Math Jams

MaryM 2021-03-12 18:23:25
The Math Jam will start at 6:30 PM Eastern / 3:30 PM Pacific.

MaryM 2021-03-12 18:23:32
The classroom is moderated, meaning that you can type into the classroom, but these comments will not go directly into the room.

MaryM 2021-03-12 18:23:47
Please do not ask about administrative aspects of the contests or ask me to speculate about the results. We do not know what the index will be for the USAJMO or the USAMO.

MaryM 2021-03-12 18:31:18
Welcome to the 2021 AIME I Math Jam!

MathWizard10 2021-03-12 18:31:30
hello

Lilavigne 2021-03-12 18:31:30
Hi !

rjeagle2019 2021-03-12 18:31:30
yay

jessiewang28 2021-03-12 18:31:30
hello!

MaryM 2021-03-12 18:31:33
I'm Maria Mendes, and I'll be leading our discussion today. Here's a little bit about me:

MaryM 2021-03-12 18:31:41
I realized math was fun as a teenager when I saw a problem about spider webs and multiples of 8. I continued to participate in Mathematical Olympiads during high school and eventually made it onto the Brazilian IMO team. Later, I graduated in Applied Mathematics and got a Master’s degree in probability theory. I joined AoPS in 2017, as a grader, finding a passion for teaching. When not working, I am likely to be found studying art, concert chasing, or playing with my two beloved kittens.

MaryM 2021-03-12 18:31:51
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.

MaryM 2021-03-12 18:31:58
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.

MaryM 2021-03-12 18:32:04
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.

MaryM 2021-03-12 18:32:11
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time.

MaryM 2021-03-12 18:32:28
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We always to try do so in our regular online classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!

BeefChickling34 2021-03-12 18:32:40
got that

GabeW1234 2021-03-12 18:32:40
LOL same number of you

MaryM 2021-03-12 18:32:43
This jam will probably be quite long, so we will take a 5 minutes break after solving problem 9.

sugar_rush 2021-03-12 18:32:54
okay

Blueclay 2021-03-12 18:32:54
ok

MaryM 2021-03-12 18:33:10
Helping out today are our two assistants Eli Brottman (goveganddomath) and Detelin Dosev (OlympiadPro2000).

MaryM 2021-03-12 18:33:19
Eli has always had a passion for math. He just graduated from Northern Illinois University, with a mathematics major and minors in computer science and statistics. In the future, he hopes to pursue a PhD in computer science, applied mathematics, or a related field, in order to fulfill his dream of becoming a computational cancer researcher. Eli has participated in MATHCOUNTS, USAMTS, AMC 10/12, and AIME. In his spare time, Eli enjoys listening to music, volunteering in his community, and using math to make the world a better place.

MaryM 2021-03-12 18:33:25
Detelin 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.

Blueclay 2021-03-12 18:33:36
Hi.

D

Blueclay 2021-03-12 18:33:36
Nice to meet you both!

goveganddomath 2021-03-12 18:33:38
Hi everyone! I look forward to working with you tonight!

OlympiadPro2000 2021-03-12 18:34:03
Hello everyone!

MaryM 2021-03-12 18:34:06
They can answer questions by whispering messages to you that will show up on your screen. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over will not get it answered faster (in fact, it is less likely to get answered), so please, just ask your question once, be patient, and please understand that we may not be able to answer all the questions tonight.

MaryM 2021-03-12 18:34:13
Also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be acknowledged.

MaryM 2021-03-12 18:34:34
Let's get started! We're going to work through all 15 problems from the 2021 AIME I, in order.

Poki 2021-03-12 18:34:58
yeayy!

mathisverynice 2021-03-12 18:34:58
alright!!

sugar_rush 2021-03-12 18:34:58
lesgo

MaryM 2021-03-12 18:35:02
1. Zou and Chou are practicing their

$100$ -meter sprints by running

$6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is

$\dfrac23$ if they won the previous race but only

$\dfrac 13$ if they lost the previous race. The probability that Zou will win exactly

$5$ of the

$6$ races is

$\dfrac{m}{n},$ where

$m$ and

$n$ are relatively prime positive integers. Find

$m+n.$

MaryM 2021-03-12 18:35:18
What are the possibilities for how the six races can go?

themathboi101 2021-03-12 18:36:14
Win or lose

Hershey806 2021-03-12 18:36:14
zou will lose one of the next 5 games

Lilavigne 2021-03-12 18:36:14
he can lose game 2, 3, 4, 5, or 6

Irving1004 2021-03-12 18:36:14
win 1, win 2, win 3, win 4, win 5, lose 6

MaryM 2021-03-12 18:36:24
Z has to win 5 and C has to win 1.

MaryM 2021-03-12 18:36:30
But we already know that Z won the first race.

MaryM 2021-03-12 18:36:38
So C has to win exactly one of the 2nd through 6th races.

MaryM 2021-03-12 18:36:41
In how many ways can this happen?

Bananaman27 2021-03-12 18:36:56
5

peace09 2021-03-12 18:36:56
5 Ways

jxwis2010 2021-03-12 18:36:56
5

rockyrockrock 2021-03-12 18:36:56
5

math-Passion 2021-03-12 18:36:56

$5$ WAYS

helloworld_ 2021-03-12 18:36:56
5

GabeW1234 2021-03-12 18:37:00
5

Z_Lu 2021-03-12 18:37:00
5 ways

MaryM 2021-03-12 18:37:03
Since there are 5 choices for which race C wins, this can happen in 5 different ways.

MaryM 2021-03-12 18:37:14
And do these 5 different ways all have equal probability?

coolotter 2021-03-12 18:37:32
nope

ryanfu2008 2021-03-12 18:37:32
NO

Lilavigne 2021-03-12 18:37:32
no

functionalmath 2021-03-12 18:37:32
no

Benranger 2021-03-12 18:37:32
yes

yayy 2021-03-12 18:37:32
no

jxwis2010 2021-03-12 18:37:32
no

Blueclay 2021-03-12 18:37:32
no

cowsrule 2021-03-12 18:37:32
no

peace09 2021-03-12 18:37:32
No!

Irving1004 2021-03-12 18:37:32
No

math-Passion 2021-03-12 18:37:32
No

GabeW1234 2021-03-12 18:37:32
no

MaryM 2021-03-12 18:37:35
No. It depends on how many times the winner "flips" from Z to C or from C to Z.

MaryM 2021-03-12 18:37:39
How many times does this flip happen?

Irving1004 2021-03-12 18:38:20
2 or 1

yayy 2021-03-12 18:38:20
1 or 2

peace09 2021-03-12 18:38:20
Either Once or Twice

jxwis2010 2021-03-12 18:38:20
once for all, twice for when C wins the last race

math-Passion 2021-03-12 18:38:20
Once or Twice

dbasu1 2021-03-12 18:38:20
twice for C-win = 2nd, 3rd, 4th, or 5th, and once for C-win = 6th

cowsrule 2021-03-12 18:38:20
1 or 2?

helloworld_ 2021-03-12 18:38:20
1 or 2

MaryM 2021-03-12 18:38:23
If C wins one of the 4 middle races, such as ZZCZZZ, then a flip happens twice: once from Z to C and then once from C back to Z.

MaryM 2021-03-12 18:38:29
Now that the probability is the same for any of these sequences of races; that is

$P(\text{ZCZZZZ}) = P(\text{ZZCZZZ}) = P(\text{ZZZCZZ}) = P(\text{ZZZZCZ}).$

MaryM 2021-03-12 18:38:33
But if C wins the last race, as in ZZZZZC, then a flip only happens once: from Z to C for the last race (and it doesn't flip back, because the races are over).

MaryM 2021-03-12 18:38:41
So our total probability is

$4P(\text{ZCZZZZ}) + P(\text{ZZZZZC})$ .

MaryM 2021-03-12 18:38:46
Now we have to compute these probabilities.

MaryM 2021-03-12 18:38:53
What is

$P(\text{ZCZZZZ})$ ?

functionalmath 2021-03-12 18:39:49
8/243 sorry

gorefeebuddie 2021-03-12 18:39:49
8/243

peace09 2021-03-12 18:39:49

$1 \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^3 = \frac{8}{243}.$

dbasu1 2021-03-12 18:39:49
=1/3 * 1/3 * 2/3 * 2/3 * 2/3

= 8/243

jxwis2010 2021-03-12 18:39:49
8/243

coolotter 2021-03-12 18:39:49
8/243

Z_Lu 2021-03-12 18:39:49
8/243

RedFireTruck 2021-03-12 18:39:49
8/243

Bananaman27 2021-03-12 18:39:49
(2/3)^3(1/3)^2

MaryM 2021-03-12 18:39:56
We have

$2$ flips and

$3$ non-flips.

MaryM 2021-03-12 18:40:01
So the probability is

$\left(\dfrac13\right)^2 \cdot \left(\dfrac23\right)^3$ .

MaryM 2021-03-12 18:40:09
And the other case? What is

$P(\text{ZZZZZC})$ ?

sugar_rush 2021-03-12 18:40:56

$\tfrac{16}{243}$

Irving1004 2021-03-12 18:40:56
2 times that, which is 16/243

jxwis2010 2021-03-12 18:40:56
16/243

Kesav 2021-03-12 18:40:56
16/243

rockyrockrock 2021-03-12 18:40:56
(2/3)^4*1/3

dbasu1 2021-03-12 18:40:56
(2/3)^4 * (1/3) for 4 non-flips and 1 flip

=16/243

Hershey806 2021-03-12 18:40:56

$\frac{2}{3}^4 \cdot \frac{1}{3}$

Bananaman27 2021-03-12 18:40:56
(2/3)^4(1/3)^1

yayy 2021-03-12 18:40:56

$(\frac{2}{3})^4 * (\frac{1}{3})$

gracefulharmony 2021-03-12 18:40:56
16/243

MaryM 2021-03-12 18:40:58
We have 1 flip and 4 non-flips.

MaryM 2021-03-12 18:41:03
So the probability is

$\left(\dfrac13\right) \cdot \left(\dfrac23\right)^4$ .

MaryM 2021-03-12 18:41:06
Therefore, what's the overall probability?

donguri 2021-03-12 18:42:03

$\dfrac{16}{243}$

ryanfu2008 2021-03-12 18:42:03
16/81

gorefeebuddie 2021-03-12 18:42:03
48/243 or 16/81

peace09 2021-03-12 18:42:03

$\frac{4 \cdot 8 + 16}{243} = \frac{48}{243} = \frac{16}{81}.$

Lilavigne 2021-03-12 18:42:03
16/81

gracefulharmony 2021-03-12 18:42:03
48/243=16/81

rjeagle2019 2021-03-12 18:42:03
sorry 48/243

MaryM 2021-03-12 18:42:08
It's

$4 \cdot \left(\dfrac13\right)^2 \cdot \left(\dfrac23\right)^3 + \left(\dfrac13\right) \cdot \left(\dfrac23\right)^4.$

MaryM 2021-03-12 18:42:12
As a single fraction, it's

$\dfrac{4 \cdot 2^3 + 2^4}{3^5}.$

MaryM 2021-03-12 18:42:16
This is

$\dfrac{48}{3^5}$ .

MaryM 2021-03-12 18:42:22
We can cancel a

$3$ to get

$\dfrac{16}{3^4}$ .

MaryM 2021-03-12 18:42:27
This won't simplify any further, so we have

$\dfrac{16}{3^4} = \dfrac{16}{81}$ as the fraction in lowest terms.

MaryM 2021-03-12 18:42:33
And the answer is...

Kesav 2021-03-12 18:42:54
the answer is 16 + 81 = 97 =D

Math192 2021-03-12 18:42:54
97

coolotter 2021-03-12 18:42:54
097

sl808 2021-03-12 18:42:54
97

superstar627 2021-03-12 18:42:54
97

Hershey806 2021-03-12 18:42:54

$\boxed{097}$

awesomeandy 2021-03-12 18:42:54
97

yayy 2021-03-12 18:42:54
97

ryanfu2008 2021-03-12 18:42:54
97

s-sawen-07 2021-03-12 18:42:54
97

uneducatedpotato 2021-03-12 18:42:57
097

Argonauts16 2021-03-12 18:42:57
97

MaryM 2021-03-12 18:42:58
Our final answer is

$16 + 81 = \boxed{097}$ .

MaryM 2021-03-12 18:43:07
2. In the diagram below,

$ABCD$ is a rectangle with side lengths

$AB = 3$ and

$BC = 11,$ and

$AECF$ is a rectangle with side lengths

$AF = 7$ and

$FC = 9,$ as shown. The area of the shaded region common to the interiors of both rectangles is

$\dfrac{m}{n},$ where

$m$ and

$n$ are relatively prime positive integers. Find

$m+n.$

MaryM 2021-03-12 18:43:15

MaryM 2021-03-12 18:43:27
Before we dive in, let's get some more info into the diagram. I generally think it's a good idea to neatly add as much info the the diagram as we reasonably can.

MaryM 2021-03-12 18:43:33
Specifically, let's label the other two points, and label some lengths. I'll also add variables for some of the unknown lengths.

MaryM 2021-03-12 18:43:40

MaryM 2021-03-12 18:43:50
Now what?

functionalmath 2021-03-12 18:44:51
use triangle similarity

coolotter 2021-03-12 18:44:51
SIMILAR TRIANGLES

peace09 2021-03-12 18:44:51
Similar Triangles!

RedFireTruck 2021-03-12 18:44:51
similar triangles?

helloworld_ 2021-03-12 18:44:51
Use similar triangles

awesomeandy 2021-03-12 18:44:51
similar triangles

MeepMurp5 2021-03-12 18:44:51
Triangle

$PCE$ is similar to triangle

$PAB$ by Angle-Angle Similarity.

yayy 2021-03-12 18:44:51
use similar triangles

rockyrockrock 2021-03-12 18:44:51
ABP and CEP are similar ration 3:7

peace09 2021-03-12 18:44:51

$\triangle ABP \sim \triangle CEP$ by AA Similarity.

MaryM 2021-03-12 18:44:53
I think I see similar triangles!

MaryM 2021-03-12 18:44:57
Indeed, we have

$\angle APB = \angle CPE$ because they're vertical angles.

MaryM 2021-03-12 18:45:00
So

$\triangle APB \sim \triangle CPE$ are similar right triangles.

MaryM 2021-03-12 18:45:04
How does that help?

RedFireTruck 2021-03-12 18:46:00
write ratios!

happyhari 2021-03-12 18:46:00
use ratios

Lilavigne 2021-03-12 18:46:00
we can write an equation

yayy 2021-03-12 18:46:00
make equations

XGHou 2021-03-12 18:46:00
Ratio of side lengths is known to be 3/7

ryanfu2008 2021-03-12 18:46:00
we can write their ratios

Irving1004 2021-03-12 18:46:00
Well we can use ratios

functionalmath 2021-03-12 18:46:00
use AB/CE

rockyrockrock 2021-03-12 18:46:00

$\frac{9-y}{x}=\frac{7}{3}$

Irving1004 2021-03-12 18:46:00
PE/PB = 7/3

happyhari 2021-03-12 18:46:00
we could use ratios

peace09 2021-03-12 18:46:00

$\tfrac{x}{9-y}=\tfrac{3}{7}=\tfrac{y}{11-x}.$

MaryM 2021-03-12 18:46:02
We have

$\dfrac{AB}{CE} = \dfrac{BP}{EP} = \dfrac{AP}{CP}$ .

MaryM 2021-03-12 18:46:08
We can fill out the lengths that we've labeled.

MaryM 2021-03-12 18:46:12
We get

$\dfrac37 = \dfrac{x}{9-y} = \dfrac{y}{11-x}$ .

MaryM 2021-03-12 18:46:19
How do we work with this?

peace09 2021-03-12 18:46:44
Split it into two equations!

MaryM 2021-03-12 18:46:47
Setting

$\dfrac37$ equal to the other two fractions gives us two equations in two variables:

$\begin{align*} \dfrac37 &= \dfrac{x}{9-y}, \\ \dfrac37 &= \dfrac{y}{11-x}. \end{align*}$

MaryM 2021-03-12 18:46:52
How we can solve this?

awesomeandy 2021-03-12 18:47:22
cross multiply

Moonshine-Dragonwing 2021-03-12 18:47:22
cross multiplication and then systems

Lilavigne 2021-03-12 18:47:22
cross multiply

superstar627 2021-03-12 18:47:22
crossmultiply

coolotter 2021-03-12 18:47:22
cross-multiply

yayy 2021-03-12 18:47:22
cross multiply

gracefulharmony 2021-03-12 18:47:22
Multiply out

math-Passion 2021-03-12 18:47:22
Cross multiply

cowsrule 2021-03-12 18:47:22
27-3y=7x and 33-3x=7y by cross multiplication

rzlng 2021-03-12 18:47:22
27-37 = 7x, 33-3x = 7y by clearing denominators.

MaryM 2021-03-12 18:47:24
Let's clear out the denominators:

$\begin{align*} 3(9-y) &= 7x, \\ 3(11-x) &= 7y. \end{align*}$

MaryM 2021-03-12 18:47:30
This looks like two linear equations in two variables. So we can definitely solve it!

MaryM 2021-03-12 18:47:37
Let's clean it up a little more:

$\begin{align*} 27 &= 7x + 3y, \\ 33 &= 3x + 7y. \end{align*}$

MaryM 2021-03-12 18:47:41
What's the most efficient way to proceed?

MaryM 2021-03-12 18:47:48
Which variable is more important for us to know,

$x$ or

$y$ ?

awesomeandy 2021-03-12 18:48:27
x

uneducatedpotato 2021-03-12 18:48:27
x

helloworld_ 2021-03-12 18:48:27
x

gracefulharmony 2021-03-12 18:48:27
x

cowsrule 2021-03-12 18:48:27
x

gorefeebuddie 2021-03-12 18:48:27
x

peace09 2021-03-12 18:48:27

$x.$

Lilavigne 2021-03-12 18:48:27
x

coolotter 2021-03-12 18:48:27
wait actually x

rzlng 2021-03-12 18:48:27
x

MaryM 2021-03-12 18:48:31
Look back at what we're trying to find.

MaryM 2021-03-12 18:48:44
We want the gray area

$[APCQ]$ . What is

$[APCQ]?$

yayy 2021-03-12 18:49:44

$(11-x) * 3$

rzlng 2021-03-12 18:49:44
(11-x) * 3

cowsrule 2021-03-12 18:49:44
(11-x)*3

Asher51 2021-03-12 18:49:44
area of a parallelogram with base 11-x and height 3

dolphindash 2021-03-12 18:49:44
33-3x

ryanfu2008 2021-03-12 18:49:44
33-3x

MaryM 2021-03-12 18:49:47
Quadrilateral

$APCQ$ is a parallelogram with base

$PC = 11 - x$ and height

$AB = 3,$ so its area is

$3(11 - x).$

MaryM 2021-03-12 18:49:54
(We can also take

$AP = y$ as the base and

$CE = 7$ as the height.)

MaryM 2021-03-12 18:49:58
So finding either variable will be helpful.

MaryM 2021-03-12 18:50:03
Most of you want to find

$x$ , so let’s do that.

MaryM 2021-03-12 18:50:15
How do we efficiently solve our system for

$x$ ?

yayy 2021-03-12 18:51:03
eliminate y

Lilavigne 2021-03-12 18:51:03
eliminate the y's

Asher51 2021-03-12 18:51:03
eliminate y

cowsrule 2021-03-12 18:51:03
multiply 2nd equation by 7 and multiply 1st equation by 3

gracefulharmony 2021-03-12 18:51:03
cancel out y

MaryM 2021-03-12 18:51:07
We can eliminate

$y$ .

MaryM 2021-03-12 18:51:12
For instance, we can multiply the top equation by

$7$ , the bottom equation by

$3$ , and then subtract.

MaryM 2021-03-12 18:51:15

$\begin{align*} 189 &= 49x + 21y, \\ 99 &= 9x + 21y. \end{align*}$

MaryM 2021-03-12 18:51:19
What do we get when we subtract?

math9990 2021-03-12 18:51:52
90 = 40x

rzlng 2021-03-12 18:51:52
90 = 40x

Kesav 2021-03-12 18:51:52
90 = 40x

coolotter 2021-03-12 18:51:52
90 = 40x

cowsrule 2021-03-12 18:51:52
90=40x

happyhari 2021-03-12 18:51:52
90 = 40x

awesomeandy 2021-03-12 18:51:52
90 = 40x

jxwis2010 2021-03-12 18:51:52
40x=90

happyhari 2021-03-12 18:51:52
x = 9/4 = 2.25

futbol3.14159 2021-03-12 18:51:52
90 = 40x, x = 9/4

skyleristhecoolest 2021-03-12 18:51:52
40x = 90

MaryM 2021-03-12 18:51:55
The

$y$ 's cancel, and we're left with

$90 = 40x$ .

