx+c
rrusczyk20:02:22
Which of these forms do we want to use for our quadratic, and why?
Wikipedian133720:02:41
h, k
THECHAMP20:02:41
the first one
bpgbcg20:02:41
Geniuscide's since it uses the vertex.
blackbelt1425320:02:41
Geniuscide's, since it gives us the vertex
james4l20:02:41
h-k form as it instantly gives us the vertex
rrusczyk20:02:53
We'll take the one that gives us the vertex in a nice form:
rrusczyk20:02:57
rrusczyk20:03:01
Now what? We have f. . .
blackbelt1425320:03:11
find g
skyhog20:03:11
find g
Yongyi78120:03:11
Now just substitute for g
rrusczyk20:03:14
OK:
rrusczyk20:03:19
rrusczyk20:03:26
Now what?
gomath88820:03:46
yes, and g(x) goes thru vertex of f(x)
Yongyi78120:03:48
And set x=h, y=k equal because we know (h, k) is on both graphs
Geniuscide20:03:48
plug in h,k
gh62520:03:48
g(h)=k
Wikipedian133720:03:48
(h,k) is on g
rrusczyk20:03:54
If we feel stuck, we have a couple guiding lights here: (1) Keep your eye on the ball -- that is, what do we care about in this problem, and (2) What information have we not used yet?
rrusczyk20:03:59
We haven't used the fact that the vertex of f is on the graph of g. So, the point (h,k) is on the graph of g.
rrusczyk20:04:09
rrusczyk20:04:19
OK. Now what?
rrusczyk20:04:58
I see a lot of you suggesting "solve", but it's not at all clear that we know what we want to solve for.
rrusczyk20:05:12
This is when it becomes very important to . . .
rrusczyk20:05:18
KEEP YOUR EYE ON THE BALL!!!
all4math20:05:33
I am still a little confused on how we got this?:-c
rrusczyk20:05:41
(h,k) is on the graph of y=g(x). That's where this equation comes from.
rrusczyk20:05:51
Keeping our eye on the ball, what do we have to think about?
Yongyi78120:06:01
To see the x-intercepts
SemonR20:06:01
solve for the x-intercepts
jvenezuela71620:06:01
x-intercepts
ThinkFlow20:06:03
Should we start looking at the x-intercepts?
rrusczyk20:06:14
We have a nasty equation that we have no idea what to do with.
rrusczyk20:06:16
We save it:
rrusczyk20:06:25
And we keep our eye on the ball.
rrusczyk20:06:32
We go after the x-intercepts.
rrusczyk20:06:40
Let's start with f(x).
rrusczyk20:06:44
What are the x-intercepts?
Yongyi78120:07:03
So solve for f(x) = 0
rrusczyk20:07:14
rrusczyk20:07:20
So, what are the x-intercepts?
Periodic20:07:34
a(x-h)^2+k = 0
rrusczyk20:07:41
So. . . .
matheoteque20:07:47
sqrt(-k/a)+h
mz9420:07:47
where a(x-h)^2=-k
Geniuscide20:07:47
sqrt(-k/a)+h
nikeballa96_920:07:47
sqrt{-k/a}+h
rrusczyk20:07:52
That's one solution. . . .
prophet88620:08:06
(+/-)sqrt(-k/a)+h
lowtopology20:08:06
+- sqrt
alpha12320:08:06
plus or minus
charles.du20:08:06
forgot plus or minus
KaneHsu20:08:06
or the negative sqrt(-k/a) +h
rrusczyk20:08:08
Exactly.
rrusczyk20:08:12
rrusczyk20:08:23
We'll stick that to the top of the room, too...
rrusczyk20:08:31
And g?
rrusczyk20:08:39
As a reminder:
rrusczyk20:08:44
107740020:09:19
100 minus those roots
Yongyi78120:09:19
So 100 - h +- sqrt(-k/a)
asdf420:09:19
100- the roots of F
blackbelt1425320:09:19
100-h -+ rt(-k/a)
rrusczyk20:09:24
rrusczyk20:09:37
rrusczyk20:09:46
Of course, we have no idea how to order them.
blackbelt1425320:09:54
which is biggest?
rrusczyk20:09:58
Good question.
VDLmath20:10:00
the difference of two of them is 150....
rrusczyk20:10:05
But we don't know which two :(
rrusczyk20:10:27
Stuck. .. Stuck . . . . Stuck . . . .
rrusczyk20:10:33
Focus on what you haven't used yet.
rrusczyk20:10:38
What have you not used yet?
asdf420:10:40
Give up and move on to the much easier #24 and 25
rrusczyk20:10:47
(That's what I did the first time around :) )
queenez0820:10:59
the differences between w intercepts
rrusczyk20:11:11
Haven't used that yet, but we can't do that until we know the order here. . .
Smartguy20:11:15
well we can go back to the ugly equation
darksigma20:11:15
The nasty equation we put aside
rrusczyk20:11:26
There's something else we haven't used. . .
rrusczyk20:11:33
We haven't used the nasty equation:
rrusczyk20:11:40
rrusczyk20:11:51
rrusczyk20:12:14
These are tough to deal with since they have 3 variables. Can we use our nasty equation in any helpful way?
bpgbcg20:12:48
sqrt(-k/a)=(100-2h)/sqrt(2)
107740020:12:48
sure, we can solve for k/a in terms of h
rrusczyk20:13:18
Our nasty equation will give us sqrt(-k/a), too. So maybe we can use the nasty equation to clean things up a good bit!
addicted2sharpiz20:14:20
Right, sqrt(-k/a) is equal to 1/sqrt(2) times (100-2h)
rrusczyk20:15:12
rrusczyk20:15:30
nikeballa96_920:15:48
let sqrt(-k/a)=z
rrusczyk20:16:01
skyhog20:16:05
hooray substitution
rrusczyk20:16:14
Indeed; it looks a lot less scary now.
rrusczyk20:16:46
When you see an algebraic form recur in a problem, sometimes substituting for it makes the problem much easier to see.
rrusczyk20:16:49
What do we do now?
matheoteque20:16:51
so then we can make it all in terms of t
rrusczyk20:16:58
rrusczyk20:17:09
(All we did was substitute for h here).
ThinkFlow20:17:13
So now can we order our x-intercepts?
rrusczyk20:17:17
Let those be A,B,C,D, respectively. Can we order them?
