Math Jams

Who Wants to Be a Mathematician, Round 2

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AoPS instructor David Patrick will discuss the problems on Round 2 of qualifying for the 2019-20 Who Wants to Be a Mathematician Championship. We will be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the Championship finals, to be held in Denver in January 2020.

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Facilitator: Dave Patrick

NerdyDude 2019-10-14 19:27:00
Weren't you on who wants to be a millionaire
DPatrick 2019-10-14 19:27:07
https://www.washington.edu/alumni/columns/march00/images/patrick.jpg

Photo Credit: Maria Melin, copyright 1999 ABC Television.
DPatrick 2019-10-14 19:27:09
Yes.
DPatrick 2019-10-14 19:27:24
I once was a contestant on ABC's Who Wants to Be a Millionaire back before I started working at AoPS, way back when Regis Philbin was still the host. Here's a picture (I'm on the left, Regis is on the right):
RagingWar 2019-10-14 19:27:40
You won a car!
pow_h_2 2019-10-14 19:27:45
you won enough to buy a car, not full 1m
DPatrick 2019-10-14 19:27:55
Some of you know the story from past Math Jams...
asdf334 2019-10-14 19:28:33
it's that equator question
DPatrick 2019-10-14 19:29:05
Yeah, you can search on the web to find all the gory details.
DPatrick 2019-10-14 19:30:05
Welcome to the 2019-20 Who Wants to Be a Mathematician Championship Round 2 Math Jam!
DPatrick 2019-10-14 19:30:16
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens (probably hundreds!) of AoPS classes over the past 15 years, and I've written or co-written a few of our textbooks.
DPatrick 2019-10-14 19:30:30
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick 2019-10-14 19:30:37
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 2019-10-14 19:30:46
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 2019-10-14 19:30:57
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the 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 2019-10-14 19:31:18
Tonight we are going to discuss the problems and solutions from Round 2 of Who Wants to Be a Mathematician, which concluded last week.
DPatrick 2019-10-14 19:31:31
Who Wants to Be a Mathematician (or WWTBAM for short) is conducted by the American Mathematical Society (AMS).

https://www.ams.org/images/content/logo.png
DPatrick 2019-10-14 19:31:50
The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society. The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)
DPatrick 2019-10-14 19:32:11
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
DPatrick 2019-10-14 19:32:22
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
mikebreen 2019-10-14 19:32:38
Hello, everyone.
DPatrick 2019-10-14 19:32:47
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician. In addition to authoring articles and presenting talks related to game-show mathematics, Bill served as mathematics consultant to the CBS television show The Price is Right from 1997 to 2009. (Hence, his username.)
TPiR 2019-10-14 19:32:58
Hi everyone. Glad to be here.
DPatrick 2019-10-14 19:33:16
And last but not least, we also have an assistant here to help out tonight: Zach Stein-Perlman (ZachSteinPerlman). Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.
ZachSteinPerlman 2019-10-14 19:33:20
Hello!
DPatrick 2019-10-14 19:33:30
Zach can try to help you if you have a question or are having some other difficulty. He may open a private window with you to chat if needed.
DPatrick 2019-10-14 19:33:55
But, as I said, 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 2019-10-14 19:34:20
Round 1 of WWTBAM was held in September. Students who scored at least 7 out of 10 in Round 1 advanced to Round 2, which was held earlier this month.
DPatrick 2019-10-14 19:34:34
Round 2 consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 90 seconds per question. (But as we'll see as we work through the problems, some of them shouldn't take you nearly that long.) No books, notes, calculators, or internet access was permitted during the contest.
DPatrick 2019-10-14 19:34:53
We'll take a bit longer than 15 minutes tonight, because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the solutions to the problems, and not to merely present the answers. "Working through the solutions" often 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.
DPatrick 2019-10-14 19:35:20
Let's get started!
DPatrick 2019-10-14 19:35:25
1. For how many primes $p < 300$ is $p^3 - 2p^2$ a positive perfect square (square of a positive integer)?
DPatrick 2019-10-14 19:35:44
What can we do with the given expression?
RagingWar 2019-10-14 19:35:56
factor
motorfinn 2019-10-14 19:35:56
Factoring p^3-2p^2 seems like a good idea
ZhaoPow 2019-10-14 19:35:56
factor out p^2 first
RagingWar 2019-10-14 19:35:56
factor $p^2$ out
AlexLikeMath 2019-10-14 19:35:56
factor p-2
t3sp 2019-10-14 19:35:56
factor?
amuthup 2019-10-14 19:35:56
factor
asdf334 2019-10-14 19:35:56
factor p^2
SoulOfHeaven 2019-10-14 19:36:03
factor out p^2?
naman12 2019-10-14 19:36:03
Factor into $p^2(p-2)$
aastha.sharma02 2019-10-14 19:36:03
for this one i factored into $p^2(p-2)$
DPatrick 2019-10-14 19:36:10
Right. We can factor out $p^2$. This makes the expression $p^2(p-2)$.
DPatrick 2019-10-14 19:36:15
When is this a perfect square?
will3145 2019-10-14 19:36:38
p-2 is a perfect squre
Qwerty71 2019-10-14 19:36:38
this means since p^2 is already a square, p-2 must be a perfect square
JustKeepRunning 2019-10-14 19:36:38
$p-2$ is perfect square
GammaZero 2019-10-14 19:36:38
when $p-2$ is a square
mathstats 2019-10-14 19:36:38
then p - 2 must be a square
Mathnerd1223334444 2019-10-14 19:36:38
When p-2 is a square
Toinfinity 2019-10-14 19:36:38
When $p-2$ is a perfect square
DPatrick 2019-10-14 19:36:56
Yes. $p^2$ is already a perfect square, so the whole thing is a perfect square exactly when $p-2$ is a perfect square.
DPatrick 2019-10-14 19:37:04
Great, we've simplified the problem!
DPatrick 2019-10-14 19:37:07
1. For how many primes $p < 300$ is $p-2$ a positive perfect square?
DPatrick 2019-10-14 19:37:17
We could list all the primes less than $300$ and check them, but what's simpler?
mathstats 2019-10-14 19:37:38
pick all the squares from 1 - 300 and see which ones work
Qwerty71 2019-10-14 19:37:38
there are 17 squares up until 300; now just add 2 and check if they are prime
asdf334 2019-10-14 19:37:38
add 2 to every square
aastha.sharma02 2019-10-14 19:37:38
so then i listed the squares below 300 because there are fewer squares then there are primes
MagnificentMathematician 2019-10-14 19:37:38
find all the squares
ZhaoPow 2019-10-14 19:37:38
list odd squares less than 300
melonder 2019-10-14 19:37:38
list perfect sqyuares and add 2
DPatrick 2019-10-14 19:37:47
Right. It's way simpler to do it in reverse: list all the positive perfect squares less than $300$ and check which are $2$ less than a prime.
DPatrick 2019-10-14 19:37:59
That is, we're answering the question:
DPatrick 2019-10-14 19:38:02
1. For how many positive integers $n$ is $p = n^2 + 2$ a prime less than $300$?
DPatrick 2019-10-14 19:38:18
And as many of you have pointed out, we only have to check it for the odd $n$. (The prime $p=2$ comes from $n=0$ but that's not positive.)
DPatrick 2019-10-14 19:38:28
So now we can make a chart:
DPatrick 2019-10-14 19:38:32
$$\begin{array}{r|r}

