| Transcript
for the Math
Jam "2007 AMC 10/12 A Math Jam"
on Feb 8. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick (18:59:04)
Welcome to the 2007 AMC 10A/12A Math Jam!
DPatrick (18:59:13)
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.
DPatrick (18:59:23)
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.
DPatrick (18:59:38)
When I post a problem, I will also post a link to the problem, like so:
DPatrick (18:59:41)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-79145746.gif
DPatrick (18:59:49)
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.
DPatrick (19:00:07)
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:
DPatrick (19:00:13)
DPatrick (19:00:15)
http://www.artofproblemsolving.com/Classes/IntroNumbers/L1/logo.gif
DPatrick (19:00:33)
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.
DPatrick (19:00:51)
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.
DPatrick (19:01:17)
The Math Jam will proceed as follows:
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.
DPatrick (19:01:35)
Let's get started with #21 from the 10A.
DPatrick (19:01:40)
DPatrick (19:01:43)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-79145746.gif
DPatrick (19:01:58)
What's the overall plan for this problem?
ragnarok23 (19:02:16)
first find the side length of the cube
math92 (19:02:17)
We should fine the length of the side of the outer cube first
Xantos C. Guin (19:02:29)
Find the length of the larger cube, find the length of the smaller cube, then find the surface area.
DPatrick (19:02:51)
Right. It's important to approach a problem like this with a plan, and not aimlessly compute.
DPatrick (19:02:54)
It seems like there is a straightforward sequence of steps that we need to do:
1. Find the side length of the large cube
2. Find the diameter of the sphere
3. Find the side length of the small cube
4. Find the surface area of the small cube (thus answering the problem)
DPatrick (19:03:08)
Step 1: what's the side length of the large cube?
EunuchOmerta (19:02:32)
s=2
EunuchOmerta (19:02:45)
from 6s^2=24
math92 (19:02:50)
First of all, you can figure out the sidelength of the cube is 2. (since 2^2*6=24).
ra5249 (19:03:12)
6s^2 = 24, so s = 2; therefore, the large cube has side length 2
ragnarok23 (19:03:26)
6s^2=24 s=2
DPatrick (19:03:39)
DPatrick (19:03:51)
Step 2: what's the diameter of the sphere?
ragnarok23 (19:04:02)
same as side length of cube
qwang (19:04:02)
The diameter of the larger sphere is equal to the cube's side length: 2
Xantos C. Guin (19:04:05)
Same as the side of the larger cube, so 2
mysmartmouth (19:04:07)
Because the side length of the cube is 2, the inscribed sphere will also have side length 2.
DPatrick (19:04:21)
The
diameter of the sphere is the same as the side length of the cube, so it's 2.
DPatrick (19:04:27)
Step 3: what's the side length of the small cube?
ragnarok23 (19:04:39)
space diagonal=diameter of sphere
DPatrick (19:04:56)
Yes, the key observation is that the
diagonal of the small cube is the same as the diameter of the sphere, so the diagonal of the cube is 2.
not_trig (19:04:47)
2/sqrt(3), because the inner diagonal of a cube is side*sqrt(3)
Xantos C. Guin (19:04:56)
The space diagonal must be 2 so the side must be 2/sqrt(3)
kostya (19:05:00)
this is equal to the diagonal of the cube, so the side length =2/sqrt(3)
DPatrick (19:05:12)
DPatrick (19:05:30)
Step 4: what's the surface area of the small cube?
math92 (19:05:29)
s^2 * 6 = 4/3 * 6 = 8
Xantos C. Guin (19:05:39)
Thus the surface area is 6(2/sqrt(3))^2=8
ragnarok23 (19:05:41)
6s^2 so 8
not_trig (19:05:46)
6*s^2 = 6*4/3 = 8
DPatrick (19:05:56)
DPatrick (19:06:11)
Note that developing the plan of attack was pretty straightforward, and that executing each step was just basic geometry.
DPatrick (19:06:25)
DPatrick (19:06:29)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-146893651.gif
DPatrick (19:06:53)
How do we begin?
not_trig (19:06:52)
let's try out some sequences
mysmartmouth (19:06:55)
Let's make a sample sequence to see if we can find a pattern.
bpms_2 (19:06:57)
we can just examine a few arbitrary sequnces and find a pattern
DPatrick (19:07:23)
One thought, for a multiple-choice test, is to try to eliminate some answers. A quick way to get rid of some choices might be to try an example or two. This may also give us a feel for the problem.
Xantos C. Guin (19:07:15)
Eliminate some answers by testing 111 as a sequence
DPatrick (19:07:39)
Sure. Note that 111, all by itself, is such a sequence, so 111 is a possible value of S.
DPatrick (19:07:57)
This already means that answer must be either (A) or (D), since 111 = 3*37.
not_trig (19:08:13)
but we still need some more examples
EunuchOmerta (19:08:16)
we're not done because 37 might not always work, right
DPatrick (19:08:32)
Right, we could try another simple example.
DPatrick (19:08:40)
Another trivial choice is a sequence of the form abc, bca, cab, where a,b,c are digits.
kostya (19:08:51)
123,231,312
not_trig (19:08:55)
123, 231, 312
frodo (19:08:58)
111(a+b+c)
DPatrick (19:09:17)
We could plug in actual numbers, or we could treat it with arbitrary digits (as frodo did).
