## 2019 USA(J)MO Discussion

Go back to the Math Jam ArchiveEvan Chen, one of the coaches of the USA IMO team, will discuss selected problems from the 2019 USA Math Olympiad.

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#### Facilitator: Evan Chen

eminentabyss
2019-04-19 19:22:22

**Welcome to the 2019 USA(J)MO Math Jam!**
eminentabyss
2019-04-19 19:32:17

In this Math Jam, Evan Chen, one of the coaches of the USA IMO team, will discuss selected problems from the 2019 USA Math Olympiad.

In this Math Jam, Evan Chen, one of the coaches of the USA IMO team, will discuss selected problems from the 2019 USA Math Olympiad.

eminentabyss
2019-04-19 19:32:30

Evan Chen

Evan Chen

**(v_Enhance)**is one of the coaches of the USA IMO team, serving as a coordinator for the team selection tests and the assistant academic director at MOP. Starting in July 2019, he will also be an editor-in-chief for the USA(J)MO editorial board. As a high school student, Evan was an IMO gold medalist and winner of the 2014 USAMO (which he took from 12:30am to 5am in Taiwan). Outside of math and teaching, Evan enjoys StarCraft and Korean pop dance.
eminentabyss
2019-04-19 19:33:11

For now, please hold your questions -- we'll let you know when you can start asking questions. Also, due to the number of people attending tonight, we may not be able to get to every question.

For now, please hold your questions -- we'll let you know when you can start asking questions. Also, due to the number of people attending tonight, we may not be able to get to every question.

eminentabyss
2019-04-19 19:33:24

We do have one amazing teaching assistant with us tonight to help answer your questions:

We do have one amazing teaching assistant with us tonight to help answer your questions:

eminentabyss
2019-04-19 19:33:34

Detelin

Detelin

**(OlympiadPro2000)**won a silver medal in the 1995 IMO and a gold medal in the 1995 Balkan Math Olympiad. He received his M.S. in Mathematics from Sofia University and his Ph.D. in Mathematics from Texas A&M.
eminentabyss
2019-04-19 19:33:56

He can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, he may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question

He can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, he may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question

**once**and be patient, and please understand that we may not be able to answer all the questions tonight.
v_Enhance
2019-04-19 19:34:31

Hello! Welcome to the 2019 USA(J)MO math jam. The last time we had a USAMO Math Jam was in 2004, as far as I can tell, so I'm super excited to be doing the next one.

Hello! Welcome to the 2019 USA(J)MO math jam. The last time we had a USAMO Math Jam was in 2004, as far as I can tell, so I'm super excited to be doing the next one.

v_Enhance
2019-04-19 19:34:42

We'll be discussing some of the problems I like from this year's contest. (Not all the problems, alas, since these questions are tough, and I can't keep the AOPS staff here forever.)

We'll be discussing some of the problems I like from this year's contest. (Not all the problems, alas, since these questions are tough, and I can't keep the AOPS staff here forever.)

v_Enhance
2019-04-19 19:34:50

To start, I'm hoping to go through the problems in this order:

To start, I'm hoping to go through the problems in this order:

v_Enhance
2019-04-19 19:34:55

**JMO5/MO4**(combinatorics with sets $S_{ij}$), proposed by Ricky Liu**JMO6/MO5**(number theory with Evan and a blackboard), proposed by Yannick Yao**JMO4**($EF$ tangent to $A$-excircle), proposed by Ankan Bhattacharya, Zack Chroman, and Anant Mudgal**MO1**(iterated functional equation), proposed by Evan Chen**JMO3/MO2**(geometry with funny condition), proposed by Ankan Bhattacharya
v_Enhance
2019-04-19 19:35:13

But I only get about two hours, and the USAMO is a nine-hour test, so we may not get to all of these. Probably, if I find myself short on time, I'll skip the MO1 functional equation. We'll see!

But I only get about two hours, and the USAMO is a nine-hour test, so we may not get to all of these. Probably, if I find myself short on time, I'll skip the MO1 functional equation. We'll see!

v_Enhance
2019-04-19 19:35:26

(Conversely, if somehow we blaze through all five of these, I'll ad-lib problems by popular request until I get cut off. )

(Conversely, if somehow we blaze through all five of these, I'll ad-lib problems by popular request until I get cut off. )

v_Enhance
2019-04-19 19:35:39

One quick favor I'd like to ask off-the-bat: I will probably not answer questions of the form "how many points do I get for X?" or "what will cutoff for X be?". The rubrics have not even been decided yet, and I prefer not to be on-record about things I'm not sure about.

One quick favor I'd like to ask off-the-bat: I will probably not answer questions of the form "how many points do I get for X?" or "what will cutoff for X be?". The rubrics have not even been decided yet, and I prefer not to be on-record about things I'm not sure about.

v_Enhance
2019-04-19 19:35:49

About the Math Jam itself: A lot of you may have already seen complete solutions to a lot of the problems, and that's fine.

About the Math Jam itself: A lot of you may have already seen complete solutions to a lot of the problems, and that's fine.

v_Enhance
2019-04-19 19:36:00

However, my goal in this Math Jam is to show you how to incrementally *find* a solution, rather than just telling you. (It's sort of like directing a movie: you only reach the climax after a lot of character development.) So in this Math Jam, I'll be building up slowly, starting from ground zero. Thus it's fine if you haven't tried the problems yet!

However, my goal in this Math Jam is to show you how to incrementally *find* a solution, rather than just telling you. (It's sort of like directing a movie: you only reach the climax after a lot of character development.) So in this Math Jam, I'll be building up slowly, starting from ground zero. Thus it's fine if you haven't tried the problems yet!

v_Enhance
2019-04-19 19:36:19

If you want to read cleaned write-ups later, without the long narrative that'll be in the Math Jam, I will upload full solutions to http://web.evanchen.cc/problems.html for this year's USA(J)MO 2019. (I also recently added all the USAMO and IMO from 2000, so some of you may be interested in that independently.).

If you want to read cleaned write-ups later, without the long narrative that'll be in the Math Jam, I will upload full solutions to http://web.evanchen.cc/problems.html for this year's USA(J)MO 2019. (I also recently added all the USAMO and IMO from 2000, so some of you may be interested in that independently.).

v_Enhance
2019-04-19 19:36:29

Alright, ready to start?

Alright, ready to start?

cj13609517288
2019-04-19 19:36:42

Let's start!

Let's start!

Brudder
2019-04-19 19:36:42

yes

yes

kevinmathz
2019-04-19 19:36:42

Yep!

Yep!

andrewwang2623
2019-04-19 19:36:42

yes

yes

AlexLikeMath
2019-04-19 19:36:42

Yes!

Yes!

cowcow
2019-04-19 19:36:42

yes

yes

Kaiser143
2019-04-19 19:36:42

yeah

yeah

azmath333
2019-04-19 19:36:42

yup!

yup!

goodball
2019-04-19 19:36:42

Yes!

Yes!

ilovemath04
2019-04-19 19:36:42

yep

yep

hliu1
2019-04-19 19:36:42

Yeah

Yeah

Awesome_guy
2019-04-19 19:36:42

yep

yep

ketul
2019-04-19 19:36:42

Yeee

Yeee

Kagebaka
2019-04-19 19:36:46

ankant wait to start

ankant wait to start

v_Enhance
2019-04-19 19:36:52

$\bullet$ for all \(0\le i,j\le n\), the set \(S_{i,j}\) has \(i+j\) elements; and

$\bullet$ \(S_{i,j}\subseteq S_{k,l}\) whenever \(0\le i\le k\le n\) and \(0\le j\le l\le n\).

**JMO5/MO4.**Let $n$ be a nonnegative integer. Determine the number of ways that one can choose \((n+1)^2\) sets \(S_{i,j}\subseteq \{1,2,\ldots, 2n\}\), for integers \(i,j\) with \(0\le i,j\le n\), such that:$\bullet$ for all \(0\le i,j\le n\), the set \(S_{i,j}\) has \(i+j\) elements; and

$\bullet$ \(S_{i,j}\subseteq S_{k,l}\) whenever \(0\le i\le k\le n\) and \(0\le j\le l\le n\).

v_Enhance
2019-04-19 19:37:08

There's a lot of notation here, so I'll start by doing some book-keeping that will help us all visualize what's happening.

There's a lot of notation here, so I'll start by doing some book-keeping that will help us all visualize what's happening.

v_Enhance
2019-04-19 19:37:16

The notation $S_{i,j}$ for $0 \le i,j \le n$ means that we have $(n+1)^2$ sets and so it'll be easier to visualize what's going on if I put them in a grid, like so.

The notation $S_{i,j}$ for $0 \le i,j \le n$ means that we have $(n+1)^2$ sets and so it'll be easier to visualize what's going on if I put them in a grid, like so.

v_Enhance
2019-04-19 19:37:23

\[

\begin{bmatrix}

S_{0n} & S_{1n} & \dots & S_{nn} \\

\vdots & \vdots & \ddots & \vdots \\

S_{01} & S_{11} & \dots & S_{1n} \\

S_{00} & S_{10} & \dots & S_{n0}

\end{bmatrix}

\]

\[

\begin{bmatrix}

S_{0n} & S_{1n} & \dots & S_{nn} \\

\vdots & \vdots & \ddots & \vdots \\

S_{01} & S_{11} & \dots & S_{1n} \\

S_{00} & S_{10} & \dots & S_{n0}

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:37:41

Let's try to figure out what that condition is saying. We'll start by filling in some of the easy steps.

Let's try to figure out what that condition is saying. We'll start by filling in some of the easy steps.

v_Enhance
2019-04-19 19:37:46

First of all, what does that $S_{00}$ in the bottom-left have to be?

First of all, what does that $S_{00}$ in the bottom-left have to be?

goodbear
2019-04-19 19:38:00

ø

ø

Brendanb4321
2019-04-19 19:38:00

empty set

empty set

oldbro
2019-04-19 19:38:00

{}

{}

boomminecraft
2019-04-19 19:38:00

0 elements, so empty set

0 elements, so empty set

Eyed
2019-04-19 19:38:00

empty set

empty set

v_Enhance
2019-04-19 19:38:04

We should have $S_{00} = \varnothing$: only set with $0$ elements.

We should have $S_{00} = \varnothing$: only set with $0$ elements.

v_Enhance
2019-04-19 19:38:09

Next, what does $S_{nn}$ have to be?

Next, what does $S_{nn}$ have to be?

Kaiser143
2019-04-19 19:38:42

all of then numbers

all of then numbers

Eyed
2019-04-19 19:38:42

{1,2,...2n}

{1,2,...2n}

Brendanb4321
2019-04-19 19:38:42

$\{1,2,\ldots,2n\}$

$\{1,2,\ldots,2n\}$

SD2014
2019-04-19 19:38:42

have all the elements

have all the elements

boomminecraft
2019-04-19 19:38:42

the whole set, as it should have 2n elements!

the whole set, as it should have 2n elements!

v_Enhance
2019-04-19 19:38:47

We should have $S_{nn} = \{1, \dots, 2n\}$; all $2n$ elements are present.

We should have $S_{nn} = \{1, \dots, 2n\}$; all $2n$ elements are present.

v_Enhance
2019-04-19 19:39:02

Now imagine we placed an ant at $S_{00}$, and it walked up this grid going right and up, ending at $S_{nn}$. How does the set change as we walk up this grid?

Now imagine we placed an ant at $S_{00}$, and it walked up this grid going right and up, ending at $S_{nn}$. How does the set change as we walk up this grid?

SD2014
2019-04-19 19:39:29

it gains one element each step

it gains one element each step

Kaiser143
2019-04-19 19:39:29

add an element

add an element

bluelinfish
2019-04-19 19:39:29

one more element each time

one more element each time

hliu1
2019-04-19 19:39:29

build up one element at a time.

build up one element at a time.

goodbear
2019-04-19 19:39:29

adds one element at a time

adds one element at a time

Brendanb4321
2019-04-19 19:39:29

keeps gaining 1 new element each time

keeps gaining 1 new element each time

v_Enhance
2019-04-19 19:39:37

Right, as you go up and to the right, you add one element each time. Because of the first condition, every time the ant makes a move, the set is one element bigger, and because of the second condition, none of the old elements go away.

Right, as you go up and to the right, you add one element each time. Because of the first condition, every time the ant makes a move, the set is one element bigger, and because of the second condition, none of the old elements go away.

v_Enhance
2019-04-19 19:39:42

Therefore, for $n = 1$, one valid example would be:

\[

\begin{bmatrix}

\{1\} & \{1,2\} \\

\varnothing & \{2\}

\end{bmatrix}

\]

Therefore, for $n = 1$, one valid example would be:

\[

\begin{bmatrix}

\{1\} & \{1,2\} \\

\varnothing & \{2\}

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:39:51

I'm going to be lazy and abbreviate this to just

\[

\begin{bmatrix}

1 & 12 \\

\varnothing & 2

\end{bmatrix}

\]

because all those commas and curly braces are going to be really hard to type once $n > 1$...

I'm going to be lazy and abbreviate this to just

\[

\begin{bmatrix}

1 & 12 \\

\varnothing & 2

\end{bmatrix}

\]

because all those commas and curly braces are going to be really hard to type once $n > 1$...

v_Enhance
2019-04-19 19:39:58

Make sense?

Make sense?

v_Enhance
2019-04-19 19:40:11

Cool.

Cool.

v_Enhance
2019-04-19 19:40:12

Are there other examples for $n = 1$?

Are there other examples for $n = 1$?

porkemon2
2019-04-19 19:40:32

yes, switch 2 and 1

yes, switch 2 and 1

RaggedQuark2
2019-04-19 19:40:32

switch the 1 and the 2

switch the 1 and the 2

goodbear
2019-04-19 19:40:32

1 12

ø 1

1 12

ø 1

ilovemath04
2019-04-19 19:40:32

the 1 and 2 are switched

the 1 and 2 are switched

v_Enhance
2019-04-19 19:40:44

Yes, we could also swap the order and have

\[

\begin{bmatrix}

2 & 12 \\

\varnothing & 1

\end{bmatrix}

\]

instead.

Yes, we could also swap the order and have

\[

\begin{bmatrix}

2 & 12 \\

\varnothing & 1

\end{bmatrix}

\]

instead.

v_Enhance
2019-04-19 19:42:03

(Hahah yeah. OK, good catch.)

(Hahah yeah. OK, good catch.)

v_Enhance
2019-04-19 19:42:06

Alright, what are the other examples?

Alright, what are the other examples?

hliu1
2019-04-19 19:42:35

1 12

ø 1

2 12

ø 2

1 12

ø 1

2 12

ø 2

porkemon2
2019-04-19 19:42:35

instead of 2 and 1 you can have 1 and 1 or 2 and 2

instead of 2 and 1 you can have 1 and 1 or 2 and 2

carzland
2019-04-19 19:42:35

There are more than 2 examples?

There are more than 2 examples?

Brendanb4321
2019-04-19 19:42:35

S_{1,0}=S_{0,1}

S_{1,0}=S_{0,1}

v_Enhance
2019-04-19 19:42:58

Yep, two mor examples:

Yep, two mor examples:

v_Enhance
2019-04-19 19:43:11

\[

\begin{bmatrix}

2 & 12 \\

\varnothing & 2

\end{bmatrix}

\]

\[

\begin{bmatrix}

2 & 12 \\

\varnothing & 2

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:43:18

\[

\begin{bmatrix}

1 & 12 \\

\varnothing & 1

\end{bmatrix}

\]

\[

\begin{bmatrix}

1 & 12 \\

\varnothing & 1

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:43:29

And those are the only four.

