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2014 AMC 10/12 B Discussion

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A discussion of problems from the AMC 10/12 B, which is administered February 19. We will cover the last 5 problems on each test, as well as requested earlier problems on the tests.

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Facilitator: Dave Patrick

DPatrick 2014-02-20 18:59:37
Welcome to the 2014 AMC 10B/12B Math Jam!
DPatrick 2014-02-20 18:59:47
I'm Dave Patrick, and I'll be leading our discussion tonight.
DPatrick 2014-02-20 19:00:00
Before we get started I would like to take a moment to quickly explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick 2014-02-20 19:00:08
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick 2014-02-20 19:00:21
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 2014-02-20 19:00:39
There are a lot of students here! As I said, only (a fraction of the) 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!!
DPatrick 2014-02-20 19:01:01
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. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick 2014-02-20 19:01:30
We have at least one teaching assistant here tonight to help out: Alyssa (aka baozhale).
DPatrick 2014-02-20 19:01:46
I think a second TA is on her way, she may be running a little late.
copeland 2014-02-20 19:02:00
Alyssa is in her third years at URI. She is in the International Engineering Program for Mechanical Engineering and Chinese, which means that she learns how to be a mechanical engineer in Chinese as well as in English. She is also studying mathematics, and began in the masters program for that this fall. Alyssa wants to do research and teach, though she has not yet decided between mathematics and nanotechnology. Right now, she is helping teach math while doing nanotechnology research, which seems a reasonable compromise. In case anyone was wondering about the username, it's the pinyin (English spelling) of the Chinese word ``exploded\" (爆�,�了.)
DPatrick 2014-02-20 19:02:14
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
DPatrick 2014-02-20 19:02:33
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 final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.
thejet 2014-02-20 19:02:48
Hi baozhale (sorry don't really know who you are :/)
sojourner1 2014-02-20 19:02:48
你好爆炸了!
DPatrick 2014-02-20 19:03:07
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B. Two of these problems are the same, 10B Problem 25 and 12B Problem 22. We'll only solve that problem once.
DPatrick 2014-02-20 19:03:31
Then we'll see how long we've been here -- we might do another problem or two if we're not all too tired.
DPatrick 2014-02-20 19:03:47
Let's get started with 10B Problem 21:
DPatrick 2014-02-20 19:03:55
21. Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length 33 and $\overline{CD}$ of length 21. The other two sides are of lengths 10 and 14. The angles at $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD?$
$\phantom{10B:21}$
$\text{(A) } 10\sqrt6 \quad \text{(B) } 25 \quad \text{(C) } 8\sqrt{10} \quad \text{(D) } 18\sqrt2 \quad \text{(E) } 26$
DPatrick 2014-02-20 19:04:20
(You'll notice the current problem will always be posted to the top. You can resize that top region by dragging the horizontal bar that separates it from the main discussion window.)
gkhsieh 2014-02-20 19:04:44
draw a picture
liuh008 2014-02-20 19:04:44
Draw a diagram;
DrMath 2014-02-20 19:04:44
Diagram always!
dli00105 2014-02-20 19:04:44
draw a picture!!
Wiggle Wam 2014-02-20 19:04:44
First draw a picture!!!
DPatrick 2014-02-20 19:04:52
Of course, we should draw a diagram:
DPatrick 2014-02-20 19:04:57
DPatrick 2014-02-20 19:05:03
Anything you think we should add to this diagram?
mathbeida 2014-02-20 19:05:21
then drop two perpendicular line from C and D
DivideBy0 2014-02-20 19:05:21
altitudes!
alex766 2014-02-20 19:05:21
heights from D and C?
math0127 2014-02-20 19:05:21
the heights from points c and d
xXpapayaXx 2014-02-20 19:05:21
Draw altitudes
brandbest1 2014-02-20 19:05:21
Draw the heights of the trapezoid.
Bomist0 2014-02-20 19:05:21
extra perpendicular lines from C and D to AB
DPatrick 2014-02-20 19:05:27
Yes, the altitudes from $C$ and $D$ to $\overline{AB}$. I'll also label some unknown edges:
DPatrick 2014-02-20 19:05:32
DPatrick 2014-02-20 19:05:46
Great -- what equations can we now write?
tanuagg13 2014-02-20 19:06:03
pythagorean theorem to find a and b
SockFoot 2014-02-20 19:06:03
pythagorean theorem stuff
codyj 2014-02-20 19:06:03
pythagorean theorem
DPatrick 2014-02-20 19:06:12
Yes, we have two equations given by the Pythagorean Theorem on the little triangles.
joshualee2000 2014-02-20 19:06:33
$a^2 + h^2 = 100$
steve123456 2014-02-20 19:06:33
a^2+h^2=100 and b^2+h^2=196
WOLFHEART 2014-02-20 19:06:33
a^2 + h^2 = 10^2
joshualee2000 2014-02-20 19:06:33
$b^2 + h^2 = 196$
ChenthuranA 2014-02-20 19:06:33
b^2+h^2=196
sparkles257 2014-02-20 19:06:33
b^2 + h^2 = 196
DPatrick 2014-02-20 19:06:41
\begin{align*}
a^2 + h^2 &= 10^2, \\
b^2 + h^2 &= 14^2.
\end{align*}
DivideBy0 2014-02-20 19:06:55
we know a+b=33-21=12
CreativeNinja 2014-02-20 19:06:55
a+b=12
2kev111 2014-02-20 19:06:55
a plus b equals 12
oink 2014-02-20 19:06:55
a+b=12
ingridzhang97 2014-02-20 19:06:55
a+b=12
hotstuffFTW 2014-02-20 19:06:55
A + B = 12
DPatrick 2014-02-20 19:07:00
We also have $a+b = 33-21 = 12$ from the length $AB$.
DPatrick 2014-02-20 19:07:09
How do we use these equations?
fprosk 2014-02-20 19:07:32
subtract the second equation from the first
mathtastic 2014-02-20 19:07:32
subtract first from second and factor as (b-a)(b+a)
mathawesomeness777 2014-02-20 19:07:32
Subtract the two top equations?
harvey2014 2014-02-20 19:07:32
b^2-a^2
syrup827 2014-02-20 19:07:32
subtract out h^2
DPatrick 2014-02-20 19:07:53
There are lots of ways to proceed. I like equating the two expressions for $h^2$ from the two Pythagorean equations.
DPatrick 2014-02-20 19:08:05
(Which is really the same thing as subtracting them to cancel the $h^2$ term.)
DPatrick 2014-02-20 19:08:17
Equating $h^2$ in our Pythagorean expressions gives us $100 - a^2 = 196 - b^2.$ So we have $b^2 - a^2 = 96.$
DaChickenInc 2014-02-20 19:08:37
b-a=8
yajaniaj 2014-02-20 19:08:37
$12(b-a)=96$
brandbest1 2014-02-20 19:08:37
difference of two squares
Bomist0 2014-02-20 19:08:37
difference of squares
IsabeltheCat 2014-02-20 19:08:37
b-a=8
Eudokia 2014-02-20 19:08:37
divide by b+a
aaa16797 2014-02-20 19:08:37
you can write it in (b-a)(b+a)
DPatrick 2014-02-20 19:08:47
Aha! Dividing this by $a+b = 12$ gives us $b-a = 96/12 = 8.$
ahaanomegas 2014-02-20 19:09:14
Therefore, a = 2, b = 10.
rraj411 2014-02-20 19:09:14
2 eqn 2 variable
dli00105 2014-02-20 19:09:14
a=2 and b=10
abishek99 2014-02-20 19:09:14
a=2, b=10
WalkerTesla 2014-02-20 19:09:14
b= 10, a=2
sojourner1 2014-02-20 19:09:14
2b is 20, b=10, a = 2
DPatrick 2014-02-20 19:09:23
Right, now we have the simple system $a+b = 12$ and $b-a = 8$, which solves to give $b = 10$ and $a= 2.$
DPatrick 2014-02-20 19:09:33
And so what is $h$?
mathmaster2012 2014-02-20 19:09:49
ws5188 2014-02-20 19:09:49
4sqrt6
ahaanomegas 2014-02-20 19:09:49
sqrt(96)
mathmaster2012 2014-02-20 19:09:49
LightningX48 2014-02-20 19:09:49
4sqrt6
acegikmoqsuwy2000 2014-02-20 19:09:49
sqrt96?
DPatrick 2014-02-20 19:09:53
We have $h^2 = 100-a^2 = 100-4 = 96,$ so $h = \sqrt{96}.$
DPatrick 2014-02-20 19:10:11
Here's the updated picture with the lengths filled in:
DPatrick 2014-02-20 19:10:21
DPatrick 2014-02-20 19:10:28
Great -- how do we finish from here?
iguana123 2014-02-20 19:10:52
AC is the shorter diagonal
flyrain 2014-02-20 19:10:52
draw the shorter diagonal
joshxiong 2014-02-20 19:10:52
Clearly AC is shorter
guilt 2014-02-20 19:10:52
find AC using pythagorean theorem
johnguolex 2014-02-20 19:10:52
draw AC and use pythag to figure out the length
masalvada12 2014-02-20 19:10:52
draw in diagonals
blueferret 2014-02-20 19:10:52
pythagorean theorem with 23 and sqrt(96)
DPatrick 2014-02-20 19:11:01
Right. We want the length $AC,$ which we can get from the right triangle below:
DPatrick 2014-02-20 19:11:05
DPatrick 2014-02-20 19:11:17
This triangle has height $h=\sqrt{96}$ and length $21+a = 21+2 = 23.$
wehac 2014-02-20 19:11:33
23^2+96=625. So the answer is B) 25
summitwei 2014-02-20 19:11:33
AC=25
hesa57 2014-02-20 19:11:33
96+23^2=x^2
ajoy 2014-02-20 19:11:33
23^2 + 96 = 625. so the square root of that is 25!
blueberry7 2014-02-20 19:11:33
so AC=25
DPatrick 2014-02-20 19:11:47
Its hypotenuse is \[\sqrt{(\sqrt{96})^2 + 23^2} = \sqrt{96 + 529} = \sqrt{625}.\]
DPatrick 2014-02-20 19:11:51
This is 25, so the answer is (B).
ingridzhang97 2014-02-20 19:11:54
wasn't b=10? not 8
DPatrick 2014-02-20 19:12:12
Oops, you're right. That "8" should be a "10" in the last two pictures.
DPatrick 2014-02-20 19:12:19
Fortunately we didn't use it!
DPatrick 2014-02-20 19:12:34
On to 10B Problem #22:
DPatrick 2014-02-20 19:12:38
22. Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?
$\phantom{10B:22}$
$\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3$
DPatrick 2014-02-20 19:12:43
DPatrick 2014-02-20 19:12:59
What segment is begging to be added to the picture?
nsd 2014-02-20 19:13:35
draw from center of semicircle to center of circle.
Darn 2014-02-20 19:13:35
Connect the center of the circle to the edge of the square
CreativeNinja 2014-02-20 19:13:35
a line going through the point of tangency
sparkles257 2014-02-20 19:13:35
the segment from center to midpt of side
starwars123 2014-02-20 19:13:35
the segment from the center to the edge of the square
fprosk 2014-02-20 19:13:35
center of circle to edge of quare
briantix 2014-02-20 19:13:35
the centers of the circles
mathbeida 2014-02-20 19:13:35
connect the center of the circle to a semicirle center
DPatrick 2014-02-20 19:13:51
Right, arguably there are two segments begging to be added.