Ignite16820:17:30
Yes we can!
rrusczyk20:17:38
Somewhere, Obama is smiling.
rrusczyk20:18:11
Put inequality signs in, so I know what direction you're listing.
darksigma20:18:30
ACBD, from greatest to least?
bpgbcg20:18:31
1+1/sqrt(2)>1-1/sqrt(2)>1/sqrt(2)-1>-1-1/sqrt(2)
bbgun3420:18:31
A>C>B>D
hongsquare20:18:31
A>C>B>D
rrusczyk20:18:35
Yes. Since t is a square root, it is positive. Clearly A is largest and D is smallest. B is negative and C is positive, so we have A > C > B > D.
rrusczyk20:18:47
Now what?
VDLmath20:18:51
So take the difference of C and B and equate that to 150
mz9420:18:51
and C-B=150
rrusczyk20:19:17
Yongyi78120:20:17
And we're left with (-2 + sqrt(2))t = 150
imaginary_person20:20:17
now solve for t
ThinkFlow20:20:23
With C-B = 150, we solve for t. We use that to find A-D, which is our answer.
alligator11220:20:28
t = 75(2-sqrt(2))
rrusczyk20:20:34
Now , we just crank algebra:
rrusczyk20:20:42
bpgbcg20:21:31
t=150/(2-sqrt(2))=150+75sqrt(2)
Yongyi78120:21:42
So t = 150/(2 - sqrt(2))?
Eugene Kevin20:21:42
t=150/(2-sqrt(2))=150+75sqrt(2)
rrusczyk20:21:46
rrusczyk20:21:57
And now, we can finally finish.
Geniuscide20:22:06
remember, we just want to find A-D
Yongyi78120:22:06
And now it's bashing x_4 - x_1
rrusczyk20:22:14
We grind through some more algebra:
rrusczyk20:22:31
(It's late, I won't make you do it!)
rrusczyk20:22:32
venkataraman20:22:51
Is there an easier way?
rrusczyk20:23:09
worthawholebean posted a nice solution here:
rrusczyk20:23:41
This is similar to the way I first did it, but I didn't think to make the nice substitution worthawholebean made to simplify the computation.
rrusczyk20:24:11
That one was quite a beast. Let's look at #24
rrusczyk20:24:30
rrusczyk20:25:32
(I won't be teaching the fundamentals of logarithms on this problem -- if you don't know logs, come back to this when you do.)
worthawholebean20:25:35
Let's try taking a couple logs, see what happens.
rrusczyk20:25:39
Scaaarry!
rrusczyk20:25:42
But when I see this, I'm not scared. I'm confident. I've seen enough AMC tests to know that problems with lots of nasty notation are almost always much, much easier than they look. So, I see this, I think, "This one's not so scary."
rrusczyk20:25:45
Let's see if I'm right.
rrusczyk20:25:51
First step, get a handle on the notation.
rrusczyk20:25:57
Look at T(n) for small values of n.
panjia12320:26:01
Replace 2009 with a smaller number and try it?
rrusczyk20:26:04
What's T(2)?
Ignite16820:26:16
2^2 = 4
VDLmath20:26:16
2^2
Mewto5555520:26:16
2^T(1)=4
rrusczyk20:26:18
rrusczyk20:26:23
What about T(3)?
Ignite16820:26:40
T(3) = 2^2^2 = 2^4 = 16
VDLmath20:26:40
2^2^2
SemonR20:26:40
2^2^2
rrusczyk20:26:41
rrusczyk20:26:46
And T(4)?
bbgun3420:26:59
2^2^2^2
mathtyro20:26:59
2^2^2^2?
MiddleSchooler20:26:59
2^2^2^2
rrusczyk20:27:05
rrusczyk20:27:09
Towers of 2s. Now we know what we're dealing with.
rrusczyk20:27:13
So, let's turn to this mess:
rrusczyk20:27:17
rrusczyk20:27:22
Where do we start with this?
Wikipedian133720:27:35
log_2 B
worthawholebean20:27:35
Take a few logs, see if it simplifies
rrusczyk20:27:55
OK, how can we write log_2 B
Yongyi78120:28:14
A log_2 T(2009)
KaneHsu20:28:14
log_2 B = A log_2 T(2009)
panjia12320:28:14
rrusczyk20:28:18
rrusczyk20:28:45
Make sure you see why we only have k-1 logs left now -- one of the original k is in the argument of the final log.
rrusczyk20:28:50
Now what?
Geniuscide20:29:05
log 2 T(2009)=T(2008)
blackbelt1425320:29:05
log_2T(2009)=T(2008) right?
Wikipedian133720:29:05
rrusczyk20:29:13
SemonR20:29:29
Check this out: log2 2^2^2 = 2^2; each log takes off a level of 2
rrusczyk20:29:40
Exactly; that's what we're doing here.
rrusczyk20:29:41
rrusczyk20:29:48
Um, OK. Now what?
Yongyi78120:29:58
We can separate A and T(2008)
worthawholebean20:29:58
take another log!
rrusczyk20:30:03
And what do we get?
Ignite16820:30:26
Write the product as a sum of logs
jvenezuela71620:30:26
log(ab)=log(a)+log(b)
panjia12320:30:26
log_2 A + log_2 T(2008)
rrusczyk20:30:31
rrusczyk20:30:45
And what do we do here?
FantasyLover20:30:57
rrusczyk20:31:01
And?
Yongyi78120:31:08
But we know A = T(2009)^T(2009)
Wikipedian133720:31:08
Break A into base & exponent
Yongyi78120:31:08
So log_2 A = T(2009) * log_2 T(2009)
Ignite16820:31:08
Note log_2 A = T(2009) * T(2008)
rrusczyk20:31:14
So, the whole thing is:
worthawholebean20:31:20
T(2008)T(2009)+T(2007)
bpgbcg20:31:20
=T(2009)T(2008)+T(2007)
KaneHsu20:31:20
T(2009) T(2008) + T (2007)
jvenezuela71620:31:20
=T(2008)T(2009)+T(2007)
rrusczyk20:31:27
Here it is again in slow motion:
rrusczyk20:31:32
rrusczyk20:31:39
rrusczyk20:31:48
rrusczyk20:32:06
Um, now what?
Eugene Kevin20:32:20
get another one!
rrusczyk20:32:28
rrusczyk20:32:36
So, what do we do?
