Who Wants to Be a Mathematician, Round 1
Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the problems on Round 1 of qualifying for the 2019 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 Baltimore in January 2019.
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Facilitator: Dave Patrick
DPatrick
2018-09-27 19:30:11
Welcome to the 2018-19 Who Wants to Be a Mathematician Championship Round 1 Math Jam!
Welcome to the 2018-19 Who Wants to Be a Mathematician Championship Round 1 Math Jam!
DPatrick
2018-09-27 19:30:23
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 14 years, and I've written or co-written a few of our textbooks.
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 14 years, and I've written or co-written a few of our textbooks.
DPatrick
2018-09-27 19:30:36
I also 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):
I also 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):
DPatrick
2018-09-27 19:30:39
DPatrick
2018-09-27 19:30:43
Photo Credit: Maria Melin, copyright 1999 ABC Television..
Photo Credit: Maria Melin, copyright 1999 ABC Television..
Bubble3
2018-09-27 19:30:59
You won a car right
You won a car right
DPatrick
2018-09-27 19:31:07
Basically, yes: I didn't win the million bucks, but I did win enough to buy a new car.
Basically, yes: I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick
2018-09-27 19:31:19
Of course, cars were cheaper back in 1999.
Of course, cars were cheaper back in 1999.
DPatrick
2018-09-27 19:31:35
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.
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
2018-09-27 19:31:52
The classroom is moderated, meaning that participants 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.
The classroom is moderated, meaning that participants 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
2018-09-27 19:32:17
This helps keep the session 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.
This helps keep the session 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
2018-09-27 19:32:34
There are a lot of students here! 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!!
There are a lot of students here! 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
2018-09-27 19:32:48
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.
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.
DPatrick
2018-09-27 19:33:01
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!
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
2018-09-27 19:33:15
We have an assistant here to help out tonight: David Stoner, or applepi2000. David is a senior at Harvard and longtime AoPSer. He plans to concentrate in theoretical mathematics, and he also enjoys studying computer science and philosophy. In his high school years, David enjoyed immersing himself in the mathematical community. He won the USAJMO in 2012, and won the USAMO in 2013, 2014, and 2015, achieving a 42 on the 2015 exam. He was a member of the winning United States team for the 2013 RMM and 2015 IMO competitions, earning a gold medal at each contest. Outside of mathematics, David enjoys playing blitz and classical chess, solving Nikoli puzzles, and eating apples.
We have an assistant here to help out tonight: David Stoner, or applepi2000. David is a senior at Harvard and longtime AoPSer. He plans to concentrate in theoretical mathematics, and he also enjoys studying computer science and philosophy. In his high school years, David enjoyed immersing himself in the mathematical community. He won the USAJMO in 2012, and won the USAMO in 2013, 2014, and 2015, achieving a 42 on the 2015 exam. He was a member of the winning United States team for the 2013 RMM and 2015 IMO competitions, earning a gold medal at each contest. Outside of mathematics, David enjoys playing blitz and classical chess, solving Nikoli puzzles, and eating apples.
DPatrick
2018-09-27 19:34:16
David 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 he's able, but again it's pretty busy tonight so please understand if he doesn't get to you quickly (or at all).
David 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 he's able, but again it's pretty busy tonight so please understand if he doesn't get to you quickly (or at all).
DPatrick
2018-09-27 19:34:39
Also joining us tonight are the co-creators of the WWTBAM contest, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
Also joining us tonight are the co-creators of the WWTBAM contest, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
DPatrick
2018-09-27 19:34:51
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.
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
2018-09-27 19:34:57
Hello, everyone. Happy to be here with you and Dave.
Hello, everyone. Happy to be here with you and Dave.
TPiR
2018-09-27 19:35:01
Hello, everyone!
Hello, everyone!
DPatrick
2018-09-27 19:35:13
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.
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.
DPatrick
2018-09-27 19:35:22
Can you guess why Bill's username is TPiR?
Can you guess why Bill's username is TPiR?
shesa
2018-09-27 19:35:29
does your name have anything to do with the price is right
does your name have anything to do with the price is right
Mathisfun04
2018-09-27 19:35:33
the price is right?
the price is right?
DPatrick
2018-09-27 19:35:40
Yes indeed! 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.)
Yes indeed! 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.)
DPatrick
2018-09-27 19:35:56
As you can see, we have a lot of game show background here tonight!
As you can see, we have a lot of game show background here tonight!
DPatrick
2018-09-27 19:36:06
On to the contest info:
On to the contest info:
DPatrick
2018-09-27 19:36:15
Who Wants to Be a Mathematician is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
Who Wants to Be a Mathematician is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick
2018-09-27 19:36:27
Tonight we'll be talking about Round 1 of the Championship contest, which concluded yesterday.
Tonight we'll be talking about Round 1 of the Championship contest, which concluded yesterday.
DPatrick
2018-09-27 19:36:46
Round 1 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 were permitted during the contest.
Round 1 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 were permitted during the contest.
DPatrick
2018-09-27 19:37:00
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.
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
2018-09-27 19:37:31
Some of the questions may have interesting sidetracks, so we may also stop and view some of the scenery along the way.
Some of the questions may have interesting sidetracks, so we may also stop and view some of the scenery along the way.
DPatrick
2018-09-27 19:37:59
This year, the qualifying round had roughly 5200 participants, of which roughly 1850 advanced to Round 2. (More about that later.)
This year, the qualifying round had roughly 5200 participants, of which roughly 1850 advanced to Round 2. (More about that later.)
DPatrick
2018-09-27 19:38:15
So let's get started on the problems!
So let's get started on the problems!
DPatrick
2018-09-27 19:38:22
1. What is the ones (units) digit of $2019^2 - 2018^2$?
$ $
(a) $1$ (b) $3$ (c) $5$ (d) $7$
1. What is the ones (units) digit of $2019^2 - 2018^2$?
$ $
(a) $1$ (b) $3$ (c) $5$ (d) $7$
DPatrick
2018-09-27 19:38:30
(All of the Round 1 questions are multiple-choice.)
(All of the Round 1 questions are multiple-choice.)
DPatrick
2018-09-27 19:38:38
Reminder: please do not simply post your answer. Instead, please post suggestion(s) as to how to solve the problem.
Reminder: please do not simply post your answer. Instead, please post suggestion(s) as to how to solve the problem.
mathman629
2018-09-27 19:39:03
just use difference of squares
just use difference of squares
shesa
2018-09-27 19:39:03
we only care about the units digit, so let's just look at the units digit of 9^2 and 8^2
we only care about the units digit, so let's just look at the units digit of 9^2 and 8^2
Allen31415
2018-09-27 19:39:03
look at the last digit: you get 9^2-8^2
look at the last digit: you get 9^2-8^2
Orangecounty2
2018-09-27 19:39:03
difference of squares
difference of squares
amuthup
2018-09-27 19:39:03
difference of squares
difference of squares
WalkerTesla
2018-09-27 19:39:03
$a^{2}-b^{2} = (a-b)(a+b)$
$a^{2}-b^{2} = (a-b)(a+b)$
aussie_math_kid
2018-09-27 19:39:03
9 squared minus 8 squared
9 squared minus 8 squared
DPatrick
2018-09-27 19:39:12
Good -- there are basically two ways to approach it.
Good -- there are basically two ways to approach it.
DPatrick
2018-09-27 19:39:25
(Possibly more ways too -- but two that I though of, and apparently most of you did too.)
(Possibly more ways too -- but two that I though of, and apparently most of you did too.)
DPatrick
2018-09-27 19:39:36
One is to note that only the units digits matter.
One is to note that only the units digits matter.
DPatrick
2018-09-27 19:39:50
That is, the units digit of $2019^2$ is the same as the units digit of $9^2$, which is $81$. So its units digit is $1$.
That is, the units digit of $2019^2$ is the same as the units digit of $9^2$, which is $81$. So its units digit is $1$.
DPatrick
2018-09-27 19:40:04
And similarly the units digit of $2018^2$ is the same as the units digit of $8^2$, which is $64$. So its units digit is $4$.
And similarly the units digit of $2018^2$ is the same as the units digit of $8^2$, which is $64$. So its units digit is $4$.
DPatrick
2018-09-27 19:40:08
And when we subtract a number ending in $4$ from a number ending in $1$, what units digit do we get?
And when we subtract a number ending in $4$ from a number ending in $1$, what units digit do we get?
hobbit16
2018-09-27 19:40:29
7
7
jcao0715
2018-09-27 19:40:29
7
7
GoodInMathEverytime
2018-09-27 19:40:29
7
7
drakecap
2018-09-27 19:40:29
7
7
Apurple
2018-09-27 19:40:29
7
7
mutinykids
2018-09-27 19:40:29
7
7
DPatrick
2018-09-27 19:40:32
We get, for example, $11-4 = 7$. So the units digit of the difference is $7$. $\boxed{\text{(d)}}$
We get, for example, $11-4 = 7$. So the units digit of the difference is $7$. $\boxed{\text{(d)}}$
DPatrick
2018-09-27 19:40:47
You could also be a little more fancy and write $1 - 4 \equiv -3 \equiv 7 \pmod{10}$. (If you're familiar with modular arithmetic, that is.)