And those are the only four.

v_Enhance
2019-04-19 19:44:04

Alright, everyone caught up on $n=1$? (Including me, I guess )

Alright, everyone caught up on $n=1$? (Including me, I guess )

cowcow
2019-04-19 19:44:16

yes

yes

P_Groudon
2019-04-19 19:44:16

Yep

Yep

AlexLikeMath
2019-04-19 19:44:16

Yes.

Yes.

rjiangbz
2019-04-19 19:44:16

yeah

yeah

v_Enhance
2019-04-19 19:44:43

Alright, let's go on to $n=2$. In that case the answer is way bigger, so we'll start by just filling in a mostly empty grid.

Alright, let's go on to $n=2$. In that case the answer is way bigger, so we'll start by just filling in a mostly empty grid.

v_Enhance
2019-04-19 19:44:54

So here's a mostly empty $3 \times 3$ grid, except I filled in the easy cases $S_{00}$ and $S_{22}$.\[ \begin{bmatrix}

\color{gray}{S_{02}} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{gray}{S_{01}} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

So here's a mostly empty $3 \times 3$ grid, except I filled in the easy cases $S_{00}$ and $S_{22}$.\[ \begin{bmatrix}

\color{gray}{S_{02}} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{gray}{S_{01}} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

v_Enhance
2019-04-19 19:45:07

Let's start throwing in some entries. Where might I put our first number?

Let's start throwing in some entries. Where might I put our first number?

RaggedQuark2
2019-04-19 19:45:29

s10 or s01

s10 or s01

chenwan2004
2019-04-19 19:45:29

top

top

carzland
2019-04-19 19:45:29

In S_10 or S_01

In S_10 or S_01

goodbear
2019-04-19 19:45:29

S_10

S_10

c4raymond
2019-04-19 19:45:29

s_01?

s_01?

Brendanb4321
2019-04-19 19:45:29

S_10

S_10

beiqiang
2019-04-19 19:45:29

S_(01)

S_(01)

annoynymous
2019-04-19 19:45:29

S_{10}

S_{10}

bluelinfish
2019-04-19 19:45:29

put 1 in S10

put 1 in S10

goodball
2019-04-19 19:45:29

S01 or S10

S01 or S10

v_Enhance
2019-04-19 19:45:39

So there's a lot of different things we could fill in. For the sake of narrative, let's start with the $S_{01}$ entry (but $S_{10}$ would have worked fine too).

So there's a lot of different things we could fill in. For the sake of narrative, let's start with the $S_{01}$ entry (but $S_{10}$ would have worked fine too).

cj13609517288
2019-04-19 19:45:47

$S_{01}=1$

$S_{01}=1$

v_Enhance
2019-04-19 19:45:50

We need to exactly one number in it, so let's just put $1$: right now all the numbers are exactly the same, so we just pick one.

We need to exactly one number in it, so let's just put $1$: right now all the numbers are exactly the same, so we just pick one.

v_Enhance
2019-04-19 19:45:52

\[ \begin{bmatrix}

\color{gray}{S_{02}} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

\[ \begin{bmatrix}

\color{gray}{S_{02}} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

v_Enhance
2019-04-19 19:45:58

Okay, good. I propose we move on to $S_{02}$ next. What could we put in there?

Okay, good. I propose we move on to $S_{02}$ next. What could we put in there?

carzland
2019-04-19 19:46:18

2

2

Brendanb4321
2019-04-19 19:46:18

12 13 14

12 13 14

snow_monkey
2019-04-19 19:46:18

12, 13, or 14

12, 13, or 14

beiqiang
2019-04-19 19:46:18

1 and any from 2 to 4

1 and any from 2 to 4

v_Enhance
2019-04-19 19:46:22

Right, we could write any of $12$, $13$, $14$, but now, the numbers $2$, $3$, $4$ all play symmetric roles, so it shouldn't matter which one we put.

Right, we could write any of $12$, $13$, $14$, but now, the numbers $2$, $3$, $4$ all play symmetric roles, so it shouldn't matter which one we put.

v_Enhance
2019-04-19 19:46:26

So let's just put $12$.

So let's just put $12$.

v_Enhance
2019-04-19 19:46:28

\[ \begin{bmatrix}

\color{blue}{12} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

\[ \begin{bmatrix}

\color{blue}{12} &

\color{gray}{S_{12}} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

v_Enhance
2019-04-19 19:46:38

And at this point you might see where this is going. What could $S_{12}$ be, and does it matter?

And at this point you might see where this is going. What could $S_{12}$ be, and does it matter?

snow_monkey
2019-04-19 19:47:02

123 or 124

123 or 124

A_drop_of_water
2019-04-19 19:47:02

123

123

AwesomeYRY
2019-04-19 19:47:02

123

123

Brendanb4321
2019-04-19 19:47:02

123 124

123 124

legolego
2019-04-19 19:47:02

123 or 124 (doesn't matter which)

123 or 124 (doesn't matter which)

ilovemath04
2019-04-19 19:47:02

123 or 124 and no it does not matter

123 or 124 and no it does not matter

hliu1
2019-04-19 19:47:02

123 or 124 but 3 and 4 are symmetris

123 or 124 but 3 and 4 are symmetris

v_Enhance
2019-04-19 19:47:04

We could put either $123$ or $1234$ in $S_{12}$, and again it doesn't matter which one, so let's put $123$.

We could put either $123$ or $1234$ in $S_{12}$, and again it doesn't matter which one, so let's put $123$.

v_Enhance
2019-04-19 19:47:07

\[ \begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

\[ \begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix} \]

v_Enhance
2019-04-19 19:47:19

At this point you'll notice that I haven't made any hard decisions yet! At every point, we filled in some entries, but then also argued that our choices didn't matter.

At this point you'll notice that I haven't made any hard decisions yet! At every point, we filled in some entries, but then also argued that our choices didn't matter.

v_Enhance
2019-04-19 19:47:25

So put another way, I propose that to find the answer for $n = 2$, we can find the number of ways to finish our partially filled grid above, and then multiply by a certain number for all our choices. What number is that?

So put another way, I propose that to find the answer for $n = 2$, we can find the number of ways to finish our partially filled grid above, and then multiply by a certain number for all our choices. What number is that?

Brendanb4321
2019-04-19 19:48:04

(2n)!

(2n)!

beiqiang
2019-04-19 19:48:04

4 * 3 * 2 = 24

4 * 3 * 2 = 24

snow_monkey
2019-04-19 19:48:04

4!=24

4!=24

rjiangbz
2019-04-19 19:48:04

24

24

AlexLikeMath
2019-04-19 19:48:04

(2n)!

(2n)!

v_Enhance
2019-04-19 19:48:13

Some of you already see what's happening for general $n$

Some of you already see what's happening for general $n$

v_Enhance
2019-04-19 19:48:20

But yes, it's just $4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24$. The $4$ comes from picking the number $1$ for $S_{10}$ (as opposed to $2$, $3$, $4$); the $3$ comes from $S_{20}$ and so on.

But yes, it's just $4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24$. The $4$ comes from picking the number $1$ for $S_{10}$ (as opposed to $2$, $3$, $4$); the $3$ comes from $S_{20}$ and so on.

v_Enhance
2019-04-19 19:48:37

Okay, so let's try to fill in the rest of that $2 \times 2$ grid now.

Okay, so let's try to fill in the rest of that $2 \times 2$ grid now.

v_Enhance
2019-04-19 19:48:38

What sets can I choose for $S_{11}$?

What sets can I choose for $S_{11}$?

AlexLikeMath
2019-04-19 19:49:25

12, or 13

12, or 13

Ttugf
2019-04-19 19:49:25

12 or 13

12 or 13

diamond606
2019-04-19 19:49:25

12, 13

12, 13

snow_monkey
2019-04-19 19:49:25

12, 13

12, 13

mathfun5
2019-04-19 19:49:25

Superset of S10 but subset of S21

Superset of S10 but subset of S21

v_Enhance
2019-04-19 19:49:35

Yeah, I can either put $S_{11} = \{1,2\}$ or $S_{11} = \{1,3\}$. For the first time, these choices are not symmetric, so I need to actually keep track of the two cases.

Yeah, I can either put $S_{11} = \{1,2\}$ or $S_{11} = \{1,3\}$. For the first time, these choices are not symmetric, so I need to actually keep track of the two cases.

v_Enhance
2019-04-19 19:49:40

So here they are.

So here they are.

v_Enhance
2019-04-19 19:49:41

\[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix}

\quad\text{ OR }\quad

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{13} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix}

\]

\[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix}

\quad\text{ OR }\quad

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{13} &

\color{gray}{S_{21}} \\

\color{blue}{\varnothing} &

\color{gray}{S_{10}} &

\color{gray}{S_{20}}

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:49:56

Okay, let's look at the left case first.

Okay, let's look at the left case first.

v_Enhance
2019-04-19 19:50:00

In the first case (left), what are the possibilities for $S_{21}$?

In the first case (left), what are the possibilities for $S_{21}$?

goodball
2019-04-19 19:50:30

123 or 124

123 or 124

bluelinfish
2019-04-19 19:50:30

123 and 124

123 and 124

porkemon2
2019-04-19 19:50:30

either add 3 or a 4, two options

either add 3 or a 4, two options

v_Enhance
2019-04-19 19:50:33

We could have $S_{21} = \{1,2,3\}$ or $S_{21} = \{1,2,4\}$.

We could have $S_{21} = \{1,2,3\}$ or $S_{21} = \{1,2,4\}$.

v_Enhance
2019-04-19 19:50:36

In the first case case (still left), what are the possibilities for $S_{10}$?

In the first case case (still left), what are the possibilities for $S_{10}$?

cowcow
2019-04-19 19:50:48

1 or 2

1 or 2

AwesomeYRY
2019-04-19 19:50:48

1 or 2

1 or 2

annoynymous
2019-04-19 19:50:51

1,2

1,2

v_Enhance
2019-04-19 19:50:54

We could have $S_{10} = \{1\}$ or $S_{10} = \{2\}$.

We could have $S_{10} = \{1\}$ or $S_{10} = \{2\}$.

v_Enhance
2019-04-19 19:51:13

So our left case right now looks like: \[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{orange}{123 \text{ or } 124} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 2} &

\color{gray}{S_{20}}

\end{bmatrix} \]

So our left case right now looks like: \[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{orange}{123 \text{ or } 124} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 2} &

\color{gray}{S_{20}}

\end{bmatrix} \]

v_Enhance
2019-04-19 19:51:19

Let's move on the the right case.

Let's move on the the right case.

v_Enhance
2019-04-19 19:51:36

(we'll deal with $S_{20}$ shortly )

(we'll deal with $S_{20}$ shortly )

cj13609517288
2019-04-19 19:51:46

THIS IS GETTING MESSIER EVERY MINUTE!!!

THIS IS GETTING MESSIER EVERY MINUTE!!!

v_Enhance
2019-04-19 19:51:49

Hang in there!

Hang in there!

v_Enhance
2019-04-19 19:52:19

Is the

Is the

*right*case, what might $S_{21}$ and $S_{10}$ be?
AwesomeYRY
2019-04-19 19:52:48

123 or 134 for S21 and S10 can be 1 or 3

123 or 134 for S21 and S10 can be 1 or 3

P_Groudon
2019-04-19 19:52:48

S_10 = 1 or 3 while S_21 = 123 or 134

S_10 = 1 or 3 while S_21 = 123 or 134

jesun
2019-04-19 19:52:48

123 or 134 for s21 and 1 or 3 for s10

123 or 134 for s21 and 1 or 3 for s10

v_Enhance
2019-04-19 19:53:00

In the right case, we have $S_{21} = \{1,2,3\}$ or $S_{21} = \{1,3,4\}$; while $S_{10} = \{1\}$ or $S_{10} = \{3\}$.

In the right case, we have $S_{21} = \{1,2,3\}$ or $S_{21} = \{1,3,4\}$; while $S_{10} = \{1\}$ or $S_{10} = \{3\}$.

v_Enhance
2019-04-19 19:53:21

OK, this is getting messy, so let's step back and see where we are before we keep diving further.

OK, this is getting messy, so let's step back and see where we are before we keep diving further.

v_Enhance
2019-04-19 19:53:36

Our map right now looks like:

\[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{orange}{123 \text{ or } 124} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 2} &

\color{gray}{S_{20}}

\end{bmatrix}

\quad\text{ OR }\quad

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{13} &

\color{orange}{123 \text{ or } 134} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 3} &

\color{gray}{S_{20}}

\end{bmatrix}

\]

Our map right now looks like:

\[

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{12} &

\color{orange}{123 \text{ or } 124} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 2} &

\color{gray}{S_{20}}

\end{bmatrix}

\quad\text{ OR }\quad

\begin{bmatrix}

\color{blue}{12} &

\color{blue}{123} &

\color{blue}{1234} \\

\color{blue}{1} &

\color{red}{13} &

\color{orange}{123 \text{ or } 134} \\

\color{blue}{\varnothing} &

\color{orange}{1 \text{ or } 3} &

\color{gray}{S_{20}}

\end{bmatrix}

\]

v_Enhance
2019-04-19 19:54:02

Now here's the elephant in the room. We have $2^3 = 8$ cases so far, and for each of them we want to think about how many ways there are to fill $S_{20}$. Well, for each case, take a guess: how many might there be?

Now here's the elephant in the room. We have $2^3 = 8$ cases so far, and for each of them we want to think about how many ways there are to fill $S_{20}$. Well, for each case, take a guess: how many might there be?

AlexLikeMath
2019-04-19 19:54:44

2

2

mathfun5
2019-04-19 19:54:44

2 each time

2 each time

Ttugf
2019-04-19 19:54:44

2 for each

2 for each

RaggedQuark2
2019-04-19 19:54:44

2 each

2 each

v_Enhance
2019-04-19 19:54:49

Yeah, the big surprise of this problem is that, no matter how we filled in the previous entries above and to the left, it seems there are always two ways to do it!

Yeah, the big surprise of this problem is that, no matter how we filled in the previous entries above and to the left, it seems there are always two ways to do it!

v_Enhance
2019-04-19 19:55:17

We're starting to see this in the binary choices for the red, and then the binary choices for the orange.

We're starting to see this in the binary choices for the red, and then the binary choices for the orange.

carzland
2019-04-19 19:55:20

So no more cases?

So no more cases?

v_Enhance
2019-04-19 19:55:24

Not if I can avoid it

Not if I can avoid it

v_Enhance
2019-04-19 19:55:30

So we have a foothold now, but we want to be careful. I haven't told you what order I'm filling in the cells yet.

So we have a foothold now, but we want to be careful. I haven't told you what order I'm filling in the cells yet.

v_Enhance
2019-04-19 19:55:38

For example, if we were just starting the $n = 2$ example, it would be silly to try and fill in $S_{20}$ (very bottom-right) first: would cause all sorts of issues.

For example, if we were just starting the $n = 2$ example, it would be silly to try and fill in $S_{20}$ (very bottom-right) first: would cause all sorts of issues.

v_Enhance
2019-04-19 19:55:59

So I want to fill in my grid (after all the blue entries) in such a way that it'll be true we always have two ways to do it.

So I want to fill in my grid (after all the blue entries) in such a way that it'll be true we always have two ways to do it.

v_Enhance
2019-04-19 19:56:13

What's a reasonable order to hit the grey cells in, in general?