MaryM 2021-03-12 18:51:59
So

$x = \dfrac94$ .

MaryM 2021-03-12 18:52:03
So what is the area of parallelogram

$APCQ?$

Kesav 2021-03-12 18:52:52
26.25 or 105/4

peace09 2021-03-12 18:52:52

$[APCQ]=\tfrac{105}{4}.$

rockyrockrock 2021-03-12 18:52:52
33-27/4

rzlng 2021-03-12 18:52:52
(11 - 9/4) * 3 = (35/4) * 3 = 105/4

rockyrockrock 2021-03-12 18:52:52
105/4

awesomeandy 2021-03-12 18:52:52
3(11-9/4)

functionalmath 2021-03-12 18:52:52
33-27/4=105/4

rockyrockrock 2021-03-12 18:52:52

$\frac{105}{4}$

coolotter 2021-03-12 18:52:52
105/4

Irving1004 2021-03-12 18:52:52
105/4

MaryM 2021-03-12 18:52:54
The area of parallelogram

$APCQ$ is

$3 (11 - x) = 3 \left( 11 - \dfrac{9}{4} \right) = \dfrac{105}{4}.$

MaryM 2021-03-12 18:52:58
So what's our final answer?

sugar_rush 2021-03-12 18:53:17

$109$

Z_Lu 2021-03-12 18:53:17
109

donguri 2021-03-12 18:53:17

$\boxed{109}$

dolphindash 2021-03-12 18:53:17
109

ryanfu2008 2021-03-12 18:53:17

$\huge{109}$

cowsrule 2021-03-12 18:53:17
109

justin6688 2021-03-12 18:53:17
109

happyhari 2021-03-12 18:53:17
109

bookworm_2019 2021-03-12 18:53:17
109

rzlng 2021-03-12 18:53:17
105 + 4 = 109

geometry6 2021-03-12 18:53:17
109

MaryM 2021-03-12 18:53:20
We get

$105 + 4 = \boxed{109}$ .

MaryM 2021-03-12 18:53:35
3. Find the number of positive integers less than

$1000$ that can be expressed as the difference of two integral powers of

$2.$

MaryM 2021-03-12 18:53:43
Just so we have some easy terminology, let's call a positive number

$n$ good if we can write it as

$n = 2^a - 2^b$ for some nonnegative integers

$a$ and

$b$ .

MaryM 2021-03-12 18:53:50
So, we want to count all the good

$n$ that are less than

$1000$ .

MaryM 2021-03-12 18:54:00
Any ideas?

MaryM 2021-03-12 18:54:03
We're working with powers of

$2$ , so what number-theory tactic does that possibly suggest?

bambithenambi 2021-03-12 18:55:03
binary!

CircleInvert 2021-03-12 18:55:03
Binary?

peace09 2021-03-12 18:55:03
Binary?

Z_Lu 2021-03-12 18:55:03
binary??

happyhari 2021-03-12 18:55:03
binary

sugar_rush 2021-03-12 18:55:03
binary representation?

MaryM 2021-03-12 18:55:09
Maybe thinking about this in base 2 will help!

MaryM 2021-03-12 18:55:13
What's

$2^a$ in base 2?

ryanfu2008 2021-03-12 18:55:52
1 followed by a 0s

rockyrockrock 2021-03-12 18:55:52
100000 with a 0's

awesomeguy856 2021-03-12 18:55:52

$1000...$ with

$a$

$0$ 's

helloworld_ 2021-03-12 18:55:52
1 followed by a zeros

yuanyuanC 2021-03-12 18:55:52
1 followed by a zeros

happyhari 2021-03-12 18:55:52
1 and a 0s after the 1

gorefeebuddie 2021-03-12 18:55:52
1 and then a 0's trailing

peace09 2021-03-12 18:55:52

$1\underbrace{000\hdots 0}_{a~\text{zeroes}}.$

bambithenambi 2021-03-12 18:55:52
1 with a zeroes

MaryM 2021-03-12 18:55:55
It's a 1 followed by

$a$ zeros.

MaryM 2021-03-12 18:56:01
(That's why thinking in base

$k$ when you're working with powers of

$k$ is often helpful: the powers

$k^n$ in base

$k$ have a nice, easy-to-work-with form.)

MaryM 2021-03-12 18:56:07
So our good number

$n$ in base 2 looks like:

$1\underbrace{0\ldots 0}_{a \text{ 0's}}- 1\underbrace{0\ldots 0}_{b \text{ 0's}}.$

MaryM 2021-03-12 18:56:11
What does it look like when we perform this subtraction?

MaryM 2021-03-12 18:56:36
Maybe it helps to write out an example. Here's

$a=8$ and

$b=5$ :

MaryM 2021-03-12 18:56:44

$\begin{array}{rccccccccc} &1&0&0&0&0&0&0&0&0 \\ - &&&&1&0&0&0&0&0 \end{array}$

MaryM 2021-03-12 18:56:55
What do we get in this particular case?

rockyrockrock 2021-03-12 18:57:42
11100000

Irving1004 2021-03-12 18:57:42
11100000

RedFireTruck 2021-03-12 18:57:42
11100000

peace09 2021-03-12 18:57:42
11100000.

BillXu21 2021-03-12 18:57:42
11100000

yayy 2021-03-12 18:57:42
11100000

ryanfu2008 2021-03-12 18:57:42
11100000

awesomeguy856 2021-03-12 18:57:45

$11100000$

MaryM 2021-03-12 18:57:48

$\begin{array}{rccccccccc} &1&0&0&0&0&0&0&0&0 \\ - &&&&1&0&0&0&0&0 \\ \hline &&1&1&1&0&0&0&0&0 \end{array}$

MaryM 2021-03-12 18:57:52
Do you see how this generalizes?

gorefeebuddie 2021-03-12 18:58:46
"a-b" 1111111 and then "b" 0's after

ryanfu2008 2021-03-12 18:58:46

$\text{11100000}$

happyhari 2021-03-12 18:58:46
a - b 1s in the beginning and b 0s in the end

dolphindash 2021-03-12 18:58:46
a-b 1's and b 0's

ryanfu2008 2021-03-12 18:58:46

$2^a-2^b$ in base 2 is [a-b] 1s followed by b 0s

yayy 2021-03-12 18:58:46

$111...10000...0$ with

$a-b$ 1's and

$b$ 0's

peace09 2021-03-12 18:58:46

$2^a-2^b$ in Base 2 is Composed of

$a-b$ Leading 1's and

$b$ Leading 0's?

Lilavigne 2021-03-12 18:58:46
b-a ones followed by b zeros

jxwis2010 2021-03-12 18:58:46
1....10...0, a-b 1s, b 0's

MaryM 2021-03-12 18:58:49
We get that a good number, written in base 2, has one or more 1's followed by zero or more 0's. Specifically,

$a-b$ is the number of 1's, and

$b$ is the number of 0's.

MaryM 2021-03-12 18:58:55
You could prove this algebraically with the identity

$2^a - 2^b = 2^{a-1} + 2^{a-2} + \cdots + 2^{b}$ . But I think the example we did is sufficiently convincing.

MaryM 2021-03-12 18:59:05
So how do we count the number of these less than

$1000$ ?

MaryM 2021-03-12 18:59:09
What do we have to decide to construct such a number?

peace09 2021-03-12 19:01:03
How Many 1's and How Many 0's

yayy 2021-03-12 19:01:03
the number of 1's and 0's

jxwis2010 2021-03-12 19:01:03
how many 0s, how many 1s

awesomeguy856 2021-03-12 19:01:03
the amount of

$1$ s and amount of

$0$ s

MaryM 2021-03-12 19:01:08
We have to decide how many

$1$ ’s and how many

$0$ ’s we have. This is the same thing as deciding in which position the first

$1$ will go, and how many

$1$ ’s we will have.

MaryM 2021-03-12 19:01:20
For example, if the first 1 is in the

$2^0$ spot, what are the choices for how many 1's we can have?

MaryM 2021-03-12 19:01:40
By first I mean leftmost, by the way.

MaryM 2021-03-12 19:03:13
Careful, we want the leftmost one to be in the units digits.

peace09 2021-03-12 19:03:47
Only 1 1!

Lilavigne 2021-03-12 19:03:47
1

MT-Math 2021-03-12 19:03:47
only one 1 possible

MaryM 2021-03-12 19:03:49
We must have

$1$ one.

MaryM 2021-03-12 19:03:58
If the leftmost 1 is in the

$2^1$ spot, how many choices?

functionalmath 2021-03-12 19:04:23
2

Irving1004 2021-03-12 19:04:23
2

rockyrockrock 2021-03-12 19:04:23
2

peace09 2021-03-12 19:04:23
One 1 and One 0, or Two 1s.

Lilavigne 2021-03-12 19:04:23
2

donguri 2021-03-12 19:04:23
2

BillXu21 2021-03-12 19:04:23
2

MaryM 2021-03-12 19:04:26
We can have

$1$ or

$2$ ones, so we have

$2$ choices.

MaryM 2021-03-12 19:04:31
How does this generalize?

MaryM 2021-03-12 19:04:35
If the first 1 is in the

$2^k$ spot, how many ones can we have?

Lilavigne 2021-03-12 19:05:33
k+1

Z_Lu 2021-03-12 19:05:33
k+1

BillXu21 2021-03-12 19:05:33
k+1

peace09 2021-03-12 19:05:33

$k+1$ 1s.

Irving1004 2021-03-12 19:05:33
k+1

happyhari 2021-03-12 19:05:33
k + 1

MT-Math 2021-03-12 19:05:33
k+1

MaryM 2021-03-12 19:05:35
We can have any number from

$1$ to

$k+1$ ones. So there are

$k+1$ choices.

MaryM 2021-03-12 19:05:39
But what else do we need to consider?

ryanfu2008 2021-03-12 19:06:35
<1000

Lilavigne 2021-03-12 19:06:35
the maximum number of 1s

yayy 2021-03-12 19:06:35
it has to be less than 1000

Z_Lu 2021-03-12 19:06:35
if it will be greater than 1000

ancientwarrior 2021-03-12 19:06:35
if the difference is >=1000

MT-Math 2021-03-12 19:06:35
what is the farthest left position we can place the first 1

MaryM 2021-03-12 19:06:38
We need the number to be less than

$1000$ .

MaryM 2021-03-12 19:06:48
What does that mean in terms of where the first

$1$ is?

Lilavigne 2021-03-12 19:07:23
the largest power of 2 smaller than 1000

ryanfu2008 2021-03-12 19:07:23
at most 2^9 position, as 2^10-1024>1000

yayy 2021-03-12 19:07:23
before the 10th spot

peace09 2021-03-12 19:07:23
...Since the positive integer must be less than 1000, the number may not have more than 10 digits.

MaryM 2021-03-12 19:07:26
If the first

$1$ is in the

$2^{10} = 1024$ position or more, then we'll never be less than

$1000$ , so that's always too big.

MaryM 2021-03-12 19:07:36
On the other hand, if the first

$1$ is in the

$2^8 = 256$ position or less, then we'll never be more than

$2^9 = 512$ , so we're OK.

MaryM 2021-03-12 19:07:43
What happens if the first

$1$ is in the

$2^9 = 512$ position?

MaryM 2021-03-12 19:07:50
How far to the right can we go without going over

$1000$ ?

sugar_rush 2021-03-12 19:09:22

$5$

peace09 2021-03-12 19:09:22
We may have at most five 1s.

yayy 2021-03-12 19:09:22
5 places

Z_Lu 2021-03-12 19:09:22
2^5

Lilavigne 2021-03-12 19:09:22
at most 5 ones

happyhari 2021-03-12 19:09:22
up to 2^5?

MaryM 2021-03-12 19:09:28

$2^9 = 512$ is OK.

MaryM 2021-03-12 19:09:31

$2^9 + 2^8 = 512 + 256 = 768$ is OK.

MaryM 2021-03-12 19:09:35

$2^9 + 2^8 + 2^7 = 512 + 256 + 128 = 896$ is OK.

MaryM 2021-03-12 19:09:39

$2^9 + \cdots + 2^6 = 896 + 64 = 960$ is OK.

MaryM 2021-03-12 19:09:45

$2^9 + \cdots + 2^5 = 960 + 32 = 992$ is OK.

MaryM 2021-03-12 19:09:49

$2^9 + \cdots + 2^4 = 992 + 16 = 1008$ is too big!

MaryM 2021-03-12 19:09:54
So there are 5 good numbers with the first one at

$2^9$ that are less than

$1000$ .

MaryM 2021-03-12 19:09:58
How do we sum all this up to get our final answer?

peace09 2021-03-12 19:11:29

$5+9+8+7+\hdots+1=5+45=\boxed{50}.$

Irving1004 2021-03-12 19:11:29
5 + 9(9+1)/2

donguri 2021-03-12 19:11:29
50

Z_Lu 2021-03-12 19:11:29
1+2+3+4+5+6+7+8+9+5=50

Lilavigne 2021-03-12 19:11:29
5+(9*10/2)

sugar_rush 2021-03-12 19:11:29
we do

$5+\dbinom{10}{2}$

BillXu21 2021-03-12 19:11:29
1+2+3+4+5+6+7+8+9+5

MaryM 2021-03-12 19:11:34
We had

$1 + 2 + \cdots + 9$ good numbers starting at

$2^0$ through

$2^8$ , plus another

$5$ starting at

$2^9$ .

MaryM 2021-03-12 19:11:42
So the number of good numbers less than

$1000$ is

$(1+2+\cdots+9) + 5$ .

MaryM 2021-03-12 19:11:46
The first term is

$(1+2+\cdots+9) = \dfrac{9(10)}{2} = 45$ .

MaryM 2021-03-12 19:11:51
So our total is

$45 + 5 = \boxed{050}$ .

BillXu21 2021-03-12 19:12:46
yay

MaryM 2021-03-12 19:12:51
4. Find the number of ways

$66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

MaryM 2021-03-12 19:13:00
Can we write an equation that we're trying to count solutions to?

Lilavigne 2021-03-12 19:14:06
x+y+z=66, such that x < y < z

ryanfu2008 2021-03-12 19:14:06
a+b+c = 66, a<b<c

BorealBear 2021-03-12 19:14:06
a+b+c=66 for a<b<c

BillXu21 2021-03-12 19:14:06
x+y+z=66 x<y<z x,y,and z are positive integers

peace09 2021-03-12 19:14:06

$a_1+a_2+a_3=66,$ where

$0<a_1<a_2<a_3,$ and

$a_i$ denotes the number of coins in the

$i$ th pile.

sugar_rush 2021-03-12 19:14:06

$a+b+c=66$ and

$0<a<b<c$

RedFireTruck 2021-03-12 19:14:06

$a+b+c=66$ ,

$0<a<b<c$

yayy 2021-03-12 19:14:06

$a + b + c = 66$ ,

$a < b < c$

ryanfu2008 2021-03-12 19:14:06

$a+b+c=60$

$a<b<c$

Z_Lu 2021-03-12 19:14:06
set x,y,z as piles 1,2 and 3, respectively. Then x+y+z=66, and x<y<z

MeepMurp5 2021-03-12 19:14:06

$x+y+z=66$ ,

$x<y<z$ ,

$x,y,z\in$ positive integers

MaryM 2021-03-12 19:14:09
We want solutions to

$x + y + z = 66$ with

$0 < x < y < z$ .

MaryM 2021-03-12 19:14:14
Any ideas?

MaryM 2021-03-12 19:14:51
There is a long bookkeeping solution where you perform casework on

$x$ and make a chart. But I'll present a more interesting solution.

MaryM 2021-03-12 19:15:01
One good problem-solving tactic is to try to solve a simpler problem.

MaryM 2021-03-12 19:15:05
What's a simpler problem?

MaryM 2021-03-12 19:15:43
Hint: can we remove one condition from the problem?

XGHou 2021-03-12 19:16:15
If we remove the restriction that x<y<z

coolotter 2021-03-12 19:16:15
remove the 0<x<y<z constraint

CircleInvert 2021-03-12 19:16:15
Remove

$a<b<c$

yalex999 2021-03-12 19:16:15
x < y < z

RedFireTruck 2021-03-12 19:16:15

$0<x<y<z$

yayy 2021-03-12 19:16:15
a b and c don't have to be ordered

b20081 2021-03-12 19:16:15
0<x<y<z, we can count the unordered and place them in order

MaryM 2021-03-12 19:16:18
Let's just count positive integer solutions to

$x + y + z = 66$ without any restrictions.

MaryM 2021-03-12 19:16:33
This is simpler because there's a commonly-used technique to solve this.

sugar_rush 2021-03-12 19:16:49
stars & bars?

peace09 2021-03-12 19:16:49
Stars-and-Bars!

volcanogobbler 2021-03-12 19:16:49
stars and bars

BillXu21 2021-03-12 19:16:51
stars and bars

MaryM 2021-03-12 19:16:53
Imaging placing the

$66$ coins in a row.

MaryM 2021-03-12 19:17:11
Then to separate them into three piles, what do we need to do?

yayy 2021-03-12 19:17:51
put dividers between them

Lilavigne 2021-03-12 19:17:51
insert 2 bars

happyhari 2021-03-12 19:17:51
make two bars

functionalmath 2021-03-12 19:17:51
draw "bars" to distinguish between a b and c

mm999aops 2021-03-12 19:17:51
place two bars except 'stars' might be equal

peace09 2021-03-12 19:17:51
Place 2 Separators Among the 66 Coins!

floradaisy136 2021-03-12 19:17:51
put in two bars

MaryM 2021-03-12 19:17:55
We need to choose two of the

$65$ "slots" between coins, to separate the coins into three piles.

MaryM 2021-03-12 19:18:01
And in how many ways can we do this?

skyleristhecoolest 2021-03-12 19:18:18
65 choose 2

coolotter 2021-03-12 19:18:18
65 choose 2 = 2080

yayy 2021-03-12 19:18:18
65 choose 2

BillXu21 2021-03-12 19:18:18
65 pick 2

sugar_rush 2021-03-12 19:18:18

$\dbinom{65}{2}=2080$

MT-Math 2021-03-12 19:18:18
65 choose 2

mm999aops 2021-03-12 19:18:18
65*64/2

functionalmath 2021-03-12 19:18:18
65C2

rockyrockrock 2021-03-12 19:18:18
65 choose 2

CircleInvert 2021-03-12 19:18:20

$\binom{65}{2}$

nathanqiu 2021-03-12 19:18:20
65 choose 2

MaryM 2021-03-12 19:18:22
We can do this in

$\dbinom{65}{2}$ ways.

MaryM 2021-03-12 19:18:27
So there are

$\dbinom{65}{2}$ solutions to

$x+y+z = 66$ , where all we require is that

$x,y,z$ are positive integers.

MaryM 2021-03-12 19:18:35
Now, which of these solutions are not solutions to our original problem?

sugar_rush 2021-03-12 19:19:31
when

$x=y$ or

$x=z$ or

$y=z$

Irving1004 2021-03-12 19:19:31
When at least two groups have the same numebr of coins

yayy 2021-03-12 19:19:31
when

$x=y$ or

$x=z$ or

$y=z$

nathanqiu 2021-03-12 19:19:31
the ones where there are two equal piles, or 3 equal piles

Montclarion6 2021-03-12 19:19:31
Any time two of the piles are equal.

uneducatedpotato 2021-03-12 19:19:31
when two or more of these are equal

MaryM 2021-03-12 19:19:34
One problem we might have is that two (or all three) of the piles might have the same number of coins.

MaryM 2021-03-12 19:19:40
Can we count these?

mm999aops 2021-03-12 19:20:24
yes

happyhari 2021-03-12 19:20:24
yes

Irving1004 2021-03-12 19:20:24
Yes

Benranger 2021-03-12 19:20:24
yes

sugar_rush 2021-03-12 19:20:24
yes!

mayowl 2021-03-12 19:20:24
yes

peace09 2021-03-12 19:20:24
Yes!

MaryM 2021-03-12 19:20:26
I think so!

MaryM 2021-03-12 19:20:32
Let's try it.

MaryM 2021-03-12 19:20:51
Let's start with solutions where

$x=y$ .