DPatrick (19:09:22)
DPatrick (19:09:39)
On a timed, multiple-choice test, I'd be awfully tempted to mark down (D) as my answer and move on, and come back and try to prove it if I had time.
mysmartmouth (19:08:49)
Well, 247, 475, 756, 568, 682, 824 is a sequence. We note that each digit used appears exactly 3 times, once in each place. So let the digits be x, y, and z. The sum of the digits would then be 111x + 111y + 111z.
ra5249 (19:09:39)
Units digits, tens digits, and hundreds digits will always be the same but in a different order, so it will be divisible by 111
EunuchOmerta (19:09:54)
notice that each digit will always appear in each of the hundreds, tens, and ones
DPatrick (19:10:05)
Exactly. Our examples actually contain the proof.
DPatrick (19:10:13)
Keep track of every digit. Each individual digit appears once in the hundreds place, once in the tens place, and once in the ones place.
DPatrick (19:10:26)
So each digit d in the series contributes 111d to the sum.
DPatrick (19:10:40)
Hence 111 divides S for any sequence, and thus so does 37. The answer is (D).
DPatrick (19:10:58)
DPatrick (19:11:02)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-37514258.gif
ragnarok23 (19:11:13)
Factor m^2-n^2 to (m-n)(m+n)
mysmartmouth (19:11:15)
DPatrick (19:11:35)
DPatrick (19:11:46)
DPatrick (19:11:50)
Now what?
Xantos C. Guin (19:12:01)
Find pairs of factors of 96
ragnarok23 (19:12:01)
factor 96 to see which pairs of factors work
worthawholebean (19:12:02)
Factor 96.
kostya (19:12:03)
factor 96
DPatrick (19:12:28)
We could do that, and it would certainly work, but there's a simplification that we can do first.
calc84maniac (19:12:10)
(m+n) and (m-n) cannot be odd
not_trig (19:12:32)
if m+n and m-n have differing parities, then m and n are not integers
DPatrick (19:12:45)
Exactly.
DPatrick (19:12:49)
There are two things to note. First, m+n must be the bigger factor, and m-n must be the smaller factor, and they're both positive.
DPatrick (19:12:56)
Second, both factors must be even, since m+n and m-n are either both even or both odd, and two odd numbers cannot multiply to give an even number.
DPatrick (19:13:07)
So we only need to count the number of ways to factor 96 into two even factors.
DPatrick (19:13:39)
Dividing two into each factor, we see that this is equal to the number of ways to factor 24 into
any two factors.
DPatrick (19:14:09)
And this is fairly easy to count!
EunuchOmerta (19:14:14)
2^3*3, so 4*2=8
DPatrick (19:14:28)
24 = 2^3 * 3 has 4*2 = 8 factors.
not_trig (19:14:43)
divide by 2
yellowlime (19:14:44)
4 pairs
DPatrick (19:14:58)
So there are 4 pairs of factors. (I thought that the AMC did you a small favor here by not listing 8 as an incorrect choice.)
DPatrick (19:15:08)
The answer is (B) 4.
DPatrick (19:15:47)
Certainly you can do this problem by listing factors of 96 and checking them, but a little simplification saves a bit of time...and time is precious on the AMCs!
DPatrick (19:16:02)
DPatrick (19:16:04)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-196567668.gif
DPatrick (19:16:10)
DPatrick (19:16:15)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/ebe0a92c27a3eab214c11bc390522364.png
DPatrick (19:16:45)
Yikes -- the picture looks a little scary. I don't know how to compute that shaded area directly.
ra5249 (19:16:36)
Connect EA and FB for 45-90-45 triangles
EunuchOmerta (19:16:36)
divide it into regions
Midnight2103 (19:16:52)
Triangle ABEF minus region EAO*2
rnwang2 (19:16:54)
its a rectangle minus two triangles minus two sectors
DPatrick (19:17:14)
Exactly. I do know how to compute a lot of other areas in the picture.
DPatrick (19:17:16)
Here's my strategy:
DPatrick (19:17:24)
DPatrick (19:17:27)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/53cae45a52f2b6241108329908fd43c9.png
DPatrick (19:17:38)
In other words, to get the black area, we compute the area of the rectangle ABEF, and subtract the areas of the red triangles AOC and BOD, and also the areas of the blue sectors ACE and BDF.
DPatrick (19:17:51)
We'll start easy: what's the area of the rectangle ABEF?
redcomet46 (19:18:01)
2*4rt2=8rt2
Midnight2103 (19:18:10)
2*squareroot 2 * 2
calc84maniac (19:18:10)
2*2sqrt(2)=4sqrt(2)
jamieding51 (19:18:12)
4 root 2 * 2 = 8 root 2
DPatrick (19:18:22)
DPatrick (19:18:51)
How about triangle AOC?
frodo (19:17:33)
ACO and BDO are 45-45-90 triangles.
DPatrick (19:19:26)
Indeed. It's a right triangle (since OC is a tangent), with one leg AC of length 2 and hypotenuse OA of length 2*sqrt(2). So it's 45-45-90.
siddarth (19:19:28)
2 & 2 are legs, os area is 2
yellowlime (19:19:39)
the side length of the triangle is 2
DPatrick (19:20:03)
The other length OC is also 2, so the area of AOC is (1/2)(2)(2) = 2.