DPatrick 2014-02-20 19:13:57
Whenever we have tangent circles, it's almost always helpful to draw the segment that connects the centers of the circles, which also passes through the point of tangency. So let's draw it:
DPatrick 2014-02-20 19:14:03
DPatrick 2014-02-20 19:14:25
And, there's a right triangle if we draw the segment from the center of the big circle straight down to the bottom base. I'll also label the edges that we know:
DPatrick 2014-02-20 19:14:32
DPatrick 2014-02-20 19:14:45
And now what do we have?
DaChickenInc 2014-02-20 19:15:00
right triangle w/ side lengths 1/2, 1, 1/2+r=$\sqrt{5}/2$
DrMath 2014-02-20 19:15:00
so hypotenuse is 1/2 root 5
Darn 2014-02-20 19:15:00
Hypotenuse is sqrt5/2
joshualee2000 2014-02-20 19:15:00
so the hypotenuse is $\sqrt{5}/2$
DPatrick 2014-02-20 19:15:27
On the one hand, by the Pythagorean Theorem, the hypotenuse of that right triangle is $\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}.$
DPatrick 2014-02-20 19:15:43
(You might recognize a 1-2-$\sqrt5$ right triangle, and this one is half that size.)
johnguolex 2014-02-20 19:16:04
a right triangle!
MATHCOUNTSmath 2014-02-20 19:16:04
and subtract 1/2
pkbehl 2014-02-20 19:16:04
the radius of the semicircle is 1/2 so we subtract
alex766 2014-02-20 19:16:04
then you have to subtract 1/2 for the radius of the smaller circle
DPatrick 2014-02-20 19:16:19
Right, on the other hand, it's $\frac12 + r$, where $r$ is the radius of the big circle.
hexagram 2014-02-20 19:16:32
mathmaster2012 2014-02-20 19:16:32
19CindyW 2014-02-20 19:16:34
so the radius of the circle is sqrt(5)-1/2
cnnwy1282 2014-02-20 19:16:38
Subtracting the radius of the semicircle, we get (sqrt5-1)/2
DPatrick 2014-02-20 19:16:46
Yep: $r = \frac{\sqrt5}{2}-\frac12 = \frac{\sqrt5-1}{2},$ answer (B).
DPatrick 2014-02-20 19:17:18
That problem was probably the easiest of all the problems we'll do tonight.
DPatrick 2014-02-20 19:17:32
On to 10B #23, which is also 12B #19:
DPatrick 2014-02-20 19:17:38
23. A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\phantom{10B:23}$
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$
DPatrick 2014-02-20 19:17:46
DPatrick 2014-02-20 19:17:58
(copeland did the diagram!)
dli00105 2014-02-20 19:18:34
cross section
Nahmid 2014-02-20 19:18:34
take a 2-d cross section!
DPatrick 2014-02-20 19:18:41
Indeed, like many 3D geometry problems, it's really a 2D problem in disguise.
DPatrick 2014-02-20 19:18:46
We can look at a cross-section of the object sliced down the middle:
DPatrick 2014-02-20 19:18:53
DPatrick 2014-02-20 19:19:04
We might find it helpful to assign variables to the lengths. How can we do that?
brandbest1 2014-02-20 19:19:39
let the top radius be 1, then let the radius be r
mapletree14 2014-02-20 19:19:39
Let the top have length 1, we can assume this WLOG.
Nahmid 2014-02-20 19:19:39
let the radius of the top be 1, the bottom be R
WalkerTesla 2014-02-20 19:19:39
We can let the top radius be 1 and solve for the bottom radius
DPatrick 2014-02-20 19:19:47
I like this idea. Since we the data that we care about (the ratio of the volumes that we're given, and the ratio of the radii that we want), we have a little flexibility in assign variables to lengths -- we can pick one length to be 1 if we want, for convenience.
DPatrick 2014-02-20 19:20:00
And since the answer we want is (radius of bottom face)/(radius of top face), it seems likely that things may turn out nice if we make the radius of the top face equal to 1, so that our answer is just the radius of the bottom face.
DPatrick 2014-02-20 19:20:17
So let's let the radius of the top face be 1 and the radius of the bottom face be $r$. (Again, note that our final answer will be just $r.$) Let's also call the radius of the sphere $s.$
DPatrick 2014-02-20 19:20:23
DPatrick 2014-02-20 19:20:35
Can we relate $r$ and $s$?
acegikmoqsuwy2000 2014-02-20 19:20:59
pythegorean theorem
joshxiong 2014-02-20 19:20:59
Use the pythagorean theorem
Arcmage101 2014-02-20 19:20:59
pythagorean theorem
hwl0304 2014-02-20 19:20:59
pythagorean theorem??
fireonice18 2014-02-20 19:20:59
pythagorean theorem
DPatrick 2014-02-20 19:21:07
How? I don't see a right triangle anywhere...
SockFoot 2014-02-20 19:21:30
make on
SirNinja 2014-02-20 19:21:30
draw one
guilt 2014-02-20 19:21:30
draw the altitude
NikhilP 2014-02-20 19:21:30
draw one
awesomethree 2014-02-20 19:21:30
draw it
brandbest1 2014-02-20 19:21:30
drop an altitude from the top vertex
medianpi 2014-02-20 19:21:30
make one
abishek99 2014-02-20 19:21:30
drop an altitiude
DPatrick 2014-02-20 19:21:43
Sure, if we need a right triangle, there's no law that says we can't just go ahead and make one!
DPatrick 2014-02-20 19:21:47
DPatrick 2014-02-20 19:22:07
It has legs $2s$ and $r-1$. But don't we also know its hypotenuse?
blueferret 2014-02-20 19:22:32
1+r
sophiazhi 2014-02-20 19:22:32
1+r
DaChickenInc 2014-02-20 19:22:32
1+r by equal tangents
urmilla 2014-02-20 19:22:32
r+1
fdas 2014-02-20 19:22:32
r+1
blueferret 2014-02-20 19:22:32
the hypotenuse is 1+r
Nahmid 2014-02-20 19:22:37
the tangents to a circle are equivalent, so the hyp is r+1
brandbest1 2014-02-20 19:22:37
by the two tangent theorem, the two segments of the hyp have length r and 1, so it's length is r+1
DPatrick 2014-02-20 19:22:51
Right! It's clearer if we add the point of tangency from the circle to the hypotenuse...
DPatrick 2014-02-20 19:22:56
DPatrick 2014-02-20 19:23:06
The distances from a point outside a circle to its two corresponding points of tangency are equal:
DPatrick 2014-02-20 19:23:14
DPatrick 2014-02-20 19:23:21
So the blue triangle has hypotenuse $r+1$.
SirNinja 2014-02-20 19:23:37
(2s)^2+(r-1)^2=(r+1)^2
hjl00 2014-02-20 19:23:37
pythagrean theorem
DPatrick 2014-02-20 19:23:42
Therefore by the Pythagorean Theorem we have \[(2s)^2 + (r-1)^2 = (r+1)^2.\]
Eudokia 2014-02-20 19:24:10
s^2=r
parley999 2014-02-20 19:24:10
s^2=r
joshualee2000 2014-02-20 19:24:10
so s^2 = r
dli00105 2014-02-20 19:24:10
which simplifies to s^2=r
LightningX48 2014-02-20 19:24:10
s^2 = r
DPatrick 2014-02-20 19:24:17
Right: expanding this out gives \[4s^2 + r^2 - 2r + 1 = r^2 + 2r + 1,\]
so $4s^2 = 4r,$ hence $r = s^2.$
DPatrick 2014-02-20 19:24:35
OK, let's file $r=s^2$ away for now.
rraj411 2014-02-20 19:24:42
volume relation now
DPatrick 2014-02-20 19:24:52
Right. We know that the volume of the sphere is $\frac43\pi s^3.$
DPatrick 2014-02-20 19:25:00
Therefore the volume of the frustum (the truncated cone) is twice that, or $\frac83\pi s^3.$
DPatrick 2014-02-20 19:25:05
How do we compute the volume of the frustum in terms of its bases and its height?
ingridzhang97 2014-02-20 19:25:38
extend the two sides of the "trapezoid" to meet at a point
Nahmid 2014-02-20 19:25:38
a bigger cone minus a smaller one
blueferret 2014-02-20 19:25:38
big cone minus little cone?
alex31415 2014-02-20 19:25:38
add a "top" and subtract
DrMath 2014-02-20 19:25:38
volume of total-chopped off pary
tanishq1 2014-02-20 19:25:38
draw the missing pyramid on top
AlcumusGuy 2014-02-20 19:25:38
volume of large cone minus volume of chopped off cone
DPatrick 2014-02-20 19:25:45
Many solutions I've seen, at this point, magically use the "formula" for volume of a right circular frustum. I suspect that most of you, like me, don't know this formula. Fortunately, it's not that hard to determine the volume of a frustum.
DPatrick 2014-02-20 19:25:55
We can "untruncate" it and draw the original cone:
DPatrick 2014-02-20 19:26:00
DPatrick 2014-02-20 19:26:14
I've labeled the additional height as $h$ -- how can we compute $h$ in terms of $r$ and/or $s$?
mathbeida 2014-02-20 19:26:31
h/(h+2s) = 1/r
VD8M9 2014-02-20 19:26:31
similar triangles
MathStudent2002 2014-02-20 19:26:31
similar triangles!
pl210741 2014-02-20 19:26:31
Similar triangles
geogirl08 2014-02-20 19:26:31
similarity
nuggetfan 2014-02-20 19:26:31
similar triangles!!!
iguana123 2014-02-20 19:26:31
similar triangles!
DPatrick 2014-02-20 19:26:38
Yes! By similar triangles, we know that \[\frac{h}{1} = \frac{h+2s}{r}.\]
DaChickenInc 2014-02-20 19:26:50
by similarity, $h=\frac{2s}{r-1}$
DPatrick 2014-02-20 19:27:00
Indeed, this simplifies to $rh = h+2s$, and hence $h = \dfrac{2s}{r-1}.$
DPatrick 2014-02-20 19:27:16
So the volume of the frustum is the volume of the entire cone minus the volume of the little cone on top.
DPatrick 2014-02-20 19:27:25
This is \[\frac13\pi r^2(2s+h) - \frac13\pi 1^2(h).\]
tennis1729 2014-02-20 19:27:58
which is equal to 8/3 s^3 pi
DrMath 2014-02-20 19:28:10
DPatrick 2014-02-20 19:28:21
Right, let's simplify a little first before making that equality though. (It doesn't matter much what order you finish up in.)
DrMath 2014-02-20 19:28:29
*s^3,my bad
DPatrick 2014-02-20 19:28:39
The frustum volume simplifies to \[\frac13\pi\left(r^2(2s+h)-h\right).\]
DPatrick 2014-02-20 19:28:54
And as we just mentioned, we need this to equal $\frac83\pi s^3$, so we have \[r^2(2s+h)-h = 8s^3.\]
DPatrick 2014-02-20 19:29:15
And this simplifies to $2r^2s + (r^2-1)h = 8s^3.$ How do we get rid of the $h$ from this?
SockFoot 2014-02-20 19:29:39
plug in h
tanishq1 2014-02-20 19:29:39
plug in 2s/(r-1)
crastybow 2014-02-20 19:29:39
plug in what we got before
DrMath 2014-02-20 19:29:39
h=2s/(r-1)
wehac 2014-02-20 19:29:39
h=2s/(r-1)
DPatrick 2014-02-20 19:29:52
Right, we had $h = \dfrac{2s}{r-1}.$
DPatrick 2014-02-20 19:30:18
So $(r^2-1)h = \frac{2s(r^2-1)}{r-1}$...
ingridzhang97 2014-02-20 19:30:26
and that r-1 cancels out with r^2-1=(r-1)(r+1)
dli00105 2014-02-20 19:30:26
it cancels nicely
DPatrick 2014-02-20 19:30:33
...indeed, it's just $2s(r+1)$.
