| Transcript
for the Math
Jam "2008 AMC 10/12 A Math Jam"
on Feb 18. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick18:56:38
Hello and welcome to the 2008 AMC 10A/12A Math Jam!
DPatrick18:56:47
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.
DPatrick18:56:57
Please hold your questions until the end of my introduction.
DPatrick18:57:10
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.
DPatrick18:57:26
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.
DPatrick18:57:34
When I post a problem, I will also post a link to the problem, like so:
DPatrick18:57:44
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.
DPatrick18:58:11
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:
DPatrick18:58:18
DPatrick18:58:37
As you can see, 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 definitely do not complain about it.
DPatrick18:58:54
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.
DPatrick18:59:51
Feel free to ask questions as we work through the solutions, but please also keep in mind that I may not be able to respond to your question due to the large number of people here.
DPatrick18:59:59
The Math Jam will proceed as follows:
DPatrick19:00:04
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. (#22 on the 12A is also #25 on the 10A, so that'll be a total of 9 problems.) After that, time permitting, I will take requests for some other problems for discussion.
kshweh19:00:30
Will you do 22/25 twice?
DPatrick19:00:41
In fact, I will. (I'll explain more when we get to it.)
Denny103819:00:51
about how long will this math jam last?
DPatrick19:01:00
We'll run until about 8:30 Eastern.
DPatrick19:01:17
Let's get started with Problem #21 from the AMC 10A:
DPatrick19:01:25
DPatrick19:01:38
DPatrick19:01:55
oops..I forgot to include the question!
DPatrick19:02:16
The question is "What is the area of ABCD?"
arkantosstevius19:02:37
is ABCD a square?
DPatrick19:02:52
That's probably the first thing we should determine. What kind of quadrilateral is ABCD?
karatemagic719:03:01
abcd is a rhombus
williamtz19:03:01
its a rhombus
jetsmath19:03:01
ABCD is a rhombus
DPatrick19:03:25
Right. By symmetry AB=AD and BC=DC, and also by symmetry AB=BC.
DPatrick19:03:32
So the quadrilateral is a rhombus.
andersonw19:03:46
it is a rhombus, so find the length of it's diagnonals
karatemagic719:03:46
The area of a rhombus is D1*D2/2
SmitSchu19:03:46
area is half the product of the diagonals
DPatrick19:04:06
Right. The area of a rhombus is half the product of the lengths of the diagonals. So if we can find AC and BD, we're set.
helenamei19:04:30
AC is a space diagonal and BD is a *face* diagonal
DPatrick19:04:48
Right. AC is a diagonal of the entire cube, and BC is congruent to a diagonal of one of the faces.
krsattack19:04:59
D1 = sqrt2, D2 = sqrt3
mathboygenius19:04:59
by the pythagorean theorem, AC=root 3 and BD=root 2
arkantosstevius19:04:59
bd is root 2
arkantosstevius19:04:59
ac is root 3
DPatrick19:05:12
DPatrick19:05:28
SmitSchu19:05:55
mathforme19:05:55
area = sqrt (6)/2
mathboygenius19:05:55
so root 3 times root 2=root 6, but you have to multiply that by 1/2
krsattack19:05:55
Answer: A) sqrt6 / 2
DPatrick19:06:04
DPatrick19:06:40
Let's continue with #22.
DPatrick19:06:45
andrewnycpops19:07:36
Draw a tree diagram
karatemagic719:07:36
Lets list all of the possibilities
williamtz19:07:36
i used brute force
pi guy19:07:36
list possibilities
karatemagic719:07:36
Can we brute force it
DPatrick19:07:48
Since there are only 8 possibilities, we might just list them all.
helenamei19:08:00
A could not be right because the denominator has to be a power of 2.
DPatrick19:08:12
That's true: one thing you might note is that there are 8 equally likely possibilities for the 3 flips that Jacob needs to get to the 4th term. So in our probability calculation, the denominator will be 8.
DPatrick19:08:20
This rules out choices (A) and (B).
jackdillon19:08:41
could you create a chart to determine what happens?
mathboygenius19:08:41
is there a systematic way to do this
DPatrick19:09:06
Right, even though we decide to use "brute force" (that is, just listing the options), we want to be systematic about it.
DPatrick19:09:31
Let's look at the first flip.
DPatrick19:09:38
If it's H, he gets 6*2-1 = 11.
If it's T, he gets 6/2-1 = 2.
DPatrick19:10:03
OK, nothing obvious there.
karatemagic719:10:05
Why dont we make a chart
DPatrick19:10:15
Good idea. Here's a chart of the first two flips.
DPatrick19:10:26
HH gives 11*2 - 1 = 21.
HT gives 11/2 - 1 = 4.5.
TH gives 2*2 - 1 = 3.
TT gives 2/2 - 1 = 0.
jamieding5119:10:33
Or a tree
baldcypress19:10:33
we can make a tree
deveaus19:10:33
We could use a sort of tree diagram, right?
DPatrick19:10:52
Right, it's essentially the same thing, but I didn't want to draw a tree on the computer. On paper, a tree is probably a good idea.