Ignite16820:32:54
The next log will get messy, because of the +T(2007), so we may need to look at bounds
Wikipedian133720:32:54
bound within two numbers
rrusczyk20:33:10
We don't need to know exactly what it is---we just need a good enough idea that we can tell how many logs base 2 we can take of it.
rrusczyk20:33:16
Can we compare it to something simpler? What makes this expression tough to deal with?
darksigma20:33:38
The +T(2007)
james4l20:33:38
T(2007)... replace with small epsilon, maybe?
rrusczyk20:33:46
That "+T(2007)" is pretty annoying.
rrusczyk20:33:57
How small an epsilon would we like to replace it with?
worthawholebean20:34:10
0!
panjia12320:34:10
zero?
rrusczyk20:34:17
We'd like it to just go away.
skyhog20:34:20
wait, what is meant by an epsilon?
rrusczyk20:34:30
(A very small number -- you can ignore it)
charles.du20:34:44
you can just do that...?
rrusczyk20:34:53
We'll see. We're doing a little wishful thinking here.
rrusczyk20:35:00
We know what we'd do without T(2007).
rrusczyk20:35:01
What?
Ignite16820:35:10
Take another log!
worthawholebean20:35:10
Take another log!
cyberspace20:35:15
product to sum again
Yongyi78120:35:15
Now product to sum again
rrusczyk20:35:19
rrusczyk20:35:32
That's what we'd do if we didn't have that annoying T(2007).
rrusczyk20:35:50
We're not sure we can do this and not affect the answer...
rrusczyk20:36:00
rrusczyk20:36:19
And this is what folks meant before by getting a bound.
rrusczyk20:36:28
(Again, make sure you see why we only have k-3 logs now -- we took one more step back.)
rrusczyk20:36:33
Now what do we do?
Wikipedian133720:36:45
get rid of T(2007) again
ThinkFlow20:36:45
but we do have the T(2007), right?
darksigma20:36:45
uh huh, but do'nt we get back in the same situation with a +T(2007) again?
Ignite16820:36:45
Another annoying +T(2007) pops up
funtwo20:36:52
its going off the side of the screen
rrusczyk20:37:02
Widen your screen. (That's what I just did.)
rrusczyk20:37:15
What do we do about that annoying +T(2007)?
bpgbcg20:37:23
Drop it.
cyberspace20:37:23
make it 0
prophet88620:37:27
replace with 0.
KaneHsu20:37:29
Can we turn it into another bound?
rrusczyk20:37:37
Well, that "compare to something simple" worked once. Let's try it again. (One of my favorite problem solving strategies is to take a tactic that worked well once in a problem and try it again immediately.)
rrusczyk20:37:59
(Again, note that what we're doing here is bounding our original expression. We aren't just replacing things with 0 willy-nilly.)
rrusczyk20:38:02
What happens now?
Yongyi78120:38:05
Now we just have log_2log_2... T(2008)
mz9420:38:05
then we just get T(2008) dont we?
rrusczyk20:38:16
And?
ThinkFlow20:38:19
so this is called replacing with an epsilon?
rrusczyk20:39:00
No - this is finding a lower bound. You can forget the epsilon thing.
Ignite16820:39:11
The T(2008) can handle 2008 log 2s, before it collapses to just 1
Wikipedian133720:39:11
now the T(2008) is smooth sailing to the end
rrusczyk20:39:15
Since T(2008) + T(2007) > T(2008), we have
rrusczyk20:39:19
rrusczyk20:39:43
This last expression is something we know how to deal with. How?
Yongyi78120:40:17
Just repeatedly do log until it reaches T(1)
cyberspace20:40:23
each log turns T(n) to T(n-1)
SemonR20:40:26
log2(T(n+1)=T(n)
skyhog20:40:26
we just use the logs to eliminate 2s in the exponent
rrusczyk20:40:32
Each time we reduce 1 log, we reduce the number inside T(k) by 1, so we ultimately get
rrusczyk20:40:36
rrusczyk20:40:41
And how many times will we get away with taking the logarithm of T(1)?
rrusczyk20:41:06
I'm seeing a lot of random guesses. Give me a reason.
Ignite16820:41:15
Twice, once to make it 1, and once more to make it 0
skyhog20:41:15
2 --> 1 --> 0
worthawholebean20:41:15
log_2 log_2 T(1)=0
rrusczyk20:41:23
rrusczyk20:41:55
So, what do we think k is?
SemonR20:42:06
so k in the last one is 2012
cyberspace20:42:06
so k-2010=2 and k=2012
VDLmath20:42:06
2012?
ytrewq20:42:06
2012
Geniuscide20:42:06
2012
rrusczyk20:42:16
We had:
rrusczyk20:42:17
rrusczyk20:42:21
Since our last string can have at most 2 logarithms, we see that k can be at most 2012.
rrusczyk20:42:24
So, is the answer 2012?
Yongyi78120:42:38
k = 2012, BUT... the + 2007's will make the final result slightly more than 0
Ignite16820:42:38
But we have to keep in mind that we it is merely a bound
cyberspace20:42:38
but this was a lower bound
TheWorstPlayer20:42:41
isnt it supposed to be greater than 2012
rrusczyk20:42:49
rrusczyk20:42:59
ThinkFlow20:43:04
Is our bound-taking valid? How do we prove it is?
rrusczyk20:43:16
We have proved this bound with all our inequalities above.
Complex_Ninja20:43:36
2012 + 1 = 2013
ZHANGWENZHONGKK20:43:36
It is 2013.
SemonR20:43:36
ic could be only 2013; is it?
rrusczyk20:43:51
Since the left side is greater than 0, we can take the logarithm one more time, for a total of 2013 logarithms on the original expression. So, the answer is (E).
ThinkFlow20:43:54
Do we know if the bound-taking results in an answer of 2013?
rrusczyk20:44:04
We have proved that the answer is at least 2013.
rrusczyk20:44:13
What have we not proved?
Eugene Kevin20:44:20
2013 is the only answer choice >2012
kthxbai20:44:20
answer choices ftw
SemonR20:44:25
thats the maximum
panjia12320:44:25
If it's more than 2013
107740020:44:25
the answer is exactly 2013
rrusczyk20:44:33
Exactly.
rrusczyk20:44:41
We are lucky here that we don't have anything larger as a choice, so we don't have to do the upper bound. How can we tackle the upper bound quickly?
nikeballa96_920:44:43
but if it WASNT a multiple choice, how wouldl we do it?
rrusczyk20:44:52
Great question. How do we tackle the upper bound?
rrusczyk20:44:56
That is, what would we compare T(2009)T(2008) + T(2007) to?
darksigma20:45:01
If it were an AIME type problem how would we do this?
rrusczyk20:45:14
Just what we're doing now. We found a lower bound. Now, we find an upper bound.
Mewto5555520:45:20
T(2010)?
ZHANGWENZHONGKK20:45:20
T(2009)*T(2008)+T(2007)<<T(2010).
skyhog20:45:20
it has to be less than T(2010)... because T(2010) is REALLY big
rrusczyk20:45:27
rrusczyk20:45:34
Does that give us our desired upper bound?