You could also be a little more fancy and write $1 - 4 \equiv -3 \equiv 7 \pmod{10}$. (If you're familiar with modular arithmetic, that is.)
DPatrick
2018-09-27 19:41:05
But also, as many of you mentioned, we can use the difference-of-squares factorization instead.
But also, as many of you mentioned, we can use the difference-of-squares factorization instead.
flyhawkeye
2018-09-27 19:41:19
$2019^2-2018^2=(2019+2018)(2019-2018)=4037$, so the answer is $7$.
$2019^2-2018^2=(2019+2018)(2019-2018)=4037$, so the answer is $7$.
RootThreeOverTwo
2018-09-27 19:41:19
(2019+2018)(2019-2018)
(2019+2018)(2019-2018)
DPatrick
2018-09-27 19:41:27
Right. We get $2019^2 - 2018^2 = (2019 + 2018)(2019 - 2018)$.
Right. We get $2019^2 - 2018^2 = (2019 + 2018)(2019 - 2018)$.
DPatrick
2018-09-27 19:41:41
This is especially convenient because the second factor is just $1$.
This is especially convenient because the second factor is just $1$.
DPatrick
2018-09-27 19:41:48
So the difference is equal to $2019 + 2018 = 4037$, and has units digit $7$.
So the difference is equal to $2019 + 2018 = 4037$, and has units digit $7$.
DPatrick
2018-09-27 19:42:02
Next up:
Next up:
DPatrick
2018-09-27 19:42:06
2. What is the area in the first quadrant bounded by the graph of $x + 2y = 4$ and the $x$- and $y$-axes?
$ $
(a) $1$ (b) $2$ (c) $4$ (d) $8$
2. What is the area in the first quadrant bounded by the graph of $x + 2y = 4$ and the $x$- and $y$-axes?
$ $
(a) $1$ (b) $2$ (c) $4$ (d) $8$
phoenix9
2018-09-27 19:42:36
I just drew a quick graph of it
I just drew a quick graph of it
khina
2018-09-27 19:42:36
just draw a triangle
just draw a triangle
a1b2
2018-09-27 19:42:36
It forms a triangle
It forms a triangle
phoenix9
2018-09-27 19:42:36
the y and x intercpt
the y and x intercpt
DPatrick
2018-09-27 19:42:58
Good idea -- let's just draw it. The graph is a line.
Good idea -- let's just draw it. The graph is a line.
DPatrick
2018-09-27 19:43:03
And we care about where it intersects the axes.
And we care about where it intersects the axes.
Orangestripe
2018-09-27 19:43:17
You could find the x and y intercepts by substituting zero for each of the variables respectively.
You could find the x and y intercepts by substituting zero for each of the variables respectively.
DPatrick
2018-09-27 19:43:29
Excellent idea.
Excellent idea.
DPatrick
2018-09-27 19:43:38
For example, the line intersects the $x$-axis where $y=0$.
For example, the line intersects the $x$-axis where $y=0$.
DPatrick
2018-09-27 19:43:55
So we just substitute $y=0$ into the equation for the line. This gives $x=4$, so the point $(4,0)$ is the intersection point.
So we just substitute $y=0$ into the equation for the line. This gives $x=4$, so the point $(4,0)$ is the intersection point.
DPatrick
2018-09-27 19:44:11
And the intersection with the $y$-axis is?
And the intersection with the $y$-axis is?
kcbhatraju
2018-09-27 19:44:32
(0, 2) for the y intercept
(0, 2) for the y intercept
bibilutza
2018-09-27 19:44:32
where x=0
where x=0
thea72
2018-09-27 19:44:32
(0,2)
(0,2)
DPatrick
2018-09-27 19:44:38
It intersects the $y$-axis where $x=0$.
It intersects the $y$-axis where $x=0$.
DPatrick
2018-09-27 19:44:43
This gives $2y = 4$, so the point $(0,2)$ is the intersection point.
This gives $2y = 4$, so the point $(0,2)$ is the intersection point.
DPatrick
2018-09-27 19:44:51
So now we can make a quick sketch:
So now we can make a quick sketch:
DPatrick
2018-09-27 19:44:54
ASnerd
2018-09-27 19:45:16
2*4/2=4
2*4/2=4
Scipio1
2018-09-27 19:45:16
2x4/2=4
2x4/2=4
CoolKidWiz
2018-09-27 19:45:16
Then we use 1/2 * h * b
Then we use 1/2 * h * b
Orangecounty2
2018-09-27 19:45:19
now its 4*2/2 = 4
now its 4*2/2 = 4
IlinoisMathlete
2018-09-27 19:45:19
1/2*b*h= (4*2)/2 which is 4
1/2*b*h= (4*2)/2 which is 4
WalkerTesla
2018-09-27 19:45:19
$\dfrac{2\cdot 4}{2} = 4$
$\dfrac{2\cdot 4}{2} = 4$
DPatrick
2018-09-27 19:45:26
Yes. This is a right triangle with legs of length $2$ and $4$.
Yes. This is a right triangle with legs of length $2$ and $4$.
DPatrick
2018-09-27 19:45:31
So its area is $\frac12(2)(4) = 4$. $\boxed{\text{(c)}}$
So its area is $\frac12(2)(4) = 4$. $\boxed{\text{(c)}}$
DPatrick
2018-09-27 19:46:11
And on to #3. (By the way, numbers 1-3 were the three easiest problems on the contest, in the sense that they had the highest percentages of correct answers.)
And on to #3. (By the way, numbers 1-3 were the three easiest problems on the contest, in the sense that they had the highest percentages of correct answers.)
DPatrick
2018-09-27 19:46:15
3. Let $f(x) = x^2 + 7$. What is $f(f(2))$?
$ $
(a) $11$ (b) $107$ (c) $121$ (d) $128$
3. Let $f(x) = x^2 + 7$. What is $f(f(2))$?
$ $
(a) $11$ (b) $107$ (c) $121$ (d) $128$
DPatrick
2018-09-27 19:46:26
How do we evaluate this?
How do we evaluate this?
abdeet
2018-09-27 19:46:43
f(2)=11
f(2)=11
mutinykids
2018-09-27 19:46:43
substitute 2 into the equation first
substitute 2 into the equation first
GoodInMathEverytime
2018-09-27 19:46:43
plug x as 2
plug x as 2
hobbit16
2018-09-27 19:46:43
First find f(2) and then plug that value into f again
First find f(2) and then plug that value into f again
Panda_dino
2018-09-27 19:46:43
substitute x for 2
substitute x for 2
kevalshah2005
2018-09-27 19:46:43
find out f(2) then find out f() of that.
find out f(2) then find out f() of that.
mathman629
2018-09-27 19:46:46
Plug in 2, and then plug in the result of f(2) into the function again. That's the answer
Plug in 2, and then plug in the result of f(2) into the function again. That's the answer
jcao0715
2018-09-27 19:46:50
substitue 2 into the function and substitute that value into the function again
substitue 2 into the function and substitute that value into the function again
DPatrick
2018-09-27 19:47:03
Exactly. First we find $f(2)$, and then whatever that is, we plug it back into $f$ to get the final answer.
Exactly. First we find $f(2)$, and then whatever that is, we plug it back into $f$ to get the final answer.
DPatrick
2018-09-27 19:47:11
So we start with $f(2) = 2^2 + 7 = 4 + 7 = 11$.
So we start with $f(2) = 2^2 + 7 = 4 + 7 = 11$.
DPatrick
2018-09-27 19:47:25
So $f(f(2)) = f(11)$...which is what?
So $f(f(2)) = f(11)$...which is what?
Robot7620_2
2018-09-27 19:47:43
128
128
Bubble3
2018-09-27 19:47:43
128
128
Mathdude102
2018-09-27 19:47:43
128
128
o99999
2018-09-27 19:47:43
128
128
Apurple
2018-09-27 19:47:43
11^2+7=128
11^2+7=128
flyhawkeye
2018-09-27 19:47:43
$11^2+7=128$
$11^2+7=128$
DPatrick
2018-09-27 19:47:46
$f(11) = 11^2 + 7 = 121 + 7 = 128$. $\boxed{\text{(d)}}$
$f(11) = 11^2 + 7 = 121 + 7 = 128$. $\boxed{\text{(d)}}$
DPatrick
2018-09-27 19:48:08
This problem turned out to be the easiest of all ten problems on the contest.
This problem turned out to be the easiest of all ten problems on the contest.
DPatrick
2018-09-27 19:48:31
Now they start to get a little harder.
Now they start to get a little harder.
DPatrick
2018-09-27 19:48:35
4. The radius of a sphere is $6$ cm. What is the sphere's volume divided by its surface area (ignore units)?
$ $
(a) $1.5$ (b) $2$ (c) $2.5$ (d) $3$
4. The radius of a sphere is $6$ cm. What is the sphere's volume divided by its surface area (ignore units)?
$ $
(a) $1.5$ (b) $2$ (c) $2.5$ (d) $3$
DPatrick
2018-09-27 19:49:01
So I think either you know these formulas or you don't. They're very hard to derive in real time.