What's a reasonable order to hit the grey cells in, in general?

hliu1
2019-04-19 19:56:58

just go in red diagonal, orange diagonal, etc. (aka just diagonals)

just go in red diagonal, orange diagonal, etc. (aka just diagonals)

snow_monkey
2019-04-19 19:56:58

S11, S21, S10, S20

S11, S21, S10, S20

annoynymous
2019-04-19 19:56:58

top left corner to bottom right corner

top left corner to bottom right corner

mathfun5
2019-04-19 19:56:58

We could fill it starting at the top and going down each row, then moving to the next column and repeating (so this way the cells to the left and above are always filled)

We could fill it starting at the top and going down each row, then moving to the next column and repeating (so this way the cells to the left and above are always filled)

Eyed
2019-04-19 19:56:58

minor diagonals

minor diagonals

v_Enhance
2019-04-19 19:57:06

All of these work (and then some)

All of these work (and then some)

v_Enhance
2019-04-19 19:57:34

So let me set up a way that will handle all of these:

So let me set up a way that will handle all of these:

v_Enhance
2019-04-19 19:57:49

\[ \begin{bmatrix}

\color{red}{ S \cup \{a\}} & \color{red}{ S \cup \{a,b\}} \\

\color{red}{ S} & \color{gray}{ ?}

\end{bmatrix} \]

There are exactly two ways to fill in $\color{gray}{ ?}$.

**MO4 KEY CLAIM**: suppose we are filling in the grid in the order described above. Consider a $2 \times 2$ square, as shown below.\[ \begin{bmatrix}

\color{red}{ S \cup \{a\}} & \color{red}{ S \cup \{a,b\}} \\

\color{red}{ S} & \color{gray}{ ?}

\end{bmatrix} \]

There are exactly two ways to fill in $\color{gray}{ ?}$.

v_Enhance
2019-04-19 19:58:07

What are they?

What are they?

Brendanb4321
2019-04-19 19:58:39

SU{a},SU{b}

SU{a},SU{b}

porkemon2
2019-04-19 19:58:39

add a or add b

add a or add b

Brudder
2019-04-19 19:58:39

s union b and s union a

s union b and s union a

snow_monkey
2019-04-19 19:58:39

S U {a}

S U {b}

S U {a}

S U {b}

AlexLikeMath
2019-04-19 19:58:39

S U a or S U b

S U a or S U b

v_Enhance
2019-04-19 19:58:51

Yes, we put either $S \cup \{a\}$ or $S \cup \{b\}$.

Yes, we put either $S \cup \{a\}$ or $S \cup \{b\}$.

v_Enhance
2019-04-19 19:59:10

And as long as we pick a sensible order, the claim will keep giving us the factors of $2$ we want. (Let's just standardize and choose minor diagonals, since that seemed to the most popular.)

And as long as we pick a sensible order, the claim will keep giving us the factors of $2$ we want. (Let's just standardize and choose minor diagonals, since that seemed to the most popular.)

v_Enhance
2019-04-19 19:59:43

OK, lots of stuff happened. Let's wrap everything up for general $n$.

OK, lots of stuff happened. Let's wrap everything up for general $n$.

v_Enhance
2019-04-19 19:59:59

First, I want to pick that far west and far north border as before: decide on $\varnothing = S_{00} \subset S_{01} \subset \dots S_{0n} \subset S_{1n} \subset \dots \subset S_{nn}$. How many ways can I do this?

First, I want to pick that far west and far north border as before: decide on $\varnothing = S_{00} \subset S_{01} \subset \dots S_{0n} \subset S_{1n} \subset \dots \subset S_{nn}$. How many ways can I do this?

v_Enhance
2019-04-19 20:00:09

(This is the generalization of the $4! = 24$ question from before.)

(This is the generalization of the $4! = 24$ question from before.)

mathfun5
2019-04-19 20:00:25

$(2n)!$

$(2n)!$

Magnitude
2019-04-19 20:00:25

(2n)!

(2n)!

cj13609517288
2019-04-19 20:00:25

$(2n)!$

$(2n)!$

ilovemath04
2019-04-19 20:00:28

$(2n)!$

$(2n)!$

v_Enhance
2019-04-19 20:00:31

It'll be $(2n)!$, like in the previous example. There are $2n$ ways for $S_{01}$, $2n-1$ for $S_{02}$, and so on.

It'll be $(2n)!$, like in the previous example. There are $2n$ ways for $S_{01}$, $2n-1$ for $S_{02}$, and so on.

v_Enhance
2019-04-19 20:00:36

For illustration, here's $n = 4$.

\[

\begin{bmatrix}

\color{blue}{ 5 \text{ choices}} &

\color{blue}{4 \text{ choices}} &

\color{blue}{3 \text{ choices}} &

\color{blue}{2 \text{ choices}} &

\color{blue}{ 12345678} \\

\color{blue}{ 6 \text{ choices}} &

\color{gray}{S_{13}} &

\color{gray}{S_{23}} &

\color{gray}{S_{33}} &

\color{gray}{S_{43}} \\

\color{blue}{ 7 \text{ choices}} &

\color{gray}{S_{12}} &

\color{gray}{S_{22}} &

\color{gray}{S_{32}} &

\color{gray}{S_{42}} \\

\color{blue}{ 8 \text{ choices}} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} &

\color{gray}{S_{31}} &

\color{gray}{S_{41}} \\

\color{blue}{ \varnothing } &

\color{gray}{S_{10}} &

\color{gray}{S_{20}} &

\color{gray}{S_{30}} &

\color{gray}{S_{40}} \\

\end{bmatrix}

\]

For illustration, here's $n = 4$.

\[

\begin{bmatrix}

\color{blue}{ 5 \text{ choices}} &

\color{blue}{4 \text{ choices}} &

\color{blue}{3 \text{ choices}} &

\color{blue}{2 \text{ choices}} &

\color{blue}{ 12345678} \\

\color{blue}{ 6 \text{ choices}} &

\color{gray}{S_{13}} &

\color{gray}{S_{23}} &

\color{gray}{S_{33}} &

\color{gray}{S_{43}} \\

\color{blue}{ 7 \text{ choices}} &

\color{gray}{S_{12}} &

\color{gray}{S_{22}} &

\color{gray}{S_{32}} &

\color{gray}{S_{42}} \\

\color{blue}{ 8 \text{ choices}} &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} &

\color{gray}{S_{31}} &

\color{gray}{S_{41}} \\

\color{blue}{ \varnothing } &

\color{gray}{S_{10}} &

\color{gray}{S_{20}} &

\color{gray}{S_{30}} &

\color{gray}{S_{40}} \\

\end{bmatrix}

\]

v_Enhance
2019-04-19 20:00:46

And like before, since the names of the numbers don't matter, I'll just pick $1$, $2$, $\ldots$, up to $2n$.

And like before, since the names of the numbers don't matter, I'll just pick $1$, $2$, $\ldots$, up to $2n$.

v_Enhance
2019-04-19 20:01:01

\[

\begin{bmatrix}

\color{blue}{ 1234 } &

\color{blue}{ 12345} &

\color{blue}{ 123456} &

\color{blue}{ 1234567} &

\color{blue}{ 12345678} \\

\color{blue}{ 123 } &

\color{gray}{S_{13}} &

\color{gray}{S_{23}} &

\color{gray}{S_{33}} &

\color{gray}{S_{43}} \\

\color{blue}{ 12 } &

\color{gray}{S_{12}} &

\color{gray}{S_{22}} &

\color{gray}{S_{32}} &

\color{gray}{S_{42}} \\

\color{blue}{ 1 } &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} &

\color{gray}{S_{31}} &

\color{gray}{S_{41}} \\

\color{blue}{ \varnothing } &

\color{gray}{S_{10}} &

\color{gray}{S_{20}} &

\color{gray}{S_{30}} &

\color{gray}{S_{40}} \\

\end{bmatrix}

\]

\[

\begin{bmatrix}

\color{blue}{ 1234 } &

\color{blue}{ 12345} &

\color{blue}{ 123456} &

\color{blue}{ 1234567} &

\color{blue}{ 12345678} \\

\color{blue}{ 123 } &

\color{gray}{S_{13}} &

\color{gray}{S_{23}} &

\color{gray}{S_{33}} &

\color{gray}{S_{43}} \\

\color{blue}{ 12 } &

\color{gray}{S_{12}} &

\color{gray}{S_{22}} &

\color{gray}{S_{32}} &

\color{gray}{S_{42}} \\

\color{blue}{ 1 } &

\color{gray}{S_{11}} &

\color{gray}{S_{21}} &

\color{gray}{S_{31}} &

\color{gray}{S_{41}} \\

\color{blue}{ \varnothing } &

\color{gray}{S_{10}} &

\color{gray}{S_{20}} &

\color{gray}{S_{30}} &

\color{gray}{S_{40}} \\

\end{bmatrix}

\]

v_Enhance
2019-04-19 20:01:09

So we've made our initial $(2n)!$ choices by picking the blue border.

So we've made our initial $(2n)!$ choices by picking the blue border.

v_Enhance
2019-04-19 20:01:31

Then again we want to fill in the remaining $n \times n$ grid, minor diagonals. (For example, when $n=4$, $S_{13}$ could be either $1234$ or $1235$, then we pick $S_{23}$ and $S_{12}$, etc.)

Then again we want to fill in the remaining $n \times n$ grid, minor diagonals. (For example, when $n=4$, $S_{13}$ could be either $1234$ or $1235$, then we pick $S_{23}$ and $S_{12}$, etc.)

v_Enhance
2019-04-19 20:01:35

And so how if we keep using our claim, how many ways to go?

And so how if we keep using our claim, how many ways to go?

RaggedQuark2
2019-04-19 20:02:02

2^(n^2)

2^(n^2)

ilovemath04
2019-04-19 20:02:02

2^(n^2)

2^(n^2)

wolfpack
2019-04-19 20:02:02

2^(n^2)

2^(n^2)

Damalone
2019-04-19 20:02:02

2^{n^2}

2^{n^2}

v_Enhance
2019-04-19 20:02:11

It's $2$ choices per cell. So $2^{n^2}$.

It's $2$ choices per cell. So $2^{n^2}$.

v_Enhance
2019-04-19 20:02:13

And so what's the final answer?

And so what's the final answer?

annoynymous
2019-04-19 20:02:37

total is (2n)!x 2^(n^2)

total is (2n)!x 2^(n^2)

Generic_Username
2019-04-19 20:02:37

$(2n)!2^{(n^2)}$

$(2n)!2^{(n^2)}$

jishu2003
2019-04-19 20:02:37

$(2n)!2^{n^2}$

$(2n)!2^{n^2}$

mathfun5
2019-04-19 20:02:37

$(2n)! \cdot 2^{n^2}$

$(2n)! \cdot 2^{n^2}$

v_Enhance
2019-04-19 20:02:45

Yep, $(2n)! \cdot 2^{n^2}$ is it. We gucci.

Yep, $(2n)! \cdot 2^{n^2}$ is it. We gucci.

Damalone
2019-04-19 20:02:52

Beautiful problem

Beautiful problem

v_Enhance
2019-04-19 20:03:09

Agreed!

Agreed!

annoynymous
2019-04-19 20:03:13

what makes you think of using this grid

what makes you think of using this grid

v_Enhance
2019-04-19 20:03:33

You can write the solution without the grid too. However, because the indices are double indexed in this way, I think it's easier to visualize this way.

You can write the solution without the grid too. However, because the indices are double indexed in this way, I think it's easier to visualize this way.

v_Enhance
2019-04-19 20:03:57

In general you're always welcome to draw pictures for problems if you're not changing the statement. I do it all the time.

In general you're always welcome to draw pictures for problems if you're not changing the statement. I do it all the time.

chesstoastmaster
2019-04-19 20:04:06

are there other ways than the grid?

are there other ways than the grid?

v_Enhance
2019-04-19 20:04:12

Yes, check the forums!

Yes, check the forums!

cj13609517288
2019-04-19 20:04:15

Consider how many ways will happen if n=100.

Consider how many ways will happen if n=100.

v_Enhance
2019-04-19 20:04:18

Bigggg.

Bigggg.

v_Enhance
2019-04-19 20:04:40

Any last questions?

Any last questions?

P_Groudon
2019-04-19 20:05:05

On to JMO 6 / AMO 5

On to JMO 6 / AMO 5

goodball
2019-04-19 20:05:05

Can we move to the next problem?

Can we move to the next problem?

v_Enhance
2019-04-19 20:05:09

I guess that works.

I guess that works.

v_Enhance
2019-04-19 20:05:16

**JMO6/MO5.**Two rational numbers \(\tfrac{m}{n}\) and \(\tfrac{n}{m}\) are written on a blackboard, where \(m\) and \(n\) are relatively prime positive integers. At any point, I may pick two of the numbers \(x\) and \(y\) written on the board and write either their arithmetic mean \(\tfrac{x+y}{2}\) or their harmonic mean \(\tfrac{2xy}{x+y}\) on the board as well. Find all pairs \((m,n)\) such that I can write 1 on the board in finitely many steps.
mathfun5
2019-04-19 20:05:42

Ooh my favorite problem

Ooh my favorite problem

CantonMathGuy
2019-04-19 20:05:50

where did Evan go?

where did Evan go?

v_Enhance
2019-04-19 20:05:53

Teaching class.

Teaching class.

v_Enhance
2019-04-19 20:06:06

House-keeping reminders: this is another "find all" problem. So it automatically has two parts: we need to prove a statement of the form "the goal is possible if and only if [answer]".

House-keeping reminders: this is another "find all" problem. So it automatically has two parts: we need to prove a statement of the form "the goal is possible if and only if [answer]".

cj13609517288
2019-04-19 20:06:41

Where do x and y come from here?

Where do x and y come from here?

v_Enhance
2019-04-19 20:07:00

$x$ and $y$ can be any two numbers on the blackboard. So after the first step, you'll have three numbers to choose from, etc.

$x$ and $y$ can be any two numbers on the blackboard. So after the first step, you'll have three numbers to choose from, etc.

v_Enhance
2019-04-19 20:07:27

In the previous problem, we broke it down by considering specific cases like $n = 2$ first. In this problem, I'm going to do the opposite: I'm going to try to talk about a more GENERAL problem, and see where it goes. We won't be able to solve it, but we'll see the right idea come out.

In the previous problem, we broke it down by considering specific cases like $n = 2$ first. In this problem, I'm going to do the opposite: I'm going to try to talk about a more GENERAL problem, and see where it goes. We won't be able to solve it, but we'll see the right idea come out.

v_Enhance
2019-04-19 20:07:41

So I'm going to for now ignore the condition that the two given rational numbers are reciprocals.

So I'm going to for now ignore the condition that the two given rational numbers are reciprocals.

v_Enhance
2019-04-19 20:07:51

Instead, let's suppose the problem started off with two ARBITRARY positive rational numbers $a$ and $b$ on the blackboard, and see what we can say there.

Instead, let's suppose the problem started off with two ARBITRARY positive rational numbers $a$ and $b$ on the blackboard, and see what we can say there.

v_Enhance
2019-04-19 20:07:58

(Zen philosophy comment for experts: this is worth thinking about because there's no reason why the reciprocal of the first number should matter; after some point we get all sorts of numbers on the board anyways.)

(Zen philosophy comment for experts: this is worth thinking about because there's no reason why the reciprocal of the first number should matter; after some point we get all sorts of numbers on the board anyways.)

v_Enhance
2019-04-19 20:08:05

**MO5, FIRST PART: IMPOSSIBILITY.**
v_Enhance
2019-04-19 20:08:09

Okay, well, let me start with a dumb case. What if $a = b$?