DPatrick19:11:04
For all of these, if we flip H, we're safe.
DPatrick19:11:19
For which of these is a T flip going to produce an integer?
helenamei19:11:30
The tree diagram should kind of look like this
tail only works for TT
adchia19:11:30
TT
krsattack19:11:30
T only works on TT
DPatrick19:11:52
Right. Only if the first two flips are TT can we flip a T on the third flip and still get an integer.
DPatrick19:11:57
So to summarize:
DPatrick19:12:00
HH gives 11*2 - 1 = 21. From here only H will give an integer.
HT gives 11/2 - 1 = 4.5. From here only H will give an integer.
TH gives 2*2 - 1 = 3. From here only H will give an integer.
TT gives 2/2 - 1 = 0. From here both H and T will give an integer.
krsattack19:12:10
That means 5 out of 8 cases work
DPatrick19:12:21
Right. There are 5 successful outcomes. The probability is thus (D) 5/8.
DPatrick19:12:49
Let's look at #23:
DPatrick19:12:53
DPatrick19:13:20
This is a good example of what is sometimes called "constructive counting". We need to count in how many ways we can construct two subsets.
baldcypress19:13:31
first find the tow in the intersection
karatemagic719:13:31
Lets figure out which ones will be odne twice
CatalystOfNostalgia19:13:31
first consider which two elements are in their intersection
DPatrick19:13:48
That's how I would begin, since that is the most severe restriction on the subsets.
DPatrick19:14:02
So to start, we can choose the two elements in the intersection.
adchia19:14:09
5c2
krsattack19:14:09
5C2 = 10
baldcypress19:14:09
5C2=10
DPatrick19:14:25
Right, this can be done in C(5,2) = 10 ways.
DPatrick19:14:51
(If you don't know the notation, think of 5 choices for the first element, then 4 choices for the second, but then we divide by 2 because their order doesn't matter. So 5*4/2 = 10 choices.)
DPatrick19:15:05
Now what about the other 3 elements?
lingomaniac8819:15:24
they go to either subset 1 or subset 2
krsattack19:15:27
3 remaining elements left two place in two subsets
DPatrick19:15:47
Each of the other 3 elements has to be in exactly one of the subsets, so that the union contains all 5 elements.
DPatrick19:16:06
But each has to be in exactly one or the other, not both, so that they don't show up in the intersection.
karatemagic719:16:16
2^3 =8 is how many ways we can put the others
desperado19:16:16
2^3 so 8
DataBox19:16:16
2^3 ways to assign them to A or B
zephyredx_619:16:16
2^3=8 ways?
DPatrick19:16:37
Right, so each remaining element has 2 choices (to be in the first subset or the second subset), so that's 2^3 = 8 more choices we have to make.
arkantosstevius19:16:46
and then 10 times 8 is 80
mathboygenius19:16:46
then 8*10=80
DPatrick19:17:13
...so that's 10*8 = 80 total choices that we have to this point.
aznjonny2419:17:20
must divide by two since order doesen't matter
Chauncy123419:17:20
but, we overcounted
mathforme19:17:22
But the sets are indistinguishable, we have to halve 80.
DPatrick19:17:39
Right. The order of the subsets doesn't matter!
DPatrick19:17:48
So we have to divide by 2 since we've counted each pair of subsets twice.
srs74719:17:59
40 possiblities
sharonh19:17:59
(b) 40
DPatrick19:18:04
Thus the answer is 10*8/2 = 40. Answer (B).
DPatrick19:18:19
My feeling is that the AMC question writers gave you a little break here...
DPatrick19:18:43
...in that 80 is a common wrong answer (forgetting to divide by 2 at the end), but it's not a choice!
DPatrick19:19:02
For some people, that was a little gift from the AMC writers :)
DPatrick19:19:17
Let's move on to #24.
DPatrick19:19:23
DPatrick19:19:37
(Note: this is also #15 on the 12A)
Moldytape19:19:55
We have to find the units digit of k.
CatalystOfNostalgia19:19:55
first find the units digit of k
cc119:19:55
first find the units digit of k
DPatrick19:20:04
I would start by finding the units digit of k.
canttouchthis2319:20:29
2008^2 is going to give you a units digit of 4
adchia19:20:29
2008^2 ends with 4
DPatrick19:20:45
The units digit of 2008^2 is the same as the units digit of 8^2, so it's 4.
cc119:21:08
make a list of powers of 2 endings
andersonw19:21:08
there is a pattern for powers of 2: 2, 4, 8, 6
zhh19:21:11
2^X repeats in its units digits
DPatrick19:21:19
Right, the pattern of units digits of powers of 2 is: 2, 4, 8, 6, 2, 4, 8, 6, 2, ...
Moldytape19:21:39
2^2008 has a pattern with the units digit, 2,4,8,6,2.... so it ends in a 6
arkantosstevius19:21:39
2008 is a multiple of 4
lingomaniac8819:21:39
the 2008th term in the pattern is 6
DPatrick19:21:45
2008 is a multiple of 4, so the units digit of 2^2008 is 6.
baldcypress19:21:55
4+6=10
jianfenl19:21:55
k's unit digit is 0
popopomnmnmn_219:21:57
add 4 and 6 so k ends in 0
DPatrick19:22:02
Therefore, the units digit of k is 4+6 = 0.