DPatrick 2014-02-20 19:30:43
So our equation becomes \[2r^2s + (r+1)2s = 8s^3.\]
ingridzhang97 2014-02-20 19:31:00
and we can also cancel out an s
blueferret 2014-02-20 19:31:00
reduce the 2s
Tommy2000 2014-02-20 19:31:00
divide by s
dli00105 2014-02-20 19:31:00
divide by 2
mikhailgromov 2014-02-20 19:31:00
Divide by s
leejp1998 2014-02-20 19:31:00
cancle s
DPatrick 2014-02-20 19:31:26
Let's divide by $2s$ and we're down to $r^2 + r + 1 = 4s^2.$
Royalreter1 2014-02-20 19:31:47
substitte r=s^2 in
geogirl08 2014-02-20 19:31:47
LightningX48 2014-02-20 19:31:47
r = s^2!
guilt 2014-02-20 19:31:47
s^2 = r
MathStudent2002 2014-02-20 19:31:47
And we have r=s^2 so the right side is 4r!
joshualee2000 2014-02-20 19:31:47
4s^2= 4r
DPatrick 2014-02-20 19:31:57
Aha, we had filed away $r = s^2$.
DPatrick 2014-02-20 19:32:07
So we just have $r^2 + r + 1 = 4r$, or $r^2 - 3r + 1 = 0.$
crastybow 2014-02-20 19:32:25
so use quadratic formula to find r to finish it off
brandbest1 2014-02-20 19:32:25
solve it for our answer
Mathgeek727 2014-02-20 19:32:25
quadritic equation?
guilt 2014-02-20 19:32:25
r^2 - 3r +1 = 0 -> r = (3+sqrt5)/2 or E
urmilla 2014-02-20 19:32:25
(E) (3+ sqrt 5) /2 by quadratic formula
DPatrick 2014-02-20 19:32:33
Yep. Solving gives \[r = \frac{3 \pm \sqrt5}{2}.\]
DPatrick 2014-02-20 19:32:40
Only the "+" is an answer choice (and only the "+" is greater than 1, which we need).
DPatrick 2014-02-20 19:32:49
So $r = \dfrac{3 + \sqrt5}{2}$ is the answer. Answer (E).
DPatrick 2014-02-20 19:33:27
This was the hardest problem of the 21-25 group on the 10B, in my opinion.
DPatrick 2014-02-20 19:33:44
Let's move on to 10B #24, which was also 12B #18:
DPatrick 2014-02-20 19:33:49
24. The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is bad if it is not true that for every $n$ from 1 to 15 one can find a subset of the numbers that appear consecutively on the circle that sum to $n.$ Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
$\phantom{10B:24}$
$\text{(A) } 1 \quad \text{(B) } 2 \quad \text{(C) } 3 \quad \text{(D) } 4 \quad \text{(E) } 5$
DPatrick 2014-02-20 19:34:07
Yikes. We need to untangle the words.
DPatrick 2014-02-20 19:34:13
Is there a simpler way to describe a bad arrangement?
brandbest1 2014-02-20 19:34:44
if there's a value of n such that you can't find consecutive numbers that sum to n
checkmate1021 2014-02-20 19:34:48
There exists an n from 1 to 15 that cannot be expressed as a sum of consecutive numbers on the circle
Tommy2000 2014-02-20 19:34:56
the sum of any group of numbers doesnt equal n for some number from 1-15
Quadratic64 2014-02-20 19:34:56
an arrangement where one cannot find consecutive numbers that sum to n
DPatrick 2014-02-20 19:35:03
A bad arrangement is an arrangement for which there's some number (between 1 and 15) that cannot be found in our circle as a sum of consecutive elements.
DPatrick 2014-02-20 19:35:45
For some reason (maybe it's just me) I find this reformulation easier that what's stated in the problem. "if it is not true that.." is harder to parse.
DPatrick 2014-02-20 19:36:19
So we need to count arrangements where some sum from 1 to 15 can't be found.
DPatrick 2014-02-20 19:36:22
How can we simplify the search?
DiscipulusBonus 2014-02-20 19:36:43
1-5 and 10-15 are automatic (for 10-14, only one number is missing), so we only need to check if an arrangement can sum to 6,7,8, and 9.
WalkerTesla 2014-02-20 19:36:43
We can always get 1,2,3,4,5,10,11,12,13,14, and 15. Therefore we only need to worry about 6,7,8, and 9
mathcool2009 2014-02-20 19:36:43
n = 1,2,3,4,5,10,11,12,13,14,15 always can be found as a sum
Joe10112 2014-02-20 19:36:43
1-5 can all be found, 10-15 can all be found
Eudokia 2014-02-20 19:36:43
We don't have to look at 1-5 and 10-15
abishek99 2014-02-20 19:36:43
1-5 and 10-15 have to be sums
DPatrick 2014-02-20 19:36:51
We can always find the sums 1 through 5 -- just take a single number.
DPatrick 2014-02-20 19:36:58
We can also always find the sums 10 through 15 -- just omit a single number, or take all five numbers to get 15.
DPatrick 2014-02-20 19:37:06
So the only way that an arrangement can be bad is if one of 6, 7, 8, or 9 is not a sum.
mssmath 2014-02-20 19:37:28
6 and 7 imply 8 and 9 so we need only check 6 and 7
DivideBy0 2014-02-20 19:37:28
Once we have found a set summing to 6, we can choose everything else and obtain a set summing to 9, and similarly for 7 and 8. Thus, we only need to check each case for weather or not we can obtain 6 or 7.
Rogman 2014-02-20 19:37:28
but no 6 implies no 9 and no 7 implies no 8
joshxiong 2014-02-20 19:37:28
also if 6 can be achieved, 9 can
mathmaster2012 2014-02-20 19:37:28
If we can find 6, we can find 15-6=9. If we can find 7, we can find 15-7=8.
DPatrick 2014-02-20 19:37:38
Right, there's a further simplification! 6 is a sum if and only if 9 is a sum: if some consecutive elements sum to 6, then the remaining elements sum to 9, and vice versa.
DPatrick 2014-02-20 19:37:43
In the same way, 7 is a sum if and only if 8 is a sum.
DPatrick 2014-02-20 19:37:48
So an arrangement is bad if and only if either 6 or 7 is not realizable as a sum.
DPatrick 2014-02-20 19:37:56
At this point we can probably just try to construct them.
DPatrick 2014-02-20 19:38:02
How do we construct a bad arrangement that doesn't contain a sum of 6?
DPatrick 2014-02-20 19:38:33
Here's a blank arrangement to start:
DPatrick 2014-02-20 19:38:37
Tuxianeer 2014-02-20 19:38:57
5 and 1 are not adjacent, same with 2 and 4
lucylai 2014-02-20 19:38:57
1 and 5 have to be across from each other
mentalgenius 2014-02-20 19:38:57
just put 5 somewhere to account for rotations
DPatrick 2014-02-20 19:39:22
Sure, let's just put the 5 in the top position -- it doesn't matter where we put it, and fixing it to the top will account for the rotations:
DPatrick 2014-02-20 19:39:27
DPatrick 2014-02-20 19:39:46
And we need the 1 not to be next to the 5. Since reflections don't matter, there's only 1 place (up to reflection) to put the 1:
DPatrick 2014-02-20 19:39:52
johnguolex 2014-02-20 19:39:59
also 4/5 can't be adjacent
jon167 2014-02-20 19:40:12
4 goes near 1
DPatrick 2014-02-20 19:40:22
Right: if we don't have a sum of 6, then we also don't have a sum of 9.
DPatrick 2014-02-20 19:40:31
So the 4 has to go in the blank next to the 1, away from the 5:
DPatrick 2014-02-20 19:40:43
blueberry7 2014-02-20 19:40:58
2 cannot be next to the 4 either
joshualee2000 2014-02-20 19:40:58
since 2 has to be away from 4, it goes on the left
alex31415 2014-02-20 19:40:58
The 2 is between the 5 and the 1.
willabc 2014-02-20 19:40:58
2 can't be next to 4
VD8M9 2014-02-20 19:40:58
2 between 5/1??
19bobhu 2014-02-20 19:40:58
2 has to go between 1 and 5
DPatrick 2014-02-20 19:41:16
At this point, we're done: the 2 can't be next to the 4, so it has to go on the left, and the 3 goes in the remaining spot on the right:
DPatrick 2014-02-20 19:41:21
DPatrick 2014-02-20 19:41:40
This is the only bad arrangement that doesn't have a sum of 6 or 9.
guilt 2014-02-20 19:41:53
now check the case of 7
19bobhu 2014-02-20 19:41:53
And now we have to make the ones that can't make 7
acegikmoqsuwy2000 2014-02-20 19:41:53
now we try doing it for 7
ssk9208 2014-02-20 19:41:53
now 7
Royalreter1 2014-02-20 19:41:53
Now we look for 7/8
DPatrick 2014-02-20 19:42:06
Right: now we have to do the same exercise for bad arrangement(s) that don't have a sum of 7 or 8.
DPatrick 2014-02-20 19:42:15
It's pretty much the same...
geogirl08 2014-02-20 19:42:27
5 is not adjacent to 2, nor is 3 adjacent to 5 for no 7
alex31415 2014-02-20 19:42:27
Place 5 at the top, and 2 at the bottom.
mathbeida 2014-02-20 19:42:33
start with 5 again, 2 can not be adjacent
blueberry7 2014-02-20 19:42:33
the 5 cannot be next to the 2... or the 3
DPatrick 2014-02-20 19:42:39
Again, we start with the 5 on top, and the 2 in the lower-left (since it can't be next to the 5):
DPatrick 2014-02-20 19:42:46
awesomethree 2014-02-20 19:43:02
and 3 in the lower right
starwars123 2014-02-20 19:43:02
3 in the lower right
guilt 2014-02-20 19:43:02
also 3 can't be next to 5
DPatrick 2014-02-20 19:43:17
Right, if we don't have a sum of 7, we also don't have a sum of 8, and that forces the 3 not to be next to the 5:
DPatrick 2014-02-20 19:43:28
elsw 2014-02-20 19:43:42
4 can't be next to 3
Eudokia 2014-02-20 19:43:42
4 can't be next to 3
wehac 2014-02-20 19:43:42
4 is in between 2 and 5
Tommy2000 2014-02-20 19:43:42
4 can't be next to 3, so 4 is between 2 and 5
hotstuffFTW 2014-02-20 19:43:42
4 between 5,2 and 1 between 3,5
DPatrick 2014-02-20 19:43:59
And now we're forced again: the 4 has to go on the left (it can't be next to the 3), and the 1 is what remains:
DPatrick 2014-02-20 19:44:03
DPatrick 2014-02-20 19:44:12
And thus this is the only bad arrangement (up to reflections and rotations) that doesn't have a sum of 7.
abishek99 2014-02-20 19:44:25
only 2 arrangements B
pinkrock 2014-02-20 19:44:25
so the answer in B
Bomist0 2014-02-20 19:44:25
2=>(B)
ericding 2014-02-20 19:44:25
so the answer is B
masalvada12 2014-02-20 19:44:25
therefore answer is 2
DPatrick 2014-02-20 19:44:33
Correct: there are only 2 bad arrangements (one with no 6s or 9s, and one with no 7s or 8s). The answer is (B).