DPatrick (19:20:11)
There are two of these triangles (the other is OBD), so the total red area that we have to subtract is 4.
DPatrick (19:20:26)
DPatrick (19:20:33)
(I know what answer I'd guess at this point!)
DPatrick (19:20:41)
How about the blue sector ACE?
jamieding51 (19:20:37)
Could you repost the picture for us?
DPatrick (19:20:49)
Sure:
DPatrick (19:20:54)
EunuchOmerta (19:20:31)
sector angle = 90-45=45
frodo (19:20:55)
Both sectors are 45 degrees.
kostya (19:20:59)
<ACE =45 degrees
DPatrick (19:21:16)
We just found out that angle CAO is 45 degrees, and we also know that EAO is 90 degrees. So EAC is 45 degrees.
kevinli2010 (19:21:15)
well ace and bdf are both 1/8 of their respective circles i think....?
redcomet46 (19:21:17)
It is 1/8 the area of the circle (which is 4pi) so its area is pi/2. So the two shaded areas sum to area pi
DPatrick (19:21:36)
Right. Each blue sector is 1/8 of the circle.
DPatrick (19:21:43)
The circles have area 4pi, so each blue sector has area (1/8)(4pi) = pi/2. There are two of them (one in each circle), so the total blue area is pi.
DPatrick (19:21:54)
now a ranger (19:21:32)
is that the only way to solve this problem? by using 45-45-90 triangles
DPatrick (19:22:35)
I wouldn't necessarily say "only", but it's by far the most straightforward.
DPatrick (19:22:52)
Last but not least...
DPatrick (19:22:56)
DPatrick (19:22:59)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-231332917.gif
DPatrick (19:23:03)
(Note: this is also question #22 on the AMC 12A, so we're also starting our discussion of the last five AMC 12A problems now, except slightly out-of-order!)
DPatrick (19:23:44)
There are several ways to approach this...I'm going to do it the first way it occured to me, which I think is the most straightforward.
DPatrick (19:23:51)
What do we know about such an n?
meamemg (19:24:03)
It is less than 2007
tjhance (19:24:03)
it is < 2007
kostya (19:24:10)
it is more than 1900
DPatrick (19:24:45)
n has to be a 4-digit number: 5-digits are too big, and 3-digits won't be big enough. In fact n < 2007 since all the terms are positive, and n > 1900 since otherwise S(n) and S(S(n)) aren't big enough.
EunuchOmerta (19:24:41)
try each set of 10s
DPatrick (19:25:16)
Yes, this suggests a simple casework approach.
DPatrick (19:25:31)
One case is n >= 2000. Write n = 200a for some units digit a with a < 7.
DPatrick (19:25:51)
What equation can we write?
Xantos C. Guin (19:25:51)
S(200a)=2+a
EunuchOmerta (19:26:03)
200a + 2+a+2+a
not_trig (19:25:58)
so a+2+a+2+a = 7
DPatrick (19:26:20)
Yes. S(n) = 2+a, a 1-digit number, and S(S(n)) = 2+a too.
DPatrick (19:26:27)
So n + S(n) + S(S(n)) = (2000+a) + (2+a) + (2+a) = 2004+3a.
ra5249 (19:26:29)
A is 1, so 2001
Xantos C. Guin (19:26:33)
a=1
rnwang2 (19:26:36)
a=1
DPatrick (19:26:47)
For this to equal 2007, a must be 1. So n = 2001 is a solution. (Check: 2001+3+3 = 2007.)
DPatrick (19:27:00)
The other case is n = 19ab, where a and b are digits.
DPatrick (19:27:17)
What are S(n) and S(S(n))?
tjhance (19:27:15)
1+9+a+b is S(n)
frodo (19:27:19)
S(n)=10+a+b
DPatrick (19:27:29)
Yes, S(n) = 10+a+b.
DPatrick (19:27:47)
But S(S(n)) depends on where S(n) begins with a 1 or a 2.
DPatrick (19:28:08)
So there are two subcases.
DPatrick (19:28:15)
What is a+b <= 9; what's S(S(n))?
DPatrick (19:28:25)
oops...What *if* a+b<=9, what's S(S(n))?
kostya (19:27:41)
a+b+1
cowpi (19:28:28)
a+b+1
DPatrick (19:29:01)
Right, a+b is the units digit of S(n), and the tens digits of S(n) is 1, so S(S(n)) = 1+a+b.
DPatrick (19:29:11)
So this gives
(1900+10a+b)+(10+a+b)+(1+a+b) = 1911+12a+3b.
DPatrick (19:29:28)
If this equals 2007, we need 12a+3b = 96.
DPatrick (19:29:38)
What are the solutions?
frodo (19:29:45)
4a+b=32
mysmartmouth (19:29:41)
Hamster1800 (19:29:53)
(8,0)
DPatrick (19:30:08)
Since a and b are digits, we see a=8,b=0 is a solution. All others (such as a=7, b=4) have a+b>9, which are not allowed in this subcase.
DPatrick (19:30:14)
This gives n=1980 as the only solution from this subcase.