DPatrick19:22:16
Of course, this means that the units digit of k^2 is 0, too.
DPatrick19:22:24
So we only need the units digit of 2^k.
andersonw19:22:42
k is also divisible by 4, so it is also 6
kshweh19:22:46
k = 0 (mod 4)
mburg19:22:50
use our list again
DPatrick19:23:04
Right, for this we'll go back to our list of units digits of powers of 2.
DPatrick19:23:11
Also, 2008^2 is certainly a multiple of 4, and so is 2^2008. So k is a multiple of 4.
DPatrick19:23:33
Therefore, the units digit of 2^k is 6.
DPatrick19:23:46
Thus, the answer is (D) 6.
mathboygenius19:24:03
how do u kno k is divisible by 4?
arkantosstevius19:24:03
how is k a multiple of 4?
popopomnmnmn_219:24:03
why is k divisible by 4?
DPatrick19:24:27
k = 2008^2 + 2^2008. 2008 is a multiple of 4, so any power is too. And 2^m is a multiple of 4 for any m>1.
DPatrick19:24:57
Finally let's go to #25 on the 10A.
DPatrick19:25:03
DPatrick19:25:18
DPatrick19:25:35
Note: this is also question #22 on the AMC 12A. Since we're in the AMC 10A portion of the Math Jam, we'll do a non-trigonometric solution here. (When we get to #22 on the AMC 12A, we'll discuss an alternate solution using trig.)
mathboygenius19:25:47
was the picture originally given with the problem?
DPatrick19:25:55
Yes, the picture was provided.
karatemagic719:26:10
why dont we make a big trinagle and use pythagoreans theorom
T3h Math Hero!19:26:10
Make a triangle
arkantosstevius19:26:17
you need to construct a really thin triangle
DPatrick19:26:31
One tool that we have is the Pythagorean Theorem. So we can try to set up a relevant right triangle.
DPatrick19:26:50
We'll probably want one side to be a radius (since then we'll know its length is 4).
adchia19:27:00
use the radius as the hypotenuse
DPatrick19:27:05
That's a good plan.
cc119:27:24
draw perpendicular to x
DPatrick19:27:30
Yeah, let's do that too.
DPatrick19:27:33
In fact, here's what I would do:
DPatrick19:27:42
DPatrick19:28:07
(There are probably other possibilities too, and as I said I'll discuss a different trig-based solution a little later.0
DPatrick19:28:08
)
DPatrick19:28:22
As already discussed, the hypotenuse of this triangle is 4.
What are the other two sides?
andersonw19:28:41
the shorter leg is x/2
karatemagic719:28:41
x/2 is one leg
cc119:28:41
1/2x
mcsquared19:28:41
x/2 and something else
DPatrick19:28:46
The short leg is half the length of the placemat, so it is x/2.
DPatrick19:28:52
That was the easy part. :)
DPatrick19:28:56
What about the longer leg?
adchia19:29:11
the other is the height of an equilateral triangle with side x
adchia19:29:11
+1
cc119:29:16
sqrt3/2x+1
arkantosstevius19:29:16
the longer leg is 1 plus root three times x over 2
andersonw19:29:19
make an equilateral triangle
DPatrick19:29:21
Right.
DPatrick19:29:29
The part that's going across the placement has length 1 (the width of the placemat).
DPatrick19:29:38
The part that extends from the center to the edge of the placemat is an altitude of an equilateral triangle of side length x:
DPatrick19:29:44
DPatrick19:29:54
So that part has length x*sqrt(3)/2.
DPatrick19:30:07
arkantosstevius19:30:20
then we use pythagoras
mathboygenius19:30:28
now u just use the pythagorean theorem
aznjonny2419:30:28
now use the pythagorean theorem
DPatrick19:30:47
Right: we have all three sides of the dark-sided triangle in the above picture, so we can apply the Pythagorean Theorem!
DPatrick19:30:56
arkantosstevius19:31:26
to construct an equation and solve for x
andrewnycpops19:31:26
Solve for x?
DPatrick19:31:35
Yes, we want to solve this for x.
DPatrick19:31:48
There are lots of ways to go from here.
DPatrick19:32:06
Me, before I did anything else, I'd multiply by 4 to get rid of the fractions:
DPatrick19:32:10
adchia19:32:31
expand and use quadratic formula
karatemagic719:32:35
x^2+sqrt(3)x-15=0
DPatrick19:32:41
We can now expand the square on the right side:
DPatrick19:32:44
DPatrick19:32:49
Simplifying, this gives us:
DPatrick19:32:53
jeffreychen19:33:13
Now use the quadratic formula
desperado19:33:13
use quadratic formula
DPatrick19:33:17
So we can apply the quadratic formula:
DPatrick19:33:24
lingomaniac8819:33:34
reject the negative
adchia19:33:37
must be positive
DPatrick19:33:40
Clearly we need to take the "+" sign since x > 0.