Okay, well, let me start with a dumb case. What if $a = b$?

eisirrational
2019-04-19 20:08:40

Only $a$ can be written

Only $a$ can be written

SomethingNeutral
2019-04-19 20:08:40

lost unless a=1=b

lost unless a=1=b

azmath333
2019-04-19 20:08:40

Good luck getting anywhere

Good luck getting anywhere

ketul
2019-04-19 20:08:40

you only get one number

you only get one number

snow_monkey
2019-04-19 20:08:40

same number over and over

same number over and over

v_Enhance
2019-04-19 20:08:43

Yeah, then you can only ever write one number on the board. So that case is very easy.

Yeah, then you can only ever write one number on the board. So that case is very easy.

v_Enhance
2019-04-19 20:08:55

Applying wishful thinking, though, I want something that works like $a = b$.

Applying wishful thinking, though, I want something that works like $a = b$.

v_Enhance
2019-04-19 20:09:05

So here's the key idea: what if we had \[ a \equiv b \pmod p \] instead of the (too-strong) $a=b$?

So here's the key idea: what if we had \[ a \equiv b \pmod p \] instead of the (too-strong) $a=b$?

porkemon2
2019-04-19 20:09:25

every number will always be the same mod p

every number will always be the same mod p

v_Enhance
2019-04-19 20:09:28

Maybe all the numbers would then be the same modulo $p$?

Maybe all the numbers would then be the same modulo $p$?

v_Enhance
2019-04-19 20:09:33

Well, okay, this is already going to cause problems because what does it mean to have a fraction modulo $p$?

Well, okay, this is already going to cause problems because what does it mean to have a fraction modulo $p$?

v_Enhance
2019-04-19 20:09:41

Well, it turns out you can do this: the short version you can remember is: "as long as you don't divide by zero, everything is okay".

Well, it turns out you can do this: the short version you can remember is: "as long as you don't divide by zero, everything is okay".

v_Enhance
2019-04-19 20:09:49

I really could dedicate a whole class to working with inverses modulo $p$, but here's the spark-notes version.

I really could dedicate a whole class to working with inverses modulo $p$, but here's the spark-notes version.

v_Enhance
2019-04-19 20:10:00

$\bullet$ $x \equiv yk \pmod p$, and

$\bullet$ $y \not\equiv 0 \pmod p$.

We don't define $a \pmod p$ if the denominator is divisible by $p$ though.

**Definition**. Let $a = \frac xy$ and let $p$ be a prime. Then we say $a \pmod p$ is congruent to $k$ if$\bullet$ $x \equiv yk \pmod p$, and

$\bullet$ $y \not\equiv 0 \pmod p$.

We don't define $a \pmod p$ if the denominator is divisible by $p$ though.

kevinmathz
2019-04-19 20:10:20

inverse

inverse

hliu1
2019-04-19 20:10:20

x*1/y i guess

x*1/y i guess

Damalone
2019-04-19 20:10:20

inverses

inverses

v_Enhance
2019-04-19 20:10:32

Yep, those of you that know about modular inverses already see where this is going.

Yep, those of you that know about modular inverses already see where this is going.

v_Enhance
2019-04-19 20:11:07

I won't use that word, but here's an example for those of you new to it. What is $\frac{13}{5} \pmod 7$?

I won't use that word, but here's an example for those of you new to it. What is $\frac{13}{5} \pmod 7$?

beiqiang
2019-04-19 20:11:42

13 * 3 = 39 = 4

13 * 3 = 39 = 4

jishu2003
2019-04-19 20:11:42

4

4

Brendanb4321
2019-04-19 20:11:42

4

4

v_Enhance
2019-04-19 20:11:46

We say $\frac{13}{5} \equiv 4 \pmod 7$, since $7 \mid 13 - 4 \cdot 5$. (I'm sweeping under the rug that there's a "unique" answer, but maybe some of you can already see why.)

We say $\frac{13}{5} \equiv 4 \pmod 7$, since $7 \mid 13 - 4 \cdot 5$. (I'm sweeping under the rug that there's a "unique" answer, but maybe some of you can already see why.)

v_Enhance
2019-04-19 20:11:59

OK, so with that definition set, here's our hope.

OK, so with that definition set, here's our hope.

v_Enhance
2019-04-19 20:12:00

**MO5 CONJECTURE 1**: If $a$ and $b$ are both $k \pmod p$, for some prime $p$, then $\frac{a+b}{2}$ is $k \pmod p$ too.
v_Enhance
2019-04-19 20:12:13

Do we think that's true?

Do we think that's true?

Brendanb4321
2019-04-19 20:12:40

iff p>2

iff p>2

mathfun5
2019-04-19 20:12:40

True for all p except 2

True for all p except 2

v_Enhance
2019-04-19 20:12:45

It's ALMOST true. Let's work it out and see what could go wrong.

It's ALMOST true. Let's work it out and see what could go wrong.

v_Enhance
2019-04-19 20:13:04

Suppose $a = \frac{x}{y}$ and $b = \frac{x'}{y'}$ are both $k \pmod p$, meaning $ky-x \equiv ky'-x' \equiv 0 \pmod p$ and $p$ does not divide $y$ or $y'$.

Suppose $a = \frac{x}{y}$ and $b = \frac{x'}{y'}$ are both $k \pmod p$, meaning $ky-x \equiv ky'-x' \equiv 0 \pmod p$ and $p$ does not divide $y$ or $y'$.

v_Enhance
2019-04-19 20:13:08

Can someone compute $\frac{a+b}{2}$ for me?

Can someone compute $\frac{a+b}{2}$ for me?

hliu1
2019-04-19 20:13:38

$\frac{xy'+yx'}{2yy'}$

$\frac{xy'+yx'}{2yy'}$

v_Enhance
2019-04-19 20:13:54

It equals \[ \frac{a+b}{2} = \frac{\frac xy + \frac{x'}{y'}}{2} = \frac{xy'+x'y}{2yy'}. \]

It equals \[ \frac{a+b}{2} = \frac{\frac xy + \frac{x'}{y'}}{2} = \frac{xy'+x'y}{2yy'}. \]

v_Enhance
2019-04-19 20:13:55

Well we wanted $xy' + x'y \equiv k \cdot 2yy' \pmod p$. As $x \equiv ky$ and $x' \equiv ky'$, that part is good.

Well we wanted $xy' + x'y \equiv k \cdot 2yy' \pmod p$. As $x \equiv ky$ and $x' \equiv ky'$, that part is good.

v_Enhance
2019-04-19 20:14:03

But we also required that the denominators were never divisible by $p$. And we have a denominator of $2yy'$.

But we also required that the denominators were never divisible by $p$. And we have a denominator of $2yy'$.

v_Enhance
2019-04-19 20:14:10

So if $p=2$ we are in trouble. But otherwise, we're okay.

So if $p=2$ we are in trouble. But otherwise, we're okay.

porkemon2
2019-04-19 20:14:28

can you do (k+k)/2 instead?

can you do (k+k)/2 instead?

v_Enhance
2019-04-19 20:14:44

Yes, if you have fully set up modular inverses well. I'm working mostly with bare hands here for the sake of people who haven't seen that yet.

Yes, if you have fully set up modular inverses well. I'm working mostly with bare hands here for the sake of people who haven't seen that yet.

v_Enhance
2019-04-19 20:14:55

So the fixed claim is:

So the fixed claim is:

v_Enhance
2019-04-19 20:14:57

**MO5 CLAIM 1**: Let $p$ be an*odd*prime. If $a$ and $b$ are rational numbers both $k \pmod p$, then $\frac{a+b}{2} \equiv k \pmod p$ too.
v_Enhance
2019-04-19 20:15:11

Just don't divide by zero. All good?

Just don't divide by zero. All good?

EricK0504
2019-04-19 20:15:26

yes mr chen

yes mr chen

v_Enhance
2019-04-19 20:15:31

You can just call me Evan.

You can just call me Evan.

beibeizhu
2019-04-19 20:15:38

why would this involve dividing by 0?

why would this involve dividing by 0?

v_Enhance
2019-04-19 20:15:58

When $p=2$, we get $2=0$, so then when we take the average we divide by zero. It's a bit tragic, really...

When $p=2$, we get $2=0$, so then when we take the average we divide by zero. It's a bit tragic, really...

v_Enhance
2019-04-19 20:16:33

Let's also try to check the harmonic mean. What we're hoping is:

Let's also try to check the harmonic mean. What we're hoping is:

v_Enhance
2019-04-19 20:16:36

**MO5 CONJECTURE 2**: If $a$ and $b$ are rational numbers both $k \pmod p$, and $p > 2$, then $\frac{2ab}{a+b}$ is $k \pmod p$ too.
v_Enhance
2019-04-19 20:16:48

This time we've learned to be afraid of $2$ already, so I already took care of that. Do you think that's true now?

This time we've learned to be afraid of $2$ already, so I already took care of that. Do you think that's true now?

azmath333
2019-04-19 20:17:19

a+b can't divide p either

a+b can't divide p either

mathfun5
2019-04-19 20:17:19

If a+b is not 0 mod p

If a+b is not 0 mod p

v_Enhance
2019-04-19 20:17:22

Good eyes.

Good eyes.

v_Enhance
2019-04-19 20:17:36

Again, it's almost true, but not quite.

Again, it's almost true, but not quite.

v_Enhance
2019-04-19 20:17:51

I'll write down the proof for completeness, for the paranoid out there.

I'll write down the proof for completeness, for the paranoid out there.

v_Enhance
2019-04-19 20:17:58

If $a = \frac{x}{y}$ and $b = \frac{x'}{y'}$, one finds \[ \frac{2ab}{a+b} = \frac{2xx'}{xy' + x'y}. \]

If $a = \frac{x}{y}$ and $b = \frac{x'}{y'}$, one finds \[ \frac{2ab}{a+b} = \frac{2xx'}{xy' + x'y}. \]

v_Enhance
2019-04-19 20:18:06

Well, if we want $2xx' \equiv k \cdot (xy' + x'y)$, then again that just follows since $x \equiv ky \pmod p$, $x' \equiv ky' \pmod p$.

Well, if we want $2xx' \equiv k \cdot (xy' + x'y)$, then again that just follows since $x \equiv ky \pmod p$, $x' \equiv ky' \pmod p$.

v_Enhance
2019-04-19 20:18:26

However, we're worried that the denominator $xy' + x'y$ might be divisible by $p$.

However, we're worried that the denominator $xy' + x'y$ might be divisible by $p$.

v_Enhance
2019-04-19 20:18:39

We have \[ xy' + x'y \equiv 2kyy' \pmod p \] and so this happens not only if $p = 2$, but also possibly when $k \equiv 0$.

We have \[ xy' + x'y \equiv 2kyy' \pmod p \] and so this happens not only if $p = 2$, but also possibly when $k \equiv 0$.

v_Enhance
2019-04-19 20:18:54

But otherwise we're okay.

But otherwise we're okay.

v_Enhance
2019-04-19 20:18:57

So here is the updated fix:

So here is the updated fix:

v_Enhance
2019-04-19 20:19:03

**MO5 CLAIM 2**: If $p > 2$ is prime, and $a$ and $b$ are rational numbers both $k \pmod p$, and $k \not\equiv 0 \pmod p$, then $\frac{2ab}{a+b}$ is $k \pmod p$ too.
carzland
2019-04-19 20:19:30

We good now?

We good now?

v_Enhance
2019-04-19 20:19:34

Thankfully.

Thankfully.

v_Enhance
2019-04-19 20:19:49

So here's the abridged version: subject to some worrisome divide-by-zero conditions, we can try to take the initial fractions on our board modulo $p$, and if they happen to both be some nonzero residue $k \pmod p$, then all the numbers on the board will forever be $k \pmod p$ as well.

So here's the abridged version: subject to some worrisome divide-by-zero conditions, we can try to take the initial fractions on our board modulo $p$, and if they happen to both be some nonzero residue $k \pmod p$, then all the numbers on the board will forever be $k \pmod p$ as well.

v_Enhance
2019-04-19 20:19:54

And that gives us a way to try and get an invariant.

And that gives us a way to try and get an invariant.

v_Enhance
2019-04-19 20:20:09

With me so far?

With me so far?

ketul
2019-04-19 20:20:23

Yes, God

Yes, God

v_Enhance
2019-04-19 20:20:29

Again, "Evan" is fine...

Again, "Evan" is fine...

v_Enhance
2019-04-19 20:20:34

Okay, now let's see how many cases this lets us rule out in the original problem where $a = \frac mn$ and $b = \frac nm$.

Okay, now let's see how many cases this lets us rule out in the original problem where $a = \frac mn$ and $b = \frac nm$.

v_Enhance
2019-04-19 20:20:44

Given relatively prime integers $m$ and $n$, for which primes $p$ do we have $\frac mn \equiv \frac nm \pmod p$?

Given relatively prime integers $m$ and $n$, for which primes $p$ do we have $\frac mn \equiv \frac nm \pmod p$?

v_Enhance
2019-04-19 20:20:59

(In terms of $m$ and $n$.)

(In terms of $m$ and $n$.)

mathfun5
2019-04-19 20:21:31

we need a/b = -1 or 1 mod p, so p divides a+b or a-b

we need a/b = -1 or 1 mod p, so p divides a+b or a-b

Damalone
2019-04-19 20:21:31

m^2-n^2 is divisible by p

m^2-n^2 is divisible by p

v_Enhance
2019-04-19 20:21:38

Yeah, by the definition it's just the primes $p \mid m^2-n^2$.

Yeah, by the definition it's just the primes $p \mid m^2-n^2$.

v_Enhance
2019-04-19 20:21:42

I'll point a different conceptual (and more natural) way of thinking about it though: if $x = \frac mn$ we are hoping for \begin{align*} x &\equiv \frac 1x \pmod p \\ \iff x^2 - 1 &\equiv 0 \pmod p \\ \iff x &\equiv 1 \pmod p \quad\text{ or }\quad x \equiv -1 \pmod p. \end{align*}

I'll point a different conceptual (and more natural) way of thinking about it though: if $x = \frac mn$ we are hoping for \begin{align*} x &\equiv \frac 1x \pmod p \\ \iff x^2 - 1 &\equiv 0 \pmod p \\ \iff x &\equiv 1 \pmod p \quad\text{ or }\quad x \equiv -1 \pmod p. \end{align*}

v_Enhance
2019-04-19 20:21:54

Now the target number is $1$.

Now the target number is $1$.

v_Enhance
2019-04-19 20:22:01

So if there's a

So if there's a

*single*prime $p > 2$ such that $\frac mn \equiv -1 \pmod p$, then all numbers on the board will always be $-1 \pmod p$, and we are all good to go.
v_Enhance
2019-04-19 20:22:24

That should narrow down things a lot! What's left at the end of the day?

That should narrow down things a lot! What's left at the end of the day?

mathfun5
2019-04-19 20:22:43

m+n is a power of 2

m+n is a power of 2

cj13609517288
2019-04-19 20:22:43

So they must be powers of $2$. Whoa!!!

So they must be powers of $2$. Whoa!!!

mathguy623
2019-04-19 20:22:58

m+n cant be divisible by any odd prime

m+n cant be divisible by any odd prime

v_Enhance
2019-04-19 20:23:00

We have $\frac mn \equiv -1 \pmod p \iff p \mid m+n$, so unless $m+n$ is a power of $2$, we're actually all set.