VDLmath20:45:54
yes
Eugene Kevin20:45:54
Yes
Smartguy20:45:54
yes
Ignite16820:45:54
Yes it does
darksigma20:45:54
Yes
rrusczyk20:45:59
rrusczyk20:46:07
As before, we can only take 2 logarithms of T(1), so when k = 2013, we have
rrusczyk20:46:12
rrusczyk20:46:22
Once we have a negative result, we know we can't take any more logarithms. . .
ThinkFlow20:46:26
which gives us 2013 as our only answer
kthxbai20:46:26
so it is less than 2014
rrusczyk20:46:36
Exactly. We have our lower bound, and our upper bound.
rrusczyk20:46:50
The answer is larger than 2012 and less than 2014. . .
rrusczyk20:46:58
So, it's 2013
SemonR20:47:08
why is it 2013 and not 2012?
rrusczyk20:47:15
We showed:
rrusczyk20:47:16
rrusczyk20:47:30
That left side is a little positive, so we can slip in one more logarithm!
ThinkFlow20:47:34
So, this is how we approach problems involving very large numbers? Taking bounds?
rrusczyk20:48:00
That's one strategy. It's also a strategy to try when eliminating one little addend gives you something you can work a lot easier with.
Wikipedian133720:48:06
I remember the 1970s IMO problem with the sum of digits.
rrusczyk20:48:11
That's another good example!
rrusczyk20:48:48
Something like "let S(n) be the sum of the digits of n, find S(S(S(4444^4444)))"
rrusczyk20:48:59
Anyway, we're ready for number 25.
rrusczyk20:49:09
rrusczyk20:49:17
Shall we start plugging and chugging?
worthawholebean20:49:25
NO!
Naonao20:49:25
no!
aggarwal20:49:25
no
rrusczyk20:49:29
Well, we could, but that looks to get ugly fast. (It will work, but it will take a loooong time.)
rrusczyk20:49:31
Do we see any clues in the problem that might be useful?
Geniuscide20:49:41
sum of tangents
VDLmath20:49:41
it's the tangent sum formula!
bpgbcg20:49:41
Sum of tangents!
Wikipedian133720:49:41
james4l20:49:41
iterative function is sum of tangents
Geniuscide20:49:41
sum of tangents recursion style
candyhyperalert20:49:41
Tangent (a + b)
aggarwal20:49:41
it is tangent sum formula
rrusczyk20:49:46
rrusczyk20:50:00
How does this help?
jvenezuela71620:50:38
Now we just need to add the angles
Smartguy20:50:38
just keep adding up angles sort of like fibonacci sequence
rrusczyk20:50:43
rrusczyk20:50:52
worthawholebean20:51:09
tan has range over all reals
Yongyi78120:51:09
Because the range of tan is all real numbers
rrusczyk20:51:14
rrusczyk20:51:17
Now, let's turn the crank.
rrusczyk20:51:20
rrusczyk20:51:25
rrusczyk20:51:27
Um, we sure don't want to do this all the way up to A_2009.
rrusczyk20:51:31
What do we do?
darkdieuguerre20:51:51
take the angles mod 180
Yongyi78120:51:51
Mod 180
Yongyi78120:51:51
Because tan(x+180) = tan(x)
rrusczyk20:52:07
Wikipedian133720:52:10
Let's divide by 15 for convenience, then it's mod 24.
rrusczyk20:52:18
That also would make things easier.
rrusczyk20:52:25
Now, it's just crank time.
Smartguy20:52:31
we look for a pattern
skodali20:52:32
the squence repeats
SemonR20:52:32
they loop back around
rrusczyk20:52:51
On your paper, you'll probably do this a lot neater than this:
rrusczyk20:52:52
rrusczyk20:53:01
(No pretty way to do that -- just crank away and be careful not to make a mistake.)
worthawholebean20:53:10
Note that any integer sequence recursively defined on a finte range will reepat eventually.
rrusczyk20:53:19
Good point - you know this is going to end somewhere.
Ignite16820:53:25
Note 2009 = 17 mod 24
Smartguy20:53:25
so the cycle repeats every 24th element
cyberspace20:53:25
so it repeats every 24 terms
ThinkFlow20:53:28
Then take 2009 mod 24.
kthxbai20:53:28
number in each part of pattern is 24
rrusczyk20:53:37
So, how do we finish?
Ignite16820:53:42
So we want A_17, which is 0
mz9420:53:42
so a_2009=a_17=0=A
Geniuscide20:53:42
a17 is 0
rrusczyk20:53:51
Yongyi78120:53:53
and arctan of 0 is 0
darksigma20:53:58
tan(0)=0
rrusczyk20:54:04
SemonR20:54:09
Is there really no easy way to do this? how would you do this on an actual test?
rrusczyk20:54:25
I'd do it just like we did above -- just list out, and be careful.
Guest20:54:27
you'd just have to see the tangent sum
rrusczyk20:54:47
That's pretty true. Going through a cycle of 24 with the radicals would be a nightmare.
107740020:54:49
dividing by 15 would have made it a lot faster
rrusczyk20:54:54
This would help, too.
rrusczyk20:55:13
We'll take a brief break (5 min or so), and then I'll take questions.
MrPotatoHead20:56:39
I'm a teensy bit confused, how does dividing by 15 help? And why 15?
rrusczyk20:56:51
All the angles in the sequence are multiples of 15.
rrusczyk20:57:04
We see that by noting that the greatest common divisor of 45 and 30 is 15.
rrusczyk20:57:13
So, divide by 15, and deal with smaller numbers.
HydraPyros20:57:41
about how long would #25 actually take on the real test?
rrusczyk20:58:07
If you see the tangent sum, you can knock this problem off in 1-2 minutes (but you'll want to spend another few minutes making sure you have your sequence right)
nikeballa96_920:58:22
wow...that was intenseee
rrusczyk20:58:45
No kidding. Teaching a room with 200 students with a collective IQ above 30,000 is scary.
mathtyro20:59:10
Thats average 150 IQ each? sure?...
rrusczyk20:59:15
Might even be higher.
Ignite16820:59:19
Do disguised trigonometric identities and formulas come up often?
rrusczyk20:59:37
Not terribly frequently, but they do show up from time to time.
ThinkFlow20:59:45
Should we keep an eye out for stuff in forms that we've seen before?
rrusczyk20:59:49
Exactly.
rrusczyk20:59:59
Pattern matching is crucial to solving problems.
rrusczyk21:00:03
Speaking of problems.
rrusczyk21:00:29
Let's do some more math problems, and then I'll take general questions about the AMC. Let's do the AMC 10 first.
rrusczyk21:00:38
Problem requests from the AMC 10?
rrusczyk21:01:01
We already did 21-25; if you missed them, you can review the transcript.
rrusczyk21:01:07
We'll start with #16:
rrusczyk21:01:12
rrusczyk21:01:42
What's a useful way to think about absolute values of differences?