DPatrick19:33:46
DPatrick19:33:52
Answer (C).
DPatrick19:34:13
In a little bit we'll do a slightly quicker solution that uses trigonometry.
DPatrick19:34:54
As I said at the beginning, if we have time at the end I'll take requests for other AMC 10A problems, but for now, let's move on to #21 on the AMC 12A.
DPatrick19:35:01
mburg19:35:41
i made a list
fragoo819:35:41
can we just list all the possible sums using two of the five numbers?
mathboygenius19:35:41
just brute force and get a pattern
jackdillon19:35:41
brute force it to look for a pattern
DPatrick19:35:44
That's an option.
DPatrick19:35:59
However, this is a clever observation that one can make that simplifies this problem enormously.
lingomaniac8819:36:12
For every "heavy-tailed" permutation (a1,a2,a3,a4,a5), there's a "light-tailed" permutation (a5,a4,a3,a2,a1)
CatalystOfNostalgia19:36:13
the number of heavy-tailed is equal to the number of 'heavy-headed'
DPatrick19:36:21
Right. A clever solution is to use symmetry.
DPatrick19:36:29
We can call a permutation "light-tailed" is a_1+a_2 > a_4 + a_5.
DPatrick19:36:41
...slight typo there...
DPatrick19:36:45
Call a permutation "light-tailed" if a_1+a_2 > a_4 + a_5.
DPatrick19:37:00
The key observation is that the number of light-tailed permutations is equal to the number of heavy-tailed permutations, by symmetry.
fragoo819:37:08
can the they never be "equal tailed"?
karatemagic719:37:08
what about the equals(a1+a2=a4+a5)
Aegor19:37:08
So we can just subtract all the " equal-tailed" permutations, then divide by two?
DPatrick19:37:14
Exactly, that's our strategy."MCP
DPatrick19:37:19
We have 5! = 120 permutations total. We can count the number of "even-tailed" permutations: those with a_1 + a_2 = a_4 + a_5. We then subtract these from 120, and take half of what's left.
sharonh19:37:34
heavy-tailed+light-tailed+"equal-tailed"=total
DPatrick19:37:51
Right, that's the way to think about it. And "heavy-tailed" = "light-tailed".
DPatrick19:38:02
So how do we count how many even-tailed permutations are there?
mburg19:38:23
brute forct
lauren1*19:38:23
list them all?
mathboygenius19:38:23
that's easy, since theyre not too many, just brute force
DPatrick19:38:34
Again we could do that too, but we can also be clever here as well.
lingomaniac8819:38:46
a1+a2+a4+a5 is even, so a3 must be odd
DPatrick19:38:56
One observation is that since the five elements sum to 15, we must have a_3 odd.
DPatrick19:39:01
So that's 3 different choices for a_3.
DPatrick19:39:21
Now we notice that no matter what we choose for a_3, the remaining 4 elements can only be paired in one way: we must take the largest and the smallest remaining element as one pair, and then the other two "middle" elements as the other pair.
DPatrick19:39:34
So what choices do we have left in our construction?
Goldey19:40:21
for any of those 3 choices, there are 2 pairs, which can be placed on either end and in any order for 2*2*2=8 ways each... a total of 24 with a1+a2=a4+a5
DPatrick19:40:32
Right. We need to choose one pair to be {a_1,a_2} and the other to be {a_4,a_5}. That gives us 2 choices.
DPatrick19:40:37
Then, we need to order each pair. That's 2*2 more choices.
DPatrick19:40:45
This gives us a total of 3*2*2*2 = 24 even-tailed permutations.
DPatrick19:41:17
So 120-24 = 96 permutations are non-even-tailed, and of those, exactly half --- 96/2 = 48 --- are heavy-tailed.
Aegor19:41:22
So the number of heavy-tailed is (120-24)/2 = 96/2 = 48
krsattack_219:41:22
Total = 5! = 120, so (120-24)/2 = 48
DPatrick19:41:25
Right. Answer (D).
DPatrick19:41:53
Next, let's revisit the placemats problem, which is #22 on the AMC12:
DPatrick19:42:00
DPatrick19:42:13
DPatrick19:42:32
There is a slightly more direct solution using trigonometry.
virtuoso19:42:37
We could use the Law of Cosines for this problem.
MCPC19:42:37
law of cosines
adchia19:42:37
law of cosines!
DPatrick19:42:53
Right, the main advantage trig gives us is that we don't have to just consider right triangles. We can use the Law of Cosines on any triangle!
lingomaniac8819:43:07
draw a line connecting an outer corner to the center, and another line connecting an inner corner to the center
adchia19:43:07
draw an obtuse triangle with sides 1, x, and 4, with angle 150
DPatrick19:43:13
That is exactly what I did:
DPatrick19:43:19
DPatrick19:43:32
The nice thing about this triangle is that we know all the sides, and they're simple.
DPatrick19:43:42
The short side is 1, the long side is 4, and the remaining side is x.
DPatrick19:43:56
Furthermore, we know that the obtuse angle is 150 degrees.