DPatrick 2014-02-20 19:44:59
OK, on to 10B #25 which is also 12B #22:
DPatrick 2014-02-20 19:45:04
25. In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. When the frog is on pad $N,$ $0<N<10,$ it will jump to pad $N-1$ with probability $\dfrac N{10}$ and to pad $N+1$ with probability $1-\dfrac N{10}.$ Each jump is independent of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
$\phantom{10B:25,12B:22}$
$\text{(A) } \dfrac{32}{79} \quad \text{(B) } \dfrac{161}{384} \quad \text{(C) } \dfrac{63}{146} \quad \text{(D) } \dfrac7{16} \quad \text{(E) } \dfrac12$
DPatrick 2014-02-20 19:45:18
Yikes again. Many many words.
DPatrick 2014-02-20 19:45:44
Let's set $p_i$ to be the probability that the frog lives starting from lily pad $i$. So we're trying to find $p_1.$
checkmate1021 2014-02-20 19:45:53
state diagram?
Rogman 2014-02-20 19:45:53
stages?
dli00105 2014-02-20 19:45:53
state diagram
DPatrick 2014-02-20 19:46:26
You could draw a diagram (often called a "state diagram") for this problem if you wanted, but I'm not sure this problem is complicated enough to need one.
DPatrick 2014-02-20 19:46:43
The diagram would basically just be the lily pads, with arrows showing the probabilities of jumping between them.
DPatrick 2014-02-20 19:46:53
Do we know any of the $p$'s right away?
joshxiong 2014-02-20 19:47:16
p_0 = 0, p_10 = 1
Bomist0 2014-02-20 19:47:16
0 and 10
checkmate1021 2014-02-20 19:47:16
p0 is 0
bli1999 2014-02-20 19:47:16
p0 is 0.
mathwizard888 2014-02-20 19:47:16
p_0=0, p_10=1
Tuxianeer 2014-02-20 19:47:16
$p_0=0$ $p_{10}=1$
DPatrick 2014-02-20 19:47:20
$p_0 = 0$ -- the frog has become snake food.
DPatrick 2014-02-20 19:47:28
$p_{10} = 1$ -- freedom!
DPatrick 2014-02-20 19:47:53
(Many of you have also noticed another key fact. I'm going to play dumb for a moment, but we'll get to your observation soon!)
DPatrick 2014-02-20 19:48:01
How are the other $p$'s related?
XxAndreixX 2014-02-20 19:48:43
p(n) = (1-n/10)p(n+1) + (n/10)p(n-1)
Arcmage101 2014-02-20 19:48:43
1-i/10 and i/10
DPatrick 2014-02-20 19:48:54
Right, we've got the data about the jumping that's given in the problem.
DPatrick 2014-02-20 19:49:00
The frog moves from pad $i$ to pad $i+1$ with probability $1-\frac{i}{10}$ and to pad $i-1$ with probability $\frac{i}{10}.$
DPatrick 2014-02-20 19:49:26
If the frog moves to pad $i+1$, his new probability of freedom is $p_{i+1}$.
DPatrick 2014-02-20 19:49:36
If the frog moves to pad $i-1$, his new probability of freedom is $p_{i-1}$.
DPatrick 2014-02-20 19:49:47
Therefore, \[ p_i = \left(1 - \frac{i}{10}\right)p_{i+1} + \frac{i}{10}p_{i-1}.\]
ingridzhang97 2014-02-20 19:50:01
so we can write some equations
DPatrick 2014-02-20 19:50:19
Indeed we can, but first, let's multiply this through by 10 to get rid of the fractions.
DPatrick 2014-02-20 19:50:26
\[ 10p_i = (10-i)p_{i+1} + ip_{i-1}.\]
DPatrick 2014-02-20 19:50:38
So now we have a system of 10 equations in 10 variables (using $p_0=0$ and $p_{10}=1$):
\begin{align*}
10p_1 &= 9p_2, \\
10p_2 &= 8p_3 + 2p_1, \\
10p_3 &= 7p_4 + 3p_2, \\
10p_4 &= 6p_5 + 4p_3, \\
10p_5 &= 5p_6 + 5p_4, \\
10p_6 &= 4p_7 + 6p_5, \\
10p_7 &= 3p_8 + 7p_6, \\
10p_8 &= 2p_9 + 8p_7, \\
10p_9 &= 1 + 9p_8.
\end{align*}
matholympiad25 2014-02-20 19:50:58
yikes
DPatrick 2014-02-20 19:51:10
Actually, I lied...it's "only" 9 equations in 9 variables.
DPatrick 2014-02-20 19:51:25
But that'll still take forever to solve. We could, mind you...
DPatrick 2014-02-20 19:51:40
...but there's an important observation that we can make to simplify things immensely.
mentalgenius 2014-02-20 19:52:01
in the middle, 1/2 chance to go either way
joshualee2000 2014-02-20 19:52:01
$p_5 = \frac{1}{2}$
pi37 2014-02-20 19:52:01
p_5=1/2
ABCDE 2014-02-20 19:52:01
p_5=1/2
Nahmid 2014-02-20 19:52:01
p5 = 1/2
joshualee2000 2014-02-20 19:52:01
but $p_5 = \frac{1}{2}$
mentalgenius 2014-02-20 19:52:01
p5 = 1/2
mathtastic 2014-02-20 19:52:09
ohhh p5=1/2 using logic!!!!!!!! (i think)
DrMath 2014-02-20 19:52:13
Remember $p_5=\frac{1}{2}$!
DPatrick 2014-02-20 19:52:38
Right! Everything in this problem is symmetric with respect to the middle lily pad. So if the frog is in the middle, it's 50-50 that he'll live or get eaten. So $p_5 = \frac12.$
DPatrick 2014-02-20 19:52:49
That's very useful, because it reduces our 9-equation system to a 4-equation system:
\begin{align*}
10p_1 &= 9p_2, \\
10p_2 &= 8p_3 + 2p_1, \\
10p_3 &= 7p_4 + 3p_2, \\
10p_4 &= 3 + 4p_3.
\end{align*}
SockFoot 2014-02-20 19:53:07
solving time
Royalreter1 2014-02-20 19:53:07
Now we can solve!
TheMaskedMagician 2014-02-20 19:53:07
Now this is something we can manage.
DPatrick 2014-02-20 19:53:16
At this point I don't think there's anything clever to do except to slog through and solve this for $p_1.$
DPatrick 2014-02-20 19:53:33
We want to end with $p_1$, so we'll work from the bottom up. Let's eliminate $p_4$ first.
DPatrick 2014-02-20 19:54:04
I'm basically going to zip through this because it's rather boring. But notice how I make choices as I go to make the simplification as painless as possible.
Tommy2000 2014-02-20 19:54:11
multiply equation 3 by 10 and equation 4 by 7
DPatrick 2014-02-20 19:54:21
Good idea: multiplying the third equation by 10 and the fourth equation by 7 gives:
\begin{align*}
100p_3 &= 70p_4 + 30p_2, \\
70p_4 &= 21 + 28p_3.
\end{align*}
DPatrick 2014-02-20 19:54:33
Then substitution gives $100p_3 = 21 + 28p_3 + 30p_2,$ or $72p_3 = 21 + 30p_2.$
DPatrick 2014-02-20 19:54:44
Now our system is:
\begin{align*}
10p_1 &= 9p_2, \\
10p_2 &= 8p_3 + 2p_1, \\
72p_3 &= 21 + 30p_2.
\end{align*}
DivideBy0 2014-02-20 19:54:59
mult eq 2 by 9
sojourner1 2014-02-20 19:54:59
multiply through by 9
DPatrick 2014-02-20 19:55:05
Multiplying the second equation by 9 gives $90p_2 = 72p_3 + 18p_1.$
DPatrick 2014-02-20 19:55:15
Now we can substitute the third equation into the second to get $90p_2 = 21 + 30p_2 + 18p_1,$ or $60p_2 = 21 + 18p_1.$
DPatrick 2014-02-20 19:55:26
Now our system is just 2 equations:
\begin{align*}
10p_1 &= 9p_2, \\
60p_2 &= 21 + 18p_1.
\end{align*}
LightningX48 2014-02-20 19:55:43
multiply first equation by 20, second by 3
blueberry7 2014-02-20 19:55:47
multiply the first by 20, the second by 3
bestwillcui1 2014-02-20 19:55:47
multiply first equation by 20, second equation by 3
DPatrick 2014-02-20 19:55:51
Multiply the first by 20 and the second by 3 to get:
\begin{align*}
200p_1 &= 180p_2, \\
180p_2 &= 63 + 54p_1.
\end{align*}
DPatrick 2014-02-20 19:56:00
So $200p_1 = 63 + 54p_1,$ and hence $p_1 = \dfrac{63}{146}.$ Answer (C).
DPatrick 2014-02-20 19:56:52
While this was an OK problem, it was somewhat less-than-elegant for a #25 in my opinion.
VD8M9 2014-02-20 19:56:58
is this the simplest possible way?
wehac 2014-02-20 19:56:58
Is there a quicker way to do this?
DPatrick 2014-02-20 19:57:15
We talked about this around the office this morning, and we didn't really see a better way.
DPatrick 2014-02-20 19:57:37
I'm going to take a 3-minute rest break...we'll resume at 8:01 Eastern.
DPatrick 2014-02-20 20:01:07
OK, we're back! On to the AMC 12B!
DPatrick 2014-02-20 20:01:13
21. In the figure, $ABCD$ is a square of side length 1. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE?$
$\phantom{12B:21}$
$\text{(A) } \dfrac12(\sqrt6-2) \quad \text{(B) } \dfrac14 \quad \text{(C) } 2-\sqrt3 \quad \text{(D) } \dfrac{\sqrt3}6 \quad \text{(E) } 1-\dfrac{\sqrt2}2$
DPatrick 2014-02-20 20:01:20
DPatrick 2014-02-20 20:01:39
Let's set $x = BE$ to be what we want. Notice that $GH = JK = x$ too.
DPatrick 2014-02-20 20:01:46
What do you notice in this picture?
guilt 2014-02-20 20:01:57
similar triangles!
XxAndreixX 2014-02-20 20:01:57
Similar triangles
HYP135peppers 2014-02-20 20:01:57
similar triangles!
Tommy2000 2014-02-20 20:01:57
similar triangle
Tuxianeer 2014-02-20 20:01:57
similar triangles
sahilp 2014-02-20 20:01:57
similar triangles
brandongeren 2014-02-20 20:01:57
similar triangles
DPatrick 2014-02-20 20:02:05
Ding ding ding! We've got similar triangles!
DPatrick 2014-02-20 20:02:22
Specifically, $AGJ \sim DHG.$ (And a couple of others too, but I'll use those two.) I'll color the corresponding sides in the diagram:
DPatrick 2014-02-20 20:02:28
DPatrick 2014-02-20 20:02:39
And since $GJ = 1$ and $GH = x$, we know that the constant of proportionality from $AGJ$ to $DGH$ is $x$ -- that is, every length in $DGH$ is $x$ times the corresponding length from $AGJ$.
DPatrick 2014-02-20 20:02:59
What can we do now?
guilt 2014-02-20 20:03:28
set up ratios
va2010 2014-02-20 20:03:28
AG+GD=1
Sesquipedalian 2014-02-20 20:03:28
DG+GA=1
tanishq1 2014-02-20 20:03:28
pythag theorem if you set a couple more variables
Larklight 2014-02-20 20:03:28
pythagorean theorem
XxAndreixX 2014-02-20 20:03:28
Set AJ = a, AG = b and write equations
DPatrick 2014-02-20 20:03:45
Yeah, let's label some more lengths so that we can write some equations using some of these facts.
DPatrick 2014-02-20 20:03:52
I set $AG = y$ and $AJ = z$. Then we know that $DH = xy$ and $DG = xz$.