DPatrick (19:30:32)
Now we go back to n = 19ab where a+b >= 10.
DPatrick (19:30:40)
We had S(n) = 10+a+b. What's S(S(n)) in this subcase?
tjhance (19:30:38)
a+b-8
not_trig (19:30:51)
S(S(n)) = a+b-8
DPatrick (19:31:08)
Right. S(S(n)) = a+b-8, since S(n) has tens digit 2 and units digit (a+b-10).
DPatrick (19:31:28)
So n+S(n)+S(S(n)) = (1900+10a+b)+(10+a+b)+(a+b-8) = 1902+12a+3b.
DPatrick (19:31:44)
This equals 2007, so 12a+3b = 105.
not_trig (19:31:51)
4a+b=35
DPatrick (19:32:20)
So what are the solutions here?
mysmartmouth (19:31:50)
now a ranger (19:32:15)
7,7 and 8,3 work for values for a and b , the only integers?
frodo (19:32:19)
(7,7) (8,3)
DPatrick (19:32:34)
This gives a=8, b=3 and a=7, b=7. Both have a+b>9, so they're valid. This gives n=1983 and n=1977.
DPatrick (19:33:21)
There are certainly other ways to have approached this problem, but as I said, this was probably the most straightforward.
DPatrick (19:33:56)
OK, let's move on to questions from the AMC 12A!
DPatrick (19:34:04)
DPatrick (19:34:08)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-65126292.gif
DPatrick (19:34:25)
How do we use the provided information?
mysmartmouth (19:34:37)
We can apply Vieta's formulas.
Rikimaru (19:34:41)
sums = -b/a product=c/a
ra5249 (19:34:44)
Sum of zeroes is -b/a, product is c/a
Xantos C. Guin (19:34:44)
sum of zeros = -b/a, product of zeros = c/a, sum of coefficients = a+b+c
DPatrick (19:35:13)
Right. Vieta's formulas relate the sums and products of a polynomial to its coefficients.
DPatrick (19:35:21)
In particular, the sum of the zeros is -b/a, and the product of the zeros is c/a.
cowpi (19:35:08)
replace c with -b
DPatrick (19:35:45)
Right, we know that b = -c.
DPatrick (19:35:53)
pianoforte (19:35:54)
a+b+c=a+b-b=a
DPatrick (19:36:15)
Indeed, now the sum of the coefficients is a.
DPatrick (19:36:23)
So we're done!
DPatrick (19:36:31)
"a" is the coefficient of x^2, so the answer is (A).
frodo (19:35:47)
a cannot equal 0.
DPatrick (19:36:53)
Indeed, if a=0 then f(x) = bx+ c, and the sum and the product of the zero(s) are trivially equal.
DPatrick (19:37:01)
Then the 3rd condition is that b+c = -c/b, which is messy, but the "common value" ends up being the x-intercept, choice (D).
DPatrick (19:37:17)
I think it's fair to implicitly assume that a is nonzero, but it probably should have been explicitly specified in the problem.
DPatrick (19:37:35)
We'll skip #22 since we already covered it as #25 from the 10A, so let's move on:
DPatrick (19:37:40)
DPatrick (19:37:45)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-29387190.gif
kostya (19:38:03)
side is 6
Potato Theory (19:38:14)
Write the vertices in terms of a reference point, a, a+6...
DPatrick (19:38:37)
OK, let's write equations based on the fact that the side length is 6.
Xantos C. Guin (19:38:37)
Let A be at (x,y), B is at (x-6,y) and C is at (x-6,y+6)
DPatrick (19:39:06)
That's what I would do, except how do we know that it's - and + and not some other combination of signs?
kostya (19:38:34)
draw a picture?
Bictor717 (19:39:14)
Make a quick sketch
DPatrick (19:39:33)
We could possibly observe it by making a sketch.
DPatrick (19:39:46)
But here's what I did...
DPatrick (19:39:57)
DPatrick (19:40:10)
DPatrick (19:40:23)
We can write the system of equations:
DPatrick (19:40:28)
DPatrick (19:40:41)
How do we work with these?
DPatrick (19:40:54)
(hopefully, if we didn't know + or -, it'll work out in the end.)
calc84maniac (19:39:43)
2loga(x)=loga(x^2)
Potato Theory (19:40:57)
Set the first two equal.
DPatrick (19:41:07)
Hamster1800 (19:41:03)
B has a - because 2log_a(x+6) > log_a x
ra5249 (19:41:05)
First two are equal, so (x+-6)^2 = x
DPatrick (19:41:50)
Right. We know x>0 (otherwise we can't take a log in the first place!).
jamieding51 (19:41:45)
root x = x +/- 6
DPatrick (19:42:06)
Yes. This means that 0 < x-6 < x. So the sign associated with the x terms is negative.
DPatrick (19:42:11)
Let's update our equations:
DPatrick (19:42:16)
frodo (19:42:08)
x^2+11x+36=0 x^2-13x+36
tjhance (19:42:29)
just solve for x
DPatrick (19:42:36)
Indeed, we can use what we just had to solve for x.