So I think either you know these formulas or you don't. They're very hard to derive in real time.
a1b2
2018-09-27 19:49:42
Use the volume and SA formula
Use the volume and SA formula
flyhawkeye
2018-09-27 19:49:42
Use $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$
Use $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$
phoenix9
2018-09-27 19:49:42
Find the volume and SA of the sphere using their formulas
Find the volume and SA of the sphere using their formulas
hobbit16
2018-09-27 19:49:42
Find the volume as 4/3 pi r^3 and surface area as 4 pi r^2 and ratio them.
Find the volume as 4/3 pi r^3 and surface area as 4 pi r^2 and ratio them.
asdf334
2018-09-27 19:49:42
(4/3 * pi * r^3) / (4 * pi * r^2)
(4/3 * pi * r^3) / (4 * pi * r^2)
Scipio1
2018-09-27 19:49:42
find the volume and surface area with formulas, 4/3pi•r^3, and 4pi•r^2?
find the volume and surface area with formulas, 4/3pi•r^3, and 4pi•r^2?
drakecap
2018-09-27 19:49:42
4/3 pi r^2 / 4 pi r^2 = 1/3 r
4/3 pi r^2 / 4 pi r^2 = 1/3 r
DPatrick
2018-09-27 19:49:55
Right -- let me work through these.
Right -- let me work through these.
DPatrick
2018-09-27 19:50:07
The volume of a sphere is $\dfrac43 \pi r^3$, where $r$ is the radius.
The volume of a sphere is $\dfrac43 \pi r^3$, where $r$ is the radius.
DPatrick
2018-09-27 19:50:14
And the surface area is $4 \pi r^2$.
And the surface area is $4 \pi r^2$.
DPatrick
2018-09-27 19:50:27
So when we divide volume by surface area, we get $$\dfrac{\frac43 \pi r^3}{4 \pi r^2}.$$
So when we divide volume by surface area, we get $$\dfrac{\frac43 \pi r^3}{4 \pi r^2}.$$
kevalshah2005
2018-09-27 19:50:38
pi cancels out
pi cancels out
abank
2018-09-27 19:50:40
r/3
r/3
DPatrick
2018-09-27 19:50:48
Hurray! This simplifies to $\dfrac{r}{3}$. (The $\pi$s cancel, which is good, since none of the answer choices contain $\pi$.)
Hurray! This simplifies to $\dfrac{r}{3}$. (The $\pi$s cancel, which is good, since none of the answer choices contain $\pi$.)
Bubble3
2018-09-27 19:50:58
Substitute $6$ for $r$
Substitute $6$ for $r$
Orangestripe
2018-09-27 19:50:58
6/3=2
6/3=2
bibilutza
2018-09-27 19:51:02
6/3=2
6/3=2
FreedomLantern
2018-09-27 19:51:02
r/3=6/3=2
r/3=6/3=2
krazykid548
2018-09-27 19:51:02
Thus the answer is 2!
Thus the answer is 2!
DPatrick
2018-09-27 19:51:11
And when $r=6$, the final answer is just $\dfrac63 = 2$. $\boxed{\text{(b)}}$
And when $r=6$, the final answer is just $\dfrac63 = 2$. $\boxed{\text{(b)}}$
o99999
2018-09-27 19:51:19
How do you derive the surface area of a sphere formula?
How do you derive the surface area of a sphere formula?
DPatrick
2018-09-27 19:51:26
Well, it's pretty hard.
Well, it's pretty hard.
DPatrick
2018-09-27 19:51:38
But there is a little historical gadget that may make them easier to remember.
But there is a little historical gadget that may make them easier to remember.
DPatrick
2018-09-27 19:51:45
The Greek mathematician Archimedes first discovered these formulas over 2000 years ago!
The Greek mathematician Archimedes first discovered these formulas over 2000 years ago!
DPatrick
2018-09-27 19:52:05
His very clever idea was to inscribe the sphere inside a cylinder. This makes the cylinder have radius $r$ and height $2r$. This is my very, very crude picture of this:
His very clever idea was to inscribe the sphere inside a cylinder. This makes the cylinder have radius $r$ and height $2r$. This is my very, very crude picture of this:
DPatrick
2018-09-27 19:52:10
DPatrick
2018-09-27 19:52:41
His further very very very clever observation was that both the volume and the surface area of the sphere are exactly $\frac23$ that of the surrounding cylinder. This requires quite a bit more work to see.
His further very very very clever observation was that both the volume and the surface area of the sphere are exactly $\frac23$ that of the surrounding cylinder. This requires quite a bit more work to see.
DPatrick
2018-09-27 19:53:15
But the nice thing is that the volume and surface area formulas for the cylinder are a lot easier.
But the nice thing is that the volume and surface area formulas for the cylinder are a lot easier.
DPatrick
2018-09-27 19:53:26
The volume of the cylinder is $V = bh$ where $b = \pi r^2$ is the area of the base and $h = 2r$ is the height. So $V = 2\pi r^3$.
The volume of the cylinder is $V = bh$ where $b = \pi r^2$ is the area of the base and $h = 2r$ is the height. So $V = 2\pi r^3$.
DPatrick
2018-09-27 19:53:42
So the volume of the sphere is $\frac23$ of this, or $\frac23(2\pi r^3) = \frac43\pi r^3$.
So the volume of the sphere is $\frac23$ of this, or $\frac23(2\pi r^3) = \frac43\pi r^3$.
DPatrick
2018-09-27 19:54:01
And, the surface area of the cylinder is twice the base area ($\pi r^2$) plus the vertical area, which is the height ($2r$) times the circumference $(2\pi r)$.
And, the surface area of the cylinder is twice the base area ($\pi r^2$) plus the vertical area, which is the height ($2r$) times the circumference $(2\pi r)$.
DPatrick
2018-09-27 19:54:12
Adding this up gives $S = 2(\pi r^2) + (2r)(2\pi r) = 6\pi r^2$.
Adding this up gives $S = 2(\pi r^2) + (2r)(2\pi r) = 6\pi r^2$.
DPatrick
2018-09-27 19:54:20
So the surface area of the sphere is $\frac23(6 \pi r^2) = 4 \pi r^2$.
So the surface area of the sphere is $\frac23(6 \pi r^2) = 4 \pi r^2$.
phoenix9
2018-09-27 19:54:30
how do we know its 2/3
how do we know its 2/3
DPatrick
2018-09-27 19:54:52
Well, it turns our Archimedes was a really clever guy! He was able to figure it out, over 2000 years ago!
Well, it turns our Archimedes was a really clever guy! He was able to figure it out, over 2000 years ago!
mathleticguyyy
2018-09-27 19:54:55
Why isn't it 1/3 like with cones
Why isn't it 1/3 like with cones
DPatrick
2018-09-27 19:55:02
These facts are definitely related.
These facts are definitely related.
DPatrick
2018-09-27 19:55:31
I encourage you to look it up sometime!
I encourage you to look it up sometime!
DPatrick
2018-09-27 19:55:48
But right now, let's move on with the contest problems.
But right now, let's move on with the contest problems.
DPatrick
2018-09-27 19:55:58
5. Which of the following is the closest integer to the cube of $\tan\left(\frac\pi3\right)$ ($\frac\pi3$ is in radians, not degrees)?
$ $
(a) $5$ (b) $6$ (c) $7$ (d) $8$
5. Which of the following is the closest integer to the cube of $\tan\left(\frac\pi3\right)$ ($\frac\pi3$ is in radians, not degrees)?
$ $
(a) $5$ (b) $6$ (c) $7$ (d) $8$
DPatrick
2018-09-27 19:56:13
First, what is $\tan\left(\frac\pi3\right)$?
First, what is $\tan\left(\frac\pi3\right)$?
amuthup
2018-09-27 19:56:25
30-60-90 triangles
30-60-90 triangles
Robot7620_2
2018-09-27 19:56:25
pi = 180 deg
pi = 180 deg
asdf334
2018-09-27 19:56:25
tan(60)
tan(60)
Robot7620_2
2018-09-27 19:56:25
tan 60
tan 60
ASnerd
2018-09-27 19:56:27
tan(60)
tan(60)
DPatrick
2018-09-27 19:56:35
True, you might prefer to think of $\frac\pi3$ radians in terms of degrees.
True, you might prefer to think of $\frac\pi3$ radians in terms of degrees.
DPatrick
2018-09-27 19:56:39
The key is to remember that $\pi$ radians is $180^\circ$.
The key is to remember that $\pi$ radians is $180^\circ$.
DPatrick
2018-09-27 19:56:44
So $\frac\pi3$ radians is $60^\circ$.