DPatrick 2014-02-20 20:03:57
DPatrick 2014-02-20 20:04:28
A couple of you mentioned that $AD = 1$, so we have the equation $y + xz = 1.$
DPatrick 2014-02-20 20:04:41
What about the other side of the square? What equation do we get from that?
SockFoot 2014-02-20 20:05:20
z+xy+x=y+xz=1
joshualee2000 2014-02-20 20:05:20
xz+y=1, z+xy+x =1
Larklight 2014-02-20 20:05:20
z+x+xy =1
Tinyyy 2014-02-20 20:05:20
xy+z+x=1
ingridzhang97 2014-02-20 20:05:20
AB=1 so x+xy+z=1
sparkles257 2014-02-20 20:05:20
xy + z = 1- BE
DPatrick 2014-02-20 20:05:44
Right: on the bottom, $BE = x$ and $EJ = xy$ (same as $DH$) and $AJ = z$, so adding these gives $x + xy + z = 1.$
DPatrick 2014-02-20 20:06:18
Therefore we have the equations
\begin{align*}
xy + x + z &= 1, \\
xz + y &= 1,
\end{align*}from the two sides of the big square.
DPatrick 2014-02-20 20:06:30
Now what?
distortedwalrus 2014-02-20 20:06:45
y^2 + z^2 = 1
XxAndreixX 2014-02-20 20:06:45
y^2 + z^2 = 1
PertBanking 2014-02-20 20:06:45
Pythagorean Theorem!
dli00105 2014-02-20 20:06:45
y^2+z^2=1
DPatrick 2014-02-20 20:06:57
Right! We also have $y^2 + z^2 = 1$ from triangle $AGJ$.
DPatrick 2014-02-20 20:07:23
So we've got 3 equations in 3 variables, and even though they're not linear, we can probably solve them.
Nahmid 2014-02-20 20:07:38
solve for x in the second equation and plug in
blueberry7 2014-02-20 20:07:38
solve for x
DPatrick 2014-02-20 20:08:03
There are lots of ways to finish from here -- one way is to substitute $x = \dfrac{1-y}{z}$ from the second equation into the first, to get
\[ \frac{y(1-y)}{z} + \frac{1-y}{z} + z = 1. \]
DPatrick 2014-02-20 20:08:26
It's just algebra from here, so again I'll kind of zip through it. Multiply through by $z$ to get \[y(1-y) = z - (1-y) - z^2,\] which simplifies to $0 = z - 1 - z^2 + y^2$ (the $y$ terms cancel).
joshualee2000 2014-02-20 20:08:59
1=z^2+y^2
DrMath 2014-02-20 20:08:59
z^2+y^2=1
DPatrick 2014-02-20 20:09:21
Right, we can replace $y^2$ with $1-z^2$ in the above equation, to get something just in terms of $z$.
DPatrick 2014-02-20 20:09:28
We substitute to get $0 = z - 1 - z^2 + (1-z^2),$ or $0 = z - 2z^2.$
matholympiad25 2014-02-20 20:09:44
z = 1/2
wehac 2014-02-20 20:09:44
z=1/2
SockFoot 2014-02-20 20:09:44
z=1/2
DrMath 2014-02-20 20:09:44
z=1/2
DPatrick 2014-02-20 20:09:49
Thus (since $z$ is clearly not zero!) we have $z = \frac12$.
DPatrick 2014-02-20 20:10:14
Now we just unravel the equations to find $x$ (our answer).
mathawesomeness777 2014-02-20 20:10:20
y=sqrt(3)/2
LightningX48 2014-02-20 20:10:20
y = sqrt3 / 2
crastybow 2014-02-20 20:10:27
JAG is 30-60-90
DPatrick 2014-02-20 20:10:31
We then have $y^2 = 1-z^2 = \frac34,$ so $y = \frac{\sqrt3}{2}.$ (In fact, notice that all the right triangles in the picture turn out to be 30-60-90.)
joshualee2000 2014-02-20 20:10:55
x= 1-y/z
DPatrick 2014-02-20 20:10:57
And to finish, we want \[x = \frac{1-y}{z} = \frac{1-\frac{\sqrt3}{2}}{\frac12} = 2-\sqrt3.\] Answer (C).
DrMath 2014-02-20 20:11:14
is there a reason to see that at first without digging thru algebra?
DPatrick 2014-02-20 20:11:24
The 30-60-90 triangles? Not that I can obviously see.
DPatrick 2014-02-20 20:11:51
There are certainly other ways to solve this problem -- one of the AMC's published solutions used trigonometry.
DPatrick 2014-02-20 20:12:11
Anyway, we already did #22 (it was 10B #25), so on to #23:
DPatrick 2014-02-20 20:12:16
23. The number 2017 is prime. Let $\displaystyle S=\sum_{k=0}^{62}\binom{2014}k.$ What is the remainder when $S$ is divided by 2017?
$\phantom{12B:23}$
$\text{(A) } 32 \quad \text{(B) } 684 \quad \text{(C) } 1024 \quad \text{(D) } 1576 \quad \text{(E) } 2016$
DPatrick 2014-02-20 20:12:36
Weird.
DPatrick 2014-02-20 20:12:55
What's a better way to write $\dbinom{2014}{k}$ mod 2017?
DPatrick 2014-02-20 20:13:28
It might help to write out the binomial coefficient: \[\dbinom{2014}{k} = \frac{(2014)(2013)\cdots(2014-k+1)}{k!}.\]
crastybow 2014-02-20 20:13:51
-3 * -4 * -5 * ...
Allan Z 2014-02-20 20:13:51
expand, write 2014=-3 mod 2017, 2013=-4 mod 2017 etc
ingridzhang97 2014-02-20 20:13:51
2014=-3 mod 2017
DPatrick 2014-02-20 20:14:00
Right: $2014 \equiv -3 \pmod{2017}.$
DPatrick 2014-02-20 20:14:12
So mod 2017 this is \[ \frac{(-3)(-4)\cdots(-3-k+1)}{k!}.\]
DPatrick 2014-02-20 20:14:29
By the way -- and this is important! -- this is the step where we use the fact that 2017 is prime.
DPatrick 2014-02-20 20:14:43
You can't replace numbers in the numerator of a fraction modulo some residue if the denominator isn't relatively prime to the residue. For example, 6/2 is 3, and 6 = -4 mod 10, but -4/2 is -2, which is not the same as 3 mod 10. It failed because 2 isn't relatively prime to 10, so we're not allowed to "divide by 2" mod 10. If the modulus is prime, however, we can divide by anything we want (provided it's not a multiple of the prime, of course.)
Allan Z 2014-02-20 20:14:56
the fact that 2017 is prime makes multiplicative inverse exist
starwars123 2014-02-20 20:14:58
it has no factors other than 1 and 2017
DPatrick 2014-02-20 20:15:04
Indeed, that's the technical fact that we're using.
DPatrick 2014-02-20 20:15:19
(If this doesn't mean anything to you, it might when you take a more advanced number theory course.)
DPatrick 2014-02-20 20:15:45
Anyway, so we've got that \[ \binom{2014}{k} \equiv \frac{(-3)(-4)\cdots(-3-k+1)}{k!} \pmod{2017}.\]
DPatrick 2014-02-20 20:15:50
What do we do with this?
TheStrangeCharm 2014-02-20 20:16:08
factor out the negatives and then cancel with the $k!$
Allan Z 2014-02-20 20:16:08
a lot of terms cancel!
lucylai 2014-02-20 20:16:08
cancel
DPatrick 2014-02-20 20:16:20
Indeed, if we pull the minus signs out, this is \[ (-1)^k\frac{(3)(4)\cdots(k+2)}{k!}.\]
DPatrick 2014-02-20 20:16:48
And then a lot of stuff cancels...
DPatrick 2014-02-20 20:17:05
This is just $$(-1)^k\frac{(k+1)(k+2)}{2}.$$
joshxiong 2014-02-20 20:17:21
This becomes $\binom{k+2}{2}$
DPatrick 2014-02-20 20:17:45
Aha, so our sum is just \[ S \equiv \sum_{k=0}^{62} (-1)^k\binom{k+2}{2} \pmod{2017}.\]
DPatrick 2014-02-20 20:18:00
That is, writing it without the $\Sigma$, our sum (mod 2017) is \[ \binom22 - \binom32 + \binom42 - \binom52 + \cdots - \binom{63}{2} + \binom{64}{2}.\]
DPatrick 2014-02-20 20:18:15
Does this sum telescope or collapse in a nice way?
DPatrick 2014-02-20 20:18:36
It may help to think of $\dbinom{n}{2}$ in another way...
awesome2478 2014-02-20 20:18:51
triangular numbers?
DrMath 2014-02-20 20:18:51
1+2+3+...+(n-2)+(n-1)
DPatrick 2014-02-20 20:18:59
Right! $\dbinom{n}{2}$ is the sum of the first $n-1$ positive integers: $\dfrac{n(n-1)}{2}.$
DPatrick 2014-02-20 20:19:16
So, for instance, what is $\binom42 - \binom32?$
sparkles257 2014-02-20 20:19:33
3
XxAndreixX 2014-02-20 20:19:33
3
guilt 2014-02-20 20:19:33
3
wehac 2014-02-20 20:19:33
3
ahaanomegas 2014-02-20 20:19:33
bestwillcui1 2014-02-20 20:19:33
3
mathwizard888 2014-02-20 20:19:33
3
TheMaskedMagician 2014-02-20 20:19:33
$3$
DPatrick 2014-02-20 20:19:37
It's just 3: $\binom42 - \binom32 = (1+2+3)-(1+2) = 3.$
DPatrick 2014-02-20 20:19:50
In the same way $\binom62 - \binom52 = (1+2+3+4+5)-(1+2+3+4) = 5$, and $\binom82 - \binom72 = 7$, and so on.
sahilp 2014-02-20 20:20:06
1+3+5+7.....+61+63
ingridzhang97 2014-02-20 20:20:11
that sum becomes the sum of the first 32 odds
DPatrick 2014-02-20 20:20:37
Right. When we pair up the terms like this (don't forget the $\binom22$ term at the front), our sum becomes: \[1+3+5+\cdots+63.\]
mjlove 2014-02-20 20:20:48
32^2
Jayjayliu 2014-02-20 20:20:48
32^2
DaChickenInc 2014-02-20 20:20:48
32^2=1024
ssk9208 2014-02-20 20:20:48
which is $32^2 = 1024$
blueferret 2014-02-20 20:20:48
32^2
dli00105 2014-02-20 20:20:48
sum=32^2=1024
1023megabytes 2014-02-20 20:20:48
32^2 = 1024 (C)
mathawesomeness777 2014-02-20 20:20:48
= 32^2
DPatrick 2014-02-20 20:20:53
That is, it's just the sum of the first 32 positive odd integers.
DPatrick 2014-02-20 20:21:01
And you should all know that this sum is just $32^2$, or $1024.$ Answer (C).
DPatrick 2014-02-20 20:21:05
(If you don't know this, try to prove it on your own.)
DPatrick 2014-02-20 20:21:39
I liked this problem, but the bit about 2017 having to be prime made it a bit technical.
DPatrick 2014-02-20 20:21:49
Anyway, on to #24:
DPatrick 2014-02-20 20:21:53
24. Let $ABCDE$ be a pentagon inscribed in a circle such that $AB=CD=3,$ $BC=DE=10,$ and $AE=14.$ The sum of the lengths of all diagonals of $ABCDE$ is equal to $\dfrac mn,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$
$\phantom{12B:24}$
$\text{(A) } 129 \quad \text{(B) } 247 \quad \text{(C) } 353 \quad \text{(D) } 391 \quad \text{(E) } 421$
xXpapayaXx 2014-02-20 20:22:12
draw a picture
ompatel99 2014-02-20 20:22:12
picture!