DPatrick (19:42:41)
Hamster1800 (19:42:58)
x-6 > 0 so x=9, not 4
DPatrick (19:43:09)
Yep. x=9 or x=4. But x=4 is bad since then x-6 = -2, and we can't take the log of that. So we must have x=9.
DPatrick (19:43:22)
Again, let's update our equations:
DPatrick (19:43:24)
DPatrick (19:43:39)
DPatrick (19:43:47)
What about the third that we haven't used yet?
cowpi (19:43:51)
subtract (3) and (2)
DPatrick (19:44:23)
worthawholebean (19:43:50)
We know that the third equation has a plus because 3log_a(3)>2log_a(3).
Hamster1800 (19:44:14)
we know that 3log_a 3 > 2log_a 3, so the +/- is a +
DPatrick (19:44:38)
How do we know this?
Hamster1800 (19:45:00)
Because log_a 3 > 0?
DPatrick (19:45:11)
How do we know this??
Bictor717 (19:45:06)
Log monotonically increases
DPatrick (19:45:32)
Only if a>1.
cowpi (19:45:33)
...none of the answer choices are <1?
DPatrick (19:45:57)
In fact, all of the answer choice are >1, so we can go ahead and assume that log_a 3 is positive.
DPatrick (19:46:12)
worthawholebean (19:45:50)
We must prove that a>1/
Rikimaru (19:46:23)
how do u prove it if the answers choices werent given
DPatrick (19:46:33)
We can't!!
DPatrick (19:46:38)
If we replace a by 1/a, and y by -y, we get an equally valid solution.
DPatrick (19:46:53)
Since all the answer choices are >1, we can "assume" a>1, but as an abstract question, a = 3^(-1/6) is a valid solution too, giving x = 9 and y = -12.
DPatrick (19:47:18)
The problem would have been better with the condition a>1 in the problem statement.
DPatrick (19:47:35)
Let's move on to #24:
DPatrick (19:47:40)
DPatrick (19:47:43)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-184068259.gif
DPatrick (19:48:12)
There seems to be an obvious question: what's the formula for F(n)?
rnwang2 (19:48:06)
try some numbers?
EunuchOmerta (19:48:07)
look at some small cases
tjhance (19:48:10)
notice that sine has a period of 2pi
calc rulz (19:48:25)
graph it!
DPatrick (19:48:37)
It's just F(n) = n+1, right?. This is I think most easily seen by simplify sketching the graphs of sinx and sin(nx) for a few small values of n, and convincing yourself of the pattern.
Potato Theory (19:48:49)
How would you know that for sure?
DPatrick (19:49:08)
We can more formally think of it this way: every period of sin(nx) gives 2 solutions (once on the way up and once on the way down), plus we get the extra solution at x=n*pi. If n is even, then sin(nx) has n/2 full periods, giving 2*(n/2)+1= n+1 solutions. If n is odd, then sin(nx) has (n-1)/2 full periods, but the half-period from ((n-1)/n)*pi to n*pi also gives 2 solutions, for a total of ((n-1)/2)*2 + 2 + 1 = n+1 solutions.
DPatrick (19:49:30)
Except...[img id=em-3]
Rikimaru (19:48:30)
tangent!!
worthawholebean (19:48:47)
F(n)=n+1 unless there are tangent intersections.
mysmartmouth (19:49:35)
what about sin 5x = sin x?
DPatrick (19:49:52)
Indeed, there's a trap. Compare sin(5x) and sin(x).
EunuchOmerta (19:49:42)
sin(5x) appeared to be tangent at pi/2
worthawholebean (19:49:43)
Watch out for tangent intersctions.
DPatrick (19:50:05)
They both pass through the point (pi/2,1), so we "lose" a solution there.
DPatrick (19:50:25)
In fact, we lose a solution at the point (pi/2,1) every time n = 1 (mod 4), meaning n is one more than a multiple of 4. The corresponding period of sin(nx) only hits sin(x) once.
DPatrick (19:50:57)
How many times does this happen between n=2 and n=2007?
EunuchOmerta (19:51:12)
501
Rikimaru (19:51:14)
501?
DPatrick (19:51:29)
Note that 2007/4 = 501.75, so we have to subtract 501.
DPatrick (19:51:43)
DPatrick (19:52:32)
tjhance (19:52:45)
2008(2009)/2
worthawholebean (19:52:56)
By Gauss, (2009)(2008/2)
lowfatsourcreme (19:52:57)
why - 504?
EunuchOmerta (19:53:10)
why minus 504, not 501
DPatrick (19:53:23)
Becuase the n=1 and n=2 terms are missing from our original sum.
DPatrick (19:53:38)
(Plus the answers choices help you out here.)
DPatrick (19:54:02)
This gives (2008)(2009)/2 - 504 = 2017036 - 504 = 2016532.
DPatrick (19:54:04)
Choice (D).
Totally Zealous (19:54:11)
What's Gauss?
DPatrick (19:54:32)
It's the formula for the sum of the first n positive numbers. 1+2+...+n = n(n+1)/2.
worthawholebean (19:54:23)
How would they expect yout to do that without a calculator?
DPatrick (19:55:05)
(2009)(2008) = (2000+8)(2000+9) = 4000000+17(2000)+72.
Rikimaru (19:55:05)
u can use calcs on amc
frodo (19:55:12)
You can use a calculator on the AMC.