MaryM 2021-03-12 19:20:58
What equation do we get in this case?

yayy 2021-03-12 19:21:28

$2x + z = 66$

peace09 2021-03-12 19:21:28

$2x+z=66.$

happyhari 2021-03-12 19:21:28
2x + z = 66

BillXu21 2021-03-12 19:21:28
2x+z=66

Lilavigne 2021-03-12 19:21:28
2x+z=66

Z_Lu 2021-03-12 19:21:28
2x+z=66

mm999aops 2021-03-12 19:21:28
2x+z=66

MaryM 2021-03-12 19:21:38
If

$x=y$ , then we have

$x + x + z = 66$ , or

$2x + z = 66$ .

MaryM 2021-03-12 19:21:39
How many solutions are there for positive

$x$ and

$z$ ?

Kesav 2021-03-12 19:22:11
32

Montclarion6 2021-03-12 19:22:11
32

functionalmath 2021-03-12 19:22:11
32

416554 2021-03-12 19:22:11
32

mm999aops 2021-03-12 19:22:11
as long as z is even- so z=2,z=4 to z=64

yayy 2021-03-12 19:22:11
32

sugar_rush 2021-03-12 19:22:11

$32$ right?

foodisgood 2021-03-12 19:22:11
32

b20081 2021-03-12 19:22:11
32

MaryM 2021-03-12 19:22:14
We must have

$0 < 2x < 66$ .

MaryM 2021-03-12 19:22:17
So we must have

$0 < x < 33$ .

MaryM 2021-03-12 19:22:21
Does every such

$x$ give us a solution?

sugar_rush 2021-03-12 19:22:45
yep

BillXu21 2021-03-12 19:22:45
yes

coolotter 2021-03-12 19:22:45
yes

peace09 2021-03-12 19:22:45
Yes!

happyhari 2021-03-12 19:22:45
yes

Lilavigne 2021-03-12 19:22:45
yes

yayy 2021-03-12 19:22:45
yes

b20081 2021-03-12 19:22:45
yes

RohanQV 2021-03-12 19:22:45
Yes!

t_ameya 2021-03-12 19:22:45
yes

MaryM 2021-03-12 19:22:49
Yes: once we have

$0 < x < 33$ , we set

$z = 66 - 2x$ .

MaryM 2021-03-12 19:22:59
There are

$32$ integers

$x$ with

$0 < x < 33$ , so there are

$32$ solutions with

$x=y$ .

MaryM 2021-03-12 19:23:05
How many solutions with

$x=z$ ?

MaryM 2021-03-12 19:23:55
(I mean, we are fixing

$x=z$ , but this time

$y$ might not the equal to

$x$ .)

t_ameya 2021-03-12 19:24:05
32

416554 2021-03-12 19:24:05
32

happyhari 2021-03-12 19:24:05
32

sugar_rush 2021-03-12 19:24:05
same

peace09 2021-03-12 19:24:05
32?

coolotter 2021-03-12 19:24:05
32 again

RohanQV 2021-03-12 19:24:05
32 also!

BillXu21 2021-03-12 19:24:05
32

s-sawen-07 2021-03-12 19:24:05
same?

MT-Math 2021-03-12 19:24:05
32

happyhari 2021-03-12 19:24:05
32, as well

volcanogobbler 2021-03-12 19:24:05

$32$ ?

ancientwarrior 2021-03-12 19:24:05
the same

MaryM 2021-03-12 19:24:07
By symmetry, it's exactly the same!

MaryM 2021-03-12 19:24:11
So there are

$32$ solutions with

$x=z$ .

MaryM 2021-03-12 19:24:15
And how many solutions with

$y=z$ ?

Kesav 2021-03-12 19:24:28
32

Lilavigne 2021-03-12 19:24:28
32

happyhari 2021-03-12 19:24:28
32

kante314 2021-03-12 19:24:28
32

PiMasterMC 2021-03-12 19:24:28
32

awesomeandy 2021-03-12 19:24:28
32

peace09 2021-03-12 19:24:28
32, As Well!

RohanQV 2021-03-12 19:24:28
32 also!

s-sawen-07 2021-03-12 19:24:28
again .. same?

Z_Lu 2021-03-12 19:24:28
32

Irving1004 2021-03-12 19:24:28
32

MaryM 2021-03-12 19:24:31
Same:

$32$ .

MaryM 2021-03-12 19:24:34
So that gives us

$32+32+32 = 96$ solutions with two piles the same.

MaryM 2021-03-12 19:24:37
Right?

Professor-Mom 2021-03-12 19:25:06
nppe

yayy 2021-03-12 19:25:06
overcount

peace09 2021-03-12 19:25:06
Wait! Some Cases Overlap!

iiRishabii 2021-03-12 19:25:06
no!

Kesav 2021-03-12 19:25:06
no we are overcounting

happyhari 2021-03-12 19:25:06
no, we are overcounting, what if they are all the same

Professor-Mom 2021-03-12 19:25:06
No

Lilavigne 2021-03-12 19:25:06
no overcount

MT-Math 2021-03-12 19:25:06
we overcounted kinda

Irving1004 2021-03-12 19:25:06
no

Z_Lu 2021-03-12 19:25:06
no, overcounting

MaryM 2021-03-12 19:25:09
Not quite.

MaryM 2021-03-12 19:25:13
What about the solutions with all three piles the same?

MaryM 2021-03-12 19:25:17
Can this happen?

mm999aops 2021-03-12 19:25:51
yes-22+22+22

peace09 2021-03-12 19:25:51
Yes...

$x=y=z=22.$

coolotter 2021-03-12 19:25:51
yes, when all are 22

yayy 2021-03-12 19:25:51
yes, (all 22)

happyhari 2021-03-12 19:25:51
yes, if x = y = z = 22

Arrowhead575 2021-03-12 19:25:51
22,22,22

sugar_rush 2021-03-12 19:25:51
when all three have 22

Lilavigne 2021-03-12 19:25:51
yes when x=y=z=22

PiMasterMC 2021-03-12 19:25:51
all are eq to 22

skyleristhecoolest 2021-03-12 19:25:51
22 22 22

Irving1004 2021-03-12 19:25:51
yes

MT-Math 2021-03-12 19:25:51
yes, if x, y, and z all have 22

MaryM 2021-03-12 19:25:53
Yes, but only in one way: all three piles must have

$22$ coins.

MaryM 2021-03-12 19:25:59
But how many times have we counted this solution in our count of

$96$ ?

PiMasterMC 2021-03-12 19:26:24
3

coolotter 2021-03-12 19:26:24
3 times

t_ameya 2021-03-12 19:26:24
3?

volcanogobbler 2021-03-12 19:26:24

$3$ times!

Arrowhead575 2021-03-12 19:26:24
3

happyhari 2021-03-12 19:26:24
3 times

s-sawen-07 2021-03-12 19:26:24
3

peace09 2021-03-12 19:26:24
3 Times

Irving1004 2021-03-12 19:26:24
3 times

BillXu21 2021-03-12 19:26:24
3 times

MaryM 2021-03-12 19:26:28
We've counted it three times: once for each pair of piles.

MaryM 2021-03-12 19:26:31
So what do we need to do?

s-sawen-07 2021-03-12 19:27:05
so we must get rid of 2 times

mm999aops 2021-03-12 19:27:05
minues two times

Summer2019 2021-03-12 19:27:05
subtract

peace09 2021-03-12 19:27:05
Subtract 2 Pairs

happyhari 2021-03-12 19:27:05
subtract 2(1) = 2

Lilavigne 2021-03-12 19:27:05
subtract it twice

awesomeandy 2021-03-12 19:27:05
subtract 2

happyhari 2021-03-12 19:27:05
subtract 2

coolotter 2021-03-12 19:27:05
add 2 back!

functionalmath 2021-03-12 19:27:05
subtract 2

PiMasterMC 2021-03-12 19:27:05
subtract 2 from the total

dolphindash 2021-03-12 19:27:05
subtract 2

happyhari 2021-03-12 19:27:05
96 - 2 = 94

MaryM 2021-03-12 19:27:07
We need to subtract

$2$ from our count of

$96$ , to account for the fact that the

$(22,22,22)$ solution was counted three times, and we only want to count it once.

MaryM 2021-03-12 19:27:11
Therefore, there are

$96 - 2 = 94$ solutions with two (or all three) piles the same.

MaryM 2021-03-12 19:27:15
How does that help?

Arrowhead575 2021-03-12 19:28:36
Subtract from original amount

mm999aops 2021-03-12 19:28:36
we can subtract that from the total number

Kesav 2021-03-12 19:28:36
we do 2080 - 94 = 1986!

sugar_rush 2021-03-12 19:28:36
subtract from 2080

mm999aops 2021-03-12 19:28:36
the total is 65*32, so we want 65*32-94?

functionalmath 2021-03-12 19:28:36
subtract 94 illegal cases from the 2080 legal ones

t_ameya 2021-03-12 19:28:36
this is the amount of cases that don't work

RohanQV 2021-03-12 19:28:36
Helps remove overcounting cases

Montclarion6 2021-03-12 19:28:36
2080-94

MaryM 2021-03-12 19:28:44
That means there are

$\dbinom{65}{2} - 94$ solutions with all three piles different.

MaryM 2021-03-12 19:28:51
And now what?

MaryM 2021-03-12 19:28:55
How many of these will have

$x < y < z$ ?

yayy 2021-03-12 19:29:27
the 6 ways of ordering

$x, y, z$ are symetrical, so we can divide by 6

nathanqiu 2021-03-12 19:29:27
then divide by 6

nathanqiu 2021-03-12 19:29:27
divide by 6 to account for overcount

biketoday 2021-03-12 19:29:27
1 out of every 6

rockyrockrock 2021-03-12 19:29:27
we subtract and divide by 6 !

Lilavigne 2021-03-12 19:29:27
1/6 of them

coolotter 2021-03-12 19:29:27
WE ARE NOT DONE, divide by 6 now since 3! = six

PiMasterMC 2021-03-12 19:29:27
the large number divide by 6

functionalmath 2021-03-12 19:29:27
divide by 3!=6 because ordering

yuanyuanC 2021-03-12 19:29:27
1/6 of them?

MaryM 2021-03-12 19:29:30
Here there's a clever observation.

MaryM 2021-03-12 19:29:49
If we have a solution with all three piles different,

$x,y,z$ could be in any order. And all the orders are equally likely!

MaryM 2021-03-12 19:29:58
That is, if we pick a random solution, we could have

$x,y,z$ in any of the

$3! = 6$ possible orders of smallest, medium, largest.

MaryM 2021-03-12 19:30:04
So the likelihood that

$x < y < z$ (the order we want) is the order we get is

$\dfrac16$ .

MaryM 2021-03-12 19:30:11
Therefore, exactly

$\dfrac16$ of our "all three piles different" solutions have

$x$ smallest,

$y$ middle, and

$z$ largest.

MaryM 2021-03-12 19:30:19
So our final count is

$\dfrac{\dbinom{65}{2}-94}{6}$ .

MaryM 2021-03-12 19:30:22
And now we compute!

MaryM 2021-03-12 19:30:26
What is

$\dbinom{65}{2}$ ?

functionalmath 2021-03-12 19:30:53
2080

kante314 2021-03-12 19:30:53
2008

ancientwarrior 2021-03-12 19:30:53
2080

b20081 2021-03-12 19:30:53
2080

skyleristhecoolest 2021-03-12 19:30:53
2080

Arrowhead575 2021-03-12 19:30:53
2080

jxwis2010 2021-03-12 19:30:53
2080

GabeW1234 2021-03-12 19:30:53
65*32=2080

yuanyuanC 2021-03-12 19:30:53
2080

PiMasterMC 2021-03-12 19:30:53
2080

MaryM 2021-03-12 19:30:55
It's

$\dfrac{65 \cdot 64}{2} = 65 \cdot 32 = 2080$ .

MaryM 2021-03-12 19:31:00
So our answer is

$\dfrac{2080-94}{6}$ .

MaryM 2021-03-12 19:31:09
What is this equal to?

coolotter 2021-03-12 19:31:32
so our final solution is (2080-96+2)/6 = 1986/6 =

$\boxed{331}$ !

Kesav 2021-03-12 19:31:32
331

yuanyuanC 2021-03-12 19:31:32
331

happyhari 2021-03-12 19:31:32
331

sugar_rush 2021-03-12 19:31:32

$\boxed{331}$ is our final answer!

PiMasterMC 2021-03-12 19:31:32
331

b20081 2021-03-12 19:31:32
331

GabeW1234 2021-03-12 19:31:32
331

mm999aops 2021-03-12 19:31:32
1986/6 = 331

uneducatedpotato 2021-03-12 19:31:32
331?

MaryM 2021-03-12 19:31:34
This simplifies to

$\dfrac{1986}{6} = \boxed{331}$ .

MaryM 2021-03-12 19:31:43
5. Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

MaryM 2021-03-12 19:32:00
You might be inclined to write a 3-term arithmetic sequence as

$a, a+d, a+2d$ for initial term

$a$ and common difference

$d$ .

MaryM 2021-03-12 19:32:03
Does anybody have another suggestion?

Mathaddict3825 2021-03-12 19:32:25
a-d, a, a+d

floradaisy136 2021-03-12 19:32:25
a-d, a, a+d

helloworld_ 2021-03-12 19:32:25
a - d, a, a + d

sugar_rush 2021-03-12 19:32:25

$x-d, x, x+d$

yayy 2021-03-12 19:32:25

$a-r, a, a+r$

mathicorn 2021-03-12 19:32:25
write as a-d, a, a+d?

Asher51 2021-03-12 19:32:25
b-d, b, and b+d

kante314 2021-03-12 19:32:25
Let the middle number be a

PiMasterMC 2021-03-12 19:32:25
a-d, a, a + d

CircleInvert 2021-03-12 19:32:27
Use

$a-d$ ,

$a$ , and

$a+d$

MaryM 2021-03-12 19:32:29
Yes, I find that when we're working with arithmetic sequences, especially of odd length, we can take advantage of some symmetry by focusing on the middle term.

MaryM 2021-03-12 19:32:32
That is, let's write our special sequence as

$a-d, a, a+d$ with middle term

$a$ and common difference

$d$ .

MaryM 2021-03-12 19:32:39
Hopefully this will provide some advantage to us later.

MaryM 2021-03-12 19:32:45
Let's write down the condition for this sequence to be special.

MaryM 2021-03-12 19:32:50
In words, we have

$\text{sum of squares of all three terms} = (\text{middle term}) \cdot (\text{square of the common difference}).$

MaryM 2021-03-12 19:32:56
What does this give us?

pinkpig 2021-03-12 19:33:38
ad^2 = 3a^2+2d^2

peace09 2021-03-12 19:33:38

$(a-d)^2+a^2+(a+d)^2=3a^2+2d^2=ad^2.$

CircleInvert 2021-03-12 19:33:38

$3a^2+2d^2=ad^2$

Irving1004 2021-03-12 19:33:38
well (a-b)^2 + (a+b)^2 = 2a^2 + 2b^2

Summer2019 2021-03-12 19:33:38
3a^2+2d^2=ad^2

kante314 2021-03-12 19:33:38
(a-d)^2 + a^2 + (a+d)^2 = a x d^2

t_ameya 2021-03-12 19:33:38

$a^2+(a-d)^2+(a+d)^2=a\cdotd^2$

Kesav 2021-03-12 19:33:38
3a^2 + 2d^2 = ad

Asher51 2021-03-12 19:33:38

$a^2 + (a-d)^2 + (a+d)^2 = ad^2$

pinkpig 2021-03-12 19:33:38
ad^2=3a^2+2d^2

MaryM 2021-03-12 19:33:41
We get

$(a-d)^2 + a^2 + (a+d)^2 = ad^2.$

MaryM 2021-03-12 19:34:06
Which simplifies to

$3a^2 + 2d^2 = ad^2.$

MaryM 2021-03-12 19:34:10
(Note that the

$+2ad$ from the first square cancels with the

$-2ad$ from the third square! This is why writing the sequence symmetrically is useful.)

MaryM 2021-03-12 19:34:15
Now what?

peace09 2021-03-12 19:35:05

$3a^2=ad^2-2d^2,$ Implying

$d^2=\tfrac{3a^2}{a-2}.$

iiRishabii 2021-03-12 19:35:05

$3a^2=ad^2-2d^2$

$3a^2=d(ad-2d)$

sugar_rush 2021-03-12 19:35:05
subtract off

$2d^{2}$ from each side

mm999aops 2021-03-12 19:35:05

$\frac{3a^2}{a-2}=d^2$

Lilavigne 2021-03-12 19:35:09
isolate d^2

MaryM 2021-03-12 19:35:12
We can try isolating the variable

$d.$

MaryM 2021-03-12 19:35:16
We can write

$3a^2 = ad^2 - 2d^2,$ so

$(a - 2) d^2 = 3a^2.$

MaryM 2021-03-12 19:35:24
What does this equation say about

$a?$

MaryM 2021-03-12 19:35:51
Hint: think about divisibility.

sugar_rush 2021-03-12 19:36:40

$a-2$ divides

$3a^{2}$ ?

pinkpig 2021-03-12 19:36:40
3/(a-2) is a perfect square because a^2 is already a perfect square

Irving1004 2021-03-12 19:36:40
a-2 divides into 3a^2 and this is a perfect square

pinkpig 2021-03-12 19:36:40
a-2 divides 3a^2

Unevenlogic 2021-03-12 19:36:40
(a-2) | 3a^2

MaryM 2021-03-12 19:36:43
This equation says that

$a - 2$ divides

$3a^2.$ That seems like it will narrow down the possible values of

$a.$

MaryM 2021-03-12 19:36:48
We could try dividing

$3a^2$ by

$a - 2$ using long division, but a faster way is to use a substitution.

MaryM 2021-03-12 19:36:54
Let

$b = a - 2.$ Then

$a = b + 2,$ so

$bd^2 = 3(b + 2)^2 = 3(b^2 + 4b + 4) = 3b^2 + 12b + 12.$

MaryM 2021-03-12 19:37:19
What does this equation say about

$b?$

MaryM 2021-03-12 19:37:58
Do you know a number that is a multiple of

$b$ by looking at that equation?

math9990 2021-03-12 19:38:17
b divides 12

pinkpig 2021-03-12 19:38:17
b divides 12

Unevenlogic 2021-03-12 19:38:17
b | 3b^2 + 12b + 12

sugar_rush 2021-03-12 19:38:17

$b$ divides

$12$ because it divides

$3b^{2}+12b+12$

pinkpig 2021-03-12 19:38:17
After you simplify it, you get that b is divisible by 12

PiMasterMC 2021-03-12 19:38:17
12?

MaryM 2021-03-12 19:38:20
Since

$b$ divides the left-hand side,

$b$ divides the right-hand side. Also,

$b$ divides

$3b^2$ and

$12b,$ so

$b$ must divide 12.

MaryM 2021-03-12 19:38:27
And from the equation

$(a - 2) d^2 = 3a^2,$

$b = a - 2$ must be positive.

MaryM 2021-03-12 19:38:31
At this point, we could go through the divisors of

$12$ , plug them into the equation

$d^2 = \frac{3(b + 2)^2}{b},$

and see which values lead to squares.

MaryM 2021-03-12 19:38:43
But we can narrow down the possible values even further, by writing the equation

$bd^2 = 3(b + 2)^2$ as

$3bd^2 = 9(b + 2)^2.$

MaryM 2021-03-12 19:38:48
What does this equation say about

$b?$

MaryM 2021-03-12 19:39:13
Hint: look at all those perfect squares!

GabeW1234 2021-03-12 19:40:11
3b is a square

Montclarion6 2021-03-12 19:40:11
b=3k^2

pinkpig 2021-03-12 19:40:11
b is three times a perfect square

math9990 2021-03-12 19:40:11
3b is a perfect square

Rubikscube3.1415 2021-03-12 19:40:11
3b is a perfect square

Lilavigne 2021-03-12 19:40:11
b=3*perfect square

MaryM 2021-03-12 19:40:17
Since

$d^2$ and

$9(b + 2)^2$ are perfect squares,

$3b$ must be a perfect square. ( This is exactly the same as saying that

$b$ is three times a perfect square).

MaryM 2021-03-12 19:40:24
Which divisors

$b$ of 12 have the property that

$3b$ is a perfect square?

pinkpig 2021-03-12 19:41:25
3 and 12

sugar_rush 2021-03-12 19:41:25

$3$ and

$12$ itself

foodisgood 2021-03-12 19:41:25
3,12

Summer2019 2021-03-12 19:41:25
b=3, 12

peace09 2021-03-12 19:41:25
3 and 12.

Arrowhead575 2021-03-12 19:41:25
3,12

yayy 2021-03-12 19:41:25
3, 12

mm999aops 2021-03-12 19:41:25
3 12

Unevenlogic 2021-03-12 19:41:25
3 and 12

MaryM 2021-03-12 19:41:27
Only

$b = 3$ and

$b = 12$ have this property.