SheilaMajumdar 2014-02-20 20:22:12
image please
dli00105 2014-02-20 20:22:12
draw a picture!!
DPatrick 2014-02-20 20:22:16
Here's a picture:
DPatrick 2014-02-20 20:22:22
DPatrick 2014-02-20 20:22:34
What do we notice about the diagonals?
joshxiong 2014-02-20 20:23:06
AC=BD=CE
Rogman 2014-02-20 20:23:06
some are congruent
mssmath 2014-02-20 20:23:06
AC=CE
1023megabytes 2014-02-20 20:23:06
DB=EC
blueferret 2014-02-20 20:23:06
some of them have equal lengths
distortedwalrus 2014-02-20 20:23:06
a lot of them are equal in length
DPatrick 2014-02-20 20:23:15
Three of them -- $\overline{AC},\,\overline{BD},\,\overline{CE}$ -- are the same length, since they all cut out the same length arc (specifically, an arc cut out by a segment of length 3 followed by a segment of length 10 in either order).
DPatrick 2014-02-20 20:23:30
So that we have a better picture of what's going on, I'll draw those in red below; the other two diagonals I'll draw in green ($\overline{AD}$) and blue ($\overline{BE}$):
DPatrick 2014-02-20 20:23:36
Arcmage101 2014-02-20 20:23:59
Do you have any techniques for quickly drawing approximate figures and/or diagrams?
mathtastic 2014-02-20 20:23:59
wow those are colorful
mathtastic 2014-02-20 20:23:59
are we allowed to use colors on the test?
DPatrick 2014-02-20 20:24:13
That's my suggestion: use colors. Bring lots of colors writing implements.
DPatrick 2014-02-20 20:24:28
Let's call the red length $r$, the green length $g$, and the blue length $b$. So we're trying to find $3r + g + b.$
DPatrick 2014-02-20 20:24:42
(That's another reason to use colors -- easier to remember which variable is which!)
DPatrick 2014-02-20 20:24:55
OK. What do we do now?
joshxiong 2014-02-20 20:25:15
we can use Ptolemy's Theorem
va2010 2014-02-20 20:25:15
Use ptolemy?
Larklight 2014-02-20 20:25:15
ptolmey's theorem?
guilt 2014-02-20 20:25:15
use ptolemy's theorem
Allan Z 2014-02-20 20:25:15
Notice cyclic quadrilaterals : ptolemy's theorem to get systems of 3 equations
distortedwalrus 2014-02-20 20:25:15
ptolemy's theorem
DPatrick 2014-02-20 20:25:26
From here, there are two major paths to the solution, depending on whether you know (or want to use) Ptolemy's Theorem or not. I'll show you the Ptolemy's Theorem solution, but then I'll also show you a solution where you don't need Ptolemy's Theorem at all.
DPatrick 2014-02-20 20:25:37
Ptolemy's Theorem states that in a cyclic quadrilateral, the sum of the product of the lengths of the opposite sides equals the product of the diagonals.
DPatrick 2014-02-20 20:26:03
And a "cyclic quadrilateral" is one whose four vertices all lie on the same circle.
DPatrick 2014-02-20 20:26:13
For example, using cyclic quadrilateral $BCDE$, we have \[(BC)(DE) + (BE)(CD) = (BD)(CE).\] Plugging in the numbers and the variables that we have, this is $100 + 3b = r^2.$
DPatrick 2014-02-20 20:26:18
Tommy2000 2014-02-20 20:26:42
MORE equations
DPatrick 2014-02-20 20:26:49
Yeah, we have 5 cyclic quadrilateral in our picture, so we have 5 equations that result from Ptolemy's Theorem. I'll list them below, and I'll label them by the point on the pentagon that's omitted (so, for example, equation ($A$) comes from quadrilateral $BCDE$, omitting $A$):
\begin{align*}
3b + 100 &= r^2 \qquad (A) \\
10r + 42 &= rg \qquad (B) \\
14r + 30 &= bg \qquad (C) \\
3r + 140 &= rb \qquad (D) \\
10g + 9 &= r^2 \qquad (E)
\end{align*}
DPatrick 2014-02-20 20:27:07
This seems like a lot of data. How can we use these equations?
DPatrick 2014-02-20 20:28:19
There are probably lots of ways to solve this system. Nobody suggested what I happened to do, but I'm sure what you might have done is just as good.
DPatrick 2014-02-20 20:28:37
I thought that (B) and (D) were easy ways to solve for $g$ and $b$ in terms of $r$.
DPatrick 2014-02-20 20:28:51
Specifically, (B) and (D) let us solve for $g$ and $b$ in terms of $r$:
\begin{align*}
g &= 10 + \frac{42}{r}, \\
b &= 3 + \frac{140}{r}.
\end{align*}
DPatrick 2014-02-20 20:29:32
(I think you could have used (A) and (E) in this first step as well -- a lot of you suggested that. But I don't have that one written up in my notes!)
blueberry7 2014-02-20 20:29:39
then we can plug into equations A and E
joshxiong 2014-02-20 20:29:39
now we can plug these into (A) and (E)
DPatrick 2014-02-20 20:29:53
Right, in fact, we really only need one of them.
DPatrick 2014-02-20 20:30:01
One way to continue is to plug the expression for $b$ into equation (A) and get an equation for $r$:
\[ 3\left(3 + \frac{140}{r}\right) + 100 = r^2.\]
Tommy2000 2014-02-20 20:30:17
solve for r
oink 2014-02-20 20:30:17
then solve for r
Rogman 2014-02-20 20:30:17
but we'll get a cubic!
DPatrick 2014-02-20 20:30:25
True: we can multiply through by $r$ to get $9r + 420 + 100r = r^3,$ or $r^3 - 109r - 420 = 0.$
mathlearning920 2014-02-20 20:30:39
It might be easier if you plugged the expression into E
DPatrick 2014-02-20 20:30:46
I bet we get the same cubic.
crastybow 2014-02-20 20:31:01
try RRT
guilt 2014-02-20 20:31:01
rrt!
va2010 2014-02-20 20:31:01
Use Rational root theorem
alex31415 2014-02-20 20:31:01
Rational Roots Theorem?
DPatrick 2014-02-20 20:31:22
It is tricky to solve cubics in general, but we have some clues.
DPatrick 2014-02-20 20:31:29
Because the problem strongly implies that $r$ is rational, we look for rational roots, which have to be factors of 420. Also because there's no quadratic term, the roots have to sum to 0.
DPatrick 2014-02-20 20:31:40
And we need two odd roots because the linear term is odd.
DPatrick 2014-02-20 20:31:52
And we're looking for the third side of a 3-10-$r$ triangle, where $r$ is the longest side, so $r$ is probably either 11 or 12, and only 12 is a factor of 420.
ABCDE 2014-02-20 20:32:08
-5, -7, 12 work
matholympiad25 2014-02-20 20:32:08
12 works
DPatrick 2014-02-20 20:32:14
Indeed, a little experimentation shows that the cubic factors as \[(r-12)(r+7)(r+5) = 0.\]
DPatrick 2014-02-20 20:32:29
The only positive root is $r=12.$
crastybow 2014-02-20 20:32:48
plug these into the equations for b and g
Tommy2000 2014-02-20 20:32:48
plug in to the equations for b and g
DPatrick 2014-02-20 20:33:00
Right, from here, we can find $b$ and $g$ using our previous equations for them:
\begin{align*}
g &= 10 + \frac{42}{r} = 10 + \frac{42}{12} = 10 + \frac72 = \frac{27}{2}, \\
b &= 3 + \frac{140}{r} = 3 + \frac{140}{12} = 3 + \frac{35}{3} = \frac{44}{3}.
\end{align*}
Rogman 2014-02-20 20:33:16
add add add
geogirl08 2014-02-20 20:33:18
3r+g+b = ?
DPatrick 2014-02-20 20:33:24
And to finish, our sum is \[3(12) + \frac{27}{2} + \frac{44}{3} = \frac{6(36)+3(27)+2(44)}{6} = \frac{216+81+88}{6} = \frac{385}{6}.\]
DPatrick 2014-02-20 20:33:31
This is in lowest terms, so our answer is $385+6=391.$ Answer (D).
DPatrick 2014-02-20 20:33:50
While this was a fine solution, it requires knowledge of Ptolemy's Theorem, which is a bit of a technical result. It's mildly unreasonable (in my opinion) to expect all contest takers to know this result.
DPatrick 2014-02-20 20:34:00
So, let me show you a solution where you don't need Ptolemy's Theorem at all.
DPatrick 2014-02-20 20:34:21
The picture is still up top, but let me post it down below again:
DPatrick 2014-02-20 20:34:30
Tommy2000 2014-02-20 20:34:41
would a law of cosines bash work as well?
DivideBy0 2014-02-20 20:34:41
law of cosines?
DPatrick 2014-02-20 20:35:10
We've got lots of lengths, so a Law of Cosines attack might work.
DPatrick 2014-02-20 20:35:33
We need a convenient triangle to get a cosine from, though.
dli00105 2014-02-20 20:35:55
ACE
Larklight 2014-02-20 20:35:55
CEA!
Allan Z 2014-02-20 20:36:00
ACE?
DPatrick 2014-02-20 20:36:13
Good idea. Triangle $ACE$ is isosceles with base 14 and two legs $r.$
DPatrick 2014-02-20 20:36:32
So if we let $\theta = \angle CAE$, what is $\cos\theta$?
DrMath 2014-02-20 20:37:04
7/AC
Tommy2000 2014-02-20 20:37:04
7/r
joshxiong 2014-02-20 20:37:04
7/r
DPatrick 2014-02-20 20:37:19
Right: $\cos\theta = \dfrac{7}{r}.$
DPatrick 2014-02-20 20:37:42
(Drop the perpendicular from $C$ to $\overline{AE}$ to get a right triangle, then $7$ is the "adjacent" side to $A$ and $r$ is the hypotenuse.)
DPatrick 2014-02-20 20:37:54
So that's helpful...because?
DPatrick 2014-02-20 20:38:12
Is there another angle that has a related cosine?
danzhi 2014-02-20 20:38:46
EDC
DPatrick 2014-02-20 20:39:00
Right! I would use angle $CDE$ on the opposite side of the circle.
DPatrick 2014-02-20 20:39:15
(We've got to use the fact that we're inside a circle somewhere.)
DPatrick 2014-02-20 20:39:44
Because $\angle CAE$ and $\angle CDE$ intercept opposite arcs that make up the whole circle, the angles are supplementary (sum to $180^\circ$.)