ThinkFlow21:01:55
distances
HydraPyros21:01:55
draw a number line
Ignite16821:01:55
Distance?
MrPotatoHead21:02:03
If it was a number line, it's the distance between the two
rrusczyk21:02:11
This is a natural way to think about it.
rrusczyk21:02:26
So, |a-b| = 2 tells us that b is 2 to the left or right of a.
rrusczyk21:02:32
Does it matter which?
worthawholebean21:02:51
No.
nikeballa96_921:02:51
nope...
ThinkFlow21:02:51
not really
blackbelt1425321:02:51
nope
MrPotatoHead21:02:51
The answer asks for abs value
rrusczyk21:03:03
No, it doesn't -- one case is a mirror image of the other, and the answer just asks for the final distance between a and d.
mathguy99921:03:09
Why doesn't it matter?
nikeballa96_921:03:21
after that it matters, but right there, it does not, becuase everything can be flipped...
rrusczyk21:03:39
Every case going left is the mirror image of a case going right, so we only need to investigate going one direction on our first step.
rrusczyk21:03:45
So, say we go right from a to b.
rrusczyk21:03:52
Then, we're 2 to the right of a.
rrusczyk21:04:04
Then what?
sagsabzi21:04:21
2-7=5 2-(-1)=3 2-1=1 2-(-7)=9 5+3+1+9=18
victorzhou21:04:21
go left or right 3
Yongyi78121:04:21
The next step could be + 3 or - 3
rrusczyk21:04:56
sagsabzi cranks right through the cases.
rrusczyk21:05:04
We start at 2.
rrusczyk21:05:14
Then, we have 4 possibilities:
rrusczyk21:05:30
right 3, right 4 (which gets us to 9)
rrusczyk21:05:38
right 3, left 4 (gets us to 1)
rrusczyk21:05:48
left 3, right 4 (gets us to 3)
rrusczyk21:06:01
left 3, left 4 (gets us to -5, which is 5 from a)
nikeballa96_921:06:16
adding these, we get 18
rrusczyk21:06:20
And then we add.
theprodigy21:06:23
5+3+1+9=18 (D)
rrusczyk21:06:32
(There are lots and lots of ways to do this problem)
rrusczyk21:06:52
Next up:
rrusczyk21:06:54
rrusczyk21:07:11
How can we make this problem a lot simpler to deal with?
Yongyi78121:07:17
OK, start by fixing the number of children to be at 100
nikeballa96_921:07:17
okay. let there be 100 kids, it makes things much esier
rrusczyk21:07:24
Yep, that makes it a lot easier.
rrusczyk21:07:29
100 children.
rrusczyk21:07:34
Then what?
AIME1521:08:03
60 kids play soccer, 30 swim, 24 swim and play soccer
victorzhou21:08:03
60 soccer, 30 swim, 24 soccer and swim
vallon2221:08:03
so 60 play soccer, 30 swim, and 24 soccer players swim
rrusczyk21:08:10
60 play soccer, 30 swim, 24 of the soccer players swim. That's given to us right in the problem.
rrusczyk21:08:18
So, how do we finish?
blackbelt1425321:08:46
of 70 non-swimmers, 36 play soccer. so 51% (D)
FantasyLover21:08:47
36/70*100=about 51
funtwo21:08:47
soccer nonswimmers/nonswimmers=36/70=~51
mathtyro21:09:00
This means 36 people ONLY play soccer, and there are 70 non-swimmers, so 36/70 = a little more than 50%
Yongyi78121:09:00
Next, we know that there are 100 - 30 = 70 non-swimmers
rrusczyk21:09:02
There are 100-30 = 70 non-swimmers, and 36 non-swimmers play soccer. 36/70 is around 51%.
rrusczyk21:09:13
(You could also have drawn a Venn Diagram.)
rrusczyk21:09:34
rrusczyk21:10:00
Where do we start with this?
math_galois_5021:10:08
find the circumfences
rrusczyk21:10:15
OK, let's start with that.
sagsabzi21:10:39
200pi 2pir
107740021:10:39
200pi and 2r*pi
Guest21:10:40
so we have 200pi and 2pir
rrusczyk21:10:42
The circumference of A is 200pi and that of B is 2r*pi
rrusczyk21:10:44
So?
sigmaforcedelta21:11:29
since the circle B makes complete revolutions, r divides 100
Yongyi78121:11:29
The quotient must be an integer
mathswimmer21:11:29
Their quotient must be an integer
daultimate121:11:29
200pi over 2r*pi must be an integer.
rrusczyk21:12:19
Exactly -- if we roll B n times, it goes 2r*pi*n. If this brings us back to where we started, we have 2r*pi*n = 200pi, which means n = (200pi)/(2r*pi).
rrusczyk21:12:27
Simplifying gives n = 100/r.
rrusczyk21:12:35
So. . . r must divide 100.
GaussianIntegral21:12:45
so, since there are 8 factors of 100 less than 100, the answer is B) 8
snowboarder32121:12:45
100=2^2*5^2 so there are (2+1)(2+1) factors minus the factor 100, so 9-1=8 (B)
nikeballa96_921:12:45
r must be a NATURAL factor of 100. 100 prime factorizes to 5^2*2^2, so 3*3=9 factors. subtracting 1 (for 100), we get an answer of 8, or (b).
rrusczyk21:12:57
We can't let r be 100, so we exclude that possibility.
rrusczyk21:13:06
100 = 2^2 * 5^2 so has 9 divisors. Throw out 100, so the answer is 8.
rrusczyk21:13:15
nikeballa96_921:13:40
we see that after 5 minutes, 5 km have been made up, so lauren is 15km away.
FantasyLover21:13:40
find their speed
rrusczyk21:14:04
After 5 minutes, they are 15km apart.
HydraPyros21:14:07
find their rate first
rrusczyk21:14:10
How?