MaryM 2021-03-12 19:41:31
So,

$a = 5$ or

$a = 14.$

MaryM 2021-03-12 19:41:36
If

$a = 5,$ then

$d^2 = \dfrac{3 \cdot 5^2}{3} = 25.$ Then what is

$d?$

BillXu21 2021-03-12 19:42:00
5

peace09 2021-03-12 19:42:00

$d=5.$

GabeW1234 2021-03-12 19:42:00
5

uneducatedpotato 2021-03-12 19:42:00
5

mm999aops 2021-03-12 19:42:00
d is 5 because it's incresaing

Euler1728 2021-03-12 19:42:00
5

coolotter 2021-03-12 19:42:00
d = 5

foodisgood 2021-03-12 19:42:00
5

MaryM 2021-03-12 19:42:02
If

$d^2 = 25,$ then

$d = 5.$ The value of

$d$ must be positive because the sequence is strictly increasing.

MaryM 2021-03-12 19:42:07
Similarly, if

$a = 14,$ then

$d^2 = \dfrac{3 \cdot 14^2}{12} = 49,$ so

$d = 7.$

MaryM 2021-03-12 19:42:11
So what is the sum of all possible third terms?

pinkpig 2021-03-12 19:42:52
031

iiRishabii 2021-03-12 19:42:52
31

Unevenlogic 2021-03-12 19:42:52
10 + 21 = 31

pinkpig 2021-03-12 19:42:52
10+21=31

Lilavigne 2021-03-12 19:42:52
31

sugar_rush 2021-03-12 19:42:52

$10+21=\boxed{31}$ !

foodisgood 2021-03-12 19:42:52
31

coolotter 2021-03-12 19:42:52
we can have 0 5 10 or 7 14 21 so we get 10 + 21 =

$\boxed{31}$ !

Kesav 2021-03-12 19:42:52
31

functionalmath 2021-03-12 19:42:52
31

Z_Lu 2021-03-12 19:42:52
31

MaryM 2021-03-12 19:42:54
The sum of all possible third terms is

$(5 + 5) + (14 + 7) = 10 + 21 = \boxed{031}.$

BillXu21 2021-03-12 19:43:02

BillXu21 2021-03-12 19:43:02
yay

MaryM 2021-03-12 19:43:05
6. Segments

$\overline{AB},\,$

$\overline{AC},\,$ and

$\overline{AD}$ are edges of a cube and segment

$\overline{AG}$ is a diagonal through the center of the cube. Point

$P$ satisfies

$BP=60\sqrt{10},\,$

$CP=60\sqrt5,\,$

$DP=120\sqrt2,\,$ and

$GP=36\sqrt7.$ Find

$AP.$

MaryM 2021-03-12 19:43:15
Any ideas?

Lilavigne 2021-03-12 19:44:03
coordinates?

Summer2019 2021-03-12 19:44:03
coordinates

functionalmath 2021-03-12 19:44:03
put it on the coordinate plane

BillXu21 2021-03-12 19:44:03
coordinate

bbmmall 2021-03-12 19:44:03
coordinate bash

b20081 2021-03-12 19:44:03
coordinate bash

math9990 2021-03-12 19:44:03
coordinates, then calculate BP CP DP GP in terms of the coordinates, then you can use that systems of equations to find AP

MaryM 2021-03-12 19:44:05
We want to compute a bunch of distances, so it may be helpful to work with coordinates.

MaryM 2021-03-12 19:44:11
How can we set the coordinates up?

GabeW1234 2021-03-12 19:44:54
a= (0,0,0)?

BillXu21 2021-03-12 19:44:54
a is the origin\

ancientwarrior 2021-03-12 19:44:54
Let A be the origin?

sugar_rush 2021-03-12 19:44:54

$A=\text{origin}$

functionalmath 2021-03-12 19:44:54
A is the origin, then set up the axis such that 3 edges are the axis

dbasu1 2021-03-12 19:44:54
A is the origin

pinkpig 2021-03-12 19:44:54
we could let a be the origin

MaryM 2021-03-12 19:44:57
I'd probably make

$A = (0,0,0)$ be the origin of 3-D space.

MaryM 2021-03-12 19:45:06
We're not told the side length of the cube.

MaryM 2021-03-12 19:45:10
So let's call it

$s$ .

MaryM 2021-03-12 19:45:14
Then what are good choices for

$B,C,D$ ?

math9990 2021-03-12 19:46:08
(s,0,0) (0,s,0) (0,0,x)

GabeW1234 2021-03-12 19:46:08
(0,0,s),(0,s,0),(s,0,0)

yayy 2021-03-12 19:46:08

$(s, 0, 0), (0, s, 0), (0, 0, s)$

ancientwarrior 2021-03-12 19:46:08
(s,0,0) (0,s,0) (0,0,s)

functionalmath 2021-03-12 19:46:08
(s,0,0), (0,s,0) and (0,0,s)

pinkpig 2021-03-12 19:46:08
B would be (s,0,0), C would be (0,s,0,) and D would be (0,0,s)

BillXu21 2021-03-12 19:46:08
b = (x,0,0) c = (0,s,0) d = (0,0,s) g = (s,s,s)

regulareagle 2021-03-12 19:46:08
(s,0,0) (0,s,0) and (0,0,s)

sugar_rush 2021-03-12 19:46:08

$B=(s, 0, 0), C=(0, s, 0), D=(0, 0, s)$

foodisgood 2021-03-12 19:46:08
(0,s,0), (s,0,0) and (0,0,s)?

MaryM 2021-03-12 19:46:15
We can set

$B = (s,0,0)$ ,

$C = (0,s,0)$ , and

$D = (0,0,s)$ .

MaryM 2021-03-12 19:46:22
Then what is

$G$ ?

sugar_rush 2021-03-12 19:46:51

$G=(s, s, s)$

pinkpig 2021-03-12 19:46:51
G would be (s,s,s)

BillXu21 2021-03-12 19:46:51
(s,s,s)

ancientwarrior 2021-03-12 19:46:51
(s, s, s)

Irving1004 2021-03-12 19:46:51
(s, s, s)

GabeW1234 2021-03-12 19:46:51
(s,s,s)

Lilavigne 2021-03-12 19:46:51
(s,s,s)

Asher51 2021-03-12 19:46:51
(s,s,s)

foodisgood 2021-03-12 19:46:51
(s,s,s,)

Summer2019 2021-03-12 19:46:51
G=(s,s,s)

dolphindash 2021-03-12 19:46:51
(s,s,s)

awesomeandy 2021-03-12 19:46:51
(s,s,s)

MaryM 2021-03-12 19:46:53
We have

$G = (s,s,s)$ .

MaryM 2021-03-12 19:46:56
And what about

$P$ ?

Irving1004 2021-03-12 19:47:25
(a, b, c)

sugar_rush 2021-03-12 19:47:25
let's call it

$(p, q, r)$

Unevenlogic 2021-03-12 19:47:25
(a, b, c)

BillXu21 2021-03-12 19:47:25
(x,y,z)

Summer2019 2021-03-12 19:47:25
(p,q,r)

functionalmath 2021-03-12 19:47:25
(x_0,y_0,z_0)

pinkpig 2021-03-12 19:47:25
we should let P be (x,y,z)

yayy 2021-03-12 19:47:25

$(x, y, z)$

MaryM 2021-03-12 19:47:28
We don't really know where

$P$ is, so let's just say

$P = (x,y,z)$ .

MaryM 2021-03-12 19:47:32
Now what?

sugar_rush 2021-03-12 19:47:57
3d distance formula bash!

BillXu21 2021-03-12 19:47:57
distance formula for 3D space

pinkpig 2021-03-12 19:47:57
set up a system of equations

Summer2019 2021-03-12 19:47:57
distance formula with coordinates

yayy 2021-03-12 19:47:57
equations

Z_Lu 2021-03-12 19:47:57
make uses of the distances to find P

Unevenlogic 2021-03-12 19:47:57
Use the 3d Pythagorean theorem to form equations

functionalmath 2021-03-12 19:47:57
use the distances given to set up a system of 4 by 4 equations and solve

MaryM 2021-03-12 19:48:01
Now we can write equations for the distances.

MaryM 2021-03-12 19:48:05
For instance, what is

$BP$ ?

MaryM 2021-03-12 19:48:52
(In terms of

$x,y,z$ and

$s$ , I mean.)

functionalmath 2021-03-12 19:49:22
(x-s)^2+y^2+z^2

sugar_rush 2021-03-12 19:49:22

$PB^{2}=p^{2}+(q-s)^{2}+r^{2}$

pinkpig 2021-03-12 19:49:22
(x-s)^2+(y^2)+z^2 = 36000

ancientwarrior 2021-03-12 19:49:22

$\sqrt{(x-s)^2+y^2+z^2}$

foodisgood 2021-03-12 19:49:22
sqrt[(x-s)^2+y^2+z^2]

Lilavigne 2021-03-12 19:49:22
sqrt{(x-s)^2+y^2+z^2}

Summer2019 2021-03-12 19:49:22
x^2+(y-s)^2+z^2=(60sqrt(10))^2

yayy 2021-03-12 19:49:22

$\sqrt[3]{(s-x)^2 + y^2 + z^2}$

Irving1004 2021-03-12 19:49:22
sqrt((x-s)^2 + y^2 + z^2)

sugar_rush 2021-03-12 19:49:22

$PB=\sqrt{(p-0)^2+(q-s)^2+(r-0)^2}$

BillXu21 2021-03-12 19:49:22
(x-s)^2+y^2+z^2 = 36000

GabeW1234 2021-03-12 19:49:22

$\sqrt{(x-s)^2+(y-0)^2+(z-0)^2}$

Asher51 2021-03-12 19:49:22

$\sqrt{(x-s)^2 + y^2 + z^2}$

MaryM 2021-03-12 19:49:26

$BP$ is the distance from

$(s,0,0)$ to

$(x,y,z)$ .

MaryM 2021-03-12 19:49:30
So

$BP = \sqrt{(x-s)^2 + y^2 + z^2}$ .

MaryM 2021-03-12 19:49:33
Hence, we have the equation

$\sqrt{(x-s)^2 + y^2 + z^2} = 60\sqrt{10}$ .

MaryM 2021-03-12 19:49:37
Ick, that's going to be hard to work with. Any ideas?

math9990 2021-03-12 19:50:09
square it

yayy 2021-03-12 19:50:09
square

Unevenlogic 2021-03-12 19:50:09
Square both sides

Summer2019 2021-03-12 19:50:09
square both sides

happyhari 2021-03-12 19:50:09
square both sides?

dolphindash 2021-03-12 19:50:09
square it

pinkpig 2021-03-12 19:50:09
we should square both sides of the equation and do the same for the rest of them to see if they can cancel out.

Z_Lu 2021-03-12 19:50:09
sqaure both sides

MaryM 2021-03-12 19:50:12
Let's square it.

MaryM 2021-03-12 19:50:15
Now we have

$(x-s)^2 + y^2 + z^2 = 60^2 \cdot 10$ .

MaryM 2021-03-12 19:50:19
Now what?

GabeW1234 2021-03-12 19:51:01
do the same for CP, DP, and GP

Irving1004 2021-03-12 19:51:01
Find CP and DP and GP and find AP

pinkpig 2021-03-12 19:51:01
do the same for the rest of them

BillXu21 2021-03-12 19:51:01
now write the rest of the equasions

yayy 2021-03-12 19:51:01
set up equations for all of the other points

Lilavigne 2021-03-12 19:51:01
do the same for all other segments

Summer2019 2021-03-12 19:51:01
same with CP, DP, and GP

sugar_rush 2021-03-12 19:51:01
similarly for

$PC, PD, PG$

happyhari 2021-03-12 19:51:01
find other equations

foodisgood 2021-03-12 19:51:01
set up the other equations?

Bananaman27 2021-03-12 19:51:01
do it for more lengths we know

MaryM 2021-03-12 19:51:04
We can write equations for the other three distances.

MaryM 2021-03-12 19:51:08
When we do, we get four equations in total:

MaryM 2021-03-12 19:51:14

$\begin{align*} (x-s)^2 + y^2 + z^2 &= 60^2 \cdot 10, \\ x^2 + (y-s)^2 + z^2 &= 60^2 \cdot 5, \\ x^2 + y^2 + (z-s)^2 &= 120^2 \cdot 2, \\ (x-s)^2 + (y-s)^2 + (z-s)^2 &= 36^2 \cdot 7. \end{align*}$

MaryM 2021-03-12 19:51:19
It's four equations in four variables, but unfortunately they're quadratics, not linear.

MaryM 2021-03-12 19:51:26
Can we solve them nonetheless?

MaryM 2021-03-12 19:51:33
Do we have to?

Z2589013 2021-03-12 19:52:02
no

math-Passion 2021-03-12 19:52:02
No

jxwis2010 2021-03-12 19:52:02
no

yayy 2021-03-12 19:52:02
we just need

$x^2 + y^2 + z^2$

gracefulharmony 2021-03-12 19:52:02
No

Z_Lu 2021-03-12 19:52:02
no, since we are trying to find AP

MaryM 2021-03-12 19:52:05
Keep you eye on the ball! What are we trying to find?

pinkpig 2021-03-12 19:52:27
AP

ancientwarrior 2021-03-12 19:52:27
we want

$\sqrt{x^2+y^2+z^2}$ , so maybe there's a way to manipulate

sugar_rush 2021-03-12 19:52:27
the distance

$PA$

awesomeandy 2021-03-12 19:52:27
AP

happyhari 2021-03-12 19:52:27
AP

Summer2019 2021-03-12 19:52:27
sqrt(x^2+y^2+z^2)

functionalmath 2021-03-12 19:52:27
sqrt(x^2+y^2+z^2)

Z_Lu 2021-03-12 19:52:27
we just need to find what x^2+y^2+z^2 will be equal to

BillXu21 2021-03-12 19:52:27
AP

MaryM 2021-03-12 19:52:28
We want

$AP$ .

MaryM 2021-03-12 19:52:33
But

$AP = \sqrt{x^2 + y^2 + z^2}$ .

MaryM 2021-03-12 19:52:37
Can we find that quantity without solving the system?

helloworld_ 2021-03-12 19:53:14
No, subtract the first 3 all from the third.

math9990 2021-03-12 19:53:14
add up the first three and subtract the fourth from the sum

b20081 2021-03-12 19:53:14
add the first 3 and subtract the 4th

GabeW1234 2021-03-12 19:53:14
yes

BillXu21 2021-03-12 19:53:14
yes

rockyrockrock 2021-03-12 19:53:14
add the first 3 and subtract the 4th and u have 2(x^2+y^2+z^2)

yayy 2021-03-12 19:53:14
add first 3 equations and subtract 4th

pinkpig 2021-03-12 19:53:14
the terms cancel out! ((first equation)+(second equation)+(third equation)-(fourth equation))/2 = AP^2

gorefeebuddie 2021-03-12 19:53:14
Yes, add everything up divide 2 and subtract the last equation

happyhari 2021-03-12 19:53:14
add all of the equations together?

BestAOPS 2021-03-12 19:53:14
add together the first three then subtract the last equation

Z2589013 2021-03-12 19:53:14
we can solve for

$AP^2$

sugar_rush 2021-03-12 19:53:14
Yes! Just add the first three equations and subtract off the fourth.

MaryM 2021-03-12 19:53:18
Yes! Add the first three equations and subtract the fourth.

MaryM 2021-03-12 19:53:24
Then all the

$(x-s)^2$ ,

$(y-s)^2$ , and

$(z-s)^2$ terms cancel!

MaryM 2021-03-12 19:53:29
And what's left on the left side?

Summer2019 2021-03-12 19:54:02
2(x^2+y^2+z^2)

GabeW1234 2021-03-12 19:54:02
2 times the square of AP

Z2589013 2021-03-12 19:54:02

$2(x^2+y^2+z^2)$

sugar_rush 2021-03-12 19:54:02

$2(x^{2}+y^{2}+z^{2}),$ which is

$2PA^{2}$

pinkpig 2021-03-12 19:54:02
2(x^2+y^2+z^2)

Z_Lu 2021-03-12 19:54:02
2x^2+2y^2+2z^2

GabeW1234 2021-03-12 19:54:02

$2x^2+2y^2+2Z^2$

MaryM 2021-03-12 19:54:05
We're left with

$2(x^2+y^2+z^2)$ .

MaryM 2021-03-12 19:54:09
But that's

$2(AP)^2$ !

MaryM 2021-03-12 19:54:14
So we get

$2(AP)^2 = 60^2 \cdot 10 + 60^2 \cdot 5 + 120^2 \cdot 2 - 36^2 \cdot 7.$

MaryM 2021-03-12 19:54:22
To finish, we have to compute

$AP$ .

pinkpig 2021-03-12 19:54:42
we can just simplify this

functionalmath 2021-03-12 19:54:42
divide and then factor out a 6^2

MaryM 2021-03-12 19:54:45
I'd probably factor out a common factor of

$12^2$ first:

$2(AP)^2 = 12^2(5^2 \cdot 10 + 5^2 \cdot 5 + 10^2 \cdot 2 - 3^2 \cdot 7).$

MaryM 2021-03-12 19:54:52
Then dividing by

$2$ and computing the squares gives

$(AP)^2 = 72(25 \cdot 10 + 25 \cdot 5 + 100 \cdot 2 - 9 \cdot 7).$

MaryM 2021-03-12 19:54:59
This is

$(AP)^2 = 72(250 + 125 + 200 - 63).$

MaryM 2021-03-12 19:55:04
And this is

$(AP)^2 = 72(512)$ .

MaryM 2021-03-12 19:55:11
Aha! This is

$(AP)^2 = 36(1024) = 6^2 \cdot 2^{10}$ .

pinkpig 2021-03-12 19:55:22
the answer is 192

sugar_rush 2021-03-12 19:55:22

$PA=\boxed{192}$

pimath 2021-03-12 19:55:22
AP = 192

MaryM 2021-03-12 19:55:24
So its square root is

$AP = 6 \cdot 2^5 = 6 \cdot 32 = \boxed{192}$ .

MaryM 2021-03-12 19:55:40
7. Find the number of pairs

$(m, n)$ of positive integers with

$1 \leq m < n \leq 30$ such that there exists a real number

$x$ satisfying

$\sin(mx) + \sin(nx) = 2.$

MaryM 2021-03-12 19:55:50
Hmmm...that's a weird equation.

MaryM 2021-03-12 19:55:54
What's really going on here?

MaryM 2021-03-12 19:56:14
How can two sines sum to

$2$ ?

rockyrockrock 2021-03-12 19:56:30
both have to be 1

ancientwarrior 2021-03-12 19:56:30
they both must be 1

sugar_rush 2021-03-12 19:56:30

$sin(mx)=sin(nx)=1$

Irving1004 2021-03-12 19:56:30
Well sin has range -1 to 1

pinkpig 2021-03-12 19:56:30
sin(x)<=1 so sin(mx)=sin(nx)=1

kante314 2021-03-12 19:56:30
algebra

GabeW1234 2021-03-12 19:56:30
1+1=2

Summer2019 2021-03-12 19:56:30
both are equal to 1

CircleInvert 2021-03-12 19:56:30

$\sin(mx)=\sin(nx)=1$

mathisfun17 2021-03-12 19:56:30
each are 1

foodisgood 2021-03-12 19:56:30
both 1

Irving1004 2021-03-12 19:56:30
and sin nx = 1

jxwis2010 2021-03-12 19:56:30
sin(mx)=1, sin(nx)=1

MaryM 2021-03-12 19:56:34
Sines are always between

$-1$ and

$1$ inclusive.

MaryM 2021-03-12 19:56:40
So the only way to satisfy that equation is to have

$\sin(mx) = \sin(nx) = 1$ .

MaryM 2021-03-12 19:56:43
But we can say more than that, right?

MaryM 2021-03-12 19:56:50
When is

$\sin\theta= 1$ ?

uneducatedpotato 2021-03-12 19:57:55
mx and nx = 90(mod 360)

sugar_rush 2021-03-12 19:57:55
when

$\theta=90\pmod{360}$

Summer2019 2021-03-12 19:57:55
theta=pi/2+2kpi

Z2589013 2021-03-12 19:57:55

$\pi/2+2\pi{n}$ where

$n$ is a positive integer

CircleInvert 2021-03-12 19:57:55
When

$\theta=k\pi+\frac{pi}{2}$

Asher51 2021-03-12 19:57:55

$\theta = \frac{\pi}{2} + 2 \pi n$

rockyrockrock 2021-03-12 19:57:55
xpi+1/2pi

mathisfun17 2021-03-12 19:57:55
theta = pi/2 + 2pi*k, k is an integer

pinkpig 2021-03-12 19:57:55
when the angle is 90 more than 360k for any constant k

jxwis2010 2021-03-12 19:57:55
$\theta = 2k\pi + \pi/2

MaryM 2021-03-12 19:58:01
Think about the graph of sine:

MaryM 2021-03-12 19:58:07

MaryM 2021-03-12 19:58:14
We have

$\sin\left(\dfrac\pi2\right) = 1$ .