DPatrick19:44:11
So we can apply the Law of Cosines!
DPatrick19:44:21
adchia19:44:32
cos 150=-root3/2
DPatrick19:44:41
Indeed, so we can plug it in and simplify.
DPatrick19:44:46
krsattack_219:44:49
We get the same quadratic equation!
CatalystOfNostalgia19:44:49
we get the same quadratic as before
DPatrick19:44:54
Yay. Mathematics is consistent. :)
DPatrick19:45:06
Now we finish just as we did before.
DPatrick19:45:10
DPatrick19:45:17
And we need to take the positive solution, so we get
DPatrick19:45:19
DPatrick19:45:22
(C) (again).
DPatrick19:46:00
Let's next look at #23:
DPatrick19:46:06
DPatrick19:46:24
What might we notice right away?
krsattack_219:46:45
Notice the binomial coefficients... Clearly connected to (z+i)^4
unimpossible19:46:45
pascal's triangle
lingomaniac8819:46:45
the coefficients: 1-4-6-4-1
mathboygenius19:46:45
1,4,6,4,1 pascal triangle
frank44_219:46:45
Similar to (z+i)^4 and (z-i)^4
Goldey19:46:45
1 4 6 4 1.. symmetrical and from pascals?
DPatrick19:46:57
The pattern of coefficients: 1,4,6,4,1. Looks like a 4th power binomial expansion (which is the 4th row of Pascal's Triangle).
DPatrick19:47:13
In fact, we note that:
DPatrick19:47:18
DPatrick19:47:32
Simplifying the i terms, we get:
DPatrick19:47:37
DPatrick19:47:44
This almost exactly matches the left side of our original equation, except the constant is 1 instead of -i.
fragoo819:47:55
add and subtract 1+i to both sides to create symmetry
cowpi19:47:55
Add i+1 to both sides then.
krsattack_219:47:55
Thus, our equation is (z+i)^4 = 1+i
DPatrick19:48:02
So we can rewrite our original equation:
DPatrick19:48:06
DPatrick19:48:14
DPatrick19:48:20
What does this mean about the solutions?
karatemagic719:48:56
they are the fourth root of i+1 an then-i
fragoo819:48:56
the points are the fourth roots of (1+i) shifted i down
DPatrick19:49:03
Right. The solutions are the four 4th roots of 1+i, shifted by -i.
adchia19:49:18
the quadrilateral is a square
frank44_219:49:21
Just solve for z+i, the shift doesnt affect the area
DPatrick19:49:36
Right. We don't have to worry about the shift by -i, since it doesn't change the shape or the area.
DPatrick19:49:52
So we have a square given by the four 4th roots of 1+i.
DPatrick19:50:09
How does this allow us to compute the area?
fragoo819:50:37
the radius of the points is 2^(1/8)
mhisno119:50:45
They are all evenly spaced
DPatrick19:50:51
Right. Note that | 1 + i | = sqrt(2) = 2^1/2.
DPatrick19:50:59
So the four solution lie evenly spaced around a circle of radius (2^1/2)^1/4 = 2^1/8.
DPatrick19:51:35
So what is the side length of the square?
adchia19:52:04
that time root2
frank44_219:52:04
2^(5/8)
lingomaniac8819:52:06
2^(1/8) * sqrt(2) = 2^(5/8)
krsattack_219:52:06
2^(5/8)
DPatrick19:52:15
Right: the four points lie on a square of side length (sqrt 2)*2^1/8 = 2^5/8.
frank44_219:52:22
area = 2^(5/4)
lingomaniac8819:52:22
so the area is 2^(5/4)
DPatrick19:52:41
To finish, we square the side length, to see that the area is (2^5/8)^2 = 2^5/4. Answer (D).
kostya19:52:56
couldn't you multiply the diagonals (2*2^1/8) and then divide by two to get 2^5/4
fragoo819:52:58
so we have a rhombus with two diagonals of length 2*2^(1/8) each
DPatrick19:53:05
Certainly, you could also finish that way.
DPatrick19:53:26
I must admit that this was, by far, my least favorite of all the problems that we're discussing today.
DPatrick19:53:44
It a bit too technical for my liking.
DPatrick19:54:10
Anyway, let's move on to what is probably the hardest problem on the contest, #24:
DPatrick19:54:16
arkantosstevius19:54:38
is there a diagram provided?
DPatrick19:54:47
No. But my first step in solving would be to draw one!
DPatrick19:54:57
DPatrick19:55:13
Note that A can lie anywhere on the shown ray.
DPatrick19:55:49
One thing we might note from our diagram is that A will always be acute, so that maximizing the angle will maximize the tangent.
DPatrick19:56:11
So how can we maximize the angle?
canttouchthis2319:56:29
move it closer to C
jackdillon19:56:29
Move A closer to C
susanzduan19:56:29
move it closer to c
lauren1*19:56:29
Make A lower down on the ray
DPatrick19:56:33
Why?
DPatrick19:56:51
Can we qualitatively (that is, in English, without formulas) describe where A should be?
DPatrick19:57:32
This is somewhat of an evil problem, because if you can't answer this question correctly, then you have very little chance of successfully solving the problem.