So $\frac\pi3$ radians is $60^\circ$.
spookyflame2
2018-09-27 19:56:53
sqrt3
sqrt3
hobbit16
2018-09-27 19:56:53
root 3
root 3
Orangecounty2
2018-09-27 19:56:53
tan(60) = sqrt(3)
tan(60) = sqrt(3)
ParadigmShift
2018-09-27 19:56:53
$\sqrt{3}$
$\sqrt{3}$
Dseg123
2018-09-27 19:56:53
sqrt(3)
sqrt(3)
Catgirl1234
2018-09-27 19:56:53
tan(60)
tan(60)
krazykid548
2018-09-27 19:56:55
You can draw a 30-60-90 triangle (because pi/3 is 60) and find the ratio of the opposite over adjacent sides
You can draw a 30-60-90 triangle (because pi/3 is 60) and find the ratio of the opposite over adjacent sides
DPatrick
2018-09-27 19:57:03
That's it: $\tan\left(\frac\pi3\right) = \sqrt3$. Think of a 30-60-90 triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$. The tangent is the ratio of the two legs.
That's it: $\tan\left(\frac\pi3\right) = \sqrt3$. Think of a 30-60-90 triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$. The tangent is the ratio of the two legs.
DPatrick
2018-09-27 19:57:08
DPatrick
2018-09-27 19:57:31
So our quantity is $(\sqrt3)^3$.
So our quantity is $(\sqrt3)^3$.
DPatrick
2018-09-27 19:57:37
What's probably a better way to write that?
What's probably a better way to write that?
DatBOIIIIIIII
2018-09-27 19:57:53
3rt3
3rt3
krazykid548
2018-09-27 19:57:53
Then to cube that, you do 3 * sqrt(3)
Then to cube that, you do 3 * sqrt(3)
jcao0715
2018-09-27 19:57:53
3sqrt3
3sqrt3
proshi
2018-09-27 19:57:53
3root 3
3root 3
r_dat_13
2018-09-27 19:57:53
3sqrt(3)
3sqrt(3)
Apurple
2018-09-27 19:57:53
3sqrt3
3sqrt3
kootrapali
2018-09-27 19:57:53
$3\sqrt{3}$
$3\sqrt{3}$
abank
2018-09-27 19:57:53
3sqrt3
3sqrt3
Panda_dino
2018-09-27 19:57:53
3 * root3
3 * root3
DPatrick
2018-09-27 19:58:06
Yeah, I'd probably write $(\sqrt3)^3 = 3\sqrt3$.
Yeah, I'd probably write $(\sqrt3)^3 = 3\sqrt3$.
DPatrick
2018-09-27 19:58:18
So we want the integer closest to $3\sqrt3$.
So we want the integer closest to $3\sqrt3$.
kootrapali
2018-09-27 19:58:38
$3\cdot1.7=5.1$
$3\cdot1.7=5.1$
coolak
2018-09-27 19:58:38
sqrt 3 is approximately 1.7
sqrt 3 is approximately 1.7
Dseg123
2018-09-27 19:58:38
about 3 * 1.7 = 5.1, answer is 5?
about 3 * 1.7 = 5.1, answer is 5?
hobbit16
2018-09-27 19:58:38
which is 5 since root 3 is 1.73 approximately
which is 5 since root 3 is 1.73 approximately
DPatrick
2018-09-27 19:58:49
Right. You may know that $\sqrt3 = 1.7\ldots$.
Right. You may know that $\sqrt3 = 1.7\ldots$.
DPatrick
2018-09-27 19:59:24
You can approximate to withing a tenth by noting that $1.7^2 = 2.89$ and $1.8^2 = 3.24$.
You can approximate to withing a tenth by noting that $1.7^2 = 2.89$ and $1.8^2 = 3.24$.
DPatrick
2018-09-27 19:59:57
So $3\sqrt3$ is between $5.1$ and $5.4$. And that's closest to $5$.
So $3\sqrt3$ is between $5.1$ and $5.4$. And that's closest to $5$.
DPatrick
2018-09-27 20:00:02
So the answer is $5$. $\boxed{\text{(a)}}$
So the answer is $5$. $\boxed{\text{(a)}}$
ChickenAgent2227-_-
2018-09-27 20:00:10
or we write it as sqrt27 and see that 5^2=25 and 6^2=36
or we write it as sqrt27 and see that 5^2=25 and 6^2=36
coolak
2018-09-27 20:00:10
if you didnt, you could think about sqrt 27
if you didnt, you could think about sqrt 27
kevalshah2005
2018-09-27 20:00:16
you can convert it to sqrt 27 which is a little greater than 5
you can convert it to sqrt 27 which is a little greater than 5
DPatrick
2018-09-27 20:00:40
Certainly, you can also use $3\sqrt3 = \sqrt{27}$, and then $\sqrt{25} < \sqrt{27} < \sqrt{36}$.
Certainly, you can also use $3\sqrt3 = \sqrt{27}$, and then $\sqrt{25} < \sqrt{27} < \sqrt{36}$.
DPatrick
2018-09-27 20:00:55
...and is a lot closer to $\sqrt{25}$.
...and is a lot closer to $\sqrt{25}$.
DPatrick
2018-09-27 20:01:09
On to #6:
On to #6:
DPatrick
2018-09-27 20:01:13
6. Let $n$ represent a positive integer greater than $1$. The number of points of the intersection of the graphs $y = x^n$ and $y = n^x$ is
$ $
(a) always odd
(b) always even
(c) odd when $n$ is even and even when $n$ is odd
(d) even when $n$ is even and odd when $n$ is odd
6. Let $n$ represent a positive integer greater than $1$. The number of points of the intersection of the graphs $y = x^n$ and $y = n^x$ is
$ $
(a) always odd
(b) always even
(c) odd when $n$ is even and even when $n$ is odd
(d) even when $n$ is even and odd when $n$ is odd
Orangestripe
2018-09-27 20:01:34
I just sketched a graph of n=2 and n=3
I just sketched a graph of n=2 and n=3
Scipio1
2018-09-27 20:01:34
try it out
try it out
Apurple
2018-09-27 20:01:34
pick sample integers to try
pick sample integers to try
phoenix9
2018-09-27 20:01:34
guess and check?
guess and check?
DPatrick
2018-09-27 20:01:42
Right: the answer choices help us out here.
Right: the answer choices help us out here.
DPatrick
2018-09-27 20:01:46
Specifically, if we can figure out what happens when $n=2$ and $n=3$, we'll know which answer is correct.
Specifically, if we can figure out what happens when $n=2$ and $n=3$, we'll know which answer is correct.
DPatrick
2018-09-27 20:02:00
So let's first look at $n=2$. We're comparing $y = x^2$ and $y = 2^x$.
So let's first look at $n=2$. We're comparing $y = x^2$ and $y = 2^x$.
DPatrick
2018-09-27 20:02:18
How many solutions does it have, and how can we be reasonably sure?
How many solutions does it have, and how can we be reasonably sure?
krazykid548
2018-09-27 20:02:47
They intersect twice, at x = 2 and x = 4
They intersect twice, at x = 2 and x = 4
DPatrick
2018-09-27 20:02:53
Certainly there are at least 2 solutions.
Certainly there are at least 2 solutions.
DPatrick
2018-09-27 20:03:01
$x=2$ gives us $2^2 = 2^2$, which I hope is true.
$x=2$ gives us $2^2 = 2^2$, which I hope is true.
DPatrick
2018-09-27 20:03:12
And $x=4$ gives us $4^2 = 2^4$, which is also true.
And $x=4$ gives us $4^2 = 2^4$, which is also true.
Mathisfun04
2018-09-27 20:03:24
after that, the graph of 2^x outgrows x^2, so they don't meet again
after that, the graph of 2^x outgrows x^2, so they don't meet again
krazykid548
2018-09-27 20:03:40
Because 2^x is an function which increases at a faster rate than x^2 after x = 4, they never intersect again.
Because 2^x is an function which increases at a faster rate than x^2 after x = 4, they never intersect again.
DPatrick
2018-09-27 20:03:44
True...
True...
a1b2
2018-09-27 20:03:52
They intersect once at a negative x
They intersect once at a negative x
mathleticguyyy
2018-09-27 20:03:52
what about negative x?
what about negative x?
spartacle
2018-09-27 20:03:52
but wait they do meet again for negative x
but wait they do meet again for negative x
DPatrick
2018-09-27 20:04:02
Hmmm...maybe we should sketch the graphs.
Hmmm...maybe we should sketch the graphs.
DPatrick
2018-09-27 20:04:14
What does the graph of $y = x^2$ look like?
What does the graph of $y = x^2$ look like?
shesa
2018-09-27 20:04:32
parabola
parabola
Allen31415
2018-09-27 20:04:32
a parabola
a parabola
jcao0715
2018-09-27 20:04:32
parabola
parabola
mathman629
2018-09-27 20:04:32
parabola
parabola
ParadigmShift
2018-09-27 20:04:32
Upwards opening parabola
Upwards opening parabola
DPatrick
2018-09-27 20:04:36
It's an upwards-opening parabola with vertex at $(0,0)$.
It's an upwards-opening parabola with vertex at $(0,0)$.
DPatrick
2018-09-27 20:04:40
DPatrick
2018-09-27 20:04:46
What does the graph of the $y = 2^x$ look like?