We have $\frac mn \equiv -1 \pmod p \iff p \mid m+n$, so unless $m+n$ is a power of $2$, we're actually all set.

v_Enhance
2019-04-19 20:23:33

Taking fractions modulo $p$ is real strong. Keep that in mind in future number theory.

Taking fractions modulo $p$ is real strong. Keep that in mind in future number theory.

porkemon2
2019-04-19 20:24:01

what if p|(x-1)?

what if p|(x-1)?

v_Enhance
2019-04-19 20:24:23

If $x \equiv 1 \pmod p$, then all the numbers on the board will indeed stay $1 \pmod p$ forever. However, that's not the invariant we want, because the target number is also $1$. That's why we switch to $-1$ instead.

If $x \equiv 1 \pmod p$, then all the numbers on the board will indeed stay $1 \pmod p$ forever. However, that's not the invariant we want, because the target number is also $1$. That's why we switch to $-1$ instead.

v_Enhance
2019-04-19 20:24:40

**MO5, SECOND PART: CONSTRUCTION.**
v_Enhance
2019-04-19 20:24:59

We've already ruled out so many cases, we might expect the construction to not be so bad.

We've already ruled out so many cases, we might expect the construction to not be so bad.

v_Enhance
2019-04-19 20:25:06

Let's go back again to having two general numbers $a$ and $b$. Here's again an easier special case: how far can we get from just using the arithmetic mean? (You'll see in a moment why this is related to powers of $2$.)

Let's go back again to having two general numbers $a$ and $b$. Here's again an easier special case: how far can we get from just using the arithmetic mean? (You'll see in a moment why this is related to powers of $2$.)

v_Enhance
2019-04-19 20:25:12

For starters, here is $a$ and $b$ and their average, $\dfrac{a+b}{2}.$

For starters, here is $a$ and $b$ and their average, $\dfrac{a+b}{2}.$

v_Enhance
2019-04-19 20:25:14

v_Enhance
2019-04-19 20:25:25

Now if use our operation again, what numbers could we get?

Now if use our operation again, what numbers could we get?

P_ksAreAllEquivalent
2019-04-19 20:25:56

a+3b/4, 3a+b/4

a+3b/4, 3a+b/4

goodball
2019-04-19 20:25:56

The quarters

The quarters

AwesomeYRY
2019-04-19 20:25:56

3a+b/4 and a+3b/4

3a+b/4 and a+3b/4

carzland
2019-04-19 20:25:56

The midpoints of the two segments

The midpoints of the two segments

v_Enhance
2019-04-19 20:25:59

We can get $\frac{3a+b}{4}$ and $\frac{a+3b}{4}$. So let's just put them all on the board.

We can get $\frac{3a+b}{4}$ and $\frac{a+3b}{4}$. So let's just put them all on the board.

v_Enhance
2019-04-19 20:26:01

v_Enhance
2019-04-19 20:26:06

Okay, where can I get if I try to get more arithmetic means?

Okay, where can I get if I try to get more arithmetic means?

P_ksAreAllEquivalent
2019-04-19 20:26:23

more midpoints!

more midpoints!

goodball
2019-04-19 20:26:23

The eighths

The eighths

mhuangw
2019-04-19 20:26:23

The midpoints of the 4 segments

The midpoints of the 4 segments

v_Enhance
2019-04-19 20:26:36

Yeah, by now you probably see the pattern. We get those fractions with denominator $8$.

Yeah, by now you probably see the pattern. We get those fractions with denominator $8$.

beiqiang
2019-04-19 20:26:55

dyadic rational sections between a and b

dyadic rational sections between a and b

kevinmathz
2019-04-19 20:26:55

you get every $2^k$th index

you get every $2^k$th index

v_Enhance
2019-04-19 20:27:12

Huh, that picture is also obviously wrong. Let's see how fast I can fix asy pictures.

Huh, that picture is also obviously wrong. Let's see how fast I can fix asy pictures.

mhuangw
2019-04-19 20:27:29

6 min later...

6 min later...

v_Enhance
2019-04-19 20:27:32

AMT1954
2019-04-19 20:27:39

there we go

there we go

azmath333
2019-04-19 20:27:51

wow that was fast

wow that was fast

budu
2019-04-19 20:27:51

how r u so good

how r u so good

goodball
2019-04-19 20:27:51

Whoa that was fast!

Whoa that was fast!

v_Enhance
2019-04-19 20:27:54

Lots of practice.

Lots of practice.

v_Enhance
2019-04-19 20:28:08

We can get any "power of $2$" combination: that is, if $u$ and $v$ are fractions with $u+v = 1$ and whose denominator is a power of $2$, then we can get $u \cdot a + v \cdot b = 1$.

We can get any "power of $2$" combination: that is, if $u$ and $v$ are fractions with $u+v = 1$ and whose denominator is a power of $2$, then we can get $u \cdot a + v \cdot b = 1$.

v_Enhance
2019-04-19 20:28:13

So now let's go back to the original problem and see how many numbers we can get out of it. Our initial numbers were $\frac mn$ and $\frac nm$. What are the weights $u$ and $v$ for which we can get our target \[ u \cdot \frac mn + v \cdot \frac nm = 1 \] and $u+v=1$?

So now let's go back to the original problem and see how many numbers we can get out of it. Our initial numbers were $\frac mn$ and $\frac nm$. What are the weights $u$ and $v$ for which we can get our target \[ u \cdot \frac mn + v \cdot \frac nm = 1 \] and $u+v=1$?

v_Enhance
2019-04-19 20:28:49

(So, I'm just asking you to solve the system of equations for $u$ and $v$.)

(So, I'm just asking you to solve the system of equations for $u$ and $v$.)

P_ksAreAllEquivalent
2019-04-19 20:29:32

n/(m+n), m/(m+n)

n/(m+n), m/(m+n)

mathfun5
2019-04-19 20:29:32

u=n/(m+n), v=m/(m+n) which works because m+n is a power of 2

u=n/(m+n), v=m/(m+n) which works because m+n is a power of 2

v_Enhance
2019-04-19 20:29:46

Yeah, it's $u = \frac{n}{m+n}$ and $v = \frac{m}{m+n}$, which has powers of $2$ in the denominator by hypothesis!

Yeah, it's $u = \frac{n}{m+n}$ and $v = \frac{m}{m+n}$, which has powers of $2$ in the denominator by hypothesis!

P_ksAreAllEquivalent
2019-04-19 20:30:06

we win

we win

v_Enhance
2019-04-19 20:30:14

Yep. That means our dyadic construction in the picture just works out of the box. No need to even use harmonic mean.

Yep. That means our dyadic construction in the picture just works out of the box. No need to even use harmonic mean.

v_Enhance
2019-04-19 20:30:20

That is game, set, match. The answer is that the goal is possible if $m+n$ is a power of $2$.

That is game, set, match. The answer is that the goal is possible if $m+n$ is a power of $2$.

SpecialBeing2017
2019-04-19 20:30:34

This is brilliant...

This is brilliant...

v_Enhance
2019-04-19 20:30:41

Phew. That was a long explanation. Hopefully some of you learned a bit of new number theory in doing this: fractions work modulo $p$ as long as you don't divide by zero. Super useful fact, so remember that if you ever run into integer equations that have fractions in them.

Phew. That was a long explanation. Hopefully some of you learned a bit of new number theory in doing this: fractions work modulo $p$ as long as you don't divide by zero. Super useful fact, so remember that if you ever run into integer equations that have fractions in them.

chesstoastmaster
2019-04-19 20:30:56

cool math

cool math

v_Enhance
2019-04-19 20:30:57

And for those of you that do know this already, hopefully you learned some more zen philosophy. In my opinion, the most unnatural thing about the entire problem was why the given involved two reciprocal fractions $\frac mn$ and $\frac nm$, when they really could have been any numbers $a$ and $b$.

And for those of you that do know this already, hopefully you learned some more zen philosophy. In my opinion, the most unnatural thing about the entire problem was why the given involved two reciprocal fractions $\frac mn$ and $\frac nm$, when they really could have been any numbers $a$ and $b$.

v_Enhance
2019-04-19 20:31:08

That's why so much of my motivation was of the form "let's think about the general situation with $a$ and $b$": because there is no clear reason that $ab = 1$ is related to either operation that's happening.

That's why so much of my motivation was of the form "let's think about the general situation with $a$ and $b$": because there is no clear reason that $ab = 1$ is related to either operation that's happening.

v_Enhance
2019-04-19 20:31:26

(Of course, I'm also telling an idealized version of the story of how I came up with this solution, since I don't have time during the Math Jam to lead you down wrong paths too far. There are other promising-looking approaches that fail, like for example trying to look at size of the numbers, that don't pan out.)

(Of course, I'm also telling an idealized version of the story of how I came up with this solution, since I don't have time during the Math Jam to lead you down wrong paths too far. There are other promising-looking approaches that fail, like for example trying to look at size of the numbers, that don't pan out.)

bluelinfish
2019-04-19 20:31:31

time for JMO 4

time for JMO 4

v_Enhance
2019-04-19 20:31:42

Yep.

Yep.

v_Enhance
2019-04-19 20:31:47

**JMO4.**Let $ABC$ be a triangle with $\angle ABC$ obtuse. The $A$-excircle is a circle in the exterior of $\triangle ABC$ that is tangent to side $BC$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respectively. Can line $EF$ be tangent to the $A$-excircle?
v_Enhance
2019-04-19 20:31:57

Yes/no geometry. Been a while since that's happened, huh?

Yes/no geometry. Been a while since that's happened, huh?

v_Enhance
2019-04-19 20:32:01

Well, how do we start every geometry problem?

Well, how do we start every geometry problem?

Math4Fun04
2019-04-19 20:32:11

diagram

diagram

carzland
2019-04-19 20:32:11

A wild geometry appeared

A wild geometry appeared

goodball
2019-04-19 20:32:11

Geometry= draw a diagram

Geometry= draw a diagram

Brendanb4321
2019-04-19 20:32:11

diagram

diagram

P_ksAreAllEquivalent
2019-04-19 20:32:11

diagram

diagram

v_Enhance
2019-04-19 20:32:20

Yep.

Yep.

Charmander3333
2019-04-19 20:32:23

coordinate bash

coordinate bash

v_Enhance
2019-04-19 20:32:25

v_Enhance
2019-04-19 20:32:31

So anyways here's a diagram.

So anyways here's a diagram.

v_Enhance
2019-04-19 20:32:43

I think that just about settles it. Ready to move on?

I think that just about settles it. Ready to move on?

P_ksAreAllEquivalent
2019-04-19 20:33:14

wow so accurate!

wow so accurate!

Brendanb4321
2019-04-19 20:33:14

no

no

molocyxu
2019-04-19 20:33:14

No but yes

No but yes

v_Enhance
2019-04-19 20:33:21

Of course that's not a proof. (And the figure is actually distorted slightly: otherwise $A$ would be way off-screen.)

Of course that's not a proof. (And the figure is actually distorted slightly: otherwise $A$ would be way off-screen.)

Eyed
2019-04-19 20:34:15

Wait aren't F and E switched?

Wait aren't F and E switched?

v_Enhance
2019-04-19 20:34:18

v_Enhance
2019-04-19 20:34:28

Didn't see nothing there. ;)

Didn't see nothing there. ;)

deadwindow
2019-04-19 20:34:35

all hail edit god

all hail edit god

molocyxu
2019-04-19 20:34:58

CALL HIM EVAN

CALL HIM EVAN

v_Enhance
2019-04-19 20:35:01

So let's start with the geometry...

So let's start with the geometry...

v_Enhance
2019-04-19 20:35:13

First of all, what do people know about quadrilateral $BECF$, even without the excircle involved?

First of all, what do people know about quadrilateral $BECF$, even without the excircle involved?

snow_monkey
2019-04-19 20:35:39

cyclic

cyclic

Magnitude
2019-04-19 20:35:39

concyclic

concyclic

epicjing333
2019-04-19 20:35:39

cyclic

cyclic

ketul
2019-04-19 20:35:39

Cyclic

Cyclic

porkemon2
2019-04-19 20:35:39

its cyclic with opposite right angles

its cyclic with opposite right angles

v_Enhance
2019-04-19 20:35:42

The quadrilateral $BFCE$ is cyclic, inscribed in a circle with diameter $BC$.

The quadrilateral $BFCE$ is cyclic, inscribed in a circle with diameter $BC$.

v_Enhance
2019-04-19 20:35:56

Now you can think of the excircle as follows: it's actually a circle tangent to lines $BC$, $EF$, $CF$, $BE$.

Now you can think of the excircle as follows: it's actually a circle tangent to lines $BC$, $EF$, $CF$, $BE$.

v_Enhance
2019-04-19 20:36:08

That is, "self-intersecting quadrilateral $BCFE$ has an excircle". So we actually don't have to think about $A$ in the problem.

That is, "self-intersecting quadrilateral $BCFE$ has an excircle". So we actually don't have to think about $A$ in the problem.

AlexLikeMath
2019-04-19 20:36:17

How is it cyclic?

How is it cyclic?

v_Enhance
2019-04-19 20:36:49

We have $\angle E = \angle F = 90^{\circ}$, so the opposite angles sum to $90^{\circ}$.

We have $\angle E = \angle F = 90^{\circ}$, so the opposite angles sum to $90^{\circ}$.

v_Enhance
2019-04-19 20:36:55

Now I realize most of you probably aren't so comfortable with self-intersecting quadrilaterals, so let's talk about a case you're more used to. Do you know anything about normal convex quadrilaterals that have incircles?

Now I realize most of you probably aren't so comfortable with self-intersecting quadrilaterals, so let's talk about a case you're more used to. Do you know anything about normal convex quadrilaterals that have incircles?

The_Turtle
2019-04-19 20:37:09

sum to 180*

sum to 180*

cj13609517288
2019-04-19 20:37:17

Um it is supposed to be 180 degrees.

Um it is supposed to be 180 degrees.

v_Enhance
2019-04-19 20:37:20

Man I'm really slipping up today.

Man I'm really slipping up today.

v_Enhance
2019-04-19 20:37:37

But anyways, as far as incircles go...

But anyways, as far as incircles go...

annoynymous
2019-04-19 20:37:49

Pitot

Pitot

mathfun5
2019-04-19 20:37:49

Pitots theorem: sum of opposite sides is equal

Pitots theorem: sum of opposite sides is equal

jj_ca888
2019-04-19 20:37:49

pitots => AB + CD = BC + AD

pitots => AB + CD = BC + AD

Harry0531
2019-04-19 20:37:49

Opposite sides add up to be equal in quadrilaterals with incircles

Opposite sides add up to be equal in quadrilaterals with incircles

Kagebaka
2019-04-19 20:37:49

pithot --> sum of opposite sides are equal

pithot --> sum of opposite sides are equal

franchester
2019-04-19 20:37:49

Pitot’s theorem

Pitot’s theorem

v_Enhance
2019-04-19 20:37:57

The so-called Pitot theorem states that in a normal convex quadrilateral with an incircle, the opposite sides have equal length. And you can see why from the picture below.

The so-called Pitot theorem states that in a normal convex quadrilateral with an incircle, the opposite sides have equal length. And you can see why from the picture below.

v_Enhance
2019-04-19 20:38:02

v_Enhance
2019-04-19 20:38:24

So now I claim you can say something similar here. Anyone able to work out what side relations we get?

So now I claim you can say something similar here. Anyone able to work out what side relations we get?

v_Enhance
2019-04-19 20:38:43

(For our quadrilateral $BECF$.)