DPatrick (19:55:31)
Yes, but none of the problems require them, including this one.
DPatrick (19:55:42)
Let's finish the AMC 12A:
DPatrick (19:55:44)
DPatrick (19:55:49)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-34276633.gif
DPatrick (19:56:17)
There are many ways to approach this...some solutions different than the one I'm about to do are on the message board.
DPatrick (19:56:24)
But when I saw this, it screamed out one thing to me.
DPatrick (19:56:40)
This seems like a classic case of where we'd like to set up a recursion.
Rikimaru (19:56:48)
whats a recursion
DPatrick (19:57:02)
I'm glad somebody asked! :)
DPatrick (19:57:16)
We're dealing with an operation on the set {1,2,...,n}, and we can build bigger spacy sets from smaller ones.
DPatrick (19:57:36)
For example, suppose we have a spacy subset of {1,2,...,n} that doesn't have n in it. What can we say about it?
Hamster1800 (19:57:57)
It's a spacy subset of {1,2,...,n-1}
DPatrick (19:58:20)
Exactly! That's what I mean by "recursion" -- we can build bigger spacy subsets from smaller ones.
DPatrick (19:58:40)
On the other hand, what if we have a spacy subset of {1,2,...,n} that *does* have n in it? What can we say about it?
mdk (19:59:00)
can you elaborate on that?
DPatrick (19:59:38)
Sure...but I think somebody else just beat me to it...
Hamster1800 (19:59:21)
If we take out the n it's a spacy subset of {1,2,...,n-3} since it can't have an n-1 or an n-2
Bictor717 (19:59:34)
Everything except n is a spacy subset for {1,2... n-3}
DPatrick (20:00:08)
Right. If we have a spacy subset of {1,2,...,n} with n in it, and we throw away the n, then what's left is a spacy subset of {1,2,...,n-3}.
DPatrick (20:00:31)
Xantos C. Guin (20:00:51)
s_{n}=s_{n-1}+s_{n-3}
worthawholebean (20:00:52)
s(n-3)+s(n-1)=s(n)
DPatrick (20:01:31)
Precisely. As we just discussed, every spacy subset is either a spacy subset of n-1 (if it doesn't have n), or a spacy subset of n-3 (if it has n, after we throw the n away).
DPatrick (20:01:36)
Potato Theory (20:01:18)
We need to find the first sets then.
Potato Theory (20:01:27)
In order to build on them.
DPatrick (20:01:55)
Right. Now we just need to calculate the base cases and churn out the numbers.
DPatrick (20:02:06)
How many spacy subsets of {1}?
frodo (20:02:13)
2
DiscreetFourierTransform (20:02:13)
2
Xantos C. Guin (20:02:17)
2, {1} and {}
Hamster1800 (20:02:18)
2: {} and {1}
DPatrick (20:02:26)
s_1 = 2 (both the empty set and {1} are spacy)
DPatrick (20:02:33)
What's s_2?
tjhance (20:02:35)
3 for n=2
frodo (20:02:36)
3
worthawholebean (20:02:37)
3
ra5249 (20:02:37)
3
Rikimaru (20:02:37)
3
DiscreetFourierTransform (20:02:38)
3
DPatrick (20:02:51)
s_2 = 3 (any subset except {1,2} is spacy)
DPatrick (20:03:01)
Similarly s_3 = 4 (the empty set and any singleton is spacy)
DPatrick (20:03:15)
Now we just churn using our recurrence relation:
DPatrick (20:03:43)
s_4 = 4+2 = 6
s_5 = 6+3 = 9
s_6 = 9+4 = 13
s_7 = 13+6 = 19
s_8 = 19+9 = 28
s_9 = 28+13 = 41
s_10 = 41+19 = 60
s_11 = 60+28 = 88
Finally,
s_12 = 88+41 = 129. Answer (E).
DPatrick (20:04:17)
Note that if the condition is instead that we don't want any two adjacent numbers, what do we get?
Hamster1800 (20:04:33)
fibonacci?
ra5249 (20:04:37)
s_n = s_n-1 + s_n-2, right?
DPatrick (20:04:50)
Exactly, we would get the Fibonacci numbers!
DPatrick (20:05:25)
Some of these have been posted on the AMC message board.
DPatrick (20:05:41)
But the advantage of the way that I just described is that you don't have to be too clever!
DPatrick (20:05:57)
Before we move on to other problem requests (I'll go until about 8:30 Eastern), I would like to mention that 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.
DPatrick (20:06:03)
http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php
DPatrick (20:06:12)
Three books in the Introduction-level series -- Counting & Probabilty, Geometry, and Number Theory -- are currently available. The fourth in the series -- Algebra -- should be available in late March.
DPatrick (20:06:23)
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.
DPatrick (20:06:36)
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 3 & 4, 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
DPatrick (20:06:55)
And if you haven't yet qualified for the AIME, you have another chance! The AMC 10B/12B is on Wednesday, February 21. 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/2006-2007/CU2007/2007-CU-list.shtml
DPatrick (20:07:11)
The registration deadline for the B test date is February 13, so don't delay!
DPatrick (20:07:34)
OK...as I said, I'll stay until about 8:30 and take requests for other problems.