Guest21:14:19
Lauren's speed=x
Guest21:14:19
Andrea=3x
Yongyi78121:14:19
so x = 1/4 and 3x = 3/4
rrusczyk21:14:45
Exactly. Let Lauren = x, so Andrea is 3x, and we are given x + 3x = 1, so x = 1/4.
sagsabzi21:15:06
15/1/4=60
FantasyLover21:15:08
15/(1/4)=60
rrusczyk21:15:54
At a rate of 1/4 km/min, it takes 60 minutes to cover 15 km.
eragon9721:16:27
+5 min for when they were both in motion
mz9421:16:27
but dont we have to add 5 to our final answer?
nnnnmore21:16:27
then add 5 minutes
HydraPyros21:16:27
so you add 5 to 60 so your answer is 65, or (D)
rrusczyk21:16:38
They had already been riding for 5 minutes before Andrea stopped, so the answer is
smokewisps21:16:41
60+5=65
rrusczyk21:16:54
Guest21:17:10
draw the figure
rrusczyk21:17:11
rrusczyk21:17:27
And what do we see?
mathtyro21:17:39
Similar triangles
theprodigy21:17:39
similar triangles?
sigmaforcedelta21:17:39
similar triangles
rrusczyk21:17:45
What kind of similar triangles?
blackbelt1425321:17:50
3-4-5 triangles
rrusczyk21:17:56
Lots of 3-4-5 triangles.
rrusczyk21:18:02
Can we find BE?
Yongyi78121:18:36
BE = 5/3 * 4 = 20/3
alligator11221:18:36
BE is 20/3
rrusczyk21:18:38
BE is the hypotenuse of a 3-4-5 triangle with short leg 4, so BE = 4(5/3) = 20/3.
rrusczyk21:18:40
And BF?
sagsabzi21:19:00
15/4
alligator11221:19:00
BF if 15/4
helagha21:19:00
15/4
rrusczyk21:19:02
Similarly, we have BF = 3(5/4) = 15/4 from 3-4-5 right triangle BCF
HydraPyros21:19:20
just add the fractions
CubeX21:19:20
so just add those....
akessler21:19:20
adding these we get 125/12 = (C)
rrusczyk21:19:22
And we add to get 125/12, C.
ss518821:19:27
wait how do you know BE is part of a 345 triamgle?
rrusczyk21:20:08
All the triangles are similar to ABD (<CBD = 90 - <ABD = <ADB, and <BCF = 90 - <CBD = <ABD, etc)
mathtyro21:20:22
The triangle is a 3x, 4x, 5x triangle. We can see BD is the altitude, so the twice the area is 25x^2. It is also 12x^2. Therefore, x = 25/12. And EF is 5x, so 125/12.
rrusczyk21:20:29
There's another quick look at the problem.
rrusczyk21:20:53
All right, let's look at a few AMC 12 problems.
rrusczyk21:21:33
helagha21:21:45
draw
rrusczyk21:21:54
rrusczyk21:22:15
Oops. I skipped a step for you by mistake -- added all the important points and lines :(
rrusczyk21:22:40
What now?
Smartguy21:23:25
pythagorean thrm
Yongyi78121:23:25
Pythagorean Theorem
alligator11221:23:35
are C1 and C2 tangent to eachother?
joeislittle21:23:53
doesn't matter
rrusczyk21:23:54
We don't know this for sure -- but it also doesn't matter.
rrusczyk21:24:15
What do we do next? How do we set up the next step?
Complex_Ninja21:24:41
r^2 + (3-r)^2 = (r+1)^2
Illos21:24:41
$(r+1)^2 = r^2 + (3-r)^2$
rrusczyk21:24:59
Let r be the radius of one of the circles.
rrusczyk21:25:35
rrusczyk21:25:56
(Basically, what I'm doing here is showing that these two cases are essentially the same.
rrusczyk21:26:07
If you let i=1, you get the little triangle.
rrusczyk21:26:16
If you let i=2, you get the big one.)
rrusczyk21:26:27
In either case, the Pythagorean Theorem gives you the same equation:
rrusczyk21:26:44
rrusczyk21:26:48
Then what?
VDLmath21:26:58
solve for r
alligator11221:26:59
solve for r
107740021:27:09
SImplify and find the sum of the roots by Vieta's
Geniuscide21:27:09
make a quadratic out of this eqn and use vieta
sagsabzi21:27:12
sqrt (r - 3)^2 + r^2} = r + 1 (r-3)^2+r^2=r^2+2r+1 r^2-8r+8=0 8
rrusczyk21:27:27
We can simplify the equation to r^2 - 8r + 8 = 0.
rrusczyk21:27:41
What do we have to check before we just say "sum of the roots is 8"
SemonR21:27:44
what's vieta's?
Geniuscide21:27:44
vietas formulas
rrusczyk21:27:58
The way to find the sum of the roots from the coefficients -- look it up on the wiki.
107740021:28:00
both roots are positive
rrusczyk21:28:15
Right -- a quick check with the quadratic formula reveals that they are.
rrusczyk21:28:26
So, the sum of the possible values of r is 8.
sagsabzi21:28:33
sum=-b/a product=c/a vietas formula
Chopstiks21:28:33
you mean -b/a?
prophet88621:28:33
is ax^2+ bx+c, sum of roots is -b/a, so the sum is -(-8)/1=8
rrusczyk21:28:40
That's what people mean by Vieta.
rrusczyk21:28:49
Next problem?
rrusczyk21:29:07
rrusczyk21:29:28
Someone give me a 1-line solution :)
Geniuscide21:29:58
area= pi(R^2-r^2)
Geniuscide21:29:58
but R^2-r^2=1, so A=B
rrusczyk21:30:13
That'll work.
rrusczyk21:30:38
rrusczyk21:30:59
rrusczyk21:31:46
And, in each case, we can build a right triangle with R as the hypotenuse (radius to a vertex) and r as a leg (radius of small circle to midpoint of a side). Then we use the Pythagorean Theorem to see that R^2 - r^2 = 1.
rrusczyk21:31:49
And we're done.
darksigma21:32:02
oh I see, we draw the triangle from center to vertex to corresponding midpoint and then we get a right triagnle with legnths R, r, and 1
rrusczyk21:32:05
Exactly.
rrusczyk21:32:10
So, the areas are the same (C)
addicted2sharpiz21:32:12
Is this an example of the invariance concept?
rrusczyk21:32:38
Essentially, yes, we have found an invariant. If we take any regular polygon with side length 2 and do this, we get the same area:
ThinkFlow21:32:40
The area of an annulus is dependent solely on the chord tangent to the inner circle. Since in both cases, this chord is the same (both side lengths are 2), the area of the annuli are the same, so C.
Smartguy21:33:18
and #11?
rrusczyk21:33:31
Unfortunately, I don't have the diagram handy for number 11.
rrusczyk21:34:08
Let me see if I can find it.
rrusczyk21:34:27
rrusczyk21:34:50
(Special thanks to worthawholebean for putting that together.)
mathguy99921:34:53
Its an arithmetic sequence!