(For our quadrilateral $BECF$.)

Harry0531
2019-04-19 20:39:03

I heard that proving the converse of Pitot is way harder.

I heard that proving the converse of Pitot is way harder.

v_Enhance
2019-04-19 20:39:05

It is indeed more involved. However, we won't need the converse tonight, so I won't use it, but it's good to know.

It is indeed more involved. However, we won't need the converse tonight, so I won't use it, but it's good to know.

cj13609517288
2019-04-19 20:39:09

EB+CF=EC+BF

EB+CF=EC+BF

P_ksAreAllEquivalent
2019-04-19 20:39:09

EF + FB = BC + EC?

EF + FB = BC + EC?

Awesome_guy
2019-04-19 20:39:09

EB+CF=CE+BF

EB+CF=CE+BF

Ashrafuzzaman_Nafees
2019-04-19 20:39:09

EC + BF = EB + CF

EC + BF = EB + CF

P_ksAreAllEquivalent
2019-04-19 20:40:18

isn't the above for incircle?

isn't the above for incircle?

v_Enhance
2019-04-19 20:40:26

Oh geez, I think I flipped all the $E$'s and $F$'s in my notes.

Oh geez, I think I flipped all the $E$'s and $F$'s in my notes.

v_Enhance
2019-04-19 20:41:07

Well, let's just run with it, and hope I work out the right formula at the end

Well, let's just run with it, and hope I work out the right formula at the end

v_Enhance
2019-04-19 20:41:25

Let's prove this. Let $t(B)$ denote the length of the tangent from $B$ to the $A$-excircle, and define $t(C)$, $t(E)$, $t(F)$ in the same way.

Let's prove this. Let $t(B)$ denote the length of the tangent from $B$ to the $A$-excircle, and define $t(C)$, $t(E)$, $t(F)$ in the same way.

v_Enhance
2019-04-19 20:41:34

Can someone write $BF$ in terms of these tangent lengths for me?

Can someone write $BF$ in terms of these tangent lengths for me?

P_ksAreAllEquivalent
2019-04-19 20:41:38

also what if there are config issues , could it change depending on config

also what if there are config issues , could it change depending on config

v_Enhance
2019-04-19 20:41:55

Thankfully, the problem tells you $\angle B > 90^{\circ}$, so most of thm go away.

Thankfully, the problem tells you $\angle B > 90^{\circ}$, so most of thm go away.

ketul
2019-04-19 20:42:06

$BF=t(b)-t(f)$

$BF=t(b)-t(f)$

hliu1
2019-04-19 20:42:06

tB-tF

tB-tF

v_Enhance
2019-04-19 20:42:23

We have $BF = t(B) - t(F)$.

We have $BF = t(B) - t(F)$.

v_Enhance
2019-04-19 20:42:44

Now what about $EF$? In the picture, I claim $EF$ is the sum of two tangent lengths. Which?

Now what about $EF$? In the picture, I claim $EF$ is the sum of two tangent lengths. Which?

ketul
2019-04-19 20:43:11

$EF = t(E)+t(F)$

$EF = t(E)+t(F)$

Magnitude
2019-04-19 20:43:11

t(E) + t(f)

t(E) + t(f)

snow_monkey
2019-04-19 20:43:11

t(E) + t(F)

t(E) + t(F)

v_Enhance
2019-04-19 20:43:20

Great.

Great.

ketul
2019-04-19 20:43:24

$BC=t(B)+t(C)$

$BC=t(B)+t(C)$

v_Enhance
2019-04-19 20:43:27

And there's $BC$.

And there's $BC$.

ketul
2019-04-19 20:43:33

$EC=t(e)-t(c)$

$EC=t(e)-t(c)$

v_Enhance
2019-04-19 20:43:42

And there's $CE$. We have $BC = t(B) + t(C)$ and $EC = t(E) - t(C)$.

And there's $CE$. We have $BC = t(B) + t(C)$ and $EC = t(E) - t(C)$.

v_Enhance
2019-04-19 20:43:59

So if I put these four equations together, what length relation do I get?

So if I put these four equations together, what length relation do I get?

v_Enhance
2019-04-19 20:44:16

i.e. what Pitot-like theorem holds for our self-intersecting quadrilateral?

i.e. what Pitot-like theorem holds for our self-intersecting quadrilateral?

hliu1
2019-04-19 20:44:19

doesn't EF=t(E)+t(F) assume that EF is tangent to the A-excircle?

doesn't EF=t(E)+t(F) assume that EF is tangent to the A-excircle?

v_Enhance
2019-04-19 20:44:40

Yes, sorry. I'm assuming $EF$ is tangent to the $A$-excircle, and then seeing what comes out of it.

Yes, sorry. I'm assuming $EF$ is tangent to the $A$-excircle, and then seeing what comes out of it.

mathfun5
2019-04-19 20:44:52

BF+EF=BC+EC

BF+EF=BC+EC

snow_monkey
2019-04-19 20:44:52

BC + EC = BF + EF

BC + EC = BF + EF

v_Enhance
2019-04-19 20:45:24

We get \[ BC + EC = BF + EF \] as the $t$ terms now all line up: both sides are equal to $t(B) + t(E)$.

We get \[ BC + EC = BF + EF \] as the $t$ terms now all line up: both sides are equal to $t(B) + t(E)$.

cj13609517288
2019-04-19 20:45:42

Ptolemy?

Ptolemy?

v_Enhance
2019-04-19 20:45:58

Good idea. Let's try it!

Good idea. Let's try it!

awesomemaths
2019-04-19 20:46:29

on wwhich quadrilateral??

on wwhich quadrilateral??

v_Enhance
2019-04-19 20:46:37

We will apply Ptolemy on $BECF$, our cyclic quadrilateral.

We will apply Ptolemy on $BECF$, our cyclic quadrilateral.

v_Enhance
2019-04-19 20:46:50

Let me scale so that $BC = 1$, and then let $x = BF$, $y = CE$.

Let me scale so that $BC = 1$, and then let $x = BF$, $y = CE$.

v_Enhance
2019-04-19 20:47:01

First of all, I need the other two sides. Can someone tell me $BE$ and $CF$ in terms of $x$ and $y$?

First of all, I need the other two sides. Can someone tell me $BE$ and $CF$ in terms of $x$ and $y$?

v_Enhance
2019-04-19 20:47:38

(Remember: $\angle E$ and $\angle F$ are right angles, and $BC=1$ is a hypotenuse!)

(Remember: $\angle E$ and $\angle F$ are right angles, and $BC=1$ is a hypotenuse!)

Awesome_guy
2019-04-19 20:48:13

sqrt(1-x^2)

sqrt(1-x^2)

hliu1
2019-04-19 20:48:13

sqrt(1-x^2) and sqrt(1-y^2)

sqrt(1-x^2) and sqrt(1-y^2)

porkemon2
2019-04-19 20:48:13

BE = sqrt(1-y^2), CF = sqrt(1-x^2)

BE = sqrt(1-y^2), CF = sqrt(1-x^2)

rjiangbz
2019-04-19 20:48:13

sqrt(1-x^2)

sqrt(1-x^2)

Eyed
2019-04-19 20:48:13

$\sqrt{1-x^{2}}$, $\sqrt{1-y^{2}}$

$\sqrt{1-x^{2}}$, $\sqrt{1-y^{2}}$

v_Enhance
2019-04-19 20:48:31

By the Pythagorean theorem, we have $BE = \sqrt{1-y^2}$, $CF = \sqrt{1-x^2}$.

By the Pythagorean theorem, we have $BE = \sqrt{1-y^2}$, $CF = \sqrt{1-x^2}$.

cj13609517288
2019-04-19 20:48:57

Let's do some ultimate bashing now!

Let's do some ultimate bashing now!

v_Enhance
2019-04-19 20:49:05

It'll be a bit cleaner than that.

It'll be a bit cleaner than that.

v_Enhance
2019-04-19 20:49:08

So Ptolemy's theorem tells us that

So Ptolemy's theorem tells us that

v_Enhance
2019-04-19 20:49:21

\[ BE \cdot CF + BF \cdot CE = BC \cdot EF \]

\[ BE \cdot CF + BF \cdot CE = BC \cdot EF \]

v_Enhance
2019-04-19 20:49:41

We know everything on the left-hand side, so the last term not in terms of $x$ and $y$ is that $EF$.

We know everything on the left-hand side, so the last term not in terms of $x$ and $y$ is that $EF$.

v_Enhance
2019-04-19 20:49:58

But since we know $BC+EC = BF+EF$, can we write $EF$ in terms of $x$ and $y$?

But since we know $BC+EC = BF+EF$, can we write $EF$ in terms of $x$ and $y$?

snow_monkey
2019-04-19 20:50:15

1+y-x

1+y-x

v_Enhance
2019-04-19 20:50:31

The equation becomes $EF = 1+y-x$.

The equation becomes $EF = 1+y-x$.

v_Enhance
2019-04-19 20:50:34

And so now Ptolemy's theorem says that

And so now Ptolemy's theorem says that

v_Enhance
2019-04-19 20:50:56

\[ \sqrt{1-y^2} \cdot \sqrt{1-x^2} + xy = 1 \cdot (1+y-x). \]

\[ \sqrt{1-y^2} \cdot \sqrt{1-x^2} + xy = 1 \cdot (1+y-x). \]

v_Enhance
2019-04-19 20:51:26

A bit messy. Anyone want to clean it up for me?

A bit messy. Anyone want to clean it up for me?

molocyxu
2019-04-19 20:52:14

WAIT

WAIT

molocyxu
2019-04-19 20:52:14

move xy to the right to get (1+y)(1-x)

move xy to the right to get (1+y)(1-x)

beiqiang
2019-04-19 20:52:14

move xy, factor, square, cancel?

move xy, factor, square, cancel?

Generic_Username
2019-04-19 20:52:14

$(1-y^2)(1-x^2)=((1+y)(1-x))^2$

$(1-y^2)(1-x^2)=((1+y)(1-x))^2$

mathguy623
2019-04-19 20:52:14

$\sqrt{1-y^2} \cdot \sqrt{1-x^2} = (1+y)(1-x)$

$\sqrt{1-y^2} \cdot \sqrt{1-x^2} = (1+y)(1-x)$

v_Enhance
2019-04-19 20:52:27

This rearranges to give \[ (1+y)(1-x) = \sqrt{(1-x^2)(1-y^2)} \implies \frac{1+y}{1-y} = \frac{1+x}{1-x}. \]

This rearranges to give \[ (1+y)(1-x) = \sqrt{(1-x^2)(1-y^2)} \implies \frac{1+y}{1-y} = \frac{1+x}{1-x}. \]

v_Enhance
2019-04-19 20:52:41

So... what do you think that gives?

So... what do you think that gives?

awesomemaths
2019-04-19 20:52:59

X = Y

X = Y

awesomemaths
2019-04-19 20:52:59

x = y

x = y

molocyxu
2019-04-19 20:52:59

um x=y

um x=y

v_Enhance
2019-04-19 20:53:03

And that implies $x=y$.

And that implies $x=y$.

v_Enhance
2019-04-19 20:53:10

So actually, what can we say about $BFCE$?

So actually, what can we say about $BFCE$?

Awesome_guy
2019-04-19 20:53:19

its a rectangle

its a rectangle

molocyxu
2019-04-19 20:53:25

its a rectangle

its a rectangle

v_Enhance
2019-04-19 20:53:30

It means $BFCE$ is a rectangle.

It means $BFCE$ is a rectangle.

v_Enhance
2019-04-19 20:53:32

cj13609517288
2019-04-19 20:53:43

AMAZING

AMAZING

awesomemaths
2019-04-19 20:53:43

this whole time

this whole time

v_Enhance
2019-04-19 20:53:56

And the means the answer is no. Because we have run into a contradiction: $BE$ and $CF$ are parallel, so they cannot intersect at any point $A$.

And the means the answer is no. Because we have run into a contradiction: $BE$ and $CF$ are parallel, so they cannot intersect at any point $A$.

v_Enhance
2019-04-19 20:54:19

Summary: we assumed $EF$ was tangent, wrote down the Pitot theorem, and concluded that $BECF$ is a rectangle. But that causes lines $BF$ and $CE$ to not intersect.

Summary: we assumed $EF$ was tangent, wrote down the Pitot theorem, and concluded that $BECF$ is a rectangle. But that causes lines $BF$ and $CE$ to not intersect.

v_Enhance
2019-04-19 20:54:26

Thus the point $A$ we thought was useless comes back to give a contradiction at the last moment.

Thus the point $A$ we thought was useless comes back to give a contradiction at the last moment.

cocohearts
2019-04-19 20:55:11

encore

encore

jamesyhr
2019-04-19 20:55:11

wow

wow

rjiangbz
2019-04-19 20:55:11

Woohoo for point A! You don't exist, point A!

Woohoo for point A! You don't exist, point A!

v_Enhance
2019-04-19 20:55:32

Incidentally, this is not the shortest solution. I chose it because I wanted to showcase the Pitot theorem, and sort of do it without using too many ideas.

Incidentally, this is not the shortest solution. I chose it because I wanted to showcase the Pitot theorem, and sort of do it without using too many ideas.

v_Enhance
2019-04-19 20:55:33

But:

But:

jeffisepic
2019-04-19 20:55:37

I have a one liner: If ABC and AEF can't have the same excircle because AEF is similar but not congruent to ABC

I have a one liner: If ABC and AEF can't have the same excircle because AEF is similar but not congruent to ABC

budu
2019-04-19 20:55:43

you can also use AEF ~ ABC to get a nice solution

you can also use AEF ~ ABC to get a nice solution

alifenix-
2019-04-19 20:55:50

reflection over angle bisector!

reflection over angle bisector!

v_Enhance
2019-04-19 20:56:27

I'll let anyone who wants to work out the details.

I'll let anyone who wants to work out the details.

mhuangw
2019-04-19 20:57:55

how many questions left?

how many questions left?

Jupiter123
2019-04-19 20:57:55

To be exact, we have less than 30 minutes.

To be exact, we have less than 30 minutes.

v_Enhance
2019-04-19 20:57:58

Alright, I only have half an hour left, so I'll probably skip to the MO2 geometry now --- as the MO1 functional equation is really very involved.

Alright, I only have half an hour left, so I'll probably skip to the MO2 geometry now --- as the MO1 functional equation is really very involved.

v_Enhance
2019-04-19 20:58:06

Sorry for any algebra fans.

Sorry for any algebra fans.

v_Enhance
2019-04-19 20:58:20

**JMO3/MO2.**Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$. Show that line $PE$ bisects $\overline{CD}$.
v_Enhance
2019-04-19 20:58:37

Well, okay, how do we start every geometry problem?

Well, okay, how do we start every geometry problem?

Jupiter123
2019-04-19 20:58:53

DRAW

DRAW

cooljoseph
2019-04-19 20:58:53

We draw a diagram

We draw a diagram

rjiangbz
2019-04-19 20:58:53

draw a piccher!

draw a piccher!

Deegs
2019-04-19 20:58:53

DIAGRAM

DIAGRAM

hliu1
2019-04-19 20:58:53

then draw a diagram

then draw a diagram

v_Enhance
2019-04-19 20:58:56

We begin by drawing a diagram. Let me say a bit this time about how you might do it.

We begin by drawing a diagram. Let me say a bit this time about how you might do it.

v_Enhance
2019-04-19 20:59:03

Because of the condition $AD^2 + BC^2 = AB^2$, you'll have to improvise a bit. So maybe in a contest, I would draw some big enough circle, then draw a chord of length $5$, then create chords of length $3$ and $4$ to get $ABCD$.