What does the graph of the $y = 2^x$ look like?
Dseg123
2018-09-27 20:05:00
exponential curve
exponential curve
hobbit16
2018-09-27 20:05:00
exponential curve
exponential curve
DPatrick
2018-09-27 20:05:17
Yeah, it's a little hard to describe in words.
Yeah, it's a little hard to describe in words.
DPatrick
2018-09-27 20:05:24
It's steadily increasing.
It's always positive.
It crosses the $y$-axis at $y = 2^0 = 1$; that is, the point $(0,1)$.
It's steadily increasing.
It's always positive.
It crosses the $y$-axis at $y = 2^0 = 1$; that is, the point $(0,1)$.
DPatrick
2018-09-27 20:05:41
(In fact, all of the functions $y = n^x$ where $n > 1$ is a positive integer have these properties. They all look about the same.)
(In fact, all of the functions $y = n^x$ where $n > 1$ is a positive integer have these properties. They all look about the same.)
DPatrick
2018-09-27 20:05:47
shesa
2018-09-27 20:05:56
they meet at one point on the negative side, because the parabola crosses to (0,0)!!
they meet at one point on the negative side, because the parabola crosses to (0,0)!!
DPatrick
2018-09-27 20:06:09
Right: they have to meet on the $x<0$ side as well!
Right: they have to meet on the $x<0$ side as well!
DPatrick
2018-09-27 20:06:34
My graph wasn't big enough to show the $(4,16)$ intersection point, but it's to the right of what I drew.
My graph wasn't big enough to show the $(4,16)$ intersection point, but it's to the right of what I drew.
DPatrick
2018-09-27 20:06:53
And as we mentioned, after that point, the red (exponential) curve is growing way faster than the blue (polynomial) curve, so the red will stay above the blue and they won't intersect again.
And as we mentioned, after that point, the red (exponential) curve is growing way faster than the blue (polynomial) curve, so the red will stay above the blue and they won't intersect again.
Orangestripe
2018-09-27 20:07:01
So there are 3 solutions to that system
So there are 3 solutions to that system
DPatrick
2018-09-27 20:07:17
Right: there are $3$ intersection points of $y=2^x$ and $y=x^2$: one negative and two positive ($x=2$ and $x=4$).
Right: there are $3$ intersection points of $y=2^x$ and $y=x^2$: one negative and two positive ($x=2$ and $x=4$).
CoolKidWiz
2018-09-27 20:07:25
always an odd number
always an odd number
DPatrick
2018-09-27 20:07:31
So when $n$ is even, the answer must be odd. We haven't proved this for any $n > 2$, of course, but the answer choices lock us in.
So when $n$ is even, the answer must be odd. We haven't proved this for any $n > 2$, of course, but the answer choices lock us in.
DPatrick
2018-09-27 20:07:42
So the answer is going to be either (a) or (c).
So the answer is going to be either (a) or (c).
DPatrick
2018-09-27 20:07:53
How about when $n$ is odd? In particular, what about $y = 3^x$ and $y = x^3$?
How about when $n$ is odd? In particular, what about $y = 3^x$ and $y = x^3$?
Mathisfun04
2018-09-27 20:08:28
intersect once at x=3
intersect once at x=3
mathleticguyyy
2018-09-27 20:08:28
the negative part is not going to intersect
the negative part is not going to intersect
DPatrick
2018-09-27 20:08:40
We definitely get an intersection at $x=3$.
We definitely get an intersection at $x=3$.
DPatrick
2018-09-27 20:08:46
Since $3^3 = 3^3$
Since $3^3 = 3^3$
DPatrick
2018-09-27 20:08:59
Let me try to sketch these:
Let me try to sketch these:
DPatrick
2018-09-27 20:09:04
a1b2
2018-09-27 20:09:26
there will be another positive intersection because it is not tangent
there will be another positive intersection because it is not tangent
Abra_Kadabra
2018-09-27 20:09:26
they intersect twice, right?
they intersect twice, right?
DPatrick
2018-09-27 20:09:38
They do intersect twice!
They do intersect twice!
DPatrick
2018-09-27 20:09:55
Because the red (exponential) has to end up above the blue (polynomial), since the red grows faster.
Because the red (exponential) has to end up above the blue (polynomial), since the red grows faster.
DPatrick
2018-09-27 20:10:08
The red starts above the blue and ends above the blue.
The red starts above the blue and ends above the blue.
DPatrick
2018-09-27 20:10:31
So if they cross once, the red goes below the blue. They have to cross a second time for the red to get back on top!
So if they cross once, the red goes below the blue. They have to cross a second time for the red to get back on top!
IcyEmerald05
2018-09-27 20:11:09
so the answer is c
so the answer is c
kootrapali
2018-09-27 20:11:09
The answer is c
The answer is c
bibilutza
2018-09-27 20:11:09
(c) odd when $n$ is even and even when $n$ is odd
(c) odd when $n$ is even and even when $n$ is odd
DPatrick
2018-09-27 20:11:18
Right! $n=3$ has 2 intersections, which is even.
Right! $n=3$ has 2 intersections, which is even.
DPatrick
2018-09-27 20:11:32
So when $n$ is odd, we get an even number of solutions.
So when $n$ is odd, we get an even number of solutions.
DPatrick
2018-09-27 20:11:34
Therefore, the final answer is $\boxed{\text{(c)}}$.
Therefore, the final answer is $\boxed{\text{(c)}}$.
Orangestripe
2018-09-27 20:11:55
How do you find the solutions algebraically (without a graph)?
How do you find the solutions algebraically (without a graph)?
DPatrick
2018-09-27 20:12:08
I don't think we can. Exponentials are really hard to work with.
I don't think we can. Exponentials are really hard to work with.
DPatrick
2018-09-27 20:12:42
OK, that was tricky, so next is an easier one for us to catch our breath before the end.
OK, that was tricky, so next is an easier one for us to catch our breath before the end.
DPatrick
2018-09-27 20:12:49
7. On a flat surface, a bug walks 1 foot north, 2 feet west, 3 feet south, 4 feet east, 5 feet north, and 6 feet west. It then walks straight back to its original starting point. How far did the bug walk total, in feet?
$ $
(a) $21$ (b) $24$ (c) $26$ (d) $21 + \sqrt5$
7. On a flat surface, a bug walks 1 foot north, 2 feet west, 3 feet south, 4 feet east, 5 feet north, and 6 feet west. It then walks straight back to its original starting point. How far did the bug walk total, in feet?
$ $
(a) $21$ (b) $24$ (c) $26$ (d) $21 + \sqrt5$
Apurple
2018-09-27 20:13:09
draw it out
draw it out
shesa
2018-09-27 20:13:09
just draw this out on a graph!
just draw this out on a graph!
Scipio1
2018-09-27 20:13:09
draw a diagram
draw a diagram
DPatrick
2018-09-27 20:13:18
Right, this should be pretty straightforward to draw.
Right, this should be pretty straightforward to draw.
DPatrick
2018-09-27 20:13:34
It makes sense to think of the bug walking on the coordinate plane, starting at $(0,0)$. That way it'll be easy to tell where it ends up.
It makes sense to think of the bug walking on the coordinate plane, starting at $(0,0)$. That way it'll be easy to tell where it ends up.
DPatrick
2018-09-27 20:13:37
DPatrick
2018-09-27 20:13:59
The segments parallel to the axes are easy: they total to $1+2+3+4+5+6 = 21$.
The segments parallel to the axes are easy: they total to $1+2+3+4+5+6 = 21$.
krazykid548
2018-09-27 20:14:11
The diagonal has lengths 3 and 4 so its length is 5
The diagonal has lengths 3 and 4 so its length is 5
phoenix9
2018-09-27 20:14:11
use Pythagorean theorem
use Pythagorean theorem
Scipio1
2018-09-27 20:14:11
3 4 5 triangle
3 4 5 triangle
GMbob
2018-09-27 20:14:11
it's a 3-4-5 right triangle to get back to where it started.
it's a 3-4-5 right triangle to get back to where it started.
DPatrick
2018-09-27 20:14:18
Right. At the end it goes from the point $(-4,3)$ back to $(0,0)$.
Right. At the end it goes from the point $(-4,3)$ back to $(0,0)$.
DPatrick
2018-09-27 20:14:29
So it's the hypotenuse of a right triangle with legs $3$ and $4$!
So it's the hypotenuse of a right triangle with legs $3$ and $4$!
DPatrick
2018-09-27 20:14:36
Thus the last step is length $5$.
Thus the last step is length $5$.