Geniuscide21:34:53
i found out it was an arithmetic progression
rrusczyk21:34:56
Exactly.
rrusczyk21:35:29
. . . but what is an arithemetic sequence?
ThinkFlow21:35:46
the diamonds added
Duelist21:35:50
The differences
rrusczyk21:35:53
Eactly.
rrusczyk21:36:02
We add 4, then add 8, then add 12, and so on:
FantasyLover21:36:05
1+(4+8+12+...)
Geniuscide21:36:07
the diamonds in each new ring is 4 more than the last one
rrusczyk21:36:31
Basically, we solve this by just finding the relationship between each figure and the next.
rrusczyk21:36:58
Once we see that relationship, we see the pattern: 1 + 4 + 8 + 12 + 16 + . . .
rrusczyk21:37:07
And now, it's a problem we know how to handle.
rrusczyk21:37:10
How do we finish?
vallon2221:37:57
1+4(1+2+3....+19) = 1+4(190)=1+760=761
alligator11221:37:57
1 + 4(1+2+3+...+19) = 1+4*190 = 761
epic200621:37:57
1 + 4(1 + 2 + 3 + ... + 19) = 1 + 4 * 19 * 20/2 = 760 + 1 = 761
FantasyLover21:38:01
1+4(1+2+...+19)=761
rrusczyk21:38:04
We have 1 + 4 + . . . + 4*19 = 1 + 4(1+2+ . . . + 19) = 1 + 4 * 20*19/2 = 761.
rrusczyk21:38:43
I see a couple requests for #20 -- that's the same as #23 in the AMC 10, which you'll be able to find in the transcript an hour or so after we finish.
rrusczyk21:39:25
We'll do one more. . .
rrusczyk21:39:31
14. A triangle has vertices (0; 0), (1; 1), and (6m; 0), and the line y = mx divides
the triangle into two triangles of equal area. What is the sum of all possible
values of m?
yankeesrule00721:40:28
Find midpoint of (6m,0) and (1,1).
Geniuscide21:40:28
find midpoint of (1,1) (6m,0)
rrusczyk21:40:32
Why does that help?
mz9421:41:42
cuz then the line y=mx automatically ivides it into 2 equal aread pieces
nikeballa96_921:41:43
the bases must be equal because the heights will automatically be equal
Smartguy21:41:43
the mediam of the triangle devides the area in half
rrusczyk21:41:44
The line y = mx goes through one vertex. If a line through a vertex of a triangle divides the triangle into two pieces of equal area, it must go through the midpoint of the opposite side (thus dividing the triangle into two triangles that have bases that are the same length, and the same altitude from the original vertex to that base).
rrusczyk21:41:55
So, we know that our line must go through the point.
rrusczyk21:42:18
rrusczyk21:42:20
How do we finish?
dgreenb80121:42:55
Substitue the values in, then use vietas
SemonR21:42:55
plug it into y=mx
rrusczyk21:43:06
Yep; we plug it in, and we have
addicted2sharpiz21:43:12
1/(6m+1)=m
rrusczyk21:43:25
1 = (6m+1)(m). (The 2's cancel)
Geniuscide21:44:00
6m^2+m-1=0, so by vieta, it's -1/6
nikeballa96_921:44:00
1=6m^2+m
rrusczyk21:44:01
Rearranging this gives 6m^2 + m - 1 = 0.
smokewisps21:44:04
6m^2+m-1=0
nikeballa96_921:44:04
6m^2+m-1=0
rrusczyk21:44:21
From here we can factor, or just note that the sum of the roots is -1/6.
rrusczyk21:44:47
Factoring gives (3m-1)(2m+1) = 0, and let's us find m and check our work :)
epic200621:44:55
Do you mean viete instead of vieta?
rrusczyk21:45:02
Perhaps if they were in Europe they might :)
rrusczyk21:45:35
We're nearly 3 hours in, so let's take a break from the math.
rrusczyk21:46:17
I'll now take a few general questions about the AMC if you have any.
theprodigy21:46:20
how long is this?
rrusczyk21:46:36
The math is over. I'll take a few more general questions, and then wrap it up.
ThinkFlow21:46:39
How hard is the AMC compared to Mathcounts?
rrusczyk21:47:06
Less speed intensive in the AMC. The problems on the AMC-10 are often about the same as Nationals. AMC 12 is much, much harder.
Geniuscide21:47:10
i wanna take AMC 12B, but my school doesnt offer it
Geniuscide21:47:10
where can i take it again?
rrusczyk21:47:18
Contact the AMC -- they can help you find a place.
THECHAMP21:47:22
what AOPS books help the most for AMC 10?
rrusczyk21:47:34
Now that's a question I like :) Volume 1 and the Intro books.
nikeballa96_921:47:45
what should my goal on aime be if i really want to make usamo and i got a 144?
rrusczyk21:48:17
Impossible to say since the difficulty of the AIME varies so much. Do your best. You're a lock if you get a 10. You won't get there with a 3. You might with a 6.
Naonao21:48:25
Will a 99 on AMC12A make AIME?
rrusczyk21:48:41
Hard to say. Some years it does, some years it doesn't. Take the B date if you can.
SemonR21:48:44
are we really not allowed to bring calculators of any sort this year?
rrusczyk21:48:46
Correct.
sagsabzi21:48:51
what is a normal score for a 7th grader on the AMC 10
rrusczyk21:49:00
There are stats for past years on the AMC site
Brut3Forc321:49:05
When they say that USAMO participants are proportionally selected by A and B dates, and you take both, which one do you count in? (The one you scored higher on?)
rrusczyk21:49:09
The higher.
mz9421:49:15
will 126 be good enough to be relatively close to having a chance at making usamo?
rrusczyk21:49:17
Sure.
alexk21:49:29
Would you say it is easier to make it to the AIME by taking AMC 10 or 12?
rrusczyk21:49:44
Historically, it has been easier to qualify for AIME through the AMC 12.
ArtofMath21:49:50
Is Pascal harder than AMC 8
rrusczyk21:49:58
I don't know the Canadian tests, sorry!
KaneHsu21:50:06
Based on the past, are AMC 12B's harder than AMC 12A's?
rrusczyk21:50:12
No; they try to make them the same.