Because of the condition $AD^2 + BC^2 = AB^2$, you'll have to improvise a bit. So maybe in a contest, I would draw some big enough circle, then draw a chord of length $5$, then create chords of length $3$ and $4$ to get $ABCD$.

v_Enhance
2019-04-19 20:59:12

As for $P$, the condition $\angle DAP = \angle BPC$ is hard to work with. So what might we do instead?

As for $P$, the condition $\angle DAP = \angle BPC$ is hard to work with. So what might we do instead?

alifenix-
2019-04-19 20:59:35

look at other condition and phantom point

look at other condition and phantom point

cooljoseph
2019-04-19 20:59:42

Redefine P

Redefine P

v_Enhance
2019-04-19 20:59:45

Well, we can construct $E$, and the midpoint of $CD$. And we allegedly think $P$ is the intersection of $AB$ and $ME$. So that gives us a way to draw it without the angle conditions.

Well, we can construct $E$, and the midpoint of $CD$. And we allegedly think $P$ is the intersection of $AB$ and $ME$. So that gives us a way to draw it without the angle conditions.

v_Enhance
2019-04-19 20:59:52

Okay, with that said, here's the diagram I promised you all.

Okay, with that said, here's the diagram I promised you all.

v_Enhance
2019-04-19 20:59:55

awesomemaths
2019-04-19 21:00:14

wow thats an accurate diagram

wow thats an accurate diagram

v_Enhance
2019-04-19 21:00:16

Thanks.

Thanks.

v_Enhance
2019-04-19 21:00:28

But again I want to stress, if you set it up the right way, you can get an accurate one during the test with ruler and compass. No eyeball needed.

But again I want to stress, if you set it up the right way, you can get an accurate one during the test with ruler and compass. No eyeball needed.

v_Enhance
2019-04-19 21:00:49

*eyeballing. You do need eyes.

*eyeballing. You do need eyes.

v_Enhance
2019-04-19 21:00:54

Anyways that condition $AD^2 + BC^2 = AB^2$ is really weird. What are we going to do with it? (The condition $\angle DAP = \angle BPC$ is also not great.)

Anyways that condition $AD^2 + BC^2 = AB^2$ is really weird. What are we going to do with it? (The condition $\angle DAP = \angle BPC$ is also not great.)

v_Enhance
2019-04-19 21:01:02

Well, does $AD^2 + BC^2 = AB^2$ remind people of anything?

Well, does $AD^2 + BC^2 = AB^2$ remind people of anything?

awesomemaths
2019-04-19 21:01:29

pytahgeroen theorem

pytahgeroen theorem

AwesomeYRY
2019-04-19 21:01:29

Pythagorean theorem

Pythagorean theorem

kevinmathz
2019-04-19 21:01:29

pythagorean theorem

pythagorean theorem

jeffisepic
2019-04-19 21:01:29

pythagorean theorem

pythagorean theorem

mathfun5
2019-04-19 21:01:29

Pythag!

Pythag!

v_Enhance
2019-04-19 21:01:53

Yeah, it sure

Yeah, it sure

*looks*like Pythag, sort of.
v_Enhance
2019-04-19 21:01:59

So here's what I'm going to do: construct right triangles with hypotenuse $AB$ and legs of length $AD$ and $BC$.

So here's what I'm going to do: construct right triangles with hypotenuse $AB$ and legs of length $AD$ and $BC$.

v_Enhance
2019-04-19 21:02:03

If I had a compass, how would I do that in the picture I gave you?

If I had a compass, how would I do that in the picture I gave you?

budu
2019-04-19 21:02:35

The length condition motivates the construction of the intersection of the circles centered at A through D and centered at B through C (let the point be F), as AFB is a right triangle by pythagorean theorem (and this also helps to construct the diagram)

The length condition motivates the construction of the intersection of the circles centered at A through D and centered at B through C (let the point be F), as AFB is a right triangle by pythagorean theorem (and this also helps to construct the diagram)

mathfun5
2019-04-19 21:02:35

Circle centered at A with radius AD and likewise for B

Circle centered at A with radius AD and likewise for B

Awesome_guy
2019-04-19 21:02:35

circle with center A and B

circle with center A and B

alifenix-
2019-04-19 21:02:45

draw a circle centered at A through D, one at B through C, intersect

draw a circle centered at A through D, one at B through C, intersect

v_Enhance
2019-04-19 21:02:49

Let's take the circle centered at $A$ with radius $AD$, and intersect it with the circle centered at $B$ with radius $BC$.

Let's take the circle centered at $A$ with radius $AD$, and intersect it with the circle centered at $B$ with radius $BC$.

v_Enhance
2019-04-19 21:03:02

(If you followed my advice for drawing the initial diagram, you would even have these circles already.)

(If you followed my advice for drawing the initial diagram, you would even have these circles already.)

v_Enhance
2019-04-19 21:03:04

They'll meet at two points $X$ and $Y$. Here's the picture.

They'll meet at two points $X$ and $Y$. Here's the picture.

v_Enhance
2019-04-19 21:03:05

v_Enhance
2019-04-19 21:03:29

(You can pop out this diagram by double-clicking it, if it's distorted. Large diagram.)

(You can pop out this diagram by double-clicking it, if it's distorted. Large diagram.)

v_Enhance
2019-04-19 21:03:50

Now here's the hardest part of the problem: in this diagram, do you see anything that looks like it might be true, that you didn't expect before?

Now here's the hardest part of the problem: in this diagram, do you see anything that looks like it might be true, that you didn't expect before?

jeffisepic
2019-04-19 21:05:10

X,P,Y are collinear?

X,P,Y are collinear?

mira74
2019-04-19 21:05:10

P is on XY?

P is on XY?

cooljoseph
2019-04-19 21:05:10

They intersect on the same places as the intersections of DP and AC and the lines PC and BD! In fact, when these are reflected over the line AB they also hit the circle!

They intersect on the same places as the intersections of DP and AC and the lines PC and BD! In fact, when these are reflected over the line AB they also hit the circle!

budu
2019-04-19 21:05:10

XP perpendicular to AB

XP perpendicular to AB

alifenix-
2019-04-19 21:05:10

X, Y, P collinear....?

X, Y, P collinear....?

mathfun5
2019-04-19 21:05:10

Intersection of DB and CP lies on the circle

Intersection of DB and CP lies on the circle

porkemon2
2019-04-19 21:05:18

AC, PD (CXY) looks like it coincides

AC, PD (CXY) looks like it coincides

khina
2019-04-19 21:05:18

xp is perpendicular to ab

xp is perpendicular to ab

v_Enhance
2019-04-19 21:05:24

Yep, I see two big ones:

Yep, I see two big ones:

v_Enhance
2019-04-19 21:05:25

First, it looks like lines $BD$ and $CP$ meet on the red circle. Symmetrically, lines $AC$ and $DP$ meet on the orange circle.

First, it looks like lines $BD$ and $CP$ meet on the red circle. Symmetrically, lines $AC$ and $DP$ meet on the orange circle.

v_Enhance
2019-04-19 21:05:29

Second, $P$ looks like the midpoint of $XY$.

Second, $P$ looks like the midpoint of $XY$.

v_Enhance
2019-04-19 21:05:36

Let's mark these all in.

Let's mark these all in.

v_Enhance
2019-04-19 21:05:37

v_Enhance
2019-04-19 21:05:54

Let's think about the point $P$ first, since it's the main player of this problem.

Let's think about the point $P$ first, since it's the main player of this problem.

v_Enhance
2019-04-19 21:06:16

We haven't proven any of our conjectures yet, so to emphasize this, I'm going to mark everything that follows as "wishful thinking."

We haven't proven any of our conjectures yet, so to emphasize this, I'm going to mark everything that follows as "wishful thinking."

v_Enhance
2019-04-19 21:06:17

*BEGIN DREAM MODE*
v_Enhance
2019-04-19 21:06:24

If $P$ really were the foot from $X$ to $AB$, then if we consider right triangle $AXB$, can you get some similarities?

If $P$ really were the foot from $X$ to $AB$, then if we consider right triangle $AXB$, can you get some similarities?

mhuangw
2019-04-19 21:06:57

move it around

move it around

mathfun5
2019-04-19 21:06:57

AP*AB=AX^2

AP*AB=AX^2

mira74
2019-04-19 21:06:57

APX similar to XPB

APX similar to XPB

alifenix-
2019-04-19 21:06:57

AXP sim XBP sim ABX..?

AXP sim XBP sim ABX..?

hliu1
2019-04-19 21:06:57

APX and BPX

APX and BPX

v_Enhance
2019-04-19 21:07:06

We would have $\triangle XAP \sim \triangle AXB$, among many other pairs.

We would have $\triangle XAP \sim \triangle AXB$, among many other pairs.

v_Enhance
2019-04-19 21:07:11

The corollary I want to draw your attention to is then

\[ AD^2 = AX^2 = AP \cdot AB. \]

The corollary I want to draw your attention to is then

\[ AD^2 = AX^2 = AP \cdot AB. \]

v_Enhance
2019-04-19 21:07:19

Which triangles does that imply are similar?

Which triangles does that imply are similar?

mathfun5
2019-04-19 21:07:44

So APD and ADB are similar!

So APD and ADB are similar!

budu
2019-04-19 21:07:44

APD ~ ADB

APD ~ ADB

alifenix-
2019-04-19 21:07:44

ADP and ABD

ADP and ABD

v_Enhance
2019-04-19 21:07:49

It means $\triangle ADP \sim \triangle ABD$.

It means $\triangle ADP \sim \triangle ABD$.

v_Enhance
2019-04-19 21:08:01

Similarly, can we make a statement about $BC^2$?

Similarly, can we make a statement about $BC^2$?

budu
2019-04-19 21:08:10

similarly BPC ~ BCA

similarly BPC ~ BCA

v_Enhance
2019-04-19 21:08:16

By similar logic,

\[ BC^2 = BX^2 = BP \cdot BA. \]

So $\triangle BCP \sim \triangle BAC$.

By similar logic,

\[ BC^2 = BX^2 = BP \cdot BA. \]

So $\triangle BCP \sim \triangle BAC$.

v_Enhance
2019-04-19 21:08:46

Great.

Great.

v_Enhance
2019-04-19 21:09:15

We're starting to see the lengths come out. Now I actually want to draw attention to the angles, though.

We're starting to see the lengths come out. Now I actually want to draw attention to the angles, though.

v_Enhance
2019-04-19 21:09:30

Given these similar triangles, how can we play with $\angle DPA$?

Given these similar triangles, how can we play with $\angle DPA$?

budu
2019-04-19 21:09:59

$\angle DPA = \angle BDA$

$\angle DPA = \angle BDA$

v_Enhance
2019-04-19 21:10:07

Good. But $ABCD$ is cyclic, so can we chase that further?

Good. But $ABCD$ is cyclic, so can we chase that further?

mathisawesome2169
2019-04-19 21:10:30

<APD = <ADB = <ACB = <BPC yay

<APD = <ADB = <ACB = <BPC yay

alifenix-
2019-04-19 21:10:30

and BPC is BCA

and BPC is BCA

OrangeToesWizard
2019-04-19 21:10:30

DPA = BDA

DPA = BDA

jj_ca888
2019-04-19 21:10:30

BCA!

BCA!

hliu1
2019-04-19 21:10:30

=<BCA=<BPC

=<BCA=<BPC

budu
2019-04-19 21:10:30

in fact $\angle DBA = \angle BDA = \angle BCA = \angle BPC$

in fact $\angle DBA = \angle BDA = \angle BCA = \angle BPC$

v_Enhance
2019-04-19 21:10:45

We have

\[ \angle DPA = \angle ADB = \angle ACB = \angle BPC \]

where the first equality is by $\triangle ADP \sim \triangle ABD$, the second by cyclic $ABCD$, the third by $\triangle BCP \sim \triangle BAC$.

We have

\[ \angle DPA = \angle ADB = \angle ACB = \angle BPC \]

where the first equality is by $\triangle ADP \sim \triangle ABD$, the second by cyclic $ABCD$, the third by $\triangle BCP \sim \triangle BAC$.

v_Enhance
2019-04-19 21:11:26

And that explains the angle condition.

And that explains the angle condition.

v_Enhance
2019-04-19 21:11:36

*END DREAM MODE*
v_Enhance
2019-04-19 21:11:58

All of this has been wishful thinking up to now, if $P$ really did lie on line $XY$.

All of this has been wishful thinking up to now, if $P$ really did lie on line $XY$.

cj13609517288
2019-04-19 21:12:10

AW...

AW...

deadwindow
2019-04-19 21:12:10

oh no

oh no

v_Enhance
2019-04-19 21:12:23

But I like these similarities a lot.

But I like these similarities a lot.

v_Enhance
2019-04-19 21:12:33

And so let's change up the problem: we will DEFINE $P$ to be the point satisfying \[ AD^2 = AP \cdot AB \quad\text{and}\quad BC^2 = BP \cdot BA. \]

And so let's change up the problem: we will DEFINE $P$ to be the point satisfying \[ AD^2 = AP \cdot AB \quad\text{and}\quad BC^2 = BP \cdot BA. \]

westford_flying_fish
2019-04-19 21:12:44

Set it as a ghost point then

Set it as a ghost point then

v_Enhance
2019-04-19 21:12:47

Yep.

Yep.

v_Enhance
2019-04-19 21:13:00

So I claim I can actually define a point $P$ satisfying these two length conditions, and therefore all the beautiful similar triangles we found.

So I claim I can actually define a point $P$ satisfying these two length conditions, and therefore all the beautiful similar triangles we found.

v_Enhance
2019-04-19 21:13:07

Why is that?

Why is that?

algebra_star1234
2019-04-19 21:13:36

if you add them you get the length condition

if you add them you get the length condition

mathguy623
2019-04-19 21:14:07

Just let P be the foot from X to AB

Just let P be the foot from X to AB

v_Enhance
2019-04-19 21:14:15

One way to see it is from just the pure lengths:

One way to see it is from just the pure lengths:

v_Enhance
2019-04-19 21:14:27

If we let $P$ obey the first equation then

\begin{align*}

BP \cdot BA &= (BA - AP) \cdot BA \\

&= AB^2 - AP \cdot AB = AB^2 - AD^2 = BC^2.

\end{align*}

If we let $P$ obey the first equation then

\begin{align*}

BP \cdot BA &= (BA - AP) \cdot BA \\

&= AB^2 - AP \cdot AB = AB^2 - AD^2 = BC^2.

\end{align*}

v_Enhance
2019-04-19 21:14:36

Alternatively, if you define $P$ to be the foot of that point $X$,

Alternatively, if you define $P$ to be the foot of that point $X$,

v_Enhance
2019-04-19 21:14:44

then those similar triangles will do it for you.

then those similar triangles will do it for you.

v_Enhance
2019-04-19 21:15:05

After changing the problem, we now have to prove TWO things:

(I) this point $P$ is the one in the problem statement (i.e. it satisfies the angle condition $\angle BPC = \angle DPA$), and

(II) line $PE$ bisects $CD$.

After changing the problem, we now have to prove TWO things:

(I) this point $P$ is the one in the problem statement (i.e. it satisfies the angle condition $\angle BPC = \angle DPA$), and

(II) line $PE$ bisects $CD$.

khina
2019-04-19 21:15:32

don't u have to prove uniqueness? or is it citable

don't u have to prove uniqueness? or is it citable

hliu1
2019-04-19 21:15:35

we already proved the first one right?

we already proved the first one right?

cj13609517288
2019-04-19 21:15:52

Yeah we already did the first one yay

Yeah we already did the first one yay

v_Enhance
2019-04-19 21:16:02

We basically proved (I) already in our dream. We just need to add one extra remark: there's only one point $P$ satisfying the angle condition, so it coincides with our re-defined point.