Catgirl1234
2018-09-27 20:14:42
So the answer is C
So the answer is C
kevalshah2005
2018-09-27 20:14:42
so the answer is 26
so the answer is 26
bibilutza
2018-09-27 20:14:42
last line is 5, so 21+5= c) $26$
last line is 5, so 21+5= c) $26$
DPatrick
2018-09-27 20:14:44
So the total length of the path is $21 + 5 = 26$. $\boxed{\text{(c)}}$
So the total length of the path is $21 + 5 = 26$. $\boxed{\text{(c)}}$
DPatrick
2018-09-27 20:15:03
On to #8:
On to #8:
DPatrick
2018-09-27 20:15:07
8. Double the sum of the first $1{,}111$ positive integers and subtract $1{,}111$. What is the result?
$ $
(a) $616{,}605$ (b) $1{,}234{,}321$ (c) $1{,}423{,}231$ (d) $1{,}432{,}231$
8. Double the sum of the first $1{,}111$ positive integers and subtract $1{,}111$. What is the result?
$ $
(a) $616{,}605$ (b) $1{,}234{,}321$ (c) $1{,}423{,}231$ (d) $1{,}432{,}231$
mathman629
2018-09-27 20:15:29
Use n(n+1)/2 for n=1111
Use n(n+1)/2 for n=1111
asdf334
2018-09-27 20:15:29
triangular number so n(n+1)/2
triangular number so n(n+1)/2
IlinoisMathlete
2018-09-27 20:15:29
n(n+1)/2 = sum of numbers formula
n(n+1)/2 = sum of numbers formula
DPatrick
2018-09-27 20:15:35
$1{,}111$ is a big number. I don't like big numbers.
$1{,}111$ is a big number. I don't like big numbers.
DPatrick
2018-09-27 20:15:38
Let's call it $n$ instead.
Let's call it $n$ instead.
DPatrick
2018-09-27 20:16:01
And as many of you have pointed out, the sum of the first $n$ positive integers is $\dfrac{n(n+1)}{2}$.
And as many of you have pointed out, the sum of the first $n$ positive integers is $\dfrac{n(n+1)}{2}$.
DPatrick
2018-09-27 20:16:13
(If you've never seen or proved this formula before, it's a fun exercise to do so!)
(If you've never seen or proved this formula before, it's a fun exercise to do so!)
asdf334
2018-09-27 20:16:23
n(n+1)/2 but you double it so it becomes n(n+1)
n(n+1)/2 but you double it so it becomes n(n+1)
DPatrick
2018-09-27 20:16:29
Right: double that sum is just $n(n+1)$.
Right: double that sum is just $n(n+1)$.
DPatrick
2018-09-27 20:16:38
And when we subtract $n$, we get $n(n+1) - n$.
And when we subtract $n$, we get $n(n+1) - n$.
DPatrick
2018-09-27 20:16:46
And what does this simplify to?
And what does this simplify to?
spookyflame2
2018-09-27 20:16:59
n(n+1)-n=n^2
n(n+1)-n=n^2
Robot7620_2
2018-09-27 20:16:59
factor
factor
amuthup
2018-09-27 20:16:59
which is n^2
which is n^2
Robot7620_2
2018-09-27 20:16:59
n^2
n^2
CoolKidWiz
2018-09-27 20:16:59
n^2
n^2
hobbit16
2018-09-27 20:16:59
n^2
n^2
DPatrick
2018-09-27 20:17:04
We can simplify this by factoring out an $n$, to get $n((n+1) - 1)$.
We can simplify this by factoring out an $n$, to get $n((n+1) - 1)$.
DPatrick
2018-09-27 20:17:08
Hey, that's just $n(n) = n^2$!
Hey, that's just $n(n) = n^2$!
DPatrick
2018-09-27 20:17:14
So the result we're looking for is just $1{,}111^2$.
So the result we're looking for is just $1{,}111^2$.
DPatrick
2018-09-27 20:17:32
(And see how much easier that was to write instead of writing $1{,}111$ over and over! Don't use a big number until you have to!)
(And see how much easier that was to write instead of writing $1{,}111$ over and over! Don't use a big number until you have to!)
Irshad
2018-09-27 20:17:47
1234321
1234321
thebossishere
2018-09-27 20:17:47
1234321
1234321
Irshad
2018-09-27 20:17:47
its 1234321
its 1234321
aadinair
2018-09-27 20:17:47
1234321, so b
1234321, so b
jcao0715
2018-09-27 20:17:47
1234321
1234321
DatBOIIIIIIII
2018-09-27 20:17:47
1234321
1234321
IcyEmerald05
2018-09-27 20:17:47
1234321 so b
1234321 so b
thebossishere
2018-09-27 20:17:47
Which is 1234321
Which is 1234321
DPatrick
2018-09-27 20:17:53
This works out to $1{,}234{,}321$. $\boxed{\text{(b)}}$
This works out to $1{,}234{,}321$. $\boxed{\text{(b)}}$
kevalshah2005
2018-09-27 20:18:05
There's a cool trick: for 11x11, its 121 for 111x111 its 12321 and for 1111x1111 its 1234321
There's a cool trick: for 11x11, its 121 for 111x111 its 12321 and for 1111x1111 its 1234321
DPatrick
2018-09-27 20:18:09
Indeed, to quickly see why this is the answer, think about doing the multiplication the old-fashioned way:
Indeed, to quickly see why this is the answer, think about doing the multiplication the old-fashioned way:
DPatrick
2018-09-27 20:18:12
$$
\begin{array}{ccccccc}
&&&1&1&1&1 \\
&&\times&1&1&1&1 \\ \hline
&&&1&1&1&1 \\
&&1&1&1&1 \\
&1&1&1&1 \\
1&1&1&1 \\ \hline
1&2&3&4&3&2&1
\end{array}
$$
$$
\begin{array}{ccccccc}
&&&1&1&1&1 \\
&&\times&1&1&1&1 \\ \hline
&&&1&1&1&1 \\
&&1&1&1&1 \\
&1&1&1&1 \\
1&1&1&1 \\ \hline
1&2&3&4&3&2&1
\end{array}
$$
DPatrick
2018-09-27 20:18:42
OK, a couple of harder problems to finish.
OK, a couple of harder problems to finish.
DPatrick
2018-09-27 20:18:48
9. Suppose $\log_{10}8 = r$ and $\log_{10}9 = s$. What is $\log_{10}5$ in terms of $r$ and/or $s$?
$ $
(a) $\sqrt[3]{r} + \sqrt{s}$ (b) $\left(\sqrt[3]{r}\right)\left(\sqrt{s}\right)$ (c) $(r/3) + (s/2)$ (d) $1 - (r/3)$
9. Suppose $\log_{10}8 = r$ and $\log_{10}9 = s$. What is $\log_{10}5$ in terms of $r$ and/or $s$?
$ $
(a) $\sqrt[3]{r} + \sqrt{s}$ (b) $\left(\sqrt[3]{r}\right)\left(\sqrt{s}\right)$ (c) $(r/3) + (s/2)$ (d) $1 - (r/3)$
DPatrick
2018-09-27 20:19:30
So here you might run into a roadblock. If you don't know what "log" means, I don't think there's much you can do.
So here you might run into a roadblock. If you don't know what "log" means, I don't think there's much you can do.
mathman629
2018-09-27 20:19:39
Rewrite in exponential form?
Rewrite in exponential form?
DPatrick
2018-09-27 20:19:47
I almost always like to do this.
I almost always like to do this.
DPatrick
2018-09-27 20:20:05
We can rewrite the given information as $10^r = 8$ and $10^s = 9$.
We can rewrite the given information as $10^r = 8$ and $10^s = 9$.
DPatrick
2018-09-27 20:20:27
(That's what "log" means. $\log_{10} 8 = r$ by definition means $10^r = 8$.)
(That's what "log" means. $\log_{10} 8 = r$ by definition means $10^r = 8$.)
DPatrick
2018-09-27 20:20:38
And we're trying to solve for $x$ in $10^x = 5$.
And we're trying to solve for $x$ in $10^x = 5$.
DPatrick
2018-09-27 20:20:58
There's not much we can do with these except multiply them or take them to a power. How are we possibly going to get a $5$ that way, though? All I see are powers of $2$'s and $3$'s.
There's not much we can do with these except multiply them or take them to a power. How are we possibly going to get a $5$ that way, though? All I see are powers of $2$'s and $3$'s.
a1b2
2018-09-27 20:21:12
$2\times5=10$
$2\times5=10$
DPatrick
2018-09-27 20:21:22
Hmmm...how does that help?
Hmmm...how does that help?
asdf334
2018-09-27 20:21:59
cube rt(8) = 2
cube rt(8) = 2
DPatrick
2018-09-27 20:22:33
OK...so if we take the cube root of $10^r = 8$, we get $\left(10^r\right)^\frac13 = 2$.
OK...so if we take the cube root of $10^r = 8$, we get $\left(10^r\right)^\frac13 = 2$.
DPatrick
2018-09-27 20:22:45
Or $10^{\frac{r}{3}} = 2$.
Or $10^{\frac{r}{3}} = 2$.
DPatrick
2018-09-27 20:22:48
And now what?
And now what?
Stephenpiano
2018-09-27 20:22:57
divide by 10!
divide by 10!
Catgirl1234
2018-09-27 20:23:08
divide by 10
divide by 10
hobbit16
2018-09-27 20:23:10
10/2 = 5 so 10/10^r/3 = 5
10/2 = 5 so 10/10^r/3 = 5
DPatrick
2018-09-27 20:23:32
Good idea! But how about we divide it into 10 instead, it's a little cleaner.