n & n^2 + 2 \\ \hline

1 & 3 \\

3 & 11 \\

5 & 27 \\

7 & 51 \\

9 & 83 \\

11 & 123 \\

13 & 171 \\

15 & 227 \\

17 & 291

\end{array}$$
DPatrick 2019-10-14 19:38:45
$n=19$ gives $n^2 + 2 = 363$, which is too big. So this is all we have to check.
DPatrick 2019-10-14 19:38:48
Which numbers in the right column are prime?
Mathnerd1223334444 2019-10-14 19:39:26
3, 11, 83, 227 I think
SoulOfHeaven 2019-10-14 19:39:26
3, 11, 83, 227
melonder 2019-10-14 19:39:26
3,11.83.227
naman12 2019-10-14 19:39:26
$3,11,83,227$
RagingWar 2019-10-14 19:39:26
3,11,83,227
Student. 2019-10-14 19:39:26
3, 11, 83, 227
Maskie 2019-10-14 19:39:26
3 11 83 227
kvedula2004 2019-10-14 19:39:26
all others divisible by 3
DPatrick 2019-10-14 19:39:40
We can start with the easy ones: $3$ and $11$ are prime.
DPatrick 2019-10-14 19:40:02
And we can get rid of some easy ones: $27$, $51$, $123$, $171$, $291$ are all multiples of $3$ (their digits sum to a multiple of $3$), so are composite.
DPatrick 2019-10-14 19:40:14
That just leaves $83$ and $227$ to check.
DPatrick 2019-10-14 19:40:38
...which you can do, and they indeed are both prime.
bigwolfy 2019-10-14 19:40:50
4
muhandae17 2019-10-14 19:40:50
total of 4
heidinet 2019-10-14 19:41:00
4
RagingWar 2019-10-14 19:41:00
4
Ausmumof3 2019-10-14 19:41:00
4
LovingPilot 2019-10-14 19:41:00
4
DPatrick 2019-10-14 19:41:01
That means that the only primes $p$ satisfying the condition of the problem are $3$, $11$, $83$, and $227$, for a total of $\boxed{4}$ primes.
DPatrick 2019-10-14 19:41:36
Surprisingly, this turned out to be one of the harder problems on the contest! Only about 28% of students taking the contest electronically got this right.
DPatrick 2019-10-14 19:41:54
Next up:
DPatrick 2019-10-14 19:41:58
2. Compute $\cos\left(\tan^{-1}\left(\dfrac{-\sqrt{7}}{3}\right)\right)$ (where $\tan^{-1}$ denotes the inverse tangent function).
DPatrick 2019-10-14 19:42:36
Before we get too deep into this: what is the sign of our answer going to be? Will it be positive or negative?
yoyopianow 2019-10-14 19:43:05
Positive
kvedula2004 2019-10-14 19:43:05
positive, range of tan^-1 is -pi/2 to pi/2
DPatrick 2019-10-14 19:43:17
That's right. $\tan^{-1}$ always gives us a value between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ (or $-90^\circ$ and $90^\circ$, if you prefer).
DPatrick 2019-10-14 19:43:32
But cosine is positive for any angle in that range.
DPatrick 2019-10-14 19:43:50
(You can think of the graph of cosine, or of the unit circle, to see this.)
DPatrick 2019-10-14 19:44:06
So our answer must end up being positive.
DPatrick 2019-10-14 19:44:14
And now what?
Mathnerd1223334444 2019-10-14 19:44:27
draw a triangle and stuff cancels
JustKeepRunning 2019-10-14 19:44:27
draw the triangle
DPatrick 2019-10-14 19:44:43
Yes: I generally like to just draw a right triangle for problems like this.
DPatrick 2019-10-14 19:44:48
DPatrick 2019-10-14 19:45:08
That's the triangle with $\tan(\theta) = \dfrac{\sqrt7}{3}$.
DPatrick 2019-10-14 19:45:40
And since we know the answer we want is positive, we can use this triangle even though what we really want is $\tan(\theta) = -\dfrac{\sqrt7}{3}$ -- the cosine will be the same.
DPatrick 2019-10-14 19:45:52
So the quantity we want for our answer is $\cos(\theta)$.
RagingWar 2019-10-14 19:45:56
cosine is adjacent/hypotenuse
kvedula2004 2019-10-14 19:45:59
pythag thrm, hypotenuse = 4
DPatrick 2019-10-14 19:46:05
Yep. The hypotenuse is $\sqrt{7 + 9} = \sqrt{16} = 4$.
Green4Applez 2019-10-14 19:46:15
so 3/4
t3sp 2019-10-14 19:46:15
so hypotenuse is 4 and cos(theta)=3/4
Maskie 2019-10-14 19:46:15
so it would be 3/4
DPatrick 2019-10-14 19:46:20
So $\cos\theta = \boxed{\dfrac34}$ is our answer.
kvedula2004 2019-10-14 19:46:47
these are tough early questions...
DPatrick 2019-10-14 19:46:59
This turned out to be the 3rd easiest...about 54% got it right.
DPatrick 2019-10-14 19:47:17
I think the contest as a whole was pretty tough this time around.
DPatrick 2019-10-14 19:47:25
On to #3:
DPatrick 2019-10-14 19:47:31
3 What proportion of the nine-digit numbers that can be formed by permuting the digits of $123456789$ are divisible by $36$?
DPatrick 2019-10-14 19:47:49
How do we determine if a number is divisible by $36$?
asdf334 2019-10-14 19:48:01
9 and 4
SoulOfHeaven 2019-10-14 19:48:01
has to be a multiple of 4 and 9
motorfinn 2019-10-14 19:48:01
Divisible by 9 and 4
aastha.sharma02 2019-10-14 19:48:01
div by 9 and 4
ZhaoPow 2019-10-14 19:48:01
divisible by 9 and 4
varunragu23 2019-10-14 19:48:01
divisible by 9 and 4
AlexLikeMath 2019-10-14 19:48:01
divisible by 4 and 9
Edwinyc 2019-10-14 19:48:05
by checking if it's divisible by 4 and 9
DPatrick 2019-10-14 19:48:10
We can look at the prime factors of $36$ separately. That is, it has to be divisible by both $4$ and $9$.
Orangecounty2 2019-10-14 19:48:40
all are divisible by 9
ilikechocolate 2019-10-14 19:48:40
for 9 we add up digits, but no matter how we permutate, sum is always 45 so always divisible by 9
kvedula2004 2019-10-14 19:48:40
divisibility by 9 is gurenteed bc sum of digits in constant (45)
aastha.sharma02 2019-10-14 19:48:40
its automatically divisible by 9 because the sum is 45
SoulOfHeaven 2019-10-14 19:48:40
9 is taken care of though because 1+2+3....+9 is a multilpe of 9
DPatrick 2019-10-14 19:48:53
Aha...we know a number is divisible by $9$ if the sum of its digits is a multiple of $9$.
DPatrick 2019-10-14 19:48:59
But we're permuting the digits, so the sum is always the same! It's $1+2+\cdots+9 = 45$.
DPatrick 2019-10-14 19:49:10
So every permutation is divisible by $9$. We only have to worry about divisibility by $4$.
ilikechocolate 2019-10-14 19:49:32
for 4, we need to have the 2 final numbers be divisible by 4
NerdyDude 2019-10-14 19:49:32
last two digits as a number have to be divisible by 4
mathstats 2019-10-14 19:49:32
for 4 last 2 digits div by 4
GammaZero 2019-10-14 19:49:32
THe divisibility for $4$ is that the last 2 numbers of a quantity are divisible by 4
DPatrick 2019-10-14 19:49:50
And we know when is a number divisible by $4$: when its last two digits is a multiple of $4$.
DPatrick 2019-10-14 19:49:55
So we only have to look at the last two digits of our permutation.
DPatrick 2019-10-14 19:50:05
How many possible last two digits are there?
DPatrick 2019-10-14 19:50:28
(in total, not necessarily divisible by 4)
varunragu23 2019-10-14 19:50:46
9*8 = 72
AlexLikeMath 2019-10-14 19:50:46
72
atmchallenge 2019-10-14 19:50:46