ThinkFlow21:50:24
Can everyone take all levels of the AMC?
rrusczyk21:50:33
Not sure what you mean. Seniors can't take the AMC 10.
rrusczyk21:50:37
(For example)
mathemagician172921:50:39
Are there any books specifically for number theory for the AMC 10,12?
rrusczyk21:50:50
Our Introduction to Number Theory book will help with that.
mathguy99921:50:55
How should I prepare for the AIME?
rrusczyk21:51:02
Do lots of old AIME problems.
rrusczyk21:51:20
We also have a Special AIME Problem Seminar that is a bit of a crash course prep for the AIME.
joeislittle21:51:31
Why does the maa offer the AMC instead of just offering the AIME to everyone? Grading issues?
rrusczyk21:51:46
That, and they don't want 150,000 zeroes, which is what they'd get if they did that.
brightzhu21:51:53
Whats the best way to prepare for AIME ?
rrusczyk21:52:02
Our Volume 2 and Interm books will help too :)
Chopstiks21:52:10
Do you suggest any specific method of studying out of the AoPs Vol 1/Vol 2 books?
rrusczyk21:52:14
But nothing beats doing problems.
rrusczyk21:52:29
When working through a book, try solving the problems *before* reading how the book does them.
dgreenb80121:52:34
What is the best way to avoid mistakes on counting problems and ones such as the log base 2 problem, I make a lot of errors on those?
rrusczyk21:52:58
On counting problems, I usually try to find 2 ways to do the problem. If they agree, I'm happy. If they don't, then back to the drawing board.
rrusczyk21:53:35
I see several questions about scoring. To be honest, that changes so much that I don't know the answer! You can look it up on the AMC site.
ThinkFlow21:53:37
Where can we get AMC/AIME problems?
rrusczyk21:53:49
There are books of them, and there many on the AoPS site.
erw31421:53:54
where can you take the AMC 10 B?
rrusczyk21:54:09
If your school doesn't offer it, then you can contact the AMC to have them help you find a location.
brightzhu21:54:12
Which AoPS book is best for AIME preparation?
rrusczyk21:54:19
Volume 2 and the two Intermediate books.
CubeX21:54:41
Is this AMC 10 recommende for 6th graders
rrusczyk21:54:49
Depends on the 6th grader :)
rrusczyk21:54:58
Most 6th graders who are ready for AoPS should be taking the AMC 10
dgreenb80121:55:01
What can we do if we exhausted the database and mock aimes?
rrusczyk21:55:07
HMMT tests are very good.
rrusczyk21:55:16
(Harvard-MIT Math Tournament)
mathguy99921:55:23
Do you need to take the AMC 12 to have a good chance to qualify for USAMO?
rrusczyk21:55:34
Plenty of AMC 10 people make it to the USAMO every year.
SemonR21:55:48
what kind of score would you have to get on the AMC 10 to qualify to the AIME?
rrusczyk21:56:02
120 is automatic qualification. Lower scores sometimes qualify.
kevinw21:56:07
So.. if the Intro series are best for the AMC10's, and the Intermediate series are good for AIME's.. which would be good for the AMC12?
rrusczyk21:56:25
The intro series covers most of the AMC 12, but you'll want the Intermediate for the last 5-6 problems.
mathguy99921:56:32
Should a ninth grader take the AMC 10 or the AMC 12?
rrusczyk21:56:37
I recommend taking both.
doonie21:56:39
is everything in AOPS vol2 fair game on the AIME?
rrusczyk21:57:00
Nearly everything. Not the really fancy geometry or some of the linear algebra/limits stuff.
erw31421:57:08
just wondering, how would you prepare for usamo?
rrusczyk21:57:13
Again, lots of problems.
rrusczyk21:57:28
Our WOOT program is also designed for this. But nothing beats practice.
mathemagician172921:57:31
When practicing hard problems, how long do you reccommend we think about the problem before consulting the solutions?
rrusczyk21:57:53
Very good question. For olympiad problems, I would say 2 hours, tops. For AIME problems, 45-60 minutes.
rrusczyk21:58:10
Your time is scarce. You don't get much out of a problem in the third or fourth hour!
epic200621:58:12
How about for AMC 10 problems
rrusczyk21:58:17
30 minutes.
mathblitz21:58:22
is there anyway to prepare for the different amcs if you dont currently have th books and dont have the oney to?
rrusczyk21:58:40
Look on the AoPS site -- there are tons of problems there from the AMCs -- look in the Contests tab.
Just4Math21:58:46
Do you mean 30 min for one problem?
rrusczyk21:58:48
Max, yes
moose094421:58:54
Where can we get the problems for 2009 amc 10a
rrusczyk21:59:00
They're posted on our site now.
Brut3Forc321:59:03
mathblitz: (Or look on AoPSWiki)
rrusczyk21:59:08
On the forum and in the wiki.
syoo21:59:12
would i be able to qualify for usamo with a 136.5 on amc 12? (and decent (6-10) aime?)
rrusczyk21:59:14
Sure.
Geniuscide21:59:17
how hard is AIME compared to AMC 12
rrusczyk21:59:20
Much harder.
rrusczyk21:59:26
Consider -- passing the AMC 12 is 100.
rrusczyk21:59:40
The average on the AIME is usually in the 2-3 range (out of 15).
rrusczyk21:59:42
Much harder.
yankeesrule00721:59:57
Do you have any tips for AMC in terms of time restrictions. After the test I find that I know how to do a lot of thoe problems, except during the test I just run out of time and can't try them out.
rrusczyk22:00:14
The main tip I have is to practice under test conditions to get used to the time constraints.
rrusczyk22:00:22
This won't be as severe on the AIME, since you have 3 hours.
rrusczyk22:00:43
And it definitely won't be as bad on the USAMO, since you have 4.5, and will be ready to sleep for 8 hours at the end of that.
THECHAMP22:00:48
So do the books just give math facts or problem solving techniques?
rrusczyk22:01:15
Our books focus much more on the problem solving strategies (we include the facts, but try to teach them with a problem solving perspective.)
mathemagician172922:01:18
For some reason I do ok in practice but during the real competition I always do less than my best (anxiety and overthinking). How do I fix this?
rrusczyk22:01:40
Try to get involved in local contests that aren't as important. This will help you get over performance anxiety.
ThinkFlow22:01:57
How do you find local contests?
rrusczyk22:02:07
Ask on the Forum, and mention where you're from.
HydraPyros22:02:11
ask ur teacher
rrusczyk22:02:24
If you have a very good, active teach, that will work, too :)
SemonR22:02:26
are there any local AoPS meetings/conventions? I think meeting other awesome math geeks would be fun :-)
rrusczyk22:02:35
Summer camps are a great way to do that.
mathemagician172922:02:38
theres a directory on the wiki also
rrusczyk22:02:49
That's another place to find out about local contests.
rrusczyk22:02:52
And summer camps.
rrusczyk22:03:07
All right, it's 3 hours in, and time for dinner for me.