Good idea! But how about we divide it into 10 instead, it's a little cleaner.
DPatrick
2018-09-27 20:23:47
We now have $\dfrac{10}{10^{\frac{r}{3}}} = \dfrac{10}{2} = 5$.
We now have $\dfrac{10}{10^{\frac{r}{3}}} = \dfrac{10}{2} = 5$.
DPatrick
2018-09-27 20:23:53
Hey, that's what we want!
Hey, that's what we want!
DPatrick
2018-09-27 20:24:19
And the fraction on the left simplifies to $10^{\left(1 - \frac{r}{3}\right)} = 5$.
And the fraction on the left simplifies to $10^{\left(1 - \frac{r}{3}\right)} = 5$.
phoenix9
2018-09-27 20:24:33
= x
= x
aadinair
2018-09-27 20:24:36
(d)
(d)
Orangestripe
2018-09-27 20:24:36
So the answer is d
So the answer is d
Apurple
2018-09-27 20:24:36
d
d
DPatrick
2018-09-27 20:24:37
So $10^{1 - (r/3)} = 5$, and thus $x = 1 - (r/3)$ is the solution. $\boxed{\text{(d)}}$
So $10^{1 - (r/3)} = 5$, and thus $x = 1 - (r/3)$ is the solution. $\boxed{\text{(d)}}$
DPatrick
2018-09-27 20:25:00
(The inclusion of $s$ was just there to distract us!)
(The inclusion of $s$ was just there to distract us!)
DPatrick
2018-09-27 20:25:27
Like I said, when I see logs, I usually try to convert to exponentials, which are more familiar territory for me.
Like I said, when I see logs, I usually try to convert to exponentials, which are more familiar territory for me.
bibilutza
2018-09-27 20:25:33
and last.....
and last.....
DPatrick
2018-09-27 20:25:36
10. Which of the following is closest to the number of ordered pairs of points $(m,n)$, where $m$ and $n$ are both between $1$ and $100$ inclusive and relatively prime (their greatest common divisor is $1$)? [$(2,3)$ and $(3,2)$ count as two such points.]
$ $
(a) $5{,}000$ (b) $6{,}000$ (c) $7{,}000$ (d) $8{,}000$
10. Which of the following is closest to the number of ordered pairs of points $(m,n)$, where $m$ and $n$ are both between $1$ and $100$ inclusive and relatively prime (their greatest common divisor is $1$)? [$(2,3)$ and $(3,2)$ count as two such points.]
$ $
(a) $5{,}000$ (b) $6{,}000$ (c) $7{,}000$ (d) $8{,}000$
DPatrick
2018-09-27 20:26:13
There are $100^2 = 10{,}000$ ordered pairs in total. How can we estimate how many of them are relatively prime?
There are $100^2 = 10{,}000$ ordered pairs in total. How can we estimate how many of them are relatively prime?
kootrapali
2018-09-27 20:26:32
Subtract the cases with two evens
Subtract the cases with two evens
Apurple
2018-09-27 20:26:42
elimination of those that are not
elimination of those that are not
DPatrick
2018-09-27 20:26:46
That's a really good place to start.
That's a really good place to start.
DPatrick
2018-09-27 20:27:05
If $(m,n)$ are both even, then we have to throw them out, as their gcd is at least $2$.
If $(m,n)$ are both even, then we have to throw them out, as their gcd is at least $2$.
DPatrick
2018-09-27 20:27:15
How many pairs $(m,n)$ are both even?
How many pairs $(m,n)$ are both even?
bradleyguo
2018-09-27 20:27:28
50*50
50*50
AforApple
2018-09-27 20:27:28
2500
2500
Mathisfun04
2018-09-27 20:27:28
2500
2500
DPatrick
2018-09-27 20:27:46
Right $50 \cdot 50 = 2500$ of them. Or, exactly $\frac14$ of them.
Right $50 \cdot 50 = 2500$ of them. Or, exactly $\frac14$ of them.
DPatrick
2018-09-27 20:27:54
So we can eliminate $\frac14$ of our pairs, and we're left with $7500$.
So we can eliminate $\frac14$ of our pairs, and we're left with $7500$.
DPatrick
2018-09-27 20:27:58
Now what?
Now what?
asdf334
2018-09-27 20:28:10
you can just do two multiples of the same prime will be eliminated
you can just do two multiples of the same prime will be eliminated
StanDaMan
2018-09-27 20:28:10
eliminate the pairs that are both divisible by 3
eliminate the pairs that are both divisible by 3
bigmath
2018-09-27 20:28:10
now factors of 3 and ones that dont work
now factors of 3 and ones that dont work
mathman629
2018-09-27 20:28:10
Now try multiples of 3
Now try multiples of 3
DPatrick
2018-09-27 20:28:19
Sure, let's just keep going! How many of the remaining pairs have both terms multiples of $3$?
Sure, let's just keep going! How many of the remaining pairs have both terms multiples of $3$?
DPatrick
2018-09-27 20:28:36
Remember: we don't need an exact count! We're only trying to get to the nearest 1000, after all.
Remember: we don't need an exact count! We're only trying to get to the nearest 1000, after all.
mathleticguyyy
2018-09-27 20:28:56
1/9 of remaining
1/9 of remaining
a1b2
2018-09-27 20:28:56
$\frac19$ of them
$\frac19$ of them
DPatrick
2018-09-27 20:29:04
Right. About $\frac19$ of them. (About $\frac13$ have $m$ a multiple of $3$ and about $\frac13$ have $n$ a multiple of $3$.)
Right. About $\frac19$ of them. (About $\frac13$ have $m$ a multiple of $3$ and about $\frac13$ have $n$ a multiple of $3$.)
DPatrick
2018-09-27 20:29:14
That should be close enough.
That should be close enough.
DPatrick
2018-09-27 20:29:21
So we throw those away, and what's left is about $\frac89 \cdot 7500$.
So we throw those away, and what's left is about $\frac89 \cdot 7500$.
DPatrick
2018-09-27 20:29:29
This is about $6700$. And we've still got more to get rid of!
This is about $6700$. And we've still got more to get rid of!
Orangestripe
2018-09-27 20:29:37
Then we would do multiples of 5
Then we would do multiples of 5
StanDaMan
2018-09-27 20:29:43
then do multiples of 5, 7, etc
then do multiples of 5, 7, etc
mathman629
2018-09-27 20:29:43
now try the next prime, 5!
now try the next prime, 5!
thebossishere
2018-09-27 20:29:43
try 5s
try 5s
DPatrick
2018-09-27 20:29:57
Sure, let's keep going! About $\frac{1}{25}$ of them are both multiples of $5$. (About $\frac15$ have $m$ a multiple of $5$ and about $\frac15$ have $n$ a multiple of $5$.)
Sure, let's keep going! About $\frac{1}{25}$ of them are both multiples of $5$. (About $\frac15$ have $m$ a multiple of $5$ and about $\frac15$ have $n$ a multiple of $5$.)
DPatrick
2018-09-27 20:30:12
So we throw those away, and what's left is about $\frac{24}{25} \cdot 6700$. This is about $6400$.
So we throw those away, and what's left is about $\frac{24}{25} \cdot 6700$. This is about $6400$.
DPatrick
2018-09-27 20:30:24
We'll throw away another $150$ or so ($\frac{1}{49}$ of $6400$) when we look at the $7$'s. Down to about $6250$.
We'll throw away another $150$ or so ($\frac{1}{49}$ of $6400$) when we look at the $7$'s. Down to about $6250$.
DPatrick
2018-09-27 20:30:30
We'll throw away another $50$ or so ($\frac{1}{121}$ of $6250$) when we look at the $11$'s. Down to about $6200$.
We'll throw away another $50$ or so ($\frac{1}{121}$ of $6250$) when we look at the $11$'s. Down to about $6200$.
kevalshah2005
2018-09-27 20:30:39
If you keep going, it goes to 6000
If you keep going, it goes to 6000
krazykid548
2018-09-27 20:30:39
We will even more slowly approach 6000
We will even more slowly approach 6000
DPatrick
2018-09-27 20:30:52
It sure looks like we'll probably end up somewhere around $6000$, maybe a little higher since there isn't much left to throw away.
It sure looks like we'll probably end up somewhere around $6000$, maybe a little higher since there isn't much left to throw away.
DPatrick
2018-09-27 20:31:08
At least, it seems clear that $6000$ is the closest of the answer choices. $\boxed{\text{(b)}}$
At least, it seems clear that $6000$ is the closest of the answer choices. $\boxed{\text{(b)}}$
DPatrick
2018-09-27 20:31:32
If we did this calculation more carefully, we might even get the actual answer, which is $6{,}087$.
If we did this calculation more carefully, we might even get the actual answer, which is $6{,}087$.
DPatrick
2018-09-27 20:31:43
Does anybody know the general result?
Does anybody know the general result?
DPatrick
2018-09-27 20:32:10
That is, if I replaced $100$ with a larger number $N$, do you happen to know the formula?