72
Green4Applez 2019-10-14 19:50:46
9*8=72
cj13609517288 2019-10-14 19:50:46
$8\cdot9=72$
Giakki 2019-10-14 19:50:46
72
JustKeepRunning 2019-10-14 19:50:46
72
DPatrick 2019-10-14 19:51:05
There are $9 \cdot 8 = 72$ choices for the last two digits. (9 choices for the units then 8 choices for the tens.)
DPatrick 2019-10-14 19:51:09
And how many of them are multiples of $4$?
JustKeepRunning 2019-10-14 19:51:47
16
Puddles_Penguin 2019-10-14 19:51:47
16
RagingWar 2019-10-14 19:51:47
16
Giakki 2019-10-14 19:51:47
16
yoyopianow 2019-10-14 19:51:47
16
DPatrick 2019-10-14 19:51:58
We can list them: $12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96.$
DPatrick 2019-10-14 19:52:13
(These are all the two-digit multiples of $4$, except for those ending in $0$ like $20,40,60,80$ and those with a repeated digit like $44,88$.)
DPatrick 2019-10-14 19:52:34
There are $16$ numbers in our list.
DPatrick 2019-10-14 19:52:41
So what's the final answer?
MegaminxHB2 2019-10-14 19:53:01
$\frac{2}{9}$
Maskie 2019-10-14 19:53:01
2/9
RWhite 2019-10-14 19:53:01
16/72 = 2/9
Alvin54968972 2019-10-14 19:53:01
So it’s just 16/72 = 2/9?
naman12 2019-10-14 19:53:01
$\frac{2}{9}$
Student. 2019-10-14 19:53:01
2/9
DPatrick 2019-10-14 19:53:11
So the proportion of numbers that work is $\dfrac{16}{72}$.
DPatrick 2019-10-14 19:53:16
In lowest terms, this is $\boxed{\dfrac{2}{9}}$.
DPatrick 2019-10-14 19:53:27
Note that we don't have to worry about the first seven digits: they don't affect the divisibility!
DPatrick 2019-10-14 19:53:42
Only the last two digits are relevant.
Puddles_Penguin 2019-10-14 19:54:05
I assume a good majority got this. Correct?
DPatrick 2019-10-14 19:54:20
Nope. Only 26%; it was the 2nd hardest (not counting #10 as we'll see later.)
DPatrick 2019-10-14 19:54:36
On to #4:
DPatrick 2019-10-14 19:54:45
4. Let $f(x) = (2x+3)^3$ and $g(x) = x^3+x^2-x-1$. What is the sum of the coefficients of the polynomial $f(g(x))$?
DPatrick 2019-10-14 19:55:04
Note that this problem appeared differently on different versions of the contest, but the solution method is the same.
DPatrick 2019-10-14 19:55:20
We could try to write an algebraic expression for $f(g(x))$, but is there a simpler idea?
kvedula2004 2019-10-14 19:55:41
plug in 1
asdf334 2019-10-14 19:55:41
plug in 1
serichaoo 2019-10-14 19:55:41
set x=1 to find coefficients
JustKeepRunning 2019-10-14 19:55:41
plug in $1$
iamhungry 2019-10-14 19:55:41
Plug in $1$
jacoporizzo 2019-10-14 19:55:41
plug in 1 for x
Ilikeapos 2019-10-14 19:55:41
Yes, just find $f(g(1))$
AlexLikeMath 2019-10-14 19:55:41
Plug in 1
DPatrick 2019-10-14 19:55:56
Right, that's the key. If $p(x)$ is any polynomial, then $p(1)$ is the sum of its coefficients.
DPatrick 2019-10-14 19:56:07
So all we need to do is to compute $f(g(1))$.
Orangecounty2 2019-10-14 19:56:22
but g(1) = 0!!
iamhungry 2019-10-14 19:56:22
$g(1)=0$
aastha.sharma02 2019-10-14 19:56:22
g(1)=0
DPatrick 2019-10-14 19:56:30
We have $g(1) = 1^3 + 1^2 - 1 - 1$.
DPatrick 2019-10-14 19:56:36
This gives $g(1) = 0$.
DPatrick 2019-10-14 19:56:45
So our answer is $f(g(1)) = f(0)$.
iamhungry 2019-10-14 19:57:05
$f(g(1))=f(0)=(3)^3=27$
SoulOfHeaven 2019-10-14 19:57:05
so 27
JustKeepRunning 2019-10-14 19:57:05
$f(0)=27$
NerdyDude 2019-10-14 19:57:05
(0+3)^3=27
aastha.sharma02 2019-10-14 19:57:05
so $f(0)=27$
ilikechocolate 2019-10-14 19:57:05
f(0) = 27
DPatrick 2019-10-14 19:57:13
And this is just $f(0) = 3^3 = \boxed{27}$.
DPatrick 2019-10-14 19:57:40
Again, if you took the contest, you might have had a different $f(x)$ (but I believe the same $g(x)$), so you might have ended up with a different numeric answer.
RagingWar 2019-10-14 19:57:52
this one must have been easier
NerdyDude 2019-10-14 19:57:52
how many got this right
DPatrick 2019-10-14 19:57:59
This was about 34%.
DPatrick 2019-10-14 19:58:35
I should mention that these stats are the ones that the AMS sent me this morning, and only include electronic participants. If you took the contest on paper, those aren't included yet.
DPatrick 2019-10-14 19:58:45
On to #5:
DPatrick 2019-10-14 19:58:51
5. The incircle (or inscribed circle) of a triangle is the circle that is tangent to all three sides of the triangle. What is the radius of the incircle in a triangle with side lengths $13$, $14$, and $15$?
DPatrick 2019-10-14 19:59:07
There is a very useful formula for the radius of the incircle of a triangle. Do you know it?
kvedula2004 2019-10-14 19:59:30
A=sr
RagingWar 2019-10-14 19:59:30
rs=area
NerdyDude 2019-10-14 19:59:30
A=rs
xiaomage 2019-10-14 19:59:30
area=rs
serichaoo 2019-10-14 19:59:30
A=rs
Ilikeapos 2019-10-14 19:59:30
$A = sr$
asdf334 2019-10-14 19:59:30
A=rs
JustKeepRunning 2019-10-14 19:59:30
a=rs
Orangecounty2 2019-10-14 19:59:30
[ABC] = rs
Maskie 2019-10-14 19:59:30
Sr=A
aop2014 2019-10-14 19:59:30
r=a/s
Alvin54968972 2019-10-14 19:59:30
A = rs
AlexLikeMath 2019-10-14 19:59:30
rs = A
mathstats 2019-10-14 19:59:30
sr=A
DPatrick 2019-10-14 19:59:36
The formula is $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is half of the perimeter of the triangle (this is called the semiperimeter).
SoulOfHeaven 2019-10-14 19:59:44
draw lines from center of circle to points of tangency
DPatrick 2019-10-14 19:59:56
Right -- even if you don't have this formula memorized, we can recover it.
DPatrick 2019-10-14 20:00:16
Let's draw a generic acute triangle and its incircle:
DPatrick 2019-10-14 20:00:19
DPatrick 2019-10-14 20:00:26
We can draw the radii from the center of the incircle (which is, not surprisingly, called the incenter) to the three sides:
DPatrick 2019-10-14 20:00:28
DPatrick 2019-10-14 20:00:42
And how does that help us prove the formula?
serichaoo 2019-10-14 20:01:01
Draw lengths from incentre to the vertices
RWhite 2019-10-14 20:01:01
we can draw lines from the vertices to the incenter
cj13609517288 2019-10-14 20:01:09
spilt the whole are into pieces
DPatrick 2019-10-14 20:01:12
We can draw segments from the incenter to the three vertices, like so:
DPatrick 2019-10-14 20:01:14
Green4Applez 2019-10-14 20:01:23
then we have tree triangles and compute the sum of those areas
kaiclc 2019-10-14 20:01:27
connect the incenter to the three vertices, three triangles combined area of sr
AlexLikeMath 2019-10-14 20:01:27
Add 3 triangles up
DPatrick 2019-10-14 20:01:30
And now we can compute the areas of the three blue triangles and sum them up!
DPatrick 2019-10-14 20:01:37
Each blue triangle has height $r$ and a base that's a side of the triangle.