MaryM 2021-03-12 19:58:37
Also, we can add any multiple of

$2\pi$ to

$\dfrac{\pi}{2}$ , because sine is periodic.

MaryM 2021-03-12 19:58:45
So if

$\sin\theta = 1$ , then we must have

$\theta = \left(2k + \dfrac12\right)\pi$ for some integer

$k$ .

coolotter 2021-03-12 19:58:49
are we doing this in degrees or radians?

happyhari 2021-03-12 19:58:49
is pi/2 radians equivalent to 90 degrees?

MaryM 2021-03-12 19:59:01
We will work with radians, though the logic for degrees is the same.

MaryM 2021-03-12 19:59:18
We have that

$\pi/2$ radians is the same as

$90^\circ$ .

MaryM 2021-03-12 19:59:38
Okay?

happyhari 2021-03-12 19:59:54
ok

GabeW1234 2021-03-12 19:59:54
ok

coolotter 2021-03-12 19:59:54
sounds good

Irving1004 2021-03-12 19:59:54
yes

uneducatedpotato 2021-03-12 19:59:54
yep

MaryM 2021-03-12 19:59:57
Back to our equation, that means that we must find some

$x$ such that:

$\begin{align*} mx &= \left(2k + \frac12\right)\pi, \\ nx &= \left(2\ell + \frac12\right)\pi, \end{align*}$

for some integers

$k$ and

$\ell$ .

MaryM 2021-03-12 20:00:03
What can we do with these equations?

MaryM 2021-03-12 20:00:33
Hint: think about an way to eliminate

$x$ .

Kevinisawesome 2021-03-12 20:00:49
divide

Unevenlogic 2021-03-12 20:00:49
Divide?

CircleInvert 2021-03-12 20:00:49
Divide them

foodisgood 2021-03-12 20:00:49
divide?

pinkpig 2021-03-12 20:00:49
divide them

MaryM 2021-03-12 20:00:51
We can divide them, which gives us

$\frac{m}{n} = \frac{2k + 1/2}{2 \ell + 1/2} = \frac{4k + 1}{4 \ell + 1}.$

MaryM 2021-03-12 20:00:57
So we need

$\dfrac{m}{n}$ to be equivalent to a fraction of the form

$\dfrac{4k + 1}{4 \ell + 1},$ for some integers

$k$ and

$\ell.$

MaryM 2021-03-12 20:01:06
Before we proceed, since we eliminated

$x,$ let's do a quick reversibility check. If

$\frac{m}{n} = \frac{4k + 1}{4 \ell + 1},$

then we can find an

$x$ that works?

peace09 2021-03-12 20:01:39
Yes!

yayy 2021-03-12 20:01:39
yes

Unevenlogic 2021-03-12 20:01:39
Yes

pinkpig 2021-03-12 20:01:39
Yes?

Irving1004 2021-03-12 20:01:39
yes always

MaryM 2021-03-12 20:01:43
If

$\dfrac{m}{n} = \dfrac{4k + 1}{4 \ell + 1},$ then

$\frac{2k + 1/2}{m} = \frac{2 \ell + 1/2}{n}.$

MaryM 2021-03-12 20:01:48
Then we can let

$\frac{x}{\pi} = \frac{2k + 1/2}{m} = \frac{2 \ell + 1/2}{n}.$

MaryM 2021-03-12 20:01:55
This gives us

$\begin{align*} mx &= \left(2k + \frac12\right)\pi, \\ nx &= \left(2\ell + \frac12\right)\pi, \end{align*}$

so

$\sin (mx) + \sin (nx) = 2.$

MaryM 2021-03-12 20:02:02
Thus, the condition

$\frac{m}{n} = \frac{4k + 1}{4 \ell + 1}$

is both necessary and sufficient.

BillXu21 2021-03-12 20:02:04

MaryM 2021-03-12 20:02:15
One thing we can note right away is that

$4k + 1$ and

$4 \ell + 1$ are always odd. So we can think about how factors of 2 play a role.

MaryM 2021-03-12 20:02:19
If

$\frac{m}{n} = \frac{4k + 1}{4 \ell + 1},$

then what does that say about the factors of 2 in

$m$ and

$n?$

Lamboreghini 2021-03-12 20:03:09
they're the same

Unevenlogic 2021-03-12 20:03:09
They have the same powers of 2

happyhari 2021-03-12 20:03:09
they cancel out

functionalmath 2021-03-12 20:03:09
must be able to cancel out

foodisgood 2021-03-12 20:03:09
they are the same power

pinkpig 2021-03-12 20:03:09
m and n have the same amount of two's in their prime factorization

dbasu1 2021-03-12 20:03:09
m and n have the same power of 2 in their prime factorisation

MaryM 2021-03-12 20:03:17
The number of factors of 2 in

$m$ must be equal to the number of factors of 2 in

$n.$ This is the only way that

$\dfrac{m}{n}$ is equivalent to a fraction where both the numerator and denominator are odd. (The powers of 2 in

$m$ and

$n$ must cancel perfectly.)

MaryM 2021-03-12 20:03:24
That means

$m = 2^e a$ and

$n = 2^e b,$ where

$e$ is a nonnegative integer, and

$a$ and

$b$ are odd integers. (In particular, the exponents

$e$ must be the same.)

MaryM 2021-03-12 20:03:29
We can then sort all the numbers from 1 to 30 into individual "buckets", depending on the exponent

$e.$ Then

$m$ and

$n$ must be in the same bucket.

MaryM 2021-03-12 20:03:40

$\begin{array}{r|l} \text{exponent of $2$} & m,n \\ \hline 0 & 1,3,5,\ldots,29 \\ 1 & 2,6,10,\ldots,30 \\ 2 & 4,12,20,28 \\ 3 & 8,24 \\ 4 & 16 \end{array}$

MaryM 2021-03-12 20:04:05
So with

$m = 2^e a$ and

$n = 2^e b,$ the condition

$\dfrac{m}{n} = \dfrac{4k + 1}{4 \ell + 1}$ becomes

$\frac{a}{b} = \frac{4k + 1}{4 \ell + 1}.$

MaryM 2021-03-12 20:04:09
Are there any easy cases where the condition

$\dfrac{a}{b} = \dfrac{4k + 1}{4 \ell + 1}$ is satisfied?

MaryM 2021-03-12 20:04:37
Hint: think in terms of

$\mod 4$ .

coolotter 2021-03-12 20:05:29
m = n = 1 mod 4 then

dbasu1 2021-03-12 20:05:29
a and b are both congruent to 1 mod 4

Lamboreghini 2021-03-12 20:05:29
when a and b are both congruent to 1 mod 4???

ancientwarrior 2021-03-12 20:05:29
they are both 1 mod 4

yayy 2021-03-12 20:05:29
a = 1 mod 4 and b = 1 mod 4

functionalmath 2021-03-12 20:05:29
both are 1 mod 4

MaryM 2021-03-12 20:05:32
If

$a$ is 1 mod 4 and

$b$ is 1 mod 4, then the condition

$\dfrac{a}{b} = \dfrac{4k + 1}{4 \ell + 1}$

is easily satisfied: Just take

$4k + 1 = a$ and

$4 \ell + 1 = b.$

MaryM 2021-03-12 20:05:38
Are there any other cases that work?

MaryM 2021-03-12 20:06:30
Any other combination of residues

$\mod 4$ that work?

GabeW1234 2021-03-12 20:07:20
yes

rockyrockrock 2021-03-12 20:07:20
3

pinkpig 2021-03-12 20:07:20
both are equal to 3 mod 4

GabeW1234 2021-03-12 20:07:20
3 times both?

functionalmath 2021-03-12 20:07:20
would 3 work?

pinkpig 2021-03-12 20:07:20
3?

MaryM 2021-03-12 20:07:25
If

$a$ is 3 mod 4 and

$b$ is 3 mod 4, then

$3a$ is 1 mod 4 and

$3b$ is 1 mod 4, and

$\frac{a}{b} = \frac{3a}{3b}.$

MaryM 2021-03-12 20:07:33
So in this case, we can take

$4k + 1 = 3a$ and

$4k + 1 = 3b.$

MaryM 2021-03-12 20:07:38
What if

$a$ is 1 mod 4 and

$b$ is 3 mod 4?

MaryM 2021-03-12 20:08:18
Will that work?

functionalmath 2021-03-12 20:08:45
that wont work

kante314 2021-03-12 20:08:45
no

GabeW1234 2021-03-12 20:08:45
no?

pinkpig 2021-03-12 20:08:45
that case wouldn't work

dbasu1 2021-03-12 20:08:45
it will not

rockyrockrock 2021-03-12 20:08:45
no

MaryM 2021-03-12 20:08:48
It doesn't look like it will work.

MaryM 2021-03-12 20:08:53
To make sure, we can start with the equation

$\frac{a}{b} = \frac{4k + 1}{4 \ell + 1}.$

MaryM 2021-03-12 20:08:57
We can write this as

$(4 \ell + 1) b = (4k + 1) a.$

MaryM 2021-03-12 20:09:09
What happens if we take

$\mod 4$ of this equation?

pinkpig 2021-03-12 20:10:21
b=a mod 4

Unevenlogic 2021-03-12 20:10:21
a congruent to b

foodisgood 2021-03-12 20:10:21
it means b mod 4 must equal a mod 4

awesomeandy 2021-03-12 20:10:21
b=a

GabeW1234 2021-03-12 20:10:21
b=a

yayy 2021-03-12 20:10:21
b = a mod 4

happyhari 2021-03-12 20:10:21
then we get b = a

MaryM 2021-03-12 20:10:25
Taking both sides modulo 4, we get

$b \equiv a \pmod{4}.$

MaryM 2021-03-12 20:10:30
So if

$a$ is 1 mod 4 and

$b$ is 3 mod 4, then no such

$k$ and

$\ell$ exist.

MaryM 2021-03-12 20:10:35
We reach the same conclusion if

$a$ is 3 mod 4 and

$b$ is 1 mod 4.

MaryM 2021-03-12 20:10:41
(Remember that

$a$ and

$b$ are odd, so this covers all the cases.)

MaryM 2021-03-12 20:10:57
This tells us that we count the pair

$(m,n),$ where

$m < n,$ if and only if

$m = 2^e a$ and

$n = 2^e b,$ where

$a$ and

$b$ are odd, and

$a \equiv b \pmod{4}.$

MaryM 2021-03-12 20:11:07
So to count the number of pairs

$(m,n),$ we must take each of our buckets and separate further by

$a, b \pmod{4}.$

MaryM 2021-03-12 20:11:17

$\begin{array}{r|c|l} \text{exponent of $2$} & a,b \pmod{4} & m,n \\ \hline 0 & 1 & 1,5,9,13,17,21,25,29 \\ 0 & 3 & 3,7,11,15,19,23,27 \\ 1 & 1 & 2,10,18,26 \\ 1 & 3 & 6,14,22,30 \\ 2 & 1 & 4,20 \\ 2 & 3 & 12,28 \\ 3 & 1 & 8 \\ 3 & 3 & 24 \\ 4 & 1 & 16 \end{array}$

byung806 2021-03-12 20:11:24
wow

MaryM 2021-03-12 20:11:38
For each bucket, how many pairs do we have?

MaryM 2021-03-12 20:13:15
(Your answer will depend on the qnantity of items in each bucket)

MaryM 2021-03-12 20:13:34
I mean, quantity.

ancientwarrior 2021-03-12 20:14:00
choose 2

Unevenlogic 2021-03-12 20:14:00
(number of items in bucket) choose 2

GabeW1234 2021-03-12 20:14:00
(the number of numbers)(the number or numbers-1)

Summer2019 2021-03-12 20:14:00
(size choose 2)

CircleInvert 2021-03-12 20:14:00

$28$ ,

$21$ ,

$6$ ,

$1$ ,

$0$ ,

$0$ , and

$0$ respectively

MaryM 2021-03-12 20:14:03
We can pick any two numbers from our bucket.

MaryM 2021-03-12 20:14:08
So if there are

$k$ numbers, we have

$\dbinom{k}{2}$ pairs.

MaryM 2021-03-12 20:14:20
The buckets have sizes 8, 7, 4, 4, 2, and 2 (the buckets with only a single element can be thrown out).

MaryM 2021-03-12 20:14:26
So the count of pairs is

$\dbinom82 + \dbinom72 + \dbinom42 + \dbinom42 + \dbinom22 + \dbinom22$ .

MaryM 2021-03-12 20:14:30
What does this work out to?

functionalmath 2021-03-12 20:15:07
63

pinkpig 2021-03-12 20:15:07
28+21+6+6+1+1=063

happyhari 2021-03-12 20:15:07
63

mop 2021-03-12 20:15:07
63!

MaryM 2021-03-12 20:15:10
We get

$28 + 21 + 6 + 6 + 1 + 1$ .

MaryM 2021-03-12 20:15:12
And the sum is

$\boxed{063}$ .

BillXu21 2021-03-12 20:15:18
yay

Lamboreghini 2021-03-12 20:15:22
yay!

MaryM 2021-03-12 20:15:32
8. Find the number of integers

$c$ such that the equation

$\Big|\big|20|x| - x^2\big|-c\Big|=21$ has

$12$ distinct real solutions.

MaryM 2021-03-12 20:15:41
Yikes.

MaryM 2021-03-12 20:15:49
Often problems that look scary are not that bad once you get past all the notation. Hopefully that's the case here.

skyleristhecoolest 2021-03-12 20:16:05
so. many. abosouloute. values.

MaryM 2021-03-12 20:16:11
How can we start investigating this?

gorefeebuddie 2021-03-12 20:17:07
handle the left hand side?

Z_Lu 2021-03-12 20:17:07
break the absolute value in the middle?

Asher51 2021-03-12 20:17:13
from the inside out

MaryM 2021-03-12 20:17:15
Let's look at the innermost part first:

$20|x| - x^2$ .

MaryM 2021-03-12 20:17:43
What do we know about this function?

happyhari 2021-03-12 20:18:15
x^2 = |x|^2

mathisfun17 2021-03-12 20:18:15
x^2 = |x|^2

functionalmath 2021-03-12 20:18:15
x^2 is always nonngeative

CircleInvert 2021-03-12 20:18:15
It is even

Kevinisawesome 2021-03-12 20:18:15
two parabolas jammed together based on whether x is positive or negative

helloworld_ 2021-03-12 20:18:15
It is even

yayy 2021-03-12 20:18:15
its even

MaryM 2021-03-12 20:18:21
If we replace

$x$ with

$-x$ , we get the same result.

MaryM 2021-03-12 20:18:29
Maybe trying to sketch a graph will be helpful?

MaryM 2021-03-12 20:18:38
We know that the graph will be symmetric across the

$y$ -axis.

MaryM 2021-03-12 20:18:48
That is, it's the graph of

$20x - x^2$ on the right side of the

$y$ -axis, and then the mirror-image of that graph across the

$y$ -axis to give us the left side.

MaryM 2021-03-12 20:18:52
And what does the graph of

$20x - x^2$ look like?

sugar_rush 2021-03-12 20:19:32
inverted parabola

klpiguy 2021-03-12 20:19:32
quadratic

happyhari 2021-03-12 20:19:32
a parabola facing downwards

MaryM 2021-03-12 20:19:35
It's a downward-opening parabola.

MaryM 2021-03-12 20:19:48
Where is its vertex?

peace09 2021-03-12 20:20:14
An upside down parabola with center (10, 100)

peace09 2021-03-12 20:20:14
(10, 100)

functionalmath 2021-03-12 20:20:14
x=10

learning0119 2021-03-12 20:20:14
on the y axis

Kevinisawesome 2021-03-12 20:20:14
10,100

happyhari 2021-03-12 20:20:21
(10,100)

ancientwarrior 2021-03-12 20:20:21
10, 100

MaryM 2021-03-12 20:20:26
It's at

$x = -\dfrac{20}{-2} = 10$ .

MaryM 2021-03-12 20:20:29
And at this point, we have

$y = 20(10) - 10^2 = 200 - 100 = 100$ .

MaryM 2021-03-12 20:20:33
So the vertex is at

$(10,100)$ .

MaryM 2021-03-12 20:20:42
What else will be helpful to graph this?

foodisgood 2021-03-12 20:21:07
the roots

Z_Lu 2021-03-12 20:21:07
intercepts

MaryM 2021-03-12 20:21:10
Knowing the roots -- that is, knowing where it crosses the

$x$ -axis.

MaryM 2021-03-12 20:21:14
What are the roots?

functionalmath 2021-03-12 20:21:30
x=0,20

klpiguy 2021-03-12 20:21:30
0,20

justin6688 2021-03-12 20:21:30
20, 0

pinkpig 2021-03-12 20:21:30
20 and 0

happyhari 2021-03-12 20:21:30
0 and 20

klpiguy 2021-03-12 20:21:30

$0,20$

MaryM 2021-03-12 20:21:33
Since

$20x - x^2 = x(20-x)$ , the roots are at

$x=0$ and

$x=20$ .

MaryM 2021-03-12 20:21:41
So this is a downward-opening parabola, with vertex at

$(10,100)$ , and roots at

$(0,0)$ and

$(20,0)$ .

MaryM 2021-03-12 20:21:47
This is enough for us to sketch a graph of this. Remember that we're going to graph this for

$x \ge 0$ , and then take its mirror image to give us the graph for

$x \le 0$ :

MaryM 2021-03-12 20:21:53

MaryM 2021-03-12 20:22:00
Note: this is definitely not to scale! (If we drew it to scale, it'd be huge, because

$(10,100)$ is *way* above the

$x$ -axis!)

MaryM 2021-03-12 20:22:10
Now what?

yayy 2021-03-12 20:22:40
take the absolute value

RaymondZhu 2021-03-12 20:22:40
deal with the absolute value around all that

peace09 2021-03-12 20:22:40
We take the absolute value of this.

MaryM 2021-03-12 20:22:42
Let's add a layer of absolute values.

MaryM 2021-03-12 20:22:54
What does that mean in terms of our graph?

Kevinisawesome 2021-03-12 20:23:22
reflect stuff below x axis above x axis

klpiguy 2021-03-12 20:23:22
everything after |x|>20 gets inverted

Lamboreghini 2021-03-12 20:23:22
all the points below the x-axis are reflected across the x axis

happyhari 2021-03-12 20:23:29
neglect the graph below the x-axis

Benranger 2021-03-12 20:23:32
all the points below the x-axis are reflected across the x axis

MaryM 2021-03-12 20:23:35
Anything that's below the

$x$ -axis (that is, is negative) gets flipped above the

$x$ -axis (that is, becomes positive).

MaryM 2021-03-12 20:23:38
So it looks like this:

MaryM 2021-03-12 20:23:42

MaryM 2021-03-12 20:23:52
Let's give this a name so I don't have to keep typing it!

MaryM 2021-03-12 20:23:57
Let's set

$f(x) = \left|20|x| - x^2\right|$ .

MaryM 2021-03-12 20:24:02
So now we need to find the integers

$c$ such that

$|f(x) - c| = 21$ has exactly

$12$ solutions.

MaryM 2021-03-12 20:24:06
What does this mean in terms of

$f(x)$ ?

MaryM 2021-03-12 20:24:44
How can we simplify

$|f(x)-c|=21$ ?

Kevinisawesome 2021-03-12 20:25:03
its 21+c or -21+c

peace09 2021-03-12 20:25:03

$f(x)=\pm{21}+c.$

Frestho 2021-03-12 20:25:03

$f(x) = \pm 21 + c$

sugar_rush 2021-03-12 20:25:03

$f(x)=21+c$ or

$f(x)=-21+c$

rockyrockrock 2021-03-12 20:25:07
f(x)-c=21 or -21

MaryM 2021-03-12 20:25:09
We must have either

$f(x) = c+21$ or

$f(x) = c-21$ .

MaryM 2021-03-12 20:25:15
But how many solutions can

$f(x) = a$ have for any given

$a$ ?

Kevinisawesome 2021-03-12 20:26:17
6

klpiguy 2021-03-12 20:26:17
2, or

klpiguy 2021-03-12 20:26:17
6

Frestho 2021-03-12 20:26:17
6

sugar_rush 2021-03-12 20:26:17

$6$ ?