CatalystOfNostalgia19:57:48
draw circle through B and D tangent to ray CA
lokito19:57:48
draw a circle with d and b on the circumference, tangent to ray CA.
Boy Soprano II19:57:48
Well, there should be a circle passing through A, D, and B tangent to line CA.
DPatrick19:58:04
That's the right answer.
DPatrick19:58:15
The picture I've drawn might be a little reminiscent of an angle inscribed in an arc of a circle.
DPatrick19:58:31
So we draw the circle that has chord DB that is tangent to the ray:
DPatrick19:58:40
DPatrick19:58:56
If (for example) A is where it's at now, then the measure of BAD is half the difference between the subtended arcs.
DPatrick19:59:27
So anywhere on the arc, it's always strictly less than half the measure of the arc DB...except...
DPatrick19:59:33
...if A is the point of tangency, then it's exactly half the measure of the arc DB, and that's the maximum that it can be.
mathboygenius20:00:05
why do u draw the circle? this is confusing
krsattack_220:00:05
What is the inspiration for this construction?
DPatrick20:00:20
That's why this problem is really hard. You have to have enough geometric experience to know to draw the above circle.
jackdillon20:00:37
So we need to find where the circle intersects with ray AC
DPatrick20:00:56
Right. First, let's clean up our diagram with A is its proper location:
DPatrick20:01:03
krsattack_220:01:33
AC^2 = CB*CD
CatalystOfNostalgia20:01:33
power of a point to get AC
DPatrick20:01:48
We can do power-of-a-point on point C to find AC.
DPatrick20:01:56
(AC)^2 = (CD)(CB) = 8, so AC = 2*sqrt(2).
DPatrick20:02:02
Now what? How do we compute tan(BAD)?
mathboygenius20:02:31
law of cosines
CatalystOfNostalgia20:02:31
law of cosines to get AD and AB, then law of cosines again to bet <BAD, then pythagorean identity, yuck
DPatrick20:02:40
yuck indeed. (I tried that first, too.)
krsattack_220:02:48
subtraction formula?
DPatrick20:02:55
I tried that next, and it's much nicer!
skipper0920:03:27
what is the subtraction formula???
jackdillon20:03:27
what is the subtraction formula?
DPatrick20:03:37
Sorry...what I meant is the tangent angle-subtraction formula.
DPatrick20:03:52
We want tan(BAD), and we can compute it via a difference of angles.
DPatrick20:04:08
But it's a lot easier if we have a right triangle.
DPatrick20:04:24
In order to get a right triangle, I would drop a perpendicular from A down to BC. Let's call the foot of this perpendicular E.
DPatrick20:04:29
DPatrick20:04:40
Then since AEC is 30-60-90, we know that AE = sqrt(6) and CE = sqrt(2).
DPatrick20:04:52
Note also that DE = 2-sqrt(2), but this wouldn't fit on the picture.
DPatrick20:05:15
DPatrick20:05:40
The nice thing is that we know tan(BAE) and tan(DAE)!
DPatrick20:05:56
DPatrick20:05:59
DPatrick20:06:26
So we can just plug these numbers in to our formula.
DPatrick20:06:32
arkantosstevius20:06:40
that's still very complicated
DPatrick20:06:46
It is, but not as much as it looks.
DPatrick20:06:52
We can multiply by 6 to clear some denominators, and the numerator simplifies nicely:
DPatrick20:06:57
DPatrick20:07:15
The only answer this possibly looks like is (D).
krsattack_220:07:31
divide by 2sqrt2 in numerator and denominator to get that answer...
DPatrick20:07:34
Indeed if we divide numerator and denominator by 2sqrt(2), we get:
DPatrick20:07:41
DPatrick20:07:51
So the answer is (D).
DPatrick20:08:01
This problem was really hard.
DPatrick20:08:28
It takes quite a bit of geometry and trig experience to get to a solution.
Strider709220:08:36
is that the only way?
DPatrick20:08:48
Probably not...but it's what feels to me like the "optimal" way.
DPatrick20:09:11
Let's finish up with #25 (which I thought was quite a bit easier).
DPatrick20:09:27
DPatrick20:10:02
Those of you who know a little bit about linear algebra, 2x2 matrices, and/or vectors in the plane might know a slick way to solve this problem.
(I might mention it at the end if we have a little extra time.)
DPatrick20:10:19
But if we didn't go that route, what might we notice?
fragoo820:10:30
30-60-90 reminiscent quantities
DPatrick20:10:50
The sums of a_n and b_n with sqrt(3) coefficients might get you thinking about the sum/difference of sines and cosines formulas.
DPatrick20:11:15
What that might suggest is writing the point (a_n,b_n) in polar coordinates. (Or in complex numbers, which is essentially the same thing.)
DPatrick20:11:38
DPatrick20:12:06
DPatrick20:12:29
So now we can write our recursion this way too.