DPatrick 2019-10-14 20:01:56
So if the three sides of the big triangle are $a$, $b$, and $c$, the areas of the blue triangles are $\frac12ra$, $\frac12rb$, and $\frac12rc$.
DPatrick 2019-10-14 20:02:14
And when we sum them, we get the area of the whole triangle. So $A = \frac12ra + \frac12rb + \frac12rc$.
RWhite 2019-10-14 20:02:17
ra/2+rb/2+rc/2=r(a+b+c)/2
DPatrick 2019-10-14 20:02:25
But then this factors as $A = r\frac12(a+b+c) = rs$, just like we wanted!
DPatrick 2019-10-14 20:02:43
So back to our problem...the semiperimeter is $s = \frac12(13+14+15) = 21$.
DPatrick 2019-10-14 20:02:48
Do we know the area?
NerdyDude 2019-10-14 20:03:15
herons formula!
cj13609517288 2019-10-14 20:03:15
we have heron! 84
SoulOfHeaven 2019-10-14 20:03:15
Use Heron's formula
naman12 2019-10-14 20:03:15
Yes, $84$ by Heron's Formula
mathstats 2019-10-14 20:03:15
herons
Giakki 2019-10-14 20:03:15
heron formula
RWhite 2019-10-14 20:03:15
heron's formula
yoyopianow 2019-10-14 20:03:15
Heron's?
lilyba 2019-10-14 20:03:15
use heron's formula
DPatrick 2019-10-14 20:03:21
One method is to use Heron's Formula, which states that the area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter and $a$,$b$,$c$ are the three sides.
DPatrick 2019-10-14 20:03:31
If we plug our data in, we get $A = \sqrt{21(21-13)(21-14)(21-15)}$.
DPatrick 2019-10-14 20:03:45
This simplifies to $A = \sqrt{21(8)(7)(6)}$, and that works out to $A = \sqrt{3 \cdot 7 \cdot 2^3 \cdot 7 \cdot 2 \cdot 3}$ in prime factors, so $A = 2^2 \cdot 3 \cdot 7 = 84$.
DPatrick 2019-10-14 20:03:53
But we don't need Heron's Formula. 13-14-15 triangles come up often in contest problems, because there's a neat little trick for them.
JustKeepRunning 2019-10-14 20:04:01
Drop altitude to make nice pythag
Puddles_Penguin 2019-10-14 20:04:05
altitude to 14 is 12
AlexLikeMath 2019-10-14 20:04:08
Yes, drop altitude to 14 side
Maskie 2019-10-14 20:04:13
split the triangle into 9-12-15 and 5-12-13 triangles
DPatrick 2019-10-14 20:04:23
What happens when we draw the altitude to the length $14$ side?
DPatrick 2019-10-14 20:04:26
DPatrick 2019-10-14 20:04:36
If you know your basic Pythagorean triples, you might be able to guess the missing side lengths.
xiaomage 2019-10-14 20:04:45
two right triangles
Alvin54968972 2019-10-14 20:04:45
Altitude is 12
aop2014 2019-10-14 20:04:45
the altitude is 12, and the base becomes 9 and 5
DPatrick 2019-10-14 20:04:49
DPatrick 2019-10-14 20:04:58
That is: a 13-14-15 triangle is just a 9-12-15 right triangle and a 5-12-13 right triangle glued together along their common length 12 sides!
DPatrick 2019-10-14 20:05:16
And we see that the area is $\frac12(12)(14) = 84$.
SoulOfHeaven 2019-10-14 20:05:35
84/21 = 4
xiaomage 2019-10-14 20:05:35
radius=4
RWhite 2019-10-14 20:05:35
so the answer would be 84=r(21) r = 4
Edwinyc 2019-10-14 20:05:35
so r=4
DPatrick 2019-10-14 20:05:42
And finally, using $A = rs$ with $A = 84$ and $s = 21$, we get $r = \boxed{4}$.
DPatrick 2019-10-14 20:06:11
This one was 41% correct.
DPatrick 2019-10-14 20:06:19
On to #6:
DPatrick 2019-10-14 20:06:26
6. The integers $3$, $4$, $5$, $6$, $12$, and $13$ are arranged, without repetition, in a horizontal row so that the sum of any two numbers in adjacent positions is a perfect square (square of a positive integer). What is the sum of the first and last numbers in the arrangement?
aastha.sharma02 2019-10-14 20:06:34
which one was easiest?
DPatrick 2019-10-14 20:06:38
This one.
DPatrick 2019-10-14 20:07:01
I don't know any clever way to solve this except to just get our handy dirty and try to construct this arrangement.
NerdyDude 2019-10-14 20:07:13
pair the numbers to find the two that can only be paired with one othe rnumber
skyguy88 2019-10-14 20:07:18
If you check which numbers can form only one perfect square with the other integers...
DPatrick 2019-10-14 20:07:25
Yes. I'd solve it like a jigsaw puzzle: look for the end pieces first.
DPatrick 2019-10-14 20:07:39
That is, are there any numbers who only have one other number to pair with to get a perfect square?
Green4Applez 2019-10-14 20:08:08
to start, six should be first (or last) since it can only add with one other number to be a perfect square
asdf334 2019-10-14 20:08:08
6and5
varunragu23 2019-10-14 20:08:08
5 and 6
RagingWar 2019-10-14 20:08:08
6 has to be a last one, can only be paired with 3
kaiclc 2019-10-14 20:08:08
5 and 6
NerdyDude 2019-10-14 20:08:08
6 can only pair with 3
mtqchen 2019-10-14 20:08:08
5 and 6
chessking1 2019-10-14 20:08:08
6 and 5
DPatrick 2019-10-14 20:08:14
$5$ only pairs with $4$.
DPatrick 2019-10-14 20:08:18
And $6$ only pairs with $3$.
yoyopianow 2019-10-14 20:08:29
So we're done?
DPatrick 2019-10-14 20:08:35
I think so. We could actually stop here!
DPatrick 2019-10-14 20:08:41
If we have a valid arrangement, we know that numbers with just one pair must go on the ends.
DPatrick 2019-10-14 20:08:55
So since we've just determined that the ends must be $5$ and $6$, our answer is $5 + 6 = \boxed{11}$.
Puddles_Penguin 2019-10-14 20:09:19
5,4,12,13,3,6 so 11
asdf334 2019-10-14 20:09:19
it's 5,4,12,13,3,6
DPatrick 2019-10-14 20:09:29
To be safe, and to check, we could construct the entire arrangement.
DPatrick 2019-10-14 20:09:34
The arrangement is $5,4,12,13,3,6$. (Or it could be reversed as $6,3,13,12,4,5$.) These are the only possibilities.
RWhite 2019-10-14 20:09:46
what percentage got this right?
DPatrick 2019-10-14 20:09:50
About 80%.
DPatrick 2019-10-14 20:09:59
This was the easiest by a wide margin.
DPatrick 2019-10-14 20:10:15
And the next problem was the second-easiest:
DPatrick 2019-10-14 20:10:19
7. Which of the following numbers is the product of three consecutive prime numbers?
$\phantom{hi}$
a. 1223 b. 1309 c. 1989 d. 2431 e. 2717
NerdyDude 2019-10-14 20:10:48
estimate with perfect cubes??
asdf334 2019-10-14 20:10:48
find lower and upper bounds
ilikechocolate 2019-10-14 20:10:54
estimate
Bole 2019-10-14 20:10:54
look at which cubes are closest
DPatrick 2019-10-14 20:11:02
Indeed. All the numbers are slightly larger than $1000 = 10^3$.
DPatrick 2019-10-14 20:11:14
$11^3 = 1331$ is in the ballpark of the choices. So is $13^3 = 2197$.
t3sp 2019-10-14 20:11:22
can you use a calculator on this test?
DPatrick 2019-10-14 20:11:26
Nope.
mathstats 2019-10-14 20:11:34
do 7 ,11 ,13
AlexLikeMath 2019-10-14 20:11:37
7*11*13 = 1001, check 11*13*17
DPatrick 2019-10-14 20:11:58