We basically proved (I) already in our dream. We just need to add one extra remark: there's only one point $P$ satisfying the angle condition, so it coincides with our re-defined point.

v_Enhance
2019-04-19 21:16:17

(To see why there's only one point, note that if you imagine starting $P$ at $A$ and moving from left to right, then one angle increases, the other decreases.)

(To see why there's only one point, note that if you imagine starting $P$ at $A$ and moving from left to right, then one angle increases, the other decreases.)

v_Enhance
2019-04-19 21:16:37

So after waking up from our dream we already have half the problem. Just need (II).

So after waking up from our dream we already have half the problem. Just need (II).

alifenix-
2019-04-19 21:17:23

so we just have to show foot of X perp AB lies on EM

so we just have to show foot of X perp AB lies on EM

cj13609517288
2019-04-19 21:17:23

So we dreamt the whole dream just to prove some similarities?

So we dreamt the whole dream just to prove some similarities?

v_Enhance
2019-04-19 21:17:32

Yes, and those similarities are (I).

Yes, and those similarities are (I).

khina
2019-04-19 21:17:51

ummm symmedian

ummm symmedian

v_Enhance
2019-04-19 21:17:56

If you know what a "symmedian" is you can actually finish the problem quite quickly from here. However, I have enough time that I want to present the completely low-tech solution.

If you know what a "symmedian" is you can actually finish the problem quite quickly from here. However, I have enough time that I want to present the completely low-tech solution.

v_Enhance
2019-04-19 21:18:36

Specifically: what points in the diagram did I draw that I haven't used yet?

Specifically: what points in the diagram did I draw that I haven't used yet?

jj_ca888
2019-04-19 21:19:00

E

E

Jupiter123
2019-04-19 21:19:00

K and L

K and L

alifenix-
2019-04-19 21:19:00

K, L

K, L

v_Enhance
2019-04-19 21:19:13

Point $E$, and $K$ and $L$ still haven't played much yet

Point $E$, and $K$ and $L$ still haven't played much yet

v_Enhance
2019-04-19 21:20:30

This is where the points $K$ and $L$ from before come in. Let us define $K$ as the intersection of $DP$ and $AC$, and $L$ as the intersection of $CP$ and $DB$.

This is where the points $K$ and $L$ from before come in. Let us define $K$ as the intersection of $DP$ and $AC$, and $L$ as the intersection of $CP$ and $DB$.

jj_ca888
2019-04-19 21:20:55

wait is it necessarily true that KL || DC?? it seems true

wait is it necessarily true that KL || DC?? it seems true

alifenix-
2019-04-19 21:20:55

looks like $KL || DC$

looks like $KL || DC$

jishu2003
2019-04-19 21:20:55

It is enough to prove that $KL||AB$

It is enough to prove that $KL||AB$

v_Enhance
2019-04-19 21:21:10

Yeah, some of you see the last crux already!

Yeah, some of you see the last crux already!

v_Enhance
2019-04-19 21:21:29

To make it clear this is also wishful, I'll use another pair of dream tags, but this will be short.

To make it clear this is also wishful, I'll use another pair of dream tags, but this will be short.

v_Enhance
2019-04-19 21:21:38

v_Enhance
2019-04-19 21:21:48

*BEGIN DREAM MODE*
v_Enhance
2019-04-19 21:21:52

Suppose $PE$ really bisected $CD$. If we apply Ceva's theorem to $\triangle PCD$ what does that give?

Suppose $PE$ really bisected $CD$. If we apply Ceva's theorem to $\triangle PCD$ what does that give?

cooljoseph
2019-04-19 21:22:25

DK/DP = CL/CP

DK/DP = CL/CP

jj_ca888
2019-04-19 21:22:25

DK/KP * PL/LC = 1

DK/KP * PL/LC = 1

v_Enhance
2019-04-19 21:22:29

If $M$ is the midpoint of $CD$, we would have

\[ \frac{PK}{KD} \cdot \underbrace{\frac{DM}{MC}}_{=1} \cdot \frac{CL}{LP}

= 1 \implies \frac{PK}{KD} = \frac{PL}{LC}. \]

This is just the same as saying $\overline{KL} \parallel \overline{CD}$.

If $M$ is the midpoint of $CD$, we would have

\[ \frac{PK}{KD} \cdot \underbrace{\frac{DM}{MC}}_{=1} \cdot \frac{CL}{LP}

= 1 \implies \frac{PK}{KD} = \frac{PL}{LC}. \]

This is just the same as saying $\overline{KL} \parallel \overline{CD}$.

v_Enhance
2019-04-19 21:22:51

Since the converse of Ceva is true, that means our desired conclusion is also just the same as saying $\overline{KL} \parallel \overline{CD}$.

Since the converse of Ceva is true, that means our desired conclusion is also just the same as saying $\overline{KL} \parallel \overline{CD}$.

v_Enhance
2019-04-19 21:22:56

So when we wake up again, that's our goal: if we can get $\overline{KL} \parallel \overline{CD}$, we win.

So when we wake up again, that's our goal: if we can get $\overline{KL} \parallel \overline{CD}$, we win.

v_Enhance
2019-04-19 21:23:02

*END DREAM MODE*
v_Enhance
2019-04-19 21:23:17

Alright, now that naptime is over, we're in really good shape. We've now gotten rid of all the weird conditions we didn't like. And so now we just need to angle chase the rest of the problem.

Alright, now that naptime is over, we're in really good shape. We've now gotten rid of all the weird conditions we didn't like. And so now we just need to angle chase the rest of the problem.

v_Enhance
2019-04-19 21:24:25

We also have four angles we already like, from our proof of (I).

We also have four angles we already like, from our proof of (I).

v_Enhance
2019-04-19 21:24:32

I'll mark them in the diagram now and call that common angle $\theta$.

I'll mark them in the diagram now and call that common angle $\theta$.

v_Enhance
2019-04-19 21:25:30

v_Enhance
2019-04-19 21:25:36

There we go.

There we go.

v_Enhance
2019-04-19 21:25:55

Alright, i spot some cyclic quadrilaterals due to $\theta$ already. Where are they?

Alright, i spot some cyclic quadrilaterals due to $\theta$ already. Where are they?

azxc
2019-04-19 21:26:44

only need to show that A,K,L.B cyclic

only need to show that A,K,L.B cyclic

cj13609517288
2019-04-19 21:26:44

DLPA

DLPA

Jupiter123
2019-04-19 21:26:44

KPBC

KPBC

mathfun5
2019-04-19 21:26:44

PKCB,ADLP

PKCB,ADLP

Jupiter123
2019-04-19 21:26:44

ADLP

ADLP

MinecraftPlayer657
2019-04-19 21:26:44

DAPL

DAPL

cj13609517288
2019-04-19 21:26:44

KPBC and DLPA

KPBC and DLPA

hliu1
2019-04-19 21:26:44

BPCK and APLD

BPCK and APLD

v_Enhance
2019-04-19 21:26:53

Yeah, since $\angle KPA = \angle KCB = \theta$, we get $KPBC$ is cyclic. Similarly, $\angle BPL = \angle ALD$ means $ADLP$ is cyclic.

Yeah, since $\angle KPA = \angle KCB = \theta$, we get $KPBC$ is cyclic. Similarly, $\angle BPL = \angle ALD$ means $ADLP$ is cyclic.

v_Enhance
2019-04-19 21:27:00

Now, can someone find $\angle AKB$ in terms of $\theta$ for me?

Now, can someone find $\angle AKB$ in terms of $\theta$ for me?

v_Enhance
2019-04-19 21:27:26

(You'll need $CKPB$ cyclic.)

(You'll need $CKPB$ cyclic.)

mathfun5
2019-04-19 21:27:50

180 - theta

180 - theta

jj_ca888
2019-04-19 21:27:50

180 - theta!

180 - theta!

v_Enhance
2019-04-19 21:27:53

We have \[ \angle AKB = 180^{\circ} - \angle BKC

= 180^{\circ} - \angle BPC = 180^{\circ} - \theta. \]

We have \[ \angle AKB = 180^{\circ} - \angle BKC

= 180^{\circ} - \angle BPC = 180^{\circ} - \theta. \]

v_Enhance
2019-04-19 21:28:02

But similarly $\angle ALB = 180^{\circ} - \theta$.

But similarly $\angle ALB = 180^{\circ} - \theta$.

goodball
2019-04-19 21:28:19

AKLB cyclic!

AKLB cyclic!

v_Enhance
2019-04-19 21:28:23

It means $AKLB$ is cyclic too.

It means $AKLB$ is cyclic too.

v_Enhance
2019-04-19 21:28:40

That should be all the ingredients I need to get the parallelism. Anyone see how to do it? Just a bit more angle chasing.

That should be all the ingredients I need to get the parallelism. Anyone see how to do it? Just a bit more angle chasing.

mathfun5
2019-04-19 21:29:20

CDL=CAB=DLK

CDL=CAB=DLK

v_Enhance
2019-04-19 21:29:40

We have \[ \angle LKC = \angle LBA = \angle DBA = \angle DCA = \angle DCK \] which gives us the desired parallel lines. (You'll sometimes here this called "Reim's theorem", but you can basically ignore that name, since it always is the same three-step angle chase.)

We have \[ \angle LKC = \angle LBA = \angle DBA = \angle DCA = \angle DCK \] which gives us the desired parallel lines. (You'll sometimes here this called "Reim's theorem", but you can basically ignore that name, since it always is the same three-step angle chase.)

v_Enhance
2019-04-19 21:29:51

*hear

*hear

v_Enhance
2019-04-19 21:29:59

So $\overline{KL} \parallel \overline{CD}$, and thus by Ceva's theorem, it follows $\overline{PE}$ bisects $\overline{CD}$.

So $\overline{KL} \parallel \overline{CD}$, and thus by Ceva's theorem, it follows $\overline{PE}$ bisects $\overline{CD}$.

v_Enhance
2019-04-19 21:30:02

The end!

The end!

cj13609517288
2019-04-19 21:30:07

WE WIN!!!!!

WE WIN!!!!!

jishu2003
2019-04-19 21:30:20

What's the symmedian proof?

What's the symmedian proof?

Maxninga938
2019-04-19 21:30:20

yee

yee

jj_ca888
2019-04-19 21:30:26

dang that's relly pro

dang that's relly pro

v_Enhance
2019-04-19 21:30:39

Some shorter solutions on the forums with symmedian, so feel free to check those out.

Some shorter solutions on the forums with symmedian, so feel free to check those out.

carzland
2019-04-19 21:30:46

Yay!

Yay!

Maxninga938
2019-04-19 21:30:59

1000 IQ

1000 IQ

jj_ca888
2019-04-19 21:30:59

i think i just doubled my iq

i think i just doubled my iq

Jupiter123
2019-04-19 21:30:59

GREAT LECTURE!!

GREAT LECTURE!!

goodball
2019-04-19 21:30:59

nice dreaming

nice dreaming

ketul
2019-04-19 21:30:59

This is the most instructive solution I've seen in a while.

This is the most instructive solution I've seen in a while.

pandadude
2019-04-19 21:30:59

hello v_Enhance!

hello v_Enhance!

v_Enhance
2019-04-19 21:31:15

Thanks all. I'm just about out of time, but I did promise people a story about MO1 on the forums.

Thanks all. I'm just about out of time, but I did promise people a story about MO1 on the forums.

v_Enhance
2019-04-19 21:31:27

So here's the story of how I wrote the MO1 that we skipped.

So here's the story of how I wrote the MO1 that we skipped.

v_Enhance
2019-04-19 21:31:39

A couple days before the grading of USAMO 2017, I woke up from a dream (maybe nightmare) that I had been assigned to grading some functional equation, and was slogging through pages of student's $f$-riddled papers.

A couple days before the grading of USAMO 2017, I woke up from a dream (maybe nightmare) that I had been assigned to grading some functional equation, and was slogging through pages of student's $f$-riddled papers.

v_Enhance
2019-04-19 21:31:50

When I woke up, I found that I actually still remembered the "problem" I was grading, and wrote it down.

When I woke up, I found that I actually still remembered the "problem" I was grading, and wrote it down.

v_Enhance
2019-04-19 21:31:59

It was ``$f^{f(n)}(n) = \frac{n^2}{f(f(n))}$''.

It was ``$f^{f(n)}(n) = \frac{n^2}{f(f(n))}$''.

Jupiter123
2019-04-19 21:32:13

Hahahaha.

Hahahaha.

khina
2019-04-19 21:32:13

what

what

v_Enhance
2019-04-19 21:32:16

You can guess the rest of the story from there!

You can guess the rest of the story from there!

v_Enhance
2019-04-19 21:32:40

Alright, back to reality. We all done for today. Thanks for coming through!

Alright, back to reality. We all done for today. Thanks for coming through!

eminentabyss
2019-04-19 21:32:52

Holy moly, what a great ending to an excellent math jam!

Holy moly, what a great ending to an excellent math jam!

eminentabyss
2019-04-19 21:33:01

Thank you all for joining us, and huge thanks to Evan for making this possible and Detelin for fielding all of your questions!

Thank you all for joining us, and huge thanks to Evan for making this possible and Detelin for fielding all of your questions!

molocyxu
2019-04-19 21:33:24

THANK YOU!!!

THANK YOU!!!

jishu2003
2019-04-19 21:33:24

Thanks for the JAM.

Thanks for the JAM.

cj13609517288
2019-04-19 21:33:24

That is actually so cool.

That is actually so cool.

goodball
2019-04-19 21:33:24

Thanks!

Thanks!

Jupiter123
2019-04-19 21:33:24

Thank you!

Thank you!

hliu1
2019-04-19 21:33:24

thank you v_enhance

thank you v_enhance

Maxninga938
2019-04-19 21:33:24

thanks

thanks

lisali666
2019-04-19 21:33:24

*end dream*

*end dream*

cj13609517288
2019-04-19 21:33:24

THANKS!!!

THANKS!!!

Damalone
2019-04-19 21:33:24

Thanks!

Thanks!

Awesome_guy
2019-04-19 21:33:24

thanks

thanks

kevinmathz
2019-04-19 21:33:24

Thank you very much!

Thank you very much!

cooljoseph
2019-04-19 21:33:57

Thanks everyone!

Thanks everyone!

wertguk
2019-04-19 21:33:57

THANKS

THANKS

mathfun5
2019-04-19 21:33:57

Thanks so much Evan!

Thanks so much Evan!

porkemon2
2019-04-19 21:33:57

Thank you!

Thank you!

azxc
2019-04-19 21:33:57

Thank you very much!

Thank you very much!

stroller
2019-04-19 21:33:57

Thanks! Nice explanation!

Thanks! Nice explanation!

ketul
2019-04-19 21:33:57

Thanks Evan, was definitely worth staying up until 3:30 AM.

Thanks Evan, was definitely worth staying up until 3:30 AM.

Ben_Chen
2019-04-19 21:33:57

Thanks

Thanks

AlexLikeMath
2019-04-19 21:33:57

thanks!

thanks!

aopssolver2
2019-04-19 21:33:57

thanks

thanks

azxc
2019-04-19 21:33:57

THANK YOU!

THANK YOU!