DPatrick20:12:36
DPatrick20:12:56
What's notable about those trig expressions on the right side?
arkantosstevius20:13:11
divide by 2 on both sides?
skipper0920:13:15
they're sum/difference
DPatrick20:13:29
Right, when we pull a 2 out, it's more clear:
DPatrick20:13:35
krsattack_220:13:50
a_n+1 = 2rcos(30 + theta)
DPatrick20:14:02
Indeed, these are the cosine and sine sum formulas!
DPatrick20:14:07
adchia20:14:29
rotate a point around 30 degrees at a time
DPatrick20:14:51
Exactly. What's going on is that the point is rotating 30 degrees every time, and doubling its distance from the origin.
krsattack_220:15:03
a_100 is 99 such rotations...
DPatrick20:15:43
Right...so every 12 steps we rotate the full 360 degrees (360 = 12*30), so applying 99 steps is a rotation of 90 degrees.
DPatrick20:16:03
...and along the way we've moved it by a factor of 2^99 away from the origin.
DPatrick20:16:20
So, to go backwards, the 1st point is the 100th point rotated 90 degrees clockwise and then shrunk by a factor of 2^(-99).
DPatrick20:16:48
How do we compute this?
krsattack_220:17:10
(a_1, b_1) = (4/2^99, -2/2^99)
DPatrick20:17:15
Right.
DPatrick20:17:21
Rotating (2,4) 90 degrees clockwise gives us (4,-2).
DPatrick20:17:31
(if you don't see this, you can sketch a picture.)
DPatrick20:17:41
DPatrick20:17:53
So that's the first point.
DPatrick20:18:12
DPatrick20:18:21
This is answer (D).
jackdillon20:18:42
What is the slick way to do this?
DPatrick20:18:47
For those of you who know a little bit about matrix algebra and vectors, we can write the recurrence as:
DPatrick20:18:52
DPatrick20:19:18
Then you can use matrix multiplication to quickly compute the effect of applying it 99 times.
fragoo820:19:23
oh and that immediately gives us the rotation and everything
DPatrick20:19:26
Right.
DPatrick20:19:39
But if you don't know what any of this means, don't worry about it...it's fairly advanced algebra.
DPatrick20:19:53
Before we move on to other problem requests, I need to pay the bills. :)
DPatrick20:20:00
AoPS publishes a series of books with the goal of providing bright math students with an appropriately challenging and modern curriculum, focusing both on the core mathematical concepts as well as problem solving techniques.
DPatrick20:20:20
The books in the Introduction-level series are Algebra, Counting & Probabilty, Geometry, and Number Theory.
DPatrick20:20:29
The scope of these books begins with MATHCOUNTS level material and quickly moves through most all AMC and AIME material, and even a bit of Olympiad material. In particular, we have tried to write these books in a way that makes advancement through more and more difficult concepts easier.
DPatrick20:20:43
We also have a newer Intermediate-level series that covers more advanced material.
DPatrick20:20:51
Our first book in this series -- Intermediate Counting & Probability -- is currently available. We should have the second book -- Intermediate Algebra -- available in the spring.
DPatrick20:21:08
Also, for those of you who have qualified for the AIME: congratulations! We will be having our Special AIME Problem Seminar on Saturday & Sunday, March 8 & 9, from 3:30 to 6:30 (Eastern) both days. The course will review several past AIME-style problems and also discuss general test-taking strategies for the AIME. There will also be a message board exclusive to the class for students to discuss the class problems and strategies. More information is at: http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php
DPatrick20:21:24
And if you haven't yet qualified for the AIME, you have another chance! The AMC 10B/12B is on Wednesday, February 27. If your school is not offering it, you may be able to take it at a local college or university. The AMC has a list of colleges offering the AMC 10B/12B on their web site at:
http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml
sjlite20:21:32
how do we know if we qualified for the AIME?
DPatrick20:21:45
If you got 120 on the 10A or 100 on the 12A, then you've automatically qualified.
DPatrick20:22:00
However, it is possible that lower scores may qualify too (as happened last year).
DPatrick20:22:13
But we won't know this until the AMC computes all the scores, which will take a few weeks.
DPatrick20:22:42
OK, I will spend a little extra time covering some other problems.
DPatrick20:23:45
(I'm just pausing to get a feel for what are requested most, since I only have time to do a couple more.)
DPatrick20:24:23
I've got tons of requests for #20 on the 12A, so let's do that one.
DPatrick20:24:32
DPatrick20:25:05
Here's a picture:
DPatrick20:25:11
adchia20:25:32
was this diagram given?
DPatrick20:25:37
No -- my first step was drawing it.
DPatrick20:25:50
The big triangle is 3-4-5.
lingomaniac8820:25:59
i tried to work through the numbers using K=rs; is there a better way?
krsattack_220:25:59
Can we use Area = rs?
DPatrick20:26:05
This is, to my mind, by far the best way.
DPatrick20:26:20
We want to find solving involving inradii, which is probably hard to compute.
DPatrick20:26:39
But we have the nice formula A = rs, where A is area and s is the semiperimeter (half the perimeter).