Sure, we can check that $7 \cdot 11 \cdot 13 = 1001$. (This is a neat fact that comes up every once in a while in problems.)
DPatrick 2019-10-14 20:12:24
That's too small, so the next one to check is $11 \cdot 13 \cdot 17$.
skyguy88 2019-10-14 20:12:50
11 times 13 times 17 is 2431
SoulOfHeaven 2019-10-14 20:12:50
11*13*17 = 2431
Alvin54968972 2019-10-14 20:12:50
11, 13, 17 gets 2431 (D)
SoulOfHeaven 2019-10-14 20:12:50
aha it's D
RWhite 2019-10-14 20:12:50
2431=11*13*17
xiaomage 2019-10-14 20:12:50
It's 2431
Giakki 2019-10-14 20:12:59
11x13x17 = 2431
DPatrick 2019-10-14 20:13:12
$13 \cdot 17 = 221$, so $11 \cdot 13 \cdot 17 = 11 \cdot 221 = 2210 + 221 = 2431$.
DPatrick 2019-10-14 20:13:19
(That's how I did it without a calculator.)
DPatrick 2019-10-14 20:13:26
Aha! That's choice $\boxed{\text{d}}$.
DPatrick 2019-10-14 20:13:41
This was about 57% correct.
yoyopianow 2019-10-14 20:13:56
Were some of the problems multiple choice and some short answer?
DPatrick 2019-10-14 20:14:02
That was the only problem that was multiple choice.
DPatrick 2019-10-14 20:14:12
Next up is #8:
DPatrick 2019-10-14 20:14:16
8. What is the sum of all real solutions to $\sqrt{x+15-8\sqrt{x-1}} = 2$?
motorfinn 2019-10-14 20:14:53
Square it
RagingWar 2019-10-14 20:14:53
square it!!!
NerdyDude 2019-10-14 20:14:53
square both sides
Bole 2019-10-14 20:14:53
get rid of the square roots
Green4Applez 2019-10-14 20:14:53
we can start off by squaring both sides
alphaone001 2019-10-14 20:14:53
square and then isolate the radical?
montana_mathlete 2019-10-14 20:14:53
Square it.
aop2014 2019-10-14 20:14:53
first square both sides
DPatrick 2019-10-14 20:15:02
Good idea -- we can square both sides to get rid of the outside radical.
DPatrick 2019-10-14 20:15:09
That gives us $x + 15 - 8\sqrt{x-1} = 4$.
DPatrick 2019-10-14 20:15:13
Now what?
asdf334 2019-10-14 20:15:37
subtract x+15 from both
Creator 2019-10-14 20:15:37
take 8sqrtx-1 to the other side and square both sides
mwizard 2019-10-14 20:15:37
move root to one side, and square again
Orangecounty2 2019-10-14 20:15:37
isolate square root and squarwe again
DPatrick 2019-10-14 20:15:47
Right: we want to isolate the radical so that we can square again.
DPatrick 2019-10-14 20:15:52
That is, we rearrange to get $x + 11 = 8\sqrt{x-1}$.
DPatrick 2019-10-14 20:16:04
And now we can square again.
DPatrick 2019-10-14 20:16:11
We get $(x+11)^2 = 64(x-1)$.
motorfinn 2019-10-14 20:16:25
x^2+22x+121=64x-64
JayEmSea 2019-10-14 20:16:25
Move x terms to one side
DPatrick 2019-10-14 20:16:27
This gives us $x^2 + 22x + 121 = 64x - 64$.
DPatrick 2019-10-14 20:16:31
And this simplifies to $x^2 - 42x + 185 = 0$.
DPatrick 2019-10-14 20:17:32
You might think "We don't need to solve this: we know that the sum of the solutions to this is $42$ by Vieta's formulas, so that's the answer."
DPatrick 2019-10-14 20:17:36
What's a potential problem with this?
naman12 2019-10-14 20:18:08
Extranneous roots
GammaZero 2019-10-14 20:18:08
extrataneous solutions
AlexLikeMath 2019-10-14 20:18:08
erraneous solutions
mathstats 2019-10-14 20:18:08
extraneous solutions
varunragu23 2019-10-14 20:18:08
oh, square root might have negative numbers inside of it and we can't have that happen
DPatrick 2019-10-14 20:18:14
We might have introduced extraneous solutions when we squared.
DPatrick 2019-10-14 20:18:31
For example, consider the much simpler equation $\sqrt{x} = x - 1$.
DPatrick 2019-10-14 20:18:42
If we square this, we get $x = x^2 - 2x + 1$, so $0 = x^2 - 3x + 1$.
DPatrick 2019-10-14 20:18:52
And by the quadratic formula, the solutions are $x = \dfrac{3 \pm \sqrt5}{2}$.
DPatrick 2019-10-14 20:19:02
However, the solution $x = \dfrac{3 - \sqrt5}{2}$ is extraneous: if we go back to our original equation and plug this in, we get $$\sqrt{\dfrac{3-\sqrt5}{2}} = \dfrac{3-\sqrt5}{2} - 1 = \dfrac{1 - \sqrt5}{2},$$ which cannot be true because the left side is positive (by definition) but the right side is negative.
DPatrick 2019-10-14 20:19:38
So, back to our problem: to be sure the solutions to our quadratic actually work in the original equation, we should find the actual solutions and check that they actually work.
DPatrick 2019-10-14 20:19:41
What are the solutions to $x^2 - 42x + 185 = 0$?
TheTallGuy2.0 2019-10-14 20:20:05
(x-5)(x-37)
RWhite 2019-10-14 20:20:05
x = 5 or 37
GammaZero 2019-10-14 20:20:05
5 and 37
xiaomage 2019-10-14 20:20:05
(x-37)(x-5), so x=5 or 37
AlexLikeMath 2019-10-14 20:20:05
5 & 37
SoulOfHeaven 2019-10-14 20:20:11
37 and 5
Edwinyc 2019-10-14 20:20:11
x=37, x=5
DPatrick 2019-10-14 20:20:21
We could certainly use the quadratic formula here...
DPatrick 2019-10-14 20:20:30
...or notice that this factors as $(x-5)(x-37) = 0$, so the solutions are $x=5$ and $x=37$.
DPatrick 2019-10-14 20:20:44
And so let's check them.
DPatrick 2019-10-14 20:20:52
Using $x=5$ we get $\sqrt{5+15-8\sqrt{5-1}} = \sqrt{5+15-8(2)} = \sqrt{4} = 2$. OK!
DPatrick 2019-10-14 20:21:06
Using $x=37$ we get $\sqrt{37+15-8\sqrt{37-1}} = \sqrt{37+15-48} = \sqrt{4} = 2$. OK!
Maskie 2019-10-14 20:21:22
42!
kvedula2004 2019-10-14 20:21:22
now our answer is 42
NerdyDude 2019-10-14 20:21:22
so it is 42
DPatrick 2019-10-14 20:21:25
So both solutions are valid, and their sum is $\boxed{42}$.
DPatrick 2019-10-14 20:21:38
This was medium difficult: 40% got it right.
DPatrick 2019-10-14 20:22:02
Next is #9, which was the hardest of 1-9 (#10 was special as we'll soon see):
DPatrick 2019-10-14 20:22:06
9. A jar contains ten balls, numbered $1$ to $10$. Three balls are randomly drawn from the jar without replacement. What is the probability that no two of the three balls are labeled with consecutive integers?
DPatrick 2019-10-14 20:22:44
This is a tricky problem. There are a number of ways to approach it.
kvedula2004 2019-10-14 20:23:02
complmentary counting...
NerdyDude 2019-10-14 20:23:02
1-P(balls have consecutive integers)
motorfinn 2019-10-14 20:23:02
Complementary Counting
alphaone001 2019-10-14 20:23:02
complementary counting?
DPatrick 2019-10-14 20:23:12
It might be easiest to use complementary counting here: let's determine the probability that we do draw balls where two of them are congruent.
DPatrick 2019-10-14 20:23:20
How could we draw a set of balls that fail?
alphaone001 2019-10-14 20:23:50
all three are consecutive, or 2 are consecutive and one is not
asdf334 2019-10-14 20:23:50
all consecutive or 2 consecutive and one is away from those 2
DPatrick 2019-10-14 20:24:04
Right. There are two cases:

1. All three balls are consecutive (like 6, 7, 8)

2. Two of the balls are consecutive but the third isn't (like 3, 4, 9)
DPatrick 2019-10-14 20:24:28
How many sets of balls in case 1? (Let's count them without regard to order, so 6,7,8 is the same as 7,6,8.)
Giakki 2019-10-14 20:25:01
8
montana_mathlete 2019-10-14 20:25:01
$8$
naman12 2019-10-14 20:25:01
$8$
Student. 2019-10-14 20:25:01
8
Yagato 2019-10-14 20:25:01
8 cases*
DPatrick 2019-10-14 20:25:08
There are exactly $8$: the middle ball can be any of 2 through 9 (inclusive).
DPatrick 2019-10-14 20:25:37
How many sets of balls in case 2, with two balls consecutive and the third not?
DPatrick 2019-10-14 20:26:13
Well...first, how many ways to choose two consecutive balls?
varunragu23 2019-10-14 20:26:32
9 ways
Yagato 2019-10-14 20:26:32
9 ways
xiaomage 2019-10-14 20:26:32
9
SoulOfHeaven 2019-10-14 20:26:32
9
bob321326 2019-10-14 20:26:32
9 ways
AlexLikeMath 2019-10-14 20:26:32
9
NerdyDude 2019-10-14 20:26:32
9 same method as before
aop2014 2019-10-14 20:26:32
9
DPatrick 2019-10-14 20:26:47
Right: we can have any pair from 1-2 up to 9-10, so there are $9$ possibilities.
DPatrick 2019-10-14 20:26:54
Then, how many ways to choose the third ball?
Maskie 2019-10-14 20:27:23
depends on which two
asdf334 2019-10-14 20:27:23
depends
JustKeepRunning 2019-10-14 20:27:23
Is determined by first choice
Student. 2019-10-14 20:27:23
6 or 7 if the first two are 1 and 2 or 9 and 10
DPatrick 2019-10-14 20:27:28
Right: it depends!
DPatrick 2019-10-14 20:27:49
If the pair is either 1-2 or 9-10, there are $7$ choices for the single ball. (For example, if the pair is 1-2, then any of 4 through 10 could be the single ball).
DPatrick 2019-10-14 20:28:09
For any other pair (and there are $7$ others), there are $6$ choices for the single ball (we must exclude the two balls in the pair and the two balls on either side).
Student. 2019-10-14 20:28:20
7 * 6 + 2 * 7 = 56
Maskie 2019-10-14 20:28:26
7*2+7*6=56
DPatrick 2019-10-14 20:28:44
Right: adding these up, there are $2 \cdot 7 + 7 \cdot 6 = 56$ ways to choose a pair and then a third ball.
asdf334 2019-10-14 20:28:52
case 1+case 2=64
RagingWar 2019-10-14 20:28:52
+ the other 8 combinations
NerdyDude 2019-10-14 20:28:52
56+8 =64 total unfavorable cases
DPatrick 2019-10-14 20:29:07
And then adding up the cases, there are $8 + 56 = 64$ sets of three balls that fail.
DPatrick 2019-10-14 20:29:15
How do we finish from here?
knchen 2019-10-14 20:29:30
find total amount of cases
mtqchen 2019-10-14 20:29:30
Subtract from the total possibilities
abcb1030 2019-10-14 20:29:36
Find the total number of ways the 3 balls can be drawn
aop2014 2019-10-14 20:29:40
subtract from $\dbinom{10}{3}$
Yagato 2019-10-14 20:29:47
a total of 10 choose 3 cases = 120
DPatrick 2019-10-14 20:29:52
Right. There are $\dbinom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$ ways to choose any three balls.
DPatrick 2019-10-14 20:29:59
And $64$ of them fail.
DPatrick 2019-10-14 20:30:02
So $120 - 64 = 56$ of them succeed.
DPatrick 2019-10-14 20:30:21
And thus what's the probability of success?
asdf334 2019-10-14 20:30:45
56/120=7/15!
kvedula2004 2019-10-14 20:30:45
56/120 = 7/15
RWhite 2019-10-14 20:30:45
is the answer 7/15
Orangecounty2 2019-10-14 20:30:45
7/15
mathstats 2019-10-14 20:30:45
7/15
motorfinn 2019-10-14 20:30:45
56/120=7/15
gmannnnnnn 2019-10-14 20:30:45
$7/15$
GammaZero 2019-10-14 20:30:45
7/15
Puddles_Penguin 2019-10-14 20:30:45
7/15
GammaZero 2019-10-14 20:30:45
$\frac{7}{15}$
Tenthgrader 2019-10-14 20:30:45
7/15
DPatrick 2019-10-14 20:30:49
It's $\dfrac{56}{120}$, which simplifies to $\boxed{\dfrac{7}{15}}$.
DPatrick 2019-10-14 20:31:13
There is a fairly slick solution that lets us get the whole thing at once.
AlexLikeMath 2019-10-14 20:31:40
Stars and bars
Alvin54968972 2019-10-14 20:31:40
Could you use stars and bars?
DPatrick 2019-10-14 20:31:52
Right. The way to construct a set of three balls that work is as follows: imagine that the 7 balls that we don't pick are in a row, like so:
X X X X X X X
DPatrick 2019-10-14 20:32:18
Then picking 3 balls that are nonconsecutive is in 1-1 correspondence to picking 3 slots from among the 8 available slots between the X's:

_ X _ X _ X _ X _ X _ X _ X _
DPatrick 2019-10-14 20:32:31
For example, if we pick the three slots shown below:

* X _ X _ X * X * X _ X _ X _

then reading left-to right, the three balls in our set are 1, 5, and 7:

1 2 _ 3 _ 4 5 6 7 8 _ 9 _ 10 _
DPatrick 2019-10-14 20:32:58
Picking them this way ensure that there's always at least one ball between the balls we pick.
DPatrick 2019-10-14 20:33:10
There are 8 slots and we need to pick three of them, so that's $\dbinom83$ ways.
DPatrick 2019-10-14 20:33:18
And we know that there are $\dbinom{10}{3}$ total possible sets of 3 balls. So the probability is $\dfrac{\binom83}{\binom{10}{3}}$.
DPatrick 2019-10-14 20:33:25
This simplifies to $\dfrac{8 \cdot 7 \cdot 6}{10 \cdot 9 \cdot 8} = \dfrac{42}{90} = \boxed{\dfrac{7}{15}}$.
SoulOfHeaven 2019-10-14 20:33:43
slick
montana_mathlete 2019-10-14 20:33:43
Wow that's much faster
DPatrick 2019-10-14 20:33:56
If you've had a lot of experience with counting problems, you might have seen this sort of solution before.
RagingWar 2019-10-14 20:34:00
how many people got this right?
DPatrick 2019-10-14 20:34:10
About 13%. It was by far the hardest of the first nine.
DPatrick 2019-10-14 20:34:20
And at last, on to #10:
DPatrick 2019-10-14 20:34:24
10. What positive integer is closest to $(e + 2\pi)^{9/2}$?
DPatrick 2019-10-14 20:34:33
This is the tie-breaker question: it's unlikely you'll come up with the exact answer, but you want to make the best estimate that you can.
DPatrick 2019-10-14 20:34:39
And no calculators!
kvedula2004 2019-10-14 20:35:07
e+2pi is scarily close to 9
asdf334 2019-10-14 20:35:07
2.71+6.28=8.99 about 9
NerdyDude 2019-10-14 20:35:17
approx 2.71 for e and 3.14 for pi the sum is really close to 9
Alvin54968972 2019-10-14 20:35:23
e is approximately 2.718 and 2π is about 6.283
DPatrick 2019-10-14 20:35:52
Indeed, if you know a few decimal places for $e$ and $\pi$, or even just one or two decimal places, you'll see that $e + 2\pi$ is really close to $9$.
DPatrick 2019-10-14 20:36:17
So $9^{9/2}$ is already a pretty good guess.
motorfinn 2019-10-14 20:36:34
sqrt(9^9)
skyguy88 2019-10-14 20:36:34
3^9 = 19683
aop2014 2019-10-14 20:36:34
so it is about sqrt(9^9), which is 3^9 which is 19683
kvedula2004 2019-10-14 20:36:34
19683
vanilla.froyo 2019-10-14 20:36:38
its about 3^9
Creator 2019-10-14 20:36:38
(sqrt9)^9 which is 3^9
DPatrick 2019-10-14 20:36:55
And yes, we have $9^{9/2} = 3^9 = 19683$.
DPatrick 2019-10-14 20:37:02
And this is a really good guess.
DPatrick 2019-10-14 20:37:18
But let me show you what I did to get an even better guess.
Alvin54968972 2019-10-14 20:37:25
But e + 2π is a little bit more than 9
DPatrick 2019-10-14 20:37:32
Right.
DPatrick 2019-10-14 20:37:41
It turns out that I know both $e$ and $\pi$ to 5 decimal places.