DPatrick (20:08:35)
I've got a requests for 17 and 19 off of both tests. I'll do them in that order.
DPatrick (20:08:46)
Let's do 17 off the 10A:
DPatrick (20:09:05)
...hang on, I'm just looking for it.
DPatrick (20:09:28)
Actually, I don't have that one typed up, sorry! :(
DPatrick (20:09:37)
So let's do #19 from the 10A:
DPatrick (20:09:43)
DPatrick (20:09:56)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-211729524.gif
DPatrick (20:10:00)
DPatrick (20:10:02)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/847aadaec40297e602f257c2e27c2ebe.png
DPatrick (20:10:23)
The funny thing about 'half' is that both halves are equal!
DPatrick (20:10:27)
Which looks easier to calcuate, the white area or the black area?
Hamster1800 (20:10:34)
White by a long shot
tjhance (20:10:36)
white area; just 4 triangles
DPatrick (20:10:46)
I vote for the white.
DPatrick (20:10:56)
Let's make the side length of the square 2, just to have an easy number to work with.
DPatrick (20:11:09)
(I'm going to go a little faster, so we can get through as many as possible)
DPatrick (20:11:21)
So each white triangle has area 1/2, and the total white area is 2.
DPatrick (20:11:34)
What's the area of a 45-45-90 triangle, if its hypotenuse has length x?
Hamster1800 (20:11:49)
x^2/4
DPatrick (20:12:03)
The sides are length x/sqrt(2), so the area is x^2/4.
DPatrick (20:12:17)
Therefore, the total white area is x^2. So x^2 = 2, and x = sqrt(2).
DPatrick (20:12:29)
How do we then get the brush width?
DPatrick (20:13:22)
Well, the sides are length 2, and the white length on each side is sqrt(2).
DPatrick (20:13:39)
This means that the sides of the little corners of the painted area have length (2-sqrt(2))/2. So the width of the brush is this times sqrt(2):
DPatrick (20:13:47)
DPatrick (20:14:07)
DPatrick (20:14:13)
Choice (C).
DPatrick (20:14:33)
Like I said, I did it pretty quickly...you may want to reread it afterwards if you didn't follow.
DPatrick (20:14:49)
My key step was arbitrarily making the side length 2, so I didn't have to drag a variable around.
DPatrick (20:15:13)
Let's quickly go back and do #17; somebody typed it for me, and the solution is quick:
worthawholebean (20:11:33)
tjhance (20:15:37)
75 divides n, so let n=(3^a)(5^b)
EunuchOmerta_2 (20:15:41)
75=5^2*3
DPatrick (20:16:12)
Right, the left side has factors of 3 and 5, so we know n does too.
DPatrick (20:16:20)
This *strongly* suggests trying for n=15.
DPatrick (20:16:36)
...because we know it can't be smaller!
EunuchOmerta_2 (20:16:24)
so m = 5*3^2, n=5*3, m+n = 45+15=60
Potato Theory (20:16:38)
Match factors. You need a 5 and 3^2.
DPatrick (20:17:10)
Yep, that's all there is to it! Letting n=15 makes m=45, and because n must have factors of 3 and 5, there's no way we can go any smaller.
DPatrick (20:17:22)
So the answer is 15+45 = 60. (D)
DPatrick (20:17:51)
Let's do #17 off the AMC 12A (which incidentally is one of my favorites!).
DPatrick (20:18:04)
DPatrick (20:18:18)
One idea is to try to bash with trig identities. The question is, which ones?
Potato Theory (20:18:18)
Use subtraction formula. cos(a-b) = cos(a)cos(b)+sin(a)sin(b)
Rikimaru (20:18:24)
cos(a-b)=(cosa)(cosb)+(sina)(sinb), if that helps
DPatrick (20:18:37)
Sure, we can start by looking at what we need:
DPatrick (20:18:41)
DPatrick (20:18:49)
But how are we going to get (cos a)(cos b) and (sin a)(sin b)?
DiscreetFourierTransform (20:18:55)
Square these equations to get something close.
Hamster1800 (20:18:56)
square stuff!
DPatrick (20:19:08)
Square both given equations!
DPatrick (20:19:12)
DPatrick (20:19:16)
worthawholebean (20:19:16)
Add them too!@
ra5249 (20:19:21)
Add
DPatrick (20:19:35)
Now add them! sin^2+cos^2 = 1 is our friend!
DPatrick (20:19:41)
DPatrick (20:19:50)
Solving gives cos(a-b) = 1/3. Answer (B).
DPatrick (20:20:04)
The most useful trig identities are usually the simplest! :)
DPatrick (20:20:19)
Let's do #19 from the 12A (another really good problem I thought).
DPatrick (20:20:29)
DiscreetFourierTransform (20:20:18)
Can we do 19 with vector cross products?
#H34N1_ (20:20:39)
Use the Shoelace Theorem [img id=em-0]
DPatrick (20:20:53)
We could do either of those, but I wouldn't recommend it.
DPatrick (20:21:04)
There's an extremely elegant solution.
DPatrick (20:21:12)
Let's sketch a crude picture:
DPatrick (20:21:21)
DPatrick (20:21:23)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/e4695ae9aa8971e118a9a291632983ba.png
DPatrick (20:21:29)
(ok, maybe not so crude.)