ASweatyAsianBoie 2021-03-12 20:26:17
6

adsupermath 2021-03-12 20:26:17
6

rockyrockrock 2021-03-12 20:26:17
6

MaryM 2021-03-12 20:26:23
We draw the line

$y=a$ and count intersection points with our graph.

MaryM 2021-03-12 20:26:28

MaryM 2021-03-12 20:26:51
The number of solutions will be either

$6$ or

$2$ depending on how big

$a$ is.

MaryM 2021-03-12 20:27:12
Since we want

$12$ solutions in total, we need

$6$ solutions for each of the two cases.

MaryM 2021-03-12 20:27:43
(We can also have 3 solutions for

$a=0$ , or

$0$ solutions for negative

$a$ .)

MaryM 2021-03-12 20:28:07
For which

$a$ do we have

$6$ solutions?

pinkpig 2021-03-12 20:28:34
0<(c+21 and c-21)<100

Frestho 2021-03-12 20:28:34

$0<a<100$

peace09 2021-03-12 20:28:34

$0<a<100.$

pinkpig 2021-03-12 20:28:34
from 0 to 100 exclusive

happyhari 2021-03-12 20:28:34
0<a<100

Z_Lu 2021-03-12 20:28:34
0<a<100

yayy 2021-03-12 20:28:34

$0 < a < 100$

foodisgood 2021-03-12 20:28:34
(0,100)

happyhari 2021-03-12 20:28:34
0 < a < 100

MaryM 2021-03-12 20:28:38
If

$0 < a < 100$ , then we'll get exactly

$6$ intersection points, so we'll get

$6$ solutions.

MaryM 2021-03-12 20:28:46
That means we need

$0 < c+21 < 100$ and

$0 < c-21 < 100$ .

MaryM 2021-03-12 20:28:51
How do we solve these inequalities?

sugar_rush 2021-03-12 20:29:39

$21<c<79$

adsupermath 2021-03-12 20:29:39

$21<c<79$

Lamboreghini 2021-03-12 20:29:39
subtract 21, add 21

MaryM 2021-03-12 20:29:43
They combine as

$0 < c-21 < c+21 < 100$ .

MaryM 2021-03-12 20:29:47
So we must have

$0 < c-21$ and

$c+21 < 100$ .

MaryM 2021-03-12 20:29:53
This gives us

$21 < c$ and

$c < 79$ .

MaryM 2021-03-12 20:29:57
So

$c$ must be an integer strictly between

$21$ and

$79$ .

MaryM 2021-03-12 20:30:01
And how many of these are there?

pinkpig 2021-03-12 20:30:25
057 integer solutions for c.

learning0119 2021-03-12 20:30:25
57

foodisgood 2021-03-12 20:30:25
57

Poki 2021-03-12 20:30:25
79-21-1 = 57

happyhari 2021-03-12 20:30:25
57

sugar_rush 2021-03-12 20:30:25

$\fbox{57}$

adsupermath 2021-03-12 20:30:25
57

MaryM 2021-03-12 20:30:27
It's the list

$22,23,\ldots,78$ .

MaryM 2021-03-12 20:30:31
There are

$78-22+1 = 57$ integers in this list.

MaryM 2021-03-12 20:30:37
Thus there are

$\boxed{057}$ valid values for

$c$ .

MaryM 2021-03-12 20:30:52
Be careful with the counting at the end of a problem like this! It's always unfortunate to do the hard work to get to

$21 < c < 79$ and then miscount the number of

$c$ 's.

pinkpig 2021-03-12 20:31:16
another problem down, just 7 more to go.

Poki 2021-03-12 20:31:29
7 more

MaryM 2021-03-12 20:31:32
9. Let

$ABCD$ be an isosceles trapezoid with

$AD=BC$ and

$AB < CD.$ Suppose that the distances from

$A$ to the lines

$BC,$

$CD,$ and

$BD$ are

$15,$

$18,$ and

$10,$ respectively. Let

$K$ be the area of

$ABCD.$ Find

$\sqrt2\cdot K.$

MaryM 2021-03-12 20:31:38
Let's try to draw a picture of what we have here. I'll also label the feet of the perpendiculars, so we can talk about them.

MaryM 2021-03-12 20:31:48

MaryM 2021-03-12 20:31:54
I think this captures all the given info, right? Note that we have

$AB < CD$ and

$BC = AD$ .

MaryM 2021-03-12 20:32:02
OK, what can we do now?

MaryM 2021-03-12 20:32:56
Hint: We have some altitudes in this diagram.

klpiguy 2021-03-12 20:33:56
areas

peace09 2021-03-12 20:33:56
Area?

mathgeek23 2021-03-12 20:33:56
areas?

ryanli1366 2021-03-12 20:33:56
find area by bh/2?

MaryM 2021-03-12 20:33:59
Altitudes suggest that using areas might be useful.

MaryM 2021-03-12 20:34:07
For example, we notice that

$AR=10$ is the height of

$\triangle ABD$ . Does that help?

klpiguy 2021-03-12 20:35:02
yes

pinkpig 2021-03-12 20:35:02
yes

MaryM 2021-03-12 20:35:04
Do you see another height of this same triangle?

Mathaddict3825 2021-03-12 20:35:28
AQ is also an altitude?

peace09 2021-03-12 20:35:28

$AQ.$

foodisgood 2021-03-12 20:35:28
AQ=18

ASweatyAsianBoie 2021-03-12 20:35:28
AQ

adsupermath 2021-03-12 20:35:28
AQ

ancientwarrior 2021-03-12 20:35:28
AQ is the altitude of the same triangle with base AB

MaryM 2021-03-12 20:35:30

$AQ$ also a height (a different height) of that same triangle

$\triangle ABD$ !

MaryM 2021-03-12 20:35:39
If we slide it over, it's the height from

$D$ to side

$\overline{AB}$ :

MaryM 2021-03-12 20:35:44

MaryM 2021-03-12 20:35:49
Gee, it sure would be nice if the third length were also a height of that triangle...

MaryM 2021-03-12 20:36:20
Is it?

MaryM 2021-03-12 20:36:54
Hint: this figure is pretty symmetric.

floradaisy136 2021-03-12 20:37:02
good news

justin6688 2021-03-12 20:37:02
yes

ryanli1366 2021-03-12 20:37:02
AP, if you flip APB

functionalmath 2021-03-12 20:37:02
yes

CircleInvert 2021-03-12 20:37:02
It is

pinkpig 2021-03-12 20:37:02
yes

Z_Lu 2021-03-12 20:37:02
yes? AP, since it is an isoceles trapezoid

Lamboreghini 2021-03-12 20:37:06
yes it is!!

Zhaom 2021-03-12 20:37:07
yes because we reflect APB across the axis of symmetry of the trapezoid

MaryM 2021-03-12 20:37:10
It is, by symmetry!

MaryM 2021-03-12 20:37:15

MaryM 2021-03-12 20:37:23
So these are all different heights for the same triangle

$\triangle ABD$ !

MaryM 2021-03-12 20:37:33
How does that help?

Kevinisawesome 2021-03-12 20:37:58
ratio of the bases are known

ASweatyAsianBoie 2021-03-12 20:37:58
Ratios of all sides?

Z_Lu 2021-03-12 20:37:58
we now have ratios of the side lengths

peace09 2021-03-12 20:37:58

$15AD=10BD=18AB...$

Mathaddict3825 2021-03-12 20:37:58
ratio of the lengths?

Zhaom 2021-03-12 20:37:58

$15AD=10BD=18AB$

adsupermath 2021-03-12 20:37:58
Set different expressions for area equal to each other

sugar_rush 2021-03-12 20:37:58
different expressions for the same area

MaryM 2021-03-12 20:38:00
We can compute the area of this triangle in three ways, using the different heights.

MaryM 2021-03-12 20:38:03
That is,

$[ABD] = \dfrac12(10)(BD) = \dfrac12(15)(AD) = \dfrac12(18)(AB).$

MaryM 2021-03-12 20:38:08
Let's drop the

$\dfrac12$ 's, so we have

$10(BD) = 15(AD) = 18(AB).$

MaryM 2021-03-12 20:38:19
What's the least common multiple of

$10$ ,

$15$ , and

$18$ ?

peace09 2021-03-12 20:38:44
90.

klpiguy 2021-03-12 20:38:44
90

happyhari 2021-03-12 20:38:44
90

Zhaom 2021-03-12 20:38:44

$90$

ryanli1366 2021-03-12 20:38:44
90

math-Passion 2021-03-12 20:38:44

$90$

Mathaddict3825 2021-03-12 20:38:48
90

Summer2019 2021-03-12 20:38:48
90

pinkpig 2021-03-12 20:38:48
90

MaryM 2021-03-12 20:38:51
It's

$90$ .

MaryM 2021-03-12 20:38:54
So we can assume that

$10(BD) = 15(AD) = 18(AB) = 90k$ for some constant

$k$ .

MaryM 2021-03-12 20:38:58
That makes

$BD = 9k$ ,

$AD = 6k$ , and

$AB = 5k$ .

MaryM 2021-03-12 20:39:03
Let's refresh the picture with these lengths:

MaryM 2021-03-12 20:39:16

MaryM 2021-03-12 20:39:29
Too bad we don't know

$CD$ .

MaryM 2021-03-12 20:39:42
Any ideas?

MaryM 2021-03-12 20:40:17
We could use Pythagoras, but there's a more elegant way.

MaryM 2021-03-12 20:40:38
Hint: The trapezoid is a cyclic quadrilateral.

gorefeebuddie 2021-03-12 20:40:55
ptolemy theorem

uneducatedpotato 2021-03-12 20:40:55
ptolemy's?

justin6688 2021-03-12 20:40:55
PTOLEMY'S THEOREM

Mathaddict3825 2021-03-12 20:40:55
ptolemy's?

bjc 2021-03-12 20:40:55
Ptolemy

Zhaom 2021-03-12 20:40:57
Ptolemy's

MaryM 2021-03-12 20:41:00
We can use Ptolemy's Theorem!

MaryM 2021-03-12 20:41:04
Ptolemy's Theorem says that for any cyclic quadrilateral

$ABCD$ , we have

$(AB)(CD) + (AD)(BC) = (AC)(BD).$ That is, the sum of the products of opposite sides equals the product of the diagonals.

MaryM 2021-03-12 20:41:13
And an isosceles trapezoid is definitely cyclic!

MaryM 2021-03-12 20:41:16
So what do we get when we apply Ptolemy's Theorem?

Zhaom 2021-03-12 20:41:46
4kCD+36k^2=81k^2

Mathaddict3825 2021-03-12 20:41:46
DC = 9k

bjc 2021-03-12 20:41:46

$CD=9k$

adsupermath 2021-03-12 20:41:46

$9k$

peace09 2021-03-12 20:41:46

$5k(CD)+36k^2=81k^2.$

MaryM 2021-03-12 20:41:50
We get

$(5k)(CD) + (6k)(6k) = (9k)(9k)$ .

MaryM 2021-03-12 20:42:02
So

$(5k)(CD) = 45k^2$ , and hence

$CD = 9k$ .

MaryM 2021-03-12 20:42:09

MaryM 2021-03-12 20:42:27
Unfortunately, we don't know what

$k$ is yet.

MaryM 2021-03-12 20:42:42
How can we figure that out?

GabeW1234 2021-03-12 20:43:11

$DQ=2k$ !

Mathaddict3825 2021-03-12 20:43:11
pythagoras on ADQ?

nathanqiu 2021-03-12 20:43:11
Pythagorean Theorem

happyhari 2021-03-12 20:43:11
use pythagoras

Kevinisawesome 2021-03-12 20:43:11
pythag

foodisgood 2021-03-12 20:43:16
draw height from B to DC

MaryM 2021-03-12 20:43:18
How about using

$\triangle ADQ$ ?

MaryM 2021-03-12 20:43:22
Indeed, we can break up

$CD$ into parts:

MaryM 2021-03-12 20:43:26

MaryM 2021-03-12 20:43:46
What do we get when we use the Pythagorean Theorem on

$\triangle ADQ$ ?

Zhaom 2021-03-12 20:44:48

$4k^2+18^2=36k^2$

justin6688 2021-03-12 20:44:48
32k^2=324

Poki 2021-03-12 20:44:48
32k^2 = 324

Lamboreghini 2021-03-12 20:44:48
4k^2+324=36k^2, solving gives k^2=81/8

awesomeandy 2021-03-12 20:44:48
k=sqrt(324/32)

MaryM 2021-03-12 20:44:50
It gives us

$18^2 + (2k)^2 = (6k)^2$ .

MaryM 2021-03-12 20:44:54
This simplifies to

$18^2 = 32k^2$ .

MaryM 2021-03-12 20:45:03
So

$k = \sqrt{\dfrac{18^2}{32}}$ .

MaryM 2021-03-12 20:45:11
Or

$k = \dfrac{18}{\sqrt{32}} = \dfrac{18}{4\sqrt2} = \dfrac{9}{2\sqrt2}$ .

MaryM 2021-03-12 20:45:15
And now can we finish?

justin6688 2021-03-12 20:45:56
yes!

foodisgood 2021-03-12 20:45:56
yep

Lamboreghini 2021-03-12 20:45:56
now we can find the area of ABCD and multiply by sqrt2

happyhari 2021-03-12 20:45:56
find the area of the trapezoid

gorefeebuddie 2021-03-12 20:45:56
Yes, the area of a trapezoid is 7k*18 or 136 k

peace09 2021-03-12 20:45:56
Yes!

MaryM 2021-03-12 20:46:01
Yes! The area of the trapezoid is the height times the average of the bases.

rockyrockrock 2021-03-12 20:46:27
7k*18

MaryM 2021-03-12 20:46:30
Therefore,

$K = 18(7k) = 18 \cdot 7 \cdot \dfrac{9}{2\sqrt2}$ .

MaryM 2021-03-12 20:46:35
This simplifies to

$K = \dfrac{9 \cdot 7 \cdot 9}{\sqrt2} = \dfrac{567}{\sqrt2}$ .

happyhari 2021-03-12 20:46:49
multiply by sqrt(2)

Lamboreghini 2021-03-12 20:46:49

$\boxed{567}$ yay!

peace09 2021-03-12 20:46:49
Our answer is

$\boxed{567}.$

ryanli1366 2021-03-12 20:46:49
so 567

MaryM 2021-03-12 20:46:52
So our final answer is

$\sqrt2 \cdot K = \boxed{567}$ .

Irving1004 2021-03-12 20:47:01
Yay!

MaryM 2021-03-12 20:47:04
As promised, let’s take a short break. I’ll be back in 5 minutes!

MaryM 2021-03-12 20:52:02
I'm back. Are you still there?

kante314 2021-03-12 20:52:14
Yep

Zhaom 2021-03-12 20:52:14
yes

nathanqiu 2021-03-12 20:52:14
yes!

coolotter 2021-03-12 20:52:14
yes we are!

MaryM 2021-03-12 20:52:17
Great, let’s tackle the last 6 problems of the test!

kante314 2021-03-12 20:52:30
Yay

MaryM 2021-03-12 20:52:33
10. Consider the sequence

$(a_k)_{k\geq 1}$ of positive rational numbers defined by

$a_1 = \dfrac{2020}{2021}$ and for

$k \geq 1,$ if

$a_k = \dfrac{m}{n}$ for relatively prime positive integers

$m$ and

$n,$ then

$a_{k+1} = \dfrac{m+18}{n+19}.$ Determine the sum of all positive integers

$j$ such that the rational number

$a_j$ can be written in the form

$\dfrac{t}{t+1}$ for some positive integer

$t.$

MaryM 2021-03-12 20:52:43
Let's call a positive integer

$j$ "good" if is satisfies the condition that

$a_j = \dfrac{t}{t+1}$ for some positive integer

$t$ .

MaryM 2021-03-12 20:52:49
We want to find the sum of all the good

$j$ 's.

MaryM 2021-03-12 20:52:59
Hopefully there are finitely many!

Zhaom 2021-03-12 20:53:10
compute a few terms and see what happens

MaryM 2021-03-12 20:53:12
Do we know any right away?

sugar_rush 2021-03-12 20:53:59

$j=1$

Zhaom 2021-03-12 20:53:59

$a_1$ works

learning0119 2021-03-12 20:53:59
k = 1

functionalmath 2021-03-12 20:53:59
a_1

klpiguy 2021-03-12 20:53:59
2020/2021

Hector2173 2021-03-12 20:53:59
j = 1

Irving1004 2021-03-12 20:53:59
1 satisfies this

MaryM 2021-03-12 20:54:05
Yes:

$j=1$ is good, since

$a_1 = \dfrac{2020}{2021}$ has the right form.

MaryM 2021-03-12 20:54:12
Any others?

adsupermath 2021-03-12 20:54:32
1, 2

Zhaom 2021-03-12 20:54:32

$a_1$ and

$a_2$ work

nathanqiu 2021-03-12 20:54:32
yes, a_2

gorefee 2021-03-12 20:54:32
1,2 works

peace09 2021-03-12 20:54:32

$a_2=\tfrac{1019}{1020}.$

gorefee 2021-03-12 20:54:32
2

jhao23 2021-03-12 20:54:32
2 also works

adsupermath 2021-03-12 20:54:32

$j=2$

MaryM 2021-03-12 20:54:34
Following the instructions, we get

$a_2 = \dfrac{2020+18}{2021+19}$ .

MaryM 2021-03-12 20:54:39
This is

$a_2 = \dfrac{2038}{2040}$ .

MaryM 2021-03-12 20:54:44
Hey, that simplifies to

$a_2 = \dfrac{1019}{1020}$ , and that's lowest terms.

MaryM 2021-03-12 20:54:50
So

$j=2$ is good too.

MaryM 2021-03-12 20:55:03
Let's continue and compute

$a_3$ .

MaryM 2021-03-12 20:55:07
What do we get?

jhao23 2021-03-12 20:55:39
1037/1039

Zhaom 2021-03-12 20:55:39

$\frac{1037}{1039}$

learning0119 2021-03-12 20:55:39
1037/1039

sugar_rush 2021-03-12 20:55:39

$\tfrac{1037}{1039}$

adsupermath 2021-03-12 20:55:39

$\frac{1037}{1039}$

functionalmath 2021-03-12 20:55:39
1037/1039

MaryM 2021-03-12 20:55:42
We get

$a_3 = \dfrac{1019+18}{1020+19} = \dfrac{1037}{1039}$ .

MaryM 2021-03-12 20:55:46
It doesn't look like it simplifies, but how can we be sure?

functionalmath 2021-03-12 20:56:21
euclids algorithm

adsupermath 2021-03-12 20:56:21
Difference of 2, but both are odd

nathanqiu 2021-03-12 20:56:21
its odd, and has difference of 2

Zhaom 2021-03-12 20:56:21

$\gcd(1037,1039)=\gcd(1037,2)=1$

foodisgood 2021-03-12 20:56:21
they don't share two so no other factors

Irving1004 2021-03-12 20:56:21
they are consecutive odd numbers

nathanqiu 2021-03-12 20:56:21
numbers are both odd, and both have difference of 2

ancientwarrior 2021-03-12 20:56:21
their difference

ryanli1366 2021-03-12 20:56:23
their difference, 2 is only divisible by 2, and because they are both odd, there is no common divisor

MaryM 2021-03-12 20:56:25
We can use the Euclidean Algorithm!

MaryM 2021-03-12 20:56:28
It tells us that

$\gcd(1039,1037) = \gcd(1039-1037,1037) = \gcd(2,1037) = 1$ .

MaryM 2021-03-12 20:56:33
So

$1037$ and

$1039$ don't have any common factors, and the fraction can't be simplified.

MaryM 2021-03-12 20:56:41
Hmmm...does this give us any clues on how to generalize this?

MaryM 2021-03-12 20:57:20
Hint: think about how can we write the next few terms of the sequence in terms of a variable.

sugar_rush 2021-03-12 20:58:17
find

$a_{2+q}$ in terms of

$q$

Zhaom 2021-03-12 20:58:17

$\frac{1001+18(j-1)}{1001+19(j-1)}$ until simplification

peace09 2021-03-12 20:58:17
How about

$a_{k+2}=\tfrac{1019+18k}{1020+19k}?$

MaryM 2021-03-12 20:58:20
We know that for any

$k \ge 1$ , as long as we can't simplify, we'll have

$a_{2+k} = \dfrac{1019+18k}{1020+19k}$ .

MaryM 2021-03-12 20:58:42
When will this first be able to be simplified?

MaryM 2021-03-12 20:59:23
Hint: think in terms of

$k$ .