That is, if I replaced $100$ with a larger number $N$, do you happen to know the formula?
Mathisfun04
2018-09-27 20:32:17
it is 10000 * 6/pi^2?
it is 10000 * 6/pi^2?
DPatrick
2018-09-27 20:32:27
Excellent! I'm glad someone has seen this!
Excellent! I'm glad someone has seen this!
mathman629
2018-09-27 20:32:35
$10000\cdot\tfrac{6}{\pi^2}\approx\boxed{6000}$
$10000\cdot\tfrac{6}{\pi^2}\approx\boxed{6000}$
DPatrick
2018-09-27 20:32:37
It turns out that the number of pairs $(m,n)$ of relatively prime positive integers where $m$ and $n$ are less than or equal to $N$ is approximately $\dfrac{6}{\pi^2} \cdot N^2$. (This requires some calculus to prove.)
It turns out that the number of pairs $(m,n)$ of relatively prime positive integers where $m$ and $n$ are less than or equal to $N$ is approximately $\dfrac{6}{\pi^2} \cdot N^2$. (This requires some calculus to prove.)
CoolKidWiz
2018-09-27 20:32:50
how did pi get in there?
how did pi get in there?
DPatrick
2018-09-27 20:32:57
That's where the calculus comes in.
That's where the calculus comes in.
DPatrick
2018-09-27 20:33:04
And indeed, when $N = 100$, this formula gives $6079$ (rounded to the nearest integer), which is remarkably close. It's only off by $8$, which is within $0.13\%$ of the actual answer.
And indeed, when $N = 100$, this formula gives $6079$ (rounded to the nearest integer), which is remarkably close. It's only off by $8$, which is within $0.13\%$ of the actual answer.
DPatrick
2018-09-27 20:33:21
And the formula gets even better as $N$ increases.
And the formula gets even better as $N$ increases.
DPatrick
2018-09-27 20:33:28
When $N=1000$, the formula gives $607{,}927$ and the actual answer is $608{,}383$. The formula is only off by $0.075\%$.
When $N=1000$, the formula gives $607{,}927$ and the actual answer is $608{,}383$. The formula is only off by $0.075\%$.
spartacle
2018-09-27 20:33:38
It comes from 1/1 + 1/4 + 1/9 + ...
It comes from 1/1 + 1/4 + 1/9 + ...
DPatrick
2018-09-27 20:33:59
Indeed, it is related to the famous Riemann Zeta Function, which in turn is related to the (probably) still-unsolved Riemann Hypothesis.
Indeed, it is related to the famous Riemann Zeta Function, which in turn is related to the (probably) still-unsolved Riemann Hypothesis.
DPatrick
2018-09-27 20:34:13
And finally, here is a pretty picture I made. Every dot is a pair $(m,n)$ where $m$ and $n$ are relatively prime.
And finally, here is a pretty picture I made. Every dot is a pair $(m,n)$ where $m$ and $n$ are relatively prime.
DPatrick
2018-09-27 20:34:17
DPatrick
2018-09-27 20:34:30
If you count them (good luck!), you'll count $6087$ dots.
If you count them (good luck!), you'll count $6087$ dots.
asdf334
2018-09-27 20:34:52
what is that diagnoal line? is it when they are the same?
what is that diagnoal line? is it when they are the same?
DPatrick
2018-09-27 20:35:12
Yes, the white diagonal are the points $(m,m)$ that have gcd $m$, so they're not colored in (except when $m=1$).
Yes, the white diagonal are the points $(m,m)$ that have gcd $m$, so they're not colored in (except when $m=1$).
Catgirl1234
2018-09-27 20:35:23
Is the graph following a set pattern?
Is the graph following a set pattern?
mikebreen
2018-09-27 20:35:32
Symmetric?
Symmetric?
DPatrick
2018-09-27 20:35:50
Nobody really knows! (If we did, then we could probably solve the Riemann Hypothesis and we'd really be a millionaire!)
Nobody really knows! (If we did, then we could probably solve the Riemann Hypothesis and we'd really be a millionaire!)
DPatrick
2018-09-27 20:36:16
So those were the 10 problems on Round 1 of this year's Championship contest!
So those were the 10 problems on Round 1 of this year's Championship contest!
DPatrick
2018-09-27 20:36:33
If you participated in the contest earlier this month and got 7 or more correct, congratulations! -- you're moving on to Round 2. Your teacher should have your invitation (or should get it soon). Round 2 will be held in October. And come back here for our Round 2 Math Jam on Tuesday, October 23 at 7:30 pm ET / 4:30 pm PT.
If you participated in the contest earlier this month and got 7 or more correct, congratulations! -- you're moving on to Round 2. Your teacher should have your invitation (or should get it soon). Round 2 will be held in October. And come back here for our Round 2 Math Jam on Tuesday, October 23 at 7:30 pm ET / 4:30 pm PT.
Bubble3
2018-09-27 20:36:43
What was the mean score?
What was the mean score?
DPatrick
2018-09-27 20:36:48
It was about 5.5.
It was about 5.5.
bibilutza
2018-09-27 20:36:52
round 3 is Baltimore, right?
round 3 is Baltimore, right?
DPatrick
2018-09-27 20:36:58
Yes: after Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Baltimore at the 2019 Joint Mathematics Meetings on January 19. Travel costs to and from Baltimore will be covered by the AMS.
Yes: after Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Baltimore at the 2019 Joint Mathematics Meetings on January 19. Travel costs to and from Baltimore will be covered by the AMS.
DPatrick
2018-09-27 20:37:09
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:
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
2018-09-27 20:37:13
DPatrick
2018-09-27 20:37:26
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Baltimore metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 2 happens to be from Baltimore).
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Baltimore metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 2 happens to be from Baltimore).
Orangestripe
2018-09-27 20:37:32
What if there is a tie?
What if there is a tie?
DPatrick
2018-09-27 20:37:55
It's unlikely: problem #10 on Round 2 will ask you to estimate a quantity. The closest answer will break ties.
It's unlikely: problem #10 on Round 2 will ask you to estimate a quantity. The closest answer will break ties.
USA
2018-09-27 20:38:08
what's the scoring method for round 2?
what's the scoring method for round 2?
DPatrick
2018-09-27 20:38:35
1 point per question, just like Round 1. Questions will be short-answer, though, not multiple-choice (except where multiple choice is more convenient for some reason).
1 point per question, just like Round 1. Questions will be short-answer, though, not multiple-choice (except where multiple choice is more convenient for some reason).
mikebreen
2018-09-27 20:38:42
Next round one starts next September
Next round one starts next September
DPatrick
2018-09-27 20:39:19
Right: if you didn't participate in Round 1 earlier this month, you can't participate further in 2018-19. Only those who advanced in Round 1 will get to take Round 2.
Right: if you didn't participate in Round 1 earlier this month, you can't participate further in 2018-19. Only those who advanced in Round 1 will get to take Round 2.
DPatrick
2018-09-27 20:39:33
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.
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.
m22pi7
2018-09-27 20:39:54
How many questions are in round 2?
How many questions are in round 2?
DPatrick
2018-09-27 20:39:58
10 problems, same as Round 1.
10 problems, same as Round 1.
AwesomeDude86
2018-09-27 20:40:01
Are round one and two scores accumulated?
Are round one and two scores accumulated?
DPatrick
2018-09-27 20:40:12
No -- the highest scorer on Round 2 in each region advances to the finals.
No -- the highest scorer on Round 2 in each region advances to the finals.
DPatrick
2018-09-27 20:40:24
...with ties broken by the best estimate on #10, if necessary.
...with ties broken by the best estimate on #10, if necessary.
mikebreen
2018-09-27 20:41:16
Usually junior or senior, but no lower age limit
Usually junior or senior, but no lower age limit
DPatrick
2018-09-27 20:41:38
If you have questions about how the contest is administered, and to see if maybe one of the many regional contests is coming your way, please visit the AMS website.
If you have questions about how the contest is administered, and to see if maybe one of the many regional contests is coming your way, please visit the AMS website.
DPatrick
2018-09-27 20:42:26
And to answer the question that I accidentally deleted: yes, Puerto Ricans and all overseas US residents can participate!
And to answer the question that I accidentally deleted: yes, Puerto Ricans and all overseas US residents can participate!
TPiR
2018-09-27 20:42:31
Thanks, everyone. Great work!
Thanks, everyone. Great work!
UnstoppableGoddess
2018-09-27 20:42:44
How much does it cost
How much does it cost
DPatrick
2018-09-27 20:42:47
It's free!
It's free!
mikebreen
2018-09-27 20:42:52
Yes, thanks Dave and all the AoPS fans.
Yes, thanks Dave and all the AoPS fans.
DPatrick
2018-09-27 20:43:12
That's it for tonight's session -- thanks for coming!
That's it for tonight's session -- thanks for coming!
DPatrick
2018-09-27 20:43:30
Please join us again on October 23 when we discuss the Round 2 problems!
Please join us again on October 23 when we discuss the Round 2 problems!
DPatrick
2018-09-27 20:43:32
Good night!
Good night!
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