Someone (20:21:30)
the y coordinate must be + or - 18 or something like that
DPatrick (20:21:46)
If we think of BC as the base of triangle ABC, then the fact that [ABC] = 2007 and BC = 223 means that A must be distance 4014/223 = 18 from line BC. So A must lie on one of the dashed lines in the picture below, each of which are 18 from the x-axis.
DPatrick (20:21:54)
(Note: the picture is not to scale!)
DPatrick (20:21:58)
DPatrick (20:22:00)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/e02b8119ee95076d89cb93cd6b437793.png
ra5249 (20:21:54)
ED has length 9rt2
PenguinIntegral (20:22:20)
and same thing with ED?
DPatrick (20:22:28)
Similarly, since [ADE] = 7002 and DE = 9*sqrt(2), the point A must be distance 14004/9*sqrt(2) = 1556*sqrt(2) from line DE. So A must also lie on one of the two diagonal dashed lines in the picture below:
DPatrick (20:22:34)
DPatrick (20:22:36)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/9e658c7d120da6770fd352546799a252.png
DPatrick (20:22:43)
So the four possible positions of A are given by the intersections of the dashed lines.
DPatrick (20:22:52)
Note that the picture is not to scale, nor did we compute those points ... becuase we don't have to!!
DPatrick (20:23:03)
What's the sum of their x-coordinates?
Xantos C. Guin (20:23:12)
The average x coordinate is equal to the x coordinate of Y
DPatrick (20:23:30)
Exactly. The four possible positions of A form a parallelogram with center Y, which is the intersection of BC and DE. So the sum of the x-coordinates of the vertices of this parallelogram is just 4 times the x-coordinate of Y.
DPatrick (20:23:46)
So all we have to do is find Y!
DiscreetFourierTransform (20:23:40)
so you just need to find the intersectoin of the line y = x - 300 with y = 0
DPatrick (20:23:55)
It is easy to determine that Y = (300,0), so the answer is 4*300 = 1200, choice (E).
PenguinIntegral (20:23:42)
that's fairly cool
DPatrick (20:24:11)
Note that we only needed a very crude, not-to-scale picture in order to see this.
DPatrick (20:24:16)
Ugly numbers in a problem statement (like 7002 and 689) should make you suspicious that there's a simple solution that doesn't require an ugly calculation. There were lots of ugly ways to do this problem, but the above solution is, in my opinion, by far the nicest!
mdk (20:24:06)
#20 #20
worthawholebean (20:24:30)
er, 12A #20
DPatrick (20:24:49)
OK, one more -- #20 from the 12A
DPatrick (20:25:33)
DPatrick (20:25:51)
now a ranger (20:26:02)
imagine it in 2d
now a ranger (20:26:05)
to make it easier
DPatrick (20:26:17)
Often the best strategy with 3D problems!
DPatrick (20:26:22)
We obviously need to figure out the size of the sliced pieces. Let's draw a face with the pieces removed:
DPatrick (20:26:29)
DPatrick (20:26:31)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/4484c63bc471ec3df8c3bc43597b90a9.png
DPatrick (20:26:34)
If the side length of the octagon is s, how to we find s?
worthawholebean (20:26:54)
Use the 45-45-90 triangles at the corners.
DPatrick (20:27:12)
The pieces removed contribute side length s/sqrt(2).
DPatrick (20:27:26)
DPatrick (20:27:56)
Note that rationalizing denomiators here makes subsequent calculations a lot easier.
DPatrick (20:28:09)
So we can compute the side length of the tetrahedra:
DPatrick (20:28:15)
DiscreetFourierTransform (20:28:08)
Now we make use of the fact that the tetrahedrons are right angled.
DPatrick (20:28:32)
Indeed, so their volumes are easy to compute!
EunuchOmerta_2 (20:28:36)
1/3Basearea*height
kostya (20:28:35)
take that, cube it and divide by 6
Someone (20:28:42)
cube it, divide by 6, multiply by 8
DPatrick (20:29:30)
Indeed, if the side length is r, then it's just r^3/6. (Think of the 45-45-90 triangle as the base, which has area r^2/2, and then r is the height, and use 1/3*base*height.)
DPatrick (20:29:39)
DPatrick (20:30:12)
(well, it's not s, s was the side length of the octagon, but my "new" s is the side length of the tetrahedron that I found above.)
DPatrick (20:30:20)
Hence the answer is (B).
DPatrick (20:30:46)
That's all for tonight! Good luck on the AMC 10B/12B if you're taking it, and see you back here in 2 weeks for the AMC B Math Jam!
Someone (20:31:08)
about when is the latest that amc results will be out?
now a ranger (20:31:24)
is there any chance the cutoff will be 114 or less?
now a ranger (20:31:25)
for amc 10
DPatrick (20:31:30)
"official" scores usually take about 3 weeks.
DPatrick (20:31:55)
The cutoff might drop below 120 if fewer than 1% of students score 120 of higher. It has happened before.
Potato Theory (20:32:23)
What about for the AMC12? Would its cutoff drop by any chance?
DPatrick (20:32:51)
The AMC 12 cutoff might drop below 100 as well, if fewer than 5% of students score 100+.