Zhaom 2021-03-12 20:59:48
when

$\gcd(1019+18k,1020+19k) \ge 2$

peace09 2021-03-12 20:59:48
When

$\text{gcd}[1019+18k, 1020+19k] > 1.$

MaryM 2021-03-12 20:59:51
We'll be able to reduce when

$\gcd(1019+18k,1020+19k) > 1$ .

MaryM 2021-03-12 20:59:59
But now we can apply the Euclidean Algorithm!

MaryM 2021-03-12 21:00:09
We get

$\gcd(1019+18k,1020+19k) = \gcd(1019+18k,(1020+19k)-(1019+18k))$ .

MaryM 2021-03-12 21:00:16
So the gcd is

$\gcd(1019+18k,1+k)$ .

MaryM 2021-03-12 21:00:25
That is,

$a_{2+k}$ will first simplify when we have

$k \ge 1$ such that

$\gcd(1019+18k,1+k) > 1$ .

MaryM 2021-03-12 21:01:01
Now what?

peace09 2021-03-12 21:01:40
Apply the Euclidean Algorithm Again to Get

$\text{gcd}[1001, 1+k] > 1.$

klpiguy 2021-03-12 21:01:40
euclidiean algorithm again

foodisgood 2021-03-12 21:01:40
go further with Euclidean algorithm

MaryM 2021-03-12 21:01:45
We can apply the Euclidean Algorithm again! How?

foodisgood 2021-03-12 21:02:15
eliminate k one one side

MaryM 2021-03-12 21:02:18
We can repeatedly subtract

$1+k$ from

$1019+18k$ .

MaryM 2021-03-12 21:02:26
How many times do you think we'll want to subtract it?

functionalmath 2021-03-12 21:02:47
multiply (1+k)b y 18

peace09 2021-03-12 21:02:47
Apply It... 18 Times

learning0119 2021-03-12 21:02:47
18

Zhaom 2021-03-12 21:02:47

$18$ times

happyhari 2021-03-12 21:02:47
18

foodisgood 2021-03-12 21:02:47
18 times

YaoAOPS 2021-03-12 21:02:47
18

MaryM 2021-03-12 21:02:50
How about

$18$ times, to get rid of the

$k$ ?

MaryM 2021-03-12 21:02:54
That is, we get

$\gcd(1019+18k,1+k) = \gcd(1019+18k - 18(1+k),1+k) = \gcd(1001,1+k)$ .

MaryM 2021-03-12 21:03:05
And when is this gcd greater than

$1$ ?

Irving1004 2021-03-12 21:03:30
gcd(1001, k+1) > 1 therefore k+1=7 works and k=6 satisfies

sugar_rush 2021-03-12 21:03:30
This happens when

$k+1$ divides

$1001$ for the first time, so

$k=6$ .

b20081 2021-03-12 21:03:30
when k=6

Hector2173 2021-03-12 21:03:30
k = 6

foodisgood 2021-03-12 21:03:33
when k+1 is a factor of 1001

MaryM 2021-03-12 21:03:35
Note that

$1001 = 7 \cdot 11 \cdot 13$ . (This is a number whose factorization seems to come up a lot in contests.)

MaryM 2021-03-12 21:03:40
So

$k=6$ is the first time, and we get

$\gcd(1001,7) = 7$ , meaning that a

$7$ will cancel out.

MaryM 2021-03-12 21:04:10
What do we get when

$k=6$ ?

MaryM 2021-03-12 21:04:41
I mean, what value will we get for the sequence?

MaryM 2021-03-12 21:05:48
When

$k=6$ , we will be dealing with

$a_8$ . And

$a_8$ equals....

Hector2173 2021-03-12 21:06:01
161/162

Hector2173 2021-03-12 21:06:05

$\frac{161}{162}$

Zhaom 2021-03-12 21:06:05

$\frac{161}{162}$

ancientwarrior 2021-03-12 21:06:05
161/162

MaryM 2021-03-12 21:06:08
We get

$a_8 = \dfrac{1019 + 18(6)}{1020 + 19(6)} = \dfrac{1127}{1134}$ .

MaryM 2021-03-12 21:06:19
And, as expected,

$7$ is a common factor! We have

$a_8 = \dfrac{1127}{1134} = \dfrac{7 \cdot 161}{7 \cdot 162} = \dfrac{161}{162}$ .

MaryM 2021-03-12 21:06:25
So

$j=8$ is a good number!

MaryM 2021-03-12 21:06:34
We've get

$j \in \{1,2,8\}$ so far.

MaryM 2021-03-12 21:06:38
Now what?

Hector2173 2021-03-12 21:06:59
do it again

MaryM 2021-03-12 21:07:13
We can play the same game!

MaryM 2021-03-12 21:07:21
That is, we'll have

$a_{8+k} = \dfrac{161+18k}{162+19k}$ until we get a common factor.

MaryM 2021-03-12 21:07:33
So we're looking for the smallest

$k$ such that

$\gcd(161+18k,162+19k) > 1$ .

MaryM 2021-03-12 21:07:57
Time for the Euclidean Algorithm again! What do we get when we simplify that

$\gcd$ ?

Zhaom 2021-03-12 21:08:31
Simplifying, we get that

$\gcd(143,k+1) \ge 2$ .

MaryM 2021-03-12 21:08:35
We get

$\gcd(161+18k,162+19k) = \gcd(161+18k,1+k)$ .

MaryM 2021-03-12 21:08:45
Then subtract

$1+k$ 18 times to get

$\gcd(143,1+k)$ .

MaryM 2021-03-12 21:08:49
What's the smallest value of

$k$ that makes this greater than

$1$ ?

Irving1004 2021-03-12 21:09:10
then we get that k+1 divides into 143 and is 11

klpiguy 2021-03-12 21:09:10
10

foodisgood 2021-03-12 21:09:10
10

Zhaom 2021-03-12 21:09:10

$k=10$

happyhari 2021-03-12 21:09:10
10

Mathaddict3825 2021-03-12 21:09:13
10

b20081 2021-03-12 21:09:13
10

MaryM 2021-03-12 21:09:15
Since

$143 = 13 \cdot 11$ , we get

$k+1 = 11$ , or

$k=10$ .

MaryM 2021-03-12 21:09:19
Let's try it!

MaryM 2021-03-12 21:09:25
We get

$a_{18} = \dfrac{161 + 18(10)}{162 + 19(10)} = \dfrac{341}{352}$ .

klpiguy 2021-03-12 21:09:41
31/32

MaryM 2021-03-12 21:09:46
Simplifying

$a_{18} = \dfrac{11 \cdot 31}{11 \cdot 32} = \dfrac{31}{32}$ .

MaryM 2021-03-12 21:09:51
So

$j= 18$ is good!

MaryM 2021-03-12 21:09:55
And our list so far is

$j \in \{1,2,8,18\}$ .

MaryM 2021-03-12 21:10:00
Now what?

Zhaom 2021-03-12 21:10:11
DO IT AGAIN

Mathaddict3825 2021-03-12 21:10:11
again

happyhari 2021-03-12 21:10:11
again

foodisgood 2021-03-12 21:10:11
again!

Irving1004 2021-03-12 21:10:11
now we do the same thing agaibn

MaryM 2021-03-12 21:10:13
Lather, rinse, repeat! Let's do it again!

MaryM 2021-03-12 21:10:18
We have

$a_{18+k} = \dfrac{31 + 18k}{32 + 19k}$ until we get a common factor.

MaryM 2021-03-12 21:10:25
Euclidean Algorithm time!

MaryM 2021-03-12 21:10:38
We have

$\gcd(31+18k,32+19k) = \gcd(31+18k,1+k)$ .

MaryM 2021-03-12 21:10:42
And then?

Zhaom 2021-03-12 21:11:09
Simplifies to

$\gcd(13,k+1) \ge 2$ .

foodisgood 2021-03-12 21:11:09
(13,k+1)

learning0119 2021-03-12 21:11:09
gcd(13, 1+k)

happyhari 2021-03-12 21:11:09
gcd(13,1+k)

MaryM 2021-03-12 21:11:11
And subtracting

$1+k$ 18 times gives

$\gcd(13,1+k)$ .

MaryM 2021-03-12 21:11:16
So what is

$k$ ?

GabeW1234 2021-03-12 21:11:32
k=12

happyhari 2021-03-12 21:11:32
k = 12

Mathaddict3825 2021-03-12 21:11:32
12

happyhari 2021-03-12 21:11:32
12

Zhaom 2021-03-12 21:11:32

$k=12$

foodisgood 2021-03-12 21:11:32
12

learning0119 2021-03-12 21:11:32
12

MaryM 2021-03-12 21:11:34

$13$ is prime, so we need

$k=12$ to get a common factor of

$13$ .

MaryM 2021-03-12 21:11:39
Let's check it and see!

MaryM 2021-03-12 21:11:43
We have

$a_{30} = \dfrac{31 + 18(12)}{32 + 19(12)} = \dfrac{247}{260}$ .

MaryM 2021-03-12 21:11:47
And indeed, we cancel a

$13$ , and we have

$a_{30} = \dfrac{19}{20}$ .

MaryM 2021-03-12 21:11:50
So

$j=30$ is good!

MaryM 2021-03-12 21:11:58
And our list now is

$j \in \{1,2,8,18,30\}$ .

MaryM 2021-03-12 21:12:04
Now what?

learning0119 2021-03-12 21:12:15
again

MaryM 2021-03-12 21:12:20
Let's do it again!

MaryM 2021-03-12 21:12:24
We have

$a_{30+k} = \dfrac{19 + 18k}{20 + 19k}$ until there's a common factor.

MaryM 2021-03-12 21:12:31
The common factor is

$\gcd(19+18k,20+19k) = \gcd(19+18k,1+k)$ , as before.

MaryM 2021-03-12 21:12:37
And when we subtract

$1+k$ 18 times, we get

$\gcd(1,1+k)$ .

MaryM 2021-03-12 21:12:44
Hmmm...what does that tell us?

klpiguy 2021-03-12 21:12:55
there are no more good numbers

GabeW1234 2021-03-12 21:12:55
no more

rockyrockrock 2021-03-12 21:12:55
it cannot be greater than 1

ancientwarrior 2021-03-12 21:12:55
we're done!

Zhaom 2021-03-12 21:12:55
Simplifies to

$\gcd(1,1+k) \ge 2$ , which is impossible.

Hector2173 2021-03-12 21:12:55
we're done

learning0119 2021-03-12 21:12:57
we are done

MaryM 2021-03-12 21:13:00
That gcd is always

$1$ .

MaryM 2021-03-12 21:13:04
So

$19+18k$ and

$20+19k$ are always relatively prime.

MaryM 2021-03-12 21:13:08
Which means

$a_{30+k}$ will never reduce!

MaryM 2021-03-12 21:13:14
That's good, because it means we're done! We've found all the good numbers:

$j \in \{1,2,8,18,30\}$ .

happyhari 2021-03-12 21:13:18
finally!

GabeW1234 2021-03-12 21:13:23
so we stop, and add

MaryM 2021-03-12 21:13:26
So what is the final answer?

Zhaom 2021-03-12 21:13:37

$1+2+8+18+30=\boxed{059}$

GabeW1234 2021-03-12 21:13:37
59

happyhari 2021-03-12 21:13:37
59

ancientwarrior 2021-03-12 21:13:37
59

Irving1004 2021-03-12 21:13:39
059

learning0119 2021-03-12 21:13:43
$\boxed{059}

MaryM 2021-03-12 21:13:46
Their sum is

$1+2+8+18+30 = \boxed{059}$ .

MaryM 2021-03-12 21:14:15
11. Let

$ABCD$ be a cyclic quadrilateral with

$AB=4,$

$BC=5,$

$CD=6,$ and

$DA=7.$ Let

$A_1$ and

$C_1$ be the feet of the perpendiculars from

$A$ and

$C,$ respectively, to line

$BD,$ and let

$B_1$ and

$D_1$ be the feet of the perpendiculars from

$B$ and

$D,$ respectively, to line

$AC.$ The perimeter of

$A_1B_1C_1D_1$ is

$\dfrac{m}{n},$ where

$m$ and

$n$ are relatively prime positive integers. Find

$m+n.$

MaryM 2021-03-12 21:14:31
Let's see if we can sketch a picture of this. First, here's the quadrilateral:

MaryM 2021-03-12 21:14:39

CrystalFlower 2021-03-12 21:14:59
kool

MaryM 2021-03-12 21:15:01
Any observations before we clutter it further with all the perpendiculars?

functionalmath 2021-03-12 21:15:23
ptolemy?

functionalmath 2021-03-12 21:15:23
use ptolemys theorem and find the diagonals?

learning0119 2021-03-12 21:15:23
ptolemy

MaryM 2021-03-12 21:15:28
Well, we do have a cyclic quadrilateral, and we know all the side lengths.

MaryM 2021-03-12 21:15:31
So we do have Ptolemy's Theorem. (Second time this contest!)

MaryM 2021-03-12 21:15:35
What does it tell us?

Zhaom 2021-03-12 21:16:09

$AC \cdot BD=59$

GabeW1234 2021-03-12 21:16:09
diagonals x and y have a product of 59

ancientwarrior 2021-03-12 21:16:09
product of diagonals is 59

MeepMurp5 2021-03-12 21:16:09

$AC\cdotBC=59$

Mathaddict3825 2021-03-12 21:16:09
(AC)(BD) = 59

Lamboreghini 2021-03-12 21:16:09
5*7+6*4=BD*AC

MaryM 2021-03-12 21:16:11
We get

$(AB)(CD) + (AD)(BC) = (AC)(BD)$ .

MaryM 2021-03-12 21:16:24
So

$(AC)(BD) = (4)(6) + (5)(7) = 24 + 35 = 59$ .

MaryM 2021-03-12 21:16:30
Not sure if this will help, so let's set it aside for now.

MaryM 2021-03-12 21:16:37

$(AC)(BD) = 59$

rockyrockrock 2021-03-12 21:16:40
wacky number

MaryM 2021-03-12 21:16:44
OK, let's add all the other stuff. I'll make the pic a little bigger so we can see all the detail.

MaryM 2021-03-12 21:16:51

MaryM 2021-03-12 21:17:01
I've also labeled the point where the diagonals intersect as

$X$ . I'm guessing that point might be important.

Irving1004 2021-03-12 21:17:06
This is where they get hard

klpiguy 2021-03-12 21:17:06
it looks messy

MaryM 2021-03-12 21:17:13
Any thoughts?

rockyrockrock 2021-03-12 21:18:01
quadrilateral 1 looks similar to original

Lamboreghini 2021-03-12 21:18:01
maybe A1B1C1D1 is also cyclic? maybe it's similar to ABCD?

happyhari 2021-03-12 21:18:01
are the two quadrilaterals similar?

MaryM 2021-03-12 21:18:06
The two quadrilaterals do look similar. But we don’t have many tools to deal with similar quadrilaterals. Why don’t we try finding some similar triangles instead?

MaryM 2021-03-12 21:19:29
But to find similar triangles we will need equal angles, so we need to take a closer look into this picture.

MaryM 2021-03-12 21:19:35
We've got the big quadrilateral

$ABCD$ and the little one

$A_1B_1C_1D_1$ , but there are others in this picture too, right?

MaryM 2021-03-12 21:19:44
For example,

$ABA_1B_1$ . What do we know about it?

CircleInvert 2021-03-12 21:20:28
It is cyclic

ancientwarrior 2021-03-12 21:20:28
it is cyclic

Irving1004 2021-03-12 21:20:28
It is cyclic

MaryM 2021-03-12 21:20:30
It's cyclic!

MaryM 2021-03-12 21:20:43
The right angles tell us it's cyclic: specifically, its circumcircle has diameter

$\overline{AB}$ :

MaryM 2021-03-12 21:20:49

MaryM 2021-03-12 21:21:00
Note how the two right angles intercept the diameter of the red circle.

MaryM 2021-03-12 21:21:13
Great -- how can we use this cyclic quadrilateral to make progress on the problem?

MaryM 2021-03-12 21:21:43
Hint: Try to find some equal angles.

Zhaom 2021-03-12 21:22:12

$\angle C_1A_1B_1=\angle BAB_1$

MaryM 2021-03-12 21:22:17
There are lots of cyclic quadrilateral facts, but a particularly useful one here is that opposite angles of a cyclic quadrilateral sum to

$180^\circ$ .

MaryM 2021-03-12 21:22:27
It means that an angle of a cyclic quadrilateral is congruent to the exterior angle at the opposite vertex.

MaryM 2021-03-12 21:22:36
That is,

$\angle BAB_1$ is congruent to the exterior angle

$\angle B_1A_1X$ on the opposite side.

MaryM 2021-03-12 21:22:50
And in the same way,

$\angle ABA_1$ is congruent to the opposite exterior angle

$\angle A_1B_1X$ .

MaryM 2021-03-12 21:22:54
This might be easier to see if I mark these in the picture:

MaryM 2021-03-12 21:23:00

MaryM 2021-03-12 21:23:05
So what do we get?

Zhaom 2021-03-12 21:23:47

$\triangle XBA\sim triangle XA_1B_1$

happyhari 2021-03-12 21:23:47
similar triangles

Lamboreghini 2021-03-12 21:23:47
A1XB1~AXB

Irving1004 2021-03-12 21:23:47
Similar triangles that take too long to type out!

MaryM 2021-03-12 21:23:49
We get that

$\triangle ABX$ and

$\triangle A_1B_1X$ are similar! They have the same two angles away from

$X$ (and the same third angle at

$X$ ).

MaryM 2021-03-12 21:23:54
In terms of lengths, what does this give us?

GabeW1234 2021-03-12 21:24:52

$4:A_1B_1=XA_1:XB_1=XB_1:A$

Zhaom 2021-03-12 21:24:52

$\frac{XB}{XB_1}=\frac{XA}{XA_1}=\frac{AB}{A_1B_1}$

MaryM 2021-03-12 21:24:55
We get

$\dfrac{AB}{A_1B_1} = \dfrac{AX}{A_1X} = \dfrac{BX}{B_1X}$ .

MaryM 2021-03-12 21:24:59
Aha -- there's the length

$A_1B_1$ we want!

MaryM 2021-03-12 21:25:08
And we know

$AB = 4$ .

MaryM 2021-03-12 21:25:18
Furthermore, do you recognize those other ratios?

MaryM 2021-03-12 21:25:33
Hint: think about trigonometry and right triangles.

MaryM 2021-03-12 21:27:30
These ratios appear in the right triangles

$\triangle AA_1X$ and

$\triangle BB_1X$ . If we set

$\angle AXB=\theta$ , how can we write the ratios?

jxwis2010 2021-03-12 21:29:00

$/frac{1}{\cos\theta}$

MaryM 2021-03-12 21:29:06
These ratios are the secant of angle

$\angle AXA_1 = \angle BXB_1$ , using the right triangles

$\triangle AA_1X$ or

$BB_1X$ . (That is, they're the reciprocal of

$\cos\angle AXA_1$ :

$A_1X$ is the "adjacent" side and

$AX$ is the hypotenuse.)

MaryM 2021-03-12 21:29:20
So if we set

$\theta = \angle AXB$ , we get

$\dfrac{AB}{A_1B_1} = \sec\theta = \dfrac{1}{\cos\theta}$ .

MaryM 2021-03-12 21:29:25
That is,

$A_1B_1 = AB \cos \theta$ .

MaryM 2021-03-12 21:29:32
Now what?

MaryM 2021-03-12 21:30:21
We handled one of the sides of the smaller quadrilateral. How can we handle the other ones?

jxwis2010 2021-03-12 21:30:37
do the same thing for the other 3 segments

MaryM 2021-03-12 21:30:41
We can do the same thing on the other three sides!

MaryM 2021-03-12 21:30:51

$CDC_1D_1$ works exactly the same way, and since

$\angle CXD = \angle AXB = \theta$ , we get

$C_1D_1 = CD \cos \theta$ as well.

MaryM 2021-03-12 21:30:56
The other two are slightly different, because

$X$ is inside the circle given by the cyclic quadrilateral. For example, let's look at

$BCC_1B_1$ :

MaryM 2021-03-12 21:31:03

MaryM 2021-03-12 21:31:23
But we still get

$\triangle BCX \sim \triangle B_1C_1X$ .

MaryM 2021-03-12 21:31:35
So, we get

$\dfrac{BC}{B_1C_1} = \dfrac{BX}{B_1X}$ .

MaryM 2021-03-12 21:31:42
What expression do we get for

$B_1C_1$ ?

functionalmath 2021-03-12 21:32:06
same tihng

Lamboreghini 2021-03-12 21:32:06
BC cos theta???