DPatrick 2014-02-20 20:39:54
So what is $\cos(\angle CDE)?$
joshxiong 2014-02-20 20:40:07
$\cos CDE=-\frac{7}{r}$
ingridzhang97 2014-02-20 20:40:07
cos(180-theta)
mathmaster2012 2014-02-20 20:40:07
-7/r
sparkles257 2014-02-20 20:40:07
-7/r
mathmaster2012 2014-02-20 20:40:15
blueferret 2014-02-20 20:40:15
-7/r
DPatrick 2014-02-20 20:40:25
Right: $\cos(\angle CDE) = -\cos(180^\circ - \theta) = -\cos\theta = -\frac{7}{r}.$
DrMath 2014-02-20 20:40:36
now LOC to find CE?
sparkles257 2014-02-20 20:40:36
now we can do cosine bash
DPatrick 2014-02-20 20:40:54
Right, now we can use the Law of Cosine on triangle $CDE$: \[(CE)^2 = (CD)^2 + (DE)^2 - 2\cos(\angle CDE)(CD)(CE).\]
DPatrick 2014-02-20 20:41:09
Plugging in our values, we get \[r^2 = 3^2 + 10^2 + 2\left(\frac{7}{r}\right)(3)(10).\]
mathtastic 2014-02-20 20:41:22
not so ugly anymore
ABCDE 2014-02-20 20:41:26
same cubic
DPatrick 2014-02-20 20:41:36
Right, when we simplify, we get $r^3 = 109r + 420,$ or $r^3 - 109r - 420 = 0.$
guilt 2014-02-20 20:41:43
same cubic so r = 12
Tommy2000 2014-02-20 20:41:43
Its the same cubic!!!!!!!!!!1
DrMath 2014-02-20 20:41:43
wait a minute- this gives the same cubic woth ptolemy!
DPatrick 2014-02-20 20:41:48
This is the same cubic as we had in the previous solution, so $r=12$ is the positive root.
DPatrick 2014-02-20 20:42:01
The fact that it's the same cubic is of course not a coincidence!
DPatrick 2014-02-20 20:42:16
Now how do we find the green and blue lengths? (Again without using Ptolemy?)
guilt 2014-02-20 20:42:59
more trig using LOC
mathtastic 2014-02-20 20:42:59
or more law of cosigns
DPatrick 2014-02-20 20:43:19
We could keep bashing with Law of Cosines, but we could also use some of the isosceles trapezoids in the picture.
DPatrick 2014-02-20 20:43:31
For example, consider $ABCD$:
DPatrick 2014-02-20 20:43:37
DPatrick 2014-02-20 20:43:47
The red diagonals have length 12. How do we find the length of the green base?
Larklight 2014-02-20 20:44:08
drop altitudes
elsw 2014-02-20 20:44:08
Right triangles?
mathtastic 2014-02-20 20:44:08
pythag??
DPatrick 2014-02-20 20:44:22
Right. We drop a perpendicular from $C$ down to $\overline{AD}$:
DPatrick 2014-02-20 20:44:28
mathtastic 2014-02-20 20:44:37
OH WAIT!! THIS IS LIKE AMC 10B #21!!!!
DPatrick 2014-02-20 20:44:44
Indeed it is.
DPatrick 2014-02-20 20:44:49
Using the Pythagorean Theorem, \[h^2 = 9-a^2 = 144-(10+a)^2.\]
DPatrick 2014-02-20 20:44:58
This simplifies to $9-a^2 = 44-20a-a^2$, so $a = \frac{35}{20} = \frac{7}{4}.$
DPatrick 2014-02-20 20:45:05
Thus the green side is $g = 10+2a = 10+\frac{7}{2}=\frac{27}{2},$ as before.
DPatrick 2014-02-20 20:45:21
A similar technique on trapezoid $BCDE$ can be used to find $b$. (I'll skip this -- you're probably sick of this problem by now.)
DPatrick 2014-02-20 20:45:42
So the problem can be solved without knowing anything about Ptolemy's Theorem!
DPatrick 2014-02-20 20:45:57
OK, on to AMC 12B #25!
DPatrick 2014-02-20 20:46:06
25. What is the sum of all positive real solutions $x$ to the equation\[2\cos(2x)\left(\cos(2x)-\cos\left(\frac{2014\pi^2}x\right)\right)=\cos(4x)-1\,?\]
$\phantom{12B:25}$
$\text{(A) } \pi \quad \text{(B) } 810\pi \quad \text{(C) } 1008\pi \quad \text{(D) } 1080\pi \quad \text{(E) } 1800\pi$
ompatel99 2014-02-20 20:46:36
2x=z?
DPatrick 2014-02-20 20:46:45
That's one idea, but I'd take it a step further.
DPatrick 2014-02-20 20:46:53
Since $\cos(2x)$ appears in a couple of places, let's call it $y.$
DPatrick 2014-02-20 20:47:02
Can we also write the right side of the equation in terms of $y$?
Larklight 2014-02-20 20:47:25
double angle identity on cos(4x)
joshxiong 2014-02-20 20:47:25
Use the double angle formula on $\cos(4x)$
Tinyyy 2014-02-20 20:47:25
Double angle formula for cos (4x)
mathwizard888 2014-02-20 20:47:25
yes, using double angle formula
DPatrick 2014-02-20 20:47:37
By the double-angle formula for cosine, \[\cos(4x) = 2\cos^2(2x)-1 = 2y^2 - 1.\]
DPatrick 2014-02-20 20:47:53
So our equation has become \[2y\left(y - \cos\left(\frac{2014\pi^2}{x}\right)\right)=2y^2-2.\]
DPatrick 2014-02-20 20:48:11
(Of course, you don't need to use $y$, but I find it easier to keep track of things.)
tanishq1 2014-02-20 20:48:17
encouraged because there's a 2y^2 on both sides.
Wickedestjr 2014-02-20 20:48:22
The 2y^2 cancels on both sides
cheeptricks 2014-02-20 20:48:28
2y^2 cancel out
ahaanomegas 2014-02-20 20:48:35
Divide both sides by 2 because why-not.
hotstuffFTW 2014-02-20 20:48:35
divide by 2
jeffgreenfan8 2014-02-20 20:48:35
we can divide by 2
DPatrick 2014-02-20 20:48:41
Hurray, life is good. The $2y^2$ terms cancel, and after dividing by -2 this is just \[y\cos\left(\frac{2014\pi^2}{x}\right) = 1.\]
DPatrick 2014-02-20 20:48:52
$y$ isn't really helping us anymore, so let's put everything back in terms of $x$: \[\cos(2x)\cos\left(\frac{2014\pi^2}{x}\right) = 1.\]
DPatrick 2014-02-20 20:49:06
So, now I ask myself: when is a product of two cosines equal to 1?
mssmath 2014-02-20 20:49:30
both are +1 or -1
ABCDE 2014-02-20 20:49:30
both 1 or -1
blueferret 2014-02-20 20:49:30
they are either both 1 or -1
Allan Z 2014-02-20 20:49:30
both of them -1 or both of them 1
joshxiong 2014-02-20 20:49:30
so they must both be -1 or 1
wehac 2014-02-20 20:49:30
when they are both 1 or both -1
ompatel99 2014-02-20 20:49:30
they are both 1 or -1
alex31415 2014-02-20 20:49:30
Both are +/-1
mathawesomeness777 2014-02-20 20:49:30
Both 1 or -1.
DPatrick 2014-02-20 20:49:39
Righto. The two cosines themselves must both be 1 or both be -1.
DPatrick 2014-02-20 20:49:50
So it looks like we have two cases to consider.
DPatrick 2014-02-20 20:50:11
We'll start with the $+1$ case. When do we have \[\cos(2x)=\cos\left(\frac{2014\pi^2}{x}\right) = 1\,?\]
DPatrick 2014-02-20 20:50:24
When is a cosine equal to 1?
ompatel99 2014-02-20 20:50:57
2pi*n
alex31415 2014-02-20 20:50:57
2pi*n for an integer n
joshualee2000 2014-02-20 20:50:57
when it is 2pin
dli00105 2014-02-20 20:50:57
when angle multiple of 2pi
blueferret 2014-02-20 20:50:57
the angle=2n*pi
Hydroxide 2014-02-20 20:50:57
multiple of 2pi
DPatrick 2014-02-20 20:51:20
Right. $\cos(\theta)=1$ if and only if $\theta$ is an even integer multiple of $\pi.$ (Please, don't use degrees in this problem -- the presence of $\pi$ should tell you that that's a bad idea.)
DPatrick 2014-02-20 20:51:35
So we must have $x = n\pi$ for some positive integer $n$ in order for $\cos(2x) = 1.$
DPatrick 2014-02-20 20:51:51
And for the other term, this leaves \[\cos\left(\frac{2014\pi^2}{x}\right) = \cos\left(\frac{2014\pi^2}{n\pi}\right) = \cos\left(\frac{2014}{n}\pi\right) = 1.\]
DPatrick 2014-02-20 20:51:58
When does this occur?
blueferret 2014-02-20 20:52:16
when 2014/n is even
joshxiong 2014-02-20 20:52:16
n is an odd factor of 2014
joshualee2000 2014-02-20 20:52:16
when n is a odd divisor of 2014
DPatrick 2014-02-20 20:52:25
Right. We need that $\dfrac{2014}{n}$ is an even positive integer.
DPatrick 2014-02-20 20:52:46
Now is a good time to note that $2014 = 2 \cdot 19 \cdot 53.$ (A common mistake here might be to factor $2014 = 2 \cdot 1007$, and assume that $1007$ is prime.)
jeffgreenfan8 2014-02-20 20:53:15
always know the prime factors of the year you are taking the exam in
DPatrick 2014-02-20 20:53:16
MathStudent2002 2014-02-20 20:53:24
4 odd factors!
Tommy2000 2014-02-20 20:53:24
4 n's, 1, 19, 53, 1007
DPatrick 2014-02-20 20:53:34
So $n$ must be one of $1,$ $19,$ $53,$ or $19 \cdot 53$ for us to have a solution. (If you mistakenly thought 1007 was prime, you'd only get the solution $n=1.$)
alex31415 2014-02-20 20:53:43
Sum of factors of 1007=(19+1)(53+1)=1080
akshay9 2014-02-20 20:53:54
This gives us four possible values of x: pi, 19pi, 53pi, 1007pi
DPatrick 2014-02-20 20:54:05
Right: our solutions in this case are $\pi$, $19\pi$, $53\pi$, and $(19 \cdot 53)\pi.$
DPatrick 2014-02-20 20:54:12
They sum to $(1+19+53+19\cdot53)\pi = (1+19)(1+53)\pi = (20)(54)\pi = 1080\pi.$ (Note the clever trick we used to add them!)
DPatrick 2014-02-20 20:54:34
This happens to be choice (D), and we've ruled out all but (D) or (E). If you're pressed for time, you can either guess that there are no solutions in the other case, or that there are.
DPatrick 2014-02-20 20:54:57
We're not pressed for time, so let's look.
DPatrick 2014-02-20 20:55:02
When do we have \[\cos(2x)=\cos\left(\frac{2014\pi^2}{x}\right) = -1\,?\]
joshxiong 2014-02-20 20:55:32
if both are an odd multiple of $\pi$
brandongeren 2014-02-20 20:55:32
odd multiples of pi
ABCDE 2014-02-20 20:55:32
odd multiple of pi
bli1999 2014-02-20 20:55:32
odd multiple of pi
DPatrick 2014-02-20 20:55:38
Right. $\cos(\theta) = -1$ if and only if $\theta$ is an odd integer multiple of $\pi$.
DPatrick 2014-02-20 20:55:48
So we must have $x = \left(n + \frac12\right)\pi$ for some nonnegative integer $n$ in order for $\cos(2x) = -1.$
DPatrick 2014-02-20 20:56:01
(That is, $x$ must be a half-integer multiple of $\pi$.)
DPatrick 2014-02-20 20:56:16
But this leaves \[\cos\left(\frac{2014\pi^2}{x}\right) = \cos\left(\frac{2014\pi^2}{\left(n+\frac12\right)\pi}\right) = \cos\left(\frac{2014}{n+\frac12}\pi\right) = -1.\]
DPatrick 2014-02-20 20:56:35
So we must have $\dfrac{2014}{n+\frac12} = \dfrac{4028}{2n+1}$ be an odd positive integer.
Rogman 2014-02-20 20:56:57
nope
Tommy2000 2014-02-20 20:56:57
impossible
DrMath 2014-02-20 20:56:57
i dont think thats possible!