\begin{align*}

e &= 2.71828\ldots \\

\pi &= 3.14159\ldots

\end{align*}
DPatrick 2019-10-14 20:37:47
(I actually know $\pi$ to a couple more and $e$ to a few more, but let's stick with this.)
DPatrick 2019-10-14 20:37:59
That gives us $e + 2\pi = 9.00146\ldots$.
DPatrick 2019-10-14 20:38:14
We might have the wrong rounded in the rightmost digit because of the "...", so lets use $e + 2\pi = 9.0015$ going forward.
DPatrick 2019-10-14 20:38:33
So how can we estimate $(9.0015)^{9/2}$?
DPatrick 2019-10-14 20:38:45
I'll write what we're trying to compute as $(9 + \epsilon)^{9/2}$, where $\epsilon = 0.0015$. (The Greek letter $\epsilon$ is commonly used for very small numbers.)
AlexLikeMath 2019-10-14 20:38:49
binomial theorem
ForeverAPenguin47 2019-10-14 20:38:53
binom thereom
DPatrick 2019-10-14 20:39:00
Yes! The "trick" is that we'll go ahead treat this like a binomial expansion, and use the Binomial Theorem on this.
DPatrick 2019-10-14 20:39:12
If $n$ were an integer, how would we expand $(x+y)^n$?
DPatrick 2019-10-14 20:39:40
I don't need all the terms, just the first two or three...
Puddles_Penguin 2019-10-14 20:40:05
x^n+nx^(n-1)y...
JustKeepRunning 2019-10-14 20:40:13
$\binom n 0 x^n+\binom n 1x^{n-1}y+\binom n 2x^{n-2}y^2\cdots$
amuthup 2019-10-14 20:40:16
$x^n + nx^{n-1}y + (n(n-1)/2)x^{n-2}y^2$
DPatrick 2019-10-14 20:40:23
Right. It starts $x^n + nx^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots$.
DPatrick 2019-10-14 20:40:34
But if $y$ is our $\epsilon$, which is a really small number, then the terms with $y^2$, $y^3$, etc. are really really small numbers.
DPatrick 2019-10-14 20:40:47
So we can forget about those terms when we approximate.
DPatrick 2019-10-14 20:40:59
That is, $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$.
DPatrick 2019-10-14 20:41:11
And the beauty of this is that this works even if $n$ isn't an integer! (Although to rigorously prove this, you probably need a little calculus.)
DPatrick 2019-10-14 20:41:29
So we will use the approximation $$(9+\epsilon)^{9/2} \approx 9^{9/2} + \frac92 \cdot 9^{7/2} \cdot \epsilon.$$
DPatrick 2019-10-14 20:41:55
And plugging in $\epsilon$, this is just $3^9 + \frac92 \cdot 3^7 \cdot 0.0015$.
DPatrick 2019-10-14 20:42:16
Conveniently, $9 \cdot 3^7$ is the same as $3^9$, so it's $3^9 \left(1 + \dfrac12 \cdot 0.0015\right)$.
DPatrick 2019-10-14 20:42:27
So this is $19683(1.00075)$.
DPatrick 2019-10-14 20:43:04
That tells us how much we should increase our original estimate of $3^9 = 19683$ by: about $0.075\%$ of $19683$.
asdf334 2019-10-14 20:43:19
0.00075 is 3/4000
asdf334 2019-10-14 20:43:26
and 19683 is abuot 20000
DPatrick 2019-10-14 20:43:34
Right, so without a calculator I'd use $0.00075 \cdot 19683 \approx 15$ (it's about $75\%$ of 19.683).
mtqchen 2019-10-14 20:43:46
Which is about 19698
DPatrick 2019-10-14 20:43:59
Exactly. My final guess was $19683 + 15 = 19698$.
kvedula2004 2019-10-14 20:44:11
really close to exact answer of 19697
asdf334 2019-10-14 20:44:11
with calculator i got 19697
vanilla.froyo 2019-10-14 20:44:11
That's really good
DPatrick 2019-10-14 20:44:19
Turns out I'm only off by one. The actual answer is $\boxed{19697}$. (Which you can check by calculator.)
DPatrick 2019-10-14 20:44:31
I was so close because the mathematics I used was legit! In particular $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$ is a really really good approximation provided $\epsilon$ is very small in comparison to $x$.
DPatrick 2019-10-14 20:44:59
I'm told that there were a handful of people who got the exact answer.
purnimak 2019-10-14 20:45:04
so how would you get the exact answer without a calculator?
DPatrick 2019-10-14 20:45:19
If you ran this estimate out to a couple more decimal places, you'd probably get it exactly.
DPatrick 2019-10-14 20:45:33
So that's it for Round 2!
DPatrick 2019-10-14 20:45:42
After the scoring for Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.
DPatrick 2019-10-14 20:45:52
Here's how the 12 finalists will be determined: 10 of the 12 will be the top scorer from Round 2 in each of the following regions:
DPatrick 2019-10-14 20:45:57
http://www.ams.org/images/wwtbam-map-us-canada.jpg
DPatrick 2019-10-14 20:46:09
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).
DPatrick 2019-10-14 20:46:19
Question #10 will be used to break ties for high scores within a region.
DPatrick 2019-10-14 20:46:30
The Championship Finals are held live in front of an audience at the Joint Mathematics Meetings, and are also live streamed on the web. (You can watch the archives of past years' finals on the WWTBAM website.) Contestants will compete directly against each other in semi-final rounds, with the semi-final winners advancing to a Jeopardy!-style buzz-in final round to determine a champion.
DPatrick 2019-10-14 20:47:01
I believe the exact date and time have been set, but I don't have that info in front of me.
mikebreen 2019-10-14 20:47:24
Sat, Jan. 18 at noon Eastern (US)
DPatrick 2019-10-14 20:47:31
Great, thanks Mike.
DPatrick 2019-10-14 20:48:02
Though I'm not sure that's right...I thought it was afternoon Denver time, which would be late afternoon ET -- am I wrong?
DPatrick 2019-10-14 20:48:07
(Wouldn't be the first time if I were.)
DPatrick 2019-10-14 20:49:00
I'll try to look it up while we continue chatting.
DPatrick 2019-10-14 20:49:14
...though I don't have much more to say. I don't have any scoring info other than the stats I've given you.
mikebreen 2019-10-14 20:49:18
We had to switch from our normal time. It will be at 10 Denver time.
DPatrick 2019-10-14 20:49:49
Aha -- that change hasn't made it into the program for the Joint Math Meetings yet (which still shows 1 pm MT).
DPatrick 2019-10-14 20:50:43
That's all for tonight. Thanks for attending!
asdf334 2019-10-14 20:50:46
thanks so much for doing this math jam!
DPatrick 2019-10-14 20:50:49
You're welcome!
mikebreen 2019-10-14 20:50:57
Thanks to all the participants in tonight's Jam--wonderful solutions--and thanks to Dave and AoPS for hosting.
TPiR 2019-10-14 20:51:08
Thanks, everyone.
Puddles_Penguin 2019-10-14 20:52:23
Thanks
ForeverAPenguin47 2019-10-14 20:52:23
thank you!
mtqchen 2019-10-14 20:52:23
Until next year
NerdyDude 2019-10-14 20:52:23
thanks
Nmath101 2019-10-14 20:52:23
thank you
NT23 2019-10-14 20:52:23
Wow this was hard but fun
aastha.sharma02 2019-10-14 20:52:23
thank you so much!
mathstats 2019-10-14 20:52:23
thank you !!
asdf334 2019-10-14 20:53:16
when and what is the next math jam?
DPatrick 2019-10-14 20:53:37
The next Math Jam is this Thursday the 17th.
DPatrick 2019-10-14 20:53:49
We will be discussing the problems from Round 1 of the USA Math Talent Search.
DPatrick 2019-10-14 20:53:58
The full schedule is at https://artofproblemsolving.com/school/mathjams
DPatrick 2019-10-14 20:55:19
I think we're ready to shut it down
devenware 2019-10-14 20:55:29
I think you're right!
DPatrick 2019-10-14 20:55:30
Everyone have a good night!
ZachSteinPerlman 2019-10-14 20:55:53
Thanks everyone!
eminentabyss 2019-10-14 20:56:02
One last huge thanks to Dave, Mike, Bill, Zach and all of you who could make it out tonight!
eminentabyss 2019-10-14 20:56:17
This classroom will close shortly.
asdf334 2019-10-14 20:56:26
goodbye!!!
asdf334 2019-10-14 20:56:26
goodbye!
piphi 2019-10-14 20:56:26
thx guys!
aastha.sharma02 2019-10-14 20:56:26
yep, thank you!
aastha.sharma02 2019-10-14 20:56:26
byeeeee
motorfinn 2019-10-14 20:56:57
Bye <3

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