DPatrick20:26:59
So if we find the areas and semiperimeters of ACD and BCD, we'll be able to find the inradii.
lingomaniac8820:27:09
angle bisector theorem to find AD and DB
DPatrick20:27:13
Ah...good start.
krsattack_220:27:20
The ratio of Areas is given as the ratio between AD and BD, which we can find by the angle bisector theorem
DPatrick20:27:46
Indeed, the ratio of the areas of ACD and BCD is just the ratio AD/BD, since both of these triangles have the same height (from C).
DPatrick20:27:56
And by the Angle Bisector Theorem, AD/BD = AC/BC.
DPatrick20:28:05
...which is 3/4.
DPatrick20:28:17
So (Area of ACD)/(Area of BCD) = 3/4.
DPatrick20:28:30
So far, so good.
DPatrick20:28:35
Now we need the semiperimeters.
krsattack_220:28:44
AD = 15/7, BD = 20/7
DPatrick20:29:05
Right. AD/BD = 3/4 and AD+BD = 5, so solving this we get AD = 15/7 and BD = 20/7.
DPatrick20:29:17
What about CD?
DPatrick20:29:24
DPatrick20:29:31
oops...I meant to put the picture up again.
DPatrick20:29:40
ppzuan20:30:10
we can use law of cosines to calculate CD
DPatrick20:30:26
That's one option. We know cos(A) = 3/5 and we know AC and AD.
Goldey20:30:30
CD= 12(sqrt2)/7 because the lines y=x(CD) and y= -4/3X+3 ( BA) intersect at 12/7,12/7
DPatrick20:30:39
That's actually what I did.
DPatrick20:31:03
If we set it up so that C is the origin, and AC and BC are the axes, then CD is the line y=x and AB is the line y = -(4/3)x + 4.
DPatrick20:31:26
We just solve to get x =12/7, so CD = (12/7)*sqrt(2).
DPatrick20:31:44
So now we have all the lengths, so we have the semiperimeters.
krsattack_220:32:27
r_a/r_b = (K_a/s_a)/(K_b/s_b)
DPatrick20:32:43
Right, and we already figured out that the ratio of the areas was 3/4.
DPatrick20:32:56
So the ratio r_a/r_b of inradii is (3/4)(perim BCD/perim ACD).
DPatrick20:33:18
We can multiply by 7 to get rid of all those 7's in the denominator, and when we crunch the numbers, the answer is:
DPatrick20:33:23
DPatrick20:33:44
DPatrick20:33:49
Answer (E).
DPatrick20:34:32
The only other problem I had multiple requests for was #19 on the AMC12A.
DPatrick20:35:04
DPatrick20:35:57
What is that square term on the right side?
DPatrick20:36:14
...if we were to write it out?
susanzduan20:37:07
1+2x+3x^2+...15x^14+14x^15+...2x^27+x^28
lingomaniac8820:37:11
x^28 + 2x^27 + 3x^26 + ... + 14x^15+15x^14 + 14x^13 + ... + 3x^2 + 2x + 1
DPatrick20:37:14
Right.
DPatrick20:37:56
It starts 1 + 2x + 3x^2 + 4x^3 + ..., the coefficients increase up to 15x^14, then the decrease back down to ... + 2x^27 + x^28.
susanzduan20:38:01
and then we match each one up with the first part?
DPatrick20:38:23
Indeed we do. Each term (except the first "1") matches with a term from (1+x+...+x^27) to contribute to the x^28 term in the entire product.
lokito20:38:27
So, it's 2+3+4+...+14+15+14+...+3+2+1?
DPatrick20:38:34
That's exactly what it is!
DPatrick20:38:56
This is twice (1+2+...+14), plus an extra 15, minus 1.
DPatrick20:39:29
And (1+2+...+14) = 14(15)/2 = 105.
krsattack_220:39:32
1+2+...+14 = 14*15/2 = 105
DPatrick20:39:34
Right.
DPatrick20:39:51
So the answer is 2*105 + 15 - 1 = 224. (C)
DPatrick20:40:26
The slick way to do this is to notice that every term, except for 1*1, of the right term is going to get used exactly once.
DPatrick20:40:46
So we just have the sum of the coefficients of (1+x+...+x^14)^2 minus 1 as our answer.
DPatrick20:40:50
So it's 15^2 - 1 = 224.
DPatrick20:41:04
And there's a really slick solution using generating functions (if you know what those are).
DPatrick20:41:25
I think I'll end it here...I've already run 15 minutes longer than scheduled!
DPatrick20:41:50
Most of the problems are being discussed on our Forum, and the AMC has posted the official answers to every problem on their web site (just the answers though, not a complete solution).
DPatrick20:42:10
There will be a transcript posted shortly.
DPatrick20:42:24
Go to the Math Jams page of our site and click on the "Transcripts" button near the top of the page.
DPatrick20:42:54
Thanks for coming and good luck on the 10B/12B if you're taking it!
kostya20:43:04
[url href=http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions]http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions[/url]
DPatrick20:43:27
Indeed, please contribute problems and solutions to our Wiki and the link above!
DPatrick20:43:46
Many of the 12A problems are up, but we need the 10A problems too!
DPatrick20:44:08
Thanks and good night :)