BobaFett101 2014-02-20 20:56:57
not possible
lucylai 2014-02-20 20:56:57
impossible
DPatrick 2014-02-20 20:57:16
We can't divide 4028 by an odd integer and be left with an odd integer!
DPatrick 2014-02-20 20:57:39
So indeed there are no solutions in this case.
joshxiong 2014-02-20 20:57:54
which means that the final answer is $1080\pi$
thiennguyen 2014-02-20 20:57:54
D 1080 $ \pi$
LightningX48 2014-02-20 20:57:54
so the ans is (d) 1080pi
DPatrick 2014-02-20 20:58:04
Our answer is our previous sum, $1080\pi$, answer (D).
mathawesomeness777 2014-02-20 20:58:19
Wow, this was easier that #24! Pretty easy for a #25, but not saying that it was easy.
mssmath 2014-02-20 20:58:19
Easy #25 in your opinion??
DPatrick 2014-02-20 20:58:36
It was fairly easy for a #25. Lots of people find trig scary, but this was pretty straightforwards.
DPatrick 2014-02-20 20:59:02
We zipped through these problems pretty quickly, so let's do one or two more from 16-20 on either test. Please vote for the one you want to see!
DPatrick 2014-02-20 20:59:22
(you need to say which contest when you vote!)
DPatrick 2014-02-20 21:00:22
We had a lot of votes for 19 and 20 on the 10B and 20 on the 12B.
DPatrick 2014-02-20 21:00:29
They're all pretty quick so let's do 'em all!
DPatrick 2014-02-20 21:00:39
19. Two concentric circles have radii 1 and 2. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\phantom{10b:19}$
(A) $\dfrac16$ (B) $\dfrac14$ (C) $\dfrac{2-\sqrt2}{2}$ (D) $\dfrac13$ (E) $\dfrac12$
wehac 2014-02-20 21:01:03
draw a diagram
summitwei 2014-02-20 21:01:03
DIAGRAM!!!
AMN300 2014-02-20 21:01:03
diagram?
fluffyanimal 2014-02-20 21:01:03
Draw a diagram?
DPatrick 2014-02-20 21:01:09
This is basically the picture you need:
DPatrick 2014-02-20 21:01:15
DPatrick 2014-02-20 21:01:40
That's the "border" case where the chord just touches the inner circle. What do we know about those triangles?
joshualee2000 2014-02-20 21:01:57
its a 30-60-90 triangle
distortedwalrus 2014-02-20 21:01:57
the right triangles are 30-60-90
hotstuffFTW 2014-02-20 21:01:57
two 30-60-90 triangles
Dragon6point1 2014-02-20 21:01:57
30-60-90s
alex31415 2014-02-20 21:01:57
30 60 90
elsw 2014-02-20 21:01:57
30-60-90
Allan Z 2014-02-20 21:01:57
30-60-90
DPatrick 2014-02-20 21:02:16
Indeed. The vertical leg is 1 (the radius of the small circle) and the hypotenuses are 2 (the radius of the big circle), so they're 30-60-90.
DPatrick 2014-02-20 21:02:25
And in particular the central angle is 120 degrees.
DPatrick 2014-02-20 21:02:42
What does that tell us about the chord? How far about do the two points have to be to intersect the small circle?
XxAndreixX 2014-02-20 21:03:08
more than 120 degrees
joshualee2000 2014-02-20 21:03:08
they have to be more than 120 degrees apart
sparkles257 2014-02-20 21:03:08
120 degrees
cheeptricks 2014-02-20 21:03:08
if the points are farther apart, they will intersect the circle
DPatrick 2014-02-20 21:03:28
Right: if the two points are more than 120 degrees apart along the outer circle, the line will intersect the inner circle.
DPatrick 2014-02-20 21:03:43
(That is, if they're farther apart than they are in the picture above).
DivideBy0 2014-02-20 21:03:57
have to be on the opposite 120 degree arc
Tommy2000 2014-02-20 21:03:57
only 120/360 degrees work
DPatrick 2014-02-20 21:04:17
Right. If we pick the first point arbitrarily, then the second point can't be within 120 degrees of the first point on either side.
DPatrick 2014-02-20 21:04:37
That means it has to be in the 120 degree arc that lies opposite the first point (or, you can think of 240 degrees worth of arc that "fail").
acegikmoqsuwy2000 2014-02-20 21:04:44
so thats 360-240 = 120 and 120/360 is 1/3
Larklight 2014-02-20 21:04:44
(d) 1/3
mathawesomeness777 2014-02-20 21:04:44
So it's just 1/3 D!
DPatrick 2014-02-20 21:04:57
Exactly. The probability is $\dfrac{120}{360} = \dfrac13.$ Answer (D).
DPatrick 2014-02-20 21:05:13
OK, on to 10B #20:
DPatrick 2014-02-20 21:05:18
20. For how many integers $x$ is the number $x^4 - 51x^2 + 50$ negative?
$\phantom{10b:20}$
(A) 8 (B) 10 (C) 12 (D) 14 (E) 16
ompatel99 2014-02-20 21:05:56
(x^2-50)(x^2-1)
dli00105 2014-02-20 21:05:56
begging to be factored!!
mathbeida 2014-02-20 21:05:56
let y = x^2
mathtastic 2014-02-20 21:05:56
let y=x^2 then factor
ahaanomegas 2014-02-20 21:05:56
Use y = x^2.
distortedwalrus 2014-02-20 21:05:56
factor as (x^2 - 50)(x^2 - 1)
DPatrick 2014-02-20 21:06:14
Right. You might find it helpful to substitute $y = x^2$ to get it as a quadratic in $y$.
DPatrick 2014-02-20 21:06:21
Or you could just factor it straight away.
DPatrick 2014-02-20 21:06:29
We end up with $(x^2-50)(x^2-1) < 0.$
DPatrick 2014-02-20 21:06:53
This is only negative when the smaller factor $(x^2-50)$ is negative and the larger factor $(x^2-1)$ is positive.
jk23541 2014-02-20 21:07:09
x^2 has to be less than 50 but greater than 1
fluffyanimal 2014-02-20 21:07:09
1<x^2<50?
wehac 2014-02-20 21:07:09
so 1 < x^2 < 50 *it's non inclusive*
VD8M9 2014-02-20 21:07:09
1<x^2<50?
DPatrick 2014-02-20 21:07:28
Exactly. We must have $x^2-50 < 0 < x^2-1$, or $1 < x^2 < 50$.
DPatrick 2014-02-20 21:07:41
But remember that we only want to count integer solutions to this.
mathtastic 2014-02-20 21:07:59
we can find -7,-6,-5,-4,-3,-2,2,3,4,5,6,7 work so 12 values which is c
ompatel99 2014-02-20 21:07:59
2 to 7 and -2 to -7
Tommy2000 2014-02-20 21:07:59
12 solutions, 6 positive and 6 negative x's
distortedwalrus 2014-02-20 21:07:59
so there are 6 on either side of 0
DPatrick 2014-02-20 21:08:20
Right: there are 6 perfect squares of integers between 1 and 50, from $2^2 = 4$ up through $7^2 = 49$.
DPatrick 2014-02-20 21:08:34
Each perfect square $x^2$ gives two solutions for $x$ (for example, $2$ and $-2$).
DPatrick 2014-02-20 21:08:45
So there are $6 \cdot 2 = 12$ solutions. Answer (C).
DPatrick 2014-02-20 21:09:15
(I think the question would have been meaner if 6 was an answer choice.)
DPatrick 2014-02-20 21:09:26
And on to 12B #20:
DPatrick 2014-02-20 21:09:32
20. For how many positive integers $x$ is $\log_{10}(x-40)+\log_{10}(60-x) < 2$?
$\phantom{12b:20}$
(A) 10 (B) 18 (C) 19 (D) 20 (E) infinitely many
blueberry7 2014-02-20 21:10:00
use log laws...
crastybow 2014-02-20 21:10:00
use log identities
distortedwalrus 2014-02-20 21:10:00
combine then using the log rule
ingridzhang97 2014-02-20 21:10:00
use the log product sum rule thing
akshay9 2014-02-20 21:10:00
Start by combining logs and then taking it out of logarithmic form!
DPatrick 2014-02-20 21:10:16
Right: we can use the identity $\log a + \log b = \log ab$.
DPatrick 2014-02-20 21:10:41
So our inequality becomes $\log_{10}(x-40)(60-x) < 2$.
elsw 2014-02-20 21:11:00
(x-40)(60-x)<100
mathmaster2012 2014-02-20 21:11:00
(x-40)(60-x)<100
ingridzhang97 2014-02-20 21:11:00
that menas (x-40)(60-x)<100
DPatrick 2014-02-20 21:11:30
Indeed, this gives $(x-40)(60-x) < 10^2 = 100.$
DPatrick 2014-02-20 21:11:59
So we can now expand and simplify this quadratic inequality.
ompatel99 2014-02-20 21:12:08
x^2-100x+2500>0
DPatrick 2014-02-20 21:12:21
It simplifies to $x^2 - 100x + 2500 > 0$, or $(x-50)^2 > 0$.
DPatrick 2014-02-20 21:12:32
So this seems true for all $x$ except for $x=50$, so is the answer (E)??
BobaFett101 2014-02-20 21:12:47
x-40 or x-60 can't be negative
XxAndreixX 2014-02-20 21:12:47
restrictions on x
joshxiong 2014-02-20 21:12:47
the logarithm is not defined for negative numbers
joshualee2000 2014-02-20 21:12:47
the log cannot be negative
Tuxianeer 2014-02-20 21:12:47
no log domain
DPatrick 2014-02-20 21:12:54
Good catch!
DPatrick 2014-02-20 21:13:10
We can't take the log of a negative number, so we need the original values in the logs to be positive.
DPatrick 2014-02-20 21:13:17
That is, we need $x-40 > 0$ and $60-x > 0$.
DPatrick 2014-02-20 21:13:25
This restricts us to $40 < x < 60$.
DPatrick 2014-02-20 21:13:39
And we ruled out $x = 50$ as well.
ILoveYouMC 2014-02-20 21:13:58
18.
cheeptricks 2014-02-20 21:13:58
18 b
sparkles257 2014-02-20 21:13:58
18
DPatrick 2014-02-20 21:14:16
There are 19 numbers between 40 and 60 (exclusive), and then we throw away $x=50$, so we have 18 solutions. Answer (B).
DPatrick 2014-02-20 21:14:41
That's all for the 2014 AMC 10B/12B Math Jam!
DPatrick 2014-02-20 21:14:47
Next up on the calendar for many of you is the 2014 AIME. The main date for the AIME is Thursday, March 13 (3 weeks from today), and the alternate date is Wednesday, March 26.
DPatrick 2014-02-20 21:14:57
For each contest, we will have our AIME Math Jam two days after the contest:
- AIME I Math Jam on Saturday, March 15 at 7 PM ET / 4 PM PT
- AIME II Math Jam on Friday, March 28 at 7 PM ET / 4 PM PT
We will discuss all 15 problems on the contest during each Math Jam. We hope you'll be able to join us!
DPatrick 2014-02-20 21:15:08
Finally, AoPS offers a weekend Special AIME Problem Seminar for students preparing for the AIME. The seminar meets on Saturday, March 1 and Sunday, March 2, from 3:30-6:30 Eastern (12:30-3:30 Pacific) each day. The cost is $80 (which includes both days). Details at http://www.artofproblemsolving.com/School/courseinfo.php?course_id=maa:aime:special
DPatrick 2014-02-20 21:15:23
Thanks -- have a good night!

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