MaryM 2021-03-12 21:32:10
The right-hand side is still the reciprocal of

$\cos\theta$ using

$\triangle BB_1X$ .

MaryM 2021-03-12 21:32:18
So we still get

$B_1C_1 = BC \cos \theta$ .

MaryM 2021-03-12 21:32:26
And on the other side, we get

$A_1D_1 = AD \cos \theta$ in the same way.

MaryM 2021-03-12 21:32:39
So, how can we write the perimeter of the smaller quadrilateral?

Zhaom 2021-03-12 21:33:10

$(4+5+6+7)\cos(\theta)$

YaoAOPS 2021-03-12 21:33:10
cos theta of each side added

MaryM 2021-03-12 21:33:13
We get

$A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA) \cos \theta.$

MaryM 2021-03-12 21:33:17
That is, the perimeter of the small brown quadrilateral is

$\cos\theta$ times the perimeter of

$ABCD$ .

MaryM 2021-03-12 21:33:23
But we know the big perimeter!

MaryM 2021-03-12 21:33:26
It's given to us:

$4+5+6+7 = 22$ .

MaryM 2021-03-12 21:33:36
So the answer we want is

$\text{Perim}(A_1B_1C_1D_1) = 22\cos\theta$ .

MaryM 2021-03-12 21:33:42
If we can figure out

$\cos\theta$ , we're done.

MaryM 2021-03-12 21:34:01
Let's go back to the simpler picture:

MaryM 2021-03-12 21:34:10

MaryM 2021-03-12 21:34:19
How can we compute the cosine of that angle?

functionalmath 2021-03-12 21:34:42
cosine law

CircleInvert 2021-03-12 21:34:46
LoC

MaryM 2021-03-12 21:34:49
Well, we have a bunch of Law of Cosines expressions that we can write down for the four small triangles.

MaryM 2021-03-12 21:34:53
For example, what do we get for

$\triangle ABX$ ?

ancientwarrior 2021-03-12 21:35:25
16=AX^2+BX^2-2AXBXcos(theta)

Zhaom 2021-03-12 21:35:25

$BX^2+AX^2-2 \cdot BX \cdot AX\cdot\cos(\theta)$

MaryM 2021-03-12 21:35:28
We get

$4^2 = (AB)^2 = (AX)^2 + (BX)^2 - 2(AX)(BX)\cos\theta$ .

MaryM 2021-03-12 21:35:32
How about

$\triangle BCX$ ?

Zhaom 2021-03-12 21:36:14

$BX^2+CX^2-2 \cdot BX \cdot CX\cdot\cos(\theta)=25$

MaryM 2021-03-12 21:36:17
We get

$5^2 = (BC)^2 = (BX)^2 + (CX)^2 - 2(BX)(CX)\cos(180^\circ-\theta)$ .

MaryM 2021-03-12 21:36:23
But

$\cos(180^\circ - \theta) = -\cos\theta$ .

ancientwarrior 2021-03-12 21:36:31
25=BX^2+CX^2+2BXCXcos(theta)

MaryM 2021-03-12 21:36:34
So our equation is

$5^2 = (BX)^2 + (CX)^2 + 2(BX)(CX)\cos\theta$ .

MaryM 2021-03-12 21:36:38
The other two triangles are essentially the same, and we get the following system of four equations:

MaryM 2021-03-12 21:36:41

$\begin{align*} 4^2 &= (AX)^2 + (BX)^2 - 2(AX)(BX)\cos\theta, \\ 5^2 &= (BX)^2 + (CX)^2 + 2(BX)(CX)\cos\theta, \\ 6^2 &= (CX)^2 + (DX)^2 - 2(CX)(DX)\cos\theta, \\ 7^2 &= (DX)^2 + (AX)^2 + 2(DX)(AX)\cos\theta. \end{align*}$

MaryM 2021-03-12 21:36:49
What can we do with these?]

MaryM 2021-03-12 21:36:54
Can we cancel out all the length-squared terms (like

$(AX)^2$ )?

CircleInvert 2021-03-12 21:37:58
Add the first and third equations, and subtract from the obtained equation the sum of the second and fourth equations

Z_Lu 2021-03-12 21:37:58
subtract and add in alternating order?

MaryM 2021-03-12 21:38:01
Yes: if we add the two with the

$+2$ cosine term and subtract the two with the

$-2$ cosine term, those terms will all cancel out.

MaryM 2021-03-12 21:38:06
We're left with

$-4^2 + 5^2 - 6^2 + 7^2 = 2((AX)(BX)+(BX)(CX)+(CX)(DX)+(DX)(AX))\cos\theta.$

MaryM 2021-03-12 21:38:16
The left side is

$-16+25-36+49 = 22$ .

MaryM 2021-03-12 21:38:24
So after dividing by

$2$ , we get

$11 = ((AX)(BX)+(BX)(CX)+(CX)(DX)+(DX)(AX))\cos\theta.$

MaryM 2021-03-12 21:38:27
Anything we can do with that expression involving the lengths?

Zhaom 2021-03-12 21:38:49
factor

MaryM 2021-03-12 21:38:53
Aha! It factors!

MaryM 2021-03-12 21:38:57
We get

$11 = ((AX + CX)(BX + DX))\cos\theta$ .

MaryM 2021-03-12 21:39:01
But what's that?

Irving1004 2021-03-12 21:39:21
Ptolemy's theorem from earlier

Lamboreghini 2021-03-12 21:39:21
59 cos theta!

YaoAOPS 2021-03-12 21:39:21
ptolmey

rockyrockrock 2021-03-12 21:39:21
59

Zhaom 2021-03-12 21:39:21

$AC \cdot BD\cos(\theta)=59\cos(\theta)$

rockyrockrock 2021-03-12 21:39:25
ac*bd

MaryM 2021-03-12 21:39:28

$AX+CX = AC$ and

$BX+DX = BD$ .

MaryM 2021-03-12 21:39:34
So this is

$11 = (AC)(BD)\cos\theta$ .

MaryM 2021-03-12 21:39:43
But we computed

$(AC)(BD) = 59$ way back at the beginning!

MaryM 2021-03-12 21:39:50
So we have

$11 = 59\cos\theta$ , hence

$\cos\theta = \dfrac{11}{59}$ .

MaryM 2021-03-12 21:39:59
What's the answer, then?

happyhari 2021-03-12 21:40:18
22(11/59) = 242/59

Zhaom 2021-03-12 21:40:18

$242+59=\boxed{301}$

Irving1004 2021-03-12 21:40:18
242/59

happyhari 2021-03-12 21:40:18
242/59

rockyrockrock 2021-03-12 21:40:20
301

MaryM 2021-03-12 21:40:22
The perimeter we wanted was

$22\cos\theta$ .

MaryM 2021-03-12 21:40:26
So it's

$22 \cdot \dfrac{11}{59} = \dfrac{242}{59}$ .

MaryM 2021-03-12 21:40:30
This is in lowest terms, so our final answer is

$242 + 59 = \boxed{301}$ .

ancientwarrior 2021-03-12 21:40:33
cool

MaryM 2021-03-12 21:40:54
12. Let

$A_1A_2A_3 \dots A_{12}$ be a dodecagon (

$12$ -gon). Three frogs initially sit at

$A_4,$

$A_8,$ and

$A_{12}.$ At the end of each minute, simultaneously each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is

$\dfrac{m}{n},$ where

$m$ and

$n$ are relatively prime positive integers. Find

$m+n.$

MaryM 2021-03-12 21:41:03
Let's draw a diagram of the frogs:

MaryM 2021-03-12 21:41:08

MaryM 2021-03-12 21:41:21
Let's call this position S (for "Start").

MaryM 2021-03-12 21:41:28
What can happen on the first move?

ancientwarrior 2021-03-12 21:42:08
two frogs get closer to each other or they all stay relatively the same

Zhaom 2021-03-12 21:42:08
each of the frogs can either hop clockwise or counterclockwise

$1$ vertex

Irving1004 2021-03-12 21:42:08
8 things, two where essentially start over

MaryM 2021-03-12 21:42:15
They might all move in the same direction: that is, they might all move clockwise or counterclockwise.

MaryM 2021-03-12 21:42:22
That's doesn't change their relative position, so it essentially keeps us at S.

MaryM 2021-03-12 21:42:27
The only other possibility is that two of them move in one direction and the third moves in the other direction.

MaryM 2021-03-12 21:42:32
For example, if the top and left frogs both move clockwise, but the right frog moves counterclockwise, we end up here:

MaryM 2021-03-12 21:42:36

MaryM 2021-03-12 21:42:42
Does it matter which frogs move in which direction?

Zhaom 2021-03-12 21:42:56
no

ancientwarrior 2021-03-12 21:42:56
not really

Lamboreghini 2021-03-12 21:43:00
no, the relative position is the same

functionalmath 2021-03-12 21:43:01
no

MaryM 2021-03-12 21:43:04
Not really. The two frogs that move in the same direction will still be 4 vertices apart. The other frog will be distance 2 from whichever frog it moves towards (and distance 6 from the other frog). But all these positions are essentially the same by symmetry.

MaryM 2021-03-12 21:43:09
Let's call this position "A" (for, um, "Amazing position").

MaryM 2021-03-12 21:43:28
What's the probability that we move S-to-S?

Zhaom 2021-03-12 21:43:51

$\frac{1}{4}$

Irving1004 2021-03-12 21:43:51
1/4

MaryM 2021-03-12 21:43:54
Imagine they move one at a time. The second and third frogs have to copy what the first frog does.

MaryM 2021-03-12 21:43:58
They each do this with probability

$\dfrac12$ , so they both do it with probability

$\dfrac14$ .

MaryM 2021-03-12 21:44:02
So we go from S to S with probability

$\dfrac14$ .

MaryM 2021-03-12 21:44:08
And thus we go from S to A with probability

$\dfrac34$ .

MaryM 2021-03-12 21:44:13
Let's make a chart of what we have so far:

MaryM 2021-03-12 21:44:20

$\begin{array}{r|ll} \text{From }\downarrow & \text{S} & \text{A} \\ \hline \text{S} & \dfrac14 & \dfrac34 \\[1.5ex] \text{A} & \end{array}$

MaryM 2021-03-12 21:44:27
Now what can happen starting from position A?

MaryM 2021-03-12 21:44:36

MaryM 2021-03-12 21:45:05
Can we win from this position?

functionalmath 2021-03-12 21:45:19
yes

Hector2173 2021-03-12 21:45:19
yes

klpiguy 2021-03-12 21:45:19
yup

donguri 2021-03-12 21:45:19
Yes

MaryM 2021-03-12 21:45:21
We might win!

MaryM 2021-03-12 21:45:25
If the two frogs that are 2 apart move towards each other, they land on top of each other, and we win.

MaryM 2021-03-12 21:45:28
What's the probability of that happening?

Lamboreghini 2021-03-12 21:45:46
1/4

happyhari 2021-03-12 21:45:46
1/4

ancientwarrior 2021-03-12 21:45:46
1/4

Zhaom 2021-03-12 21:45:46

$\frac{1}{4}$

Hector2173 2021-03-12 21:45:48
1/4

MaryM 2021-03-12 21:45:50
Each of the bottom two frogs must pick the right direction. (It doesn't matter what the third frog does.)

MaryM 2021-03-12 21:45:59
This happens with probability

$\dfrac12 \cdot \dfrac12 = \dfrac14$ .

MaryM 2021-03-12 21:46:03
So let's update the chart:

MaryM 2021-03-12 21:46:19

$\begin{array}{r|ccc} \text{From }\downarrow & \text{S} & \text{A} & \text{Win} \\ \hline \text{S} & \dfrac14 & \dfrac34 \\[1.5ex] \text{A} & & & \dfrac14 \end{array}$

MaryM 2021-03-12 21:46:49
Can we still be at

$A$ after the frogs move?

Zhaom 2021-03-12 21:47:11
yes

Hector2173 2021-03-12 21:47:11
yes

happyhari 2021-03-12 21:47:11
yes

klpiguy 2021-03-12 21:47:11
yes

MaryM 2021-03-12 21:47:13
Yes, they could all move in the same direction. That keeps us at A.

MaryM 2021-03-12 21:47:17
And what's the probability that this happens?

klpiguy 2021-03-12 21:47:34
1/4

Zhaom 2021-03-12 21:47:34

$\frac{1}{4}$

Z_Lu 2021-03-12 21:47:34
1/4

happyhari 2021-03-12 21:47:34
1/4

MaryM 2021-03-12 21:47:36
It's

$\dfrac14$ , just like before.

MaryM 2021-03-12 21:47:52
The two bottom frogs could also move in the same direction, but the other frog move opposite.

MaryM 2021-03-12 21:47:57
What happens then?

klpiguy 2021-03-12 21:48:29
A NEW POSITION!!!

Zhaom 2021-03-12 21:48:29
new position

Lamboreghini 2021-03-12 21:48:29
then, the frogs are equally spaced with 1 place between each

MaryM 2021-03-12 21:48:32
They could move towards each other, like so:

MaryM 2021-03-12 21:48:35

MaryM 2021-03-12 21:48:41
This is a new position. Let's call it B (for, um, "Beautiful"?).

MaryM 2021-03-12 21:48:45
What's the probability of this?

Lamboreghini 2021-03-12 21:49:11
1/8

happyhari 2021-03-12 21:49:11
1/8

Zhaom 2021-03-12 21:49:11

$\frac{1}{8}$

klpiguy 2021-03-12 21:49:15
1/8

Irving1004 2021-03-12 21:49:17
1.8

Irving1004 2021-03-12 21:49:23
1/8

MaryM 2021-03-12 21:49:24
All three frogs have to move in the "right" direction to get this.

MaryM 2021-03-12 21:49:30
So it's

$\left(\dfrac12\right)^3 = \dfrac18$ .

MaryM 2021-03-12 21:49:33
What if the bottom two move in the same direction, but the third moves away from them?

Hector2173 2021-03-12 21:50:16
A

happyhari 2021-03-12 21:50:16
same position as A

Zhaom 2021-03-12 21:50:16
still

$A$

MaryM 2021-03-12 21:50:18
We end up here:

MaryM 2021-03-12 21:50:23

MaryM 2021-03-12 21:50:27
This is A again (though it's a mirror image).

MaryM 2021-03-12 21:50:37
So that's another

$\dfrac18$ that we stay in A. Combined with the

$\dfrac14$ from before, that's

$\dfrac38$ total.

MaryM 2021-03-12 21:51:03
The two close frogs in A could also move away from each other.

MaryM 2021-03-12 21:51:12
Then third frog ends up in one of the red positions here:

MaryM 2021-03-12 21:51:18

MaryM 2021-03-12 21:51:21
If it's the top red dot, what position are we in?

Zhaom 2021-03-12 21:51:36

$S$

Lamboreghini 2021-03-12 21:51:36
S

happyhari 2021-03-12 21:51:36
S

Z_Lu 2021-03-12 21:51:36
S

MaryM 2021-03-12 21:51:39
We're back to S.

MaryM 2021-03-12 21:51:43
If it's the bottom red dot, what position are we in?

Lamboreghini 2021-03-12 21:51:55
A

Z_Lu 2021-03-12 21:51:55
A

Zhaom 2021-03-12 21:51:55

$A$ again

ancientwarrior 2021-03-12 21:51:55
A

Hector2173 2021-03-12 21:51:55
AA

happyhari 2021-03-12 21:51:55
A

MaryM 2021-03-12 21:51:57
We'll still in A.

MaryM 2021-03-12 21:52:01
And that's another

$\dfrac18$ probability for each.

MaryM 2021-03-12 21:52:06
Note that's now a total of

$\dfrac12$ that we stay in A.

MaryM 2021-03-12 21:52:11
So here's the updated chart with all this data:

MaryM 2021-03-12 21:52:16

$\begin{array}{r|cccc} \text{From }\downarrow & \text{S} & \text{A} & \text{B} & \text{Win} \\ \hline \text{S} & \dfrac14 & \dfrac34 \\[2.5ex] \text{A} & \dfrac18 & \dfrac12 & \dfrac18 & \dfrac14 \\[2.5ex] \text{B} & \end{array}$

MaryM 2021-03-12 21:52:20
What's next?

Zhaom 2021-03-12 21:52:41
do cases for

$B$

klpiguy 2021-03-12 21:52:41
we use B

Z_Lu 2021-03-12 21:52:41
starting from b postion

MaryM 2021-03-12 21:52:45
We need to figure out what can happen from B:

MaryM 2021-03-12 21:52:53

MaryM 2021-03-12 21:52:59
What's the probability we win from here?

klpiguy 2021-03-12 21:53:24
1/2

happyhari 2021-03-12 21:53:24
1/2

Zhaom 2021-03-12 21:53:24

$\frac{1}{2}$

rockyrockrock 2021-03-12 21:53:24
1/2

MaryM 2021-03-12 21:53:28
The middle frog has to move one way or the other.

MaryM 2021-03-12 21:53:33
In either case, if the frog on that side moves towards the middle frog, we win!

MaryM 2021-03-12 21:53:55
The probability of this happening is

$\dfrac12$ . The outside frog has to make the right choice.

MaryM 2021-03-12 21:54:01
So we win with probability

$\dfrac12$ .

MaryM 2021-03-12 21:54:06
What else could happen?

Zhaom 2021-03-12 21:54:35
the frogs move in the same direction

ancientwarrior 2021-03-12 21:54:35
they all shift in same direction

MaryM 2021-03-12 21:54:37
As usual, all three frogs could move in the same direction with probability

$\dfrac14$ . This keeps us in B.

MaryM 2021-03-12 21:55:02
It's also possible that the middle frog moves in one direction, and the other frog in that direction also goes in that direction, but the third frog goes in the other direction.

MaryM 2021-03-12 21:55:07
Where does that take us?

ancientwarrior 2021-03-12 21:55:32
i think that's A?

Z_Lu 2021-03-12 21:55:32
A

functionalmath 2021-03-12 21:55:36
A

MaryM 2021-03-12 21:55:38
It takes us back to A.

MaryM 2021-03-12 21:55:44
And this occurs with probability

$\dfrac14$ , because the other two frogs have to move a certain way relative to the middle frog.

MaryM 2021-03-12 21:55:49
So here's the final chart:

MaryM 2021-03-12 21:56:10

$\begin{array}{r|cccc} \text{From }\downarrow & \text{S} & \text{A} & \text{B} & \text{Win} \\ \hline \text{S} & \dfrac14 & \dfrac34 \\[2.5ex] \text{A} & \dfrac18 & \dfrac12 & \dfrac18 & \dfrac14 \\[2.5ex] \text{B} & & \dfrac14 & \dfrac14 & \dfrac12 \end{array}$

MaryM 2021-03-12 21:56:18
As a check, each row sums to

$1$ , so we've captured all the possible moves from each position.

MaryM 2021-03-12 21:56:33
Now, what do we do with this chart?

Zhaom 2021-03-12 21:57:00
use states

functionalmath 2021-03-12 21:57:00
find expected value

MaryM 2021-03-12 21:57:05
We can write down equations for the expected number of moves until we win.

MaryM 2021-03-12 21:57:09
Let's set

$s,a,b$ to be the expected number of moves needed to win from S, A, B, respectively.

MaryM 2021-03-12 21:57:15
What's the equation for

$s$ ?

MaryM 2021-03-12 21:57:49
Hint: We spend 1 move, and we end up back at S (with probability

$\dfrac14$ ) or at A (with probability

$\dfrac34$ ). If we end up at S, we expect to need

$s$ more moves to win. If we end up at A, we expect to need

$a$ more moves to win.

MaryM 2021-03-12 21:58:21
Don't forget to account for the first move we make.

Zhaom 2021-03-12 21:59:21

$s=\frac{s+1}{4}+\frac{3(a+1)}{4}$

ancientwarrior 2021-03-12 21:59:21
s=1/4(s+1)+3/4(a+1)

MaryM 2021-03-12 21:59:26
We get

$s = 1 + \dfrac14s + \dfrac34a$ . The

$1$ is the first move, and then we need

$s$ or

$a$ more moves, depending on which position we land on.

MaryM 2021-03-12 21:59:41
In a similar fashion, what's the equation for

$a$ ?

ancientwarrior 2021-03-12 22:00:50
a=1/8(s+1)+1/2(a+1)+1/8(b+1)+1/4

Zhaom 2021-03-12 22:00:50

$a=\frac{s+1}{8}+\frac{a+1}{2}+\frac{b+1}{8}+\frac{1}{4}$

MaryM 2021-03-12 22:00:54
We have

$a = 1 + \dfrac18s + \dfrac12a + \dfrac18b$ . (If we win, we don't need to make any more moves -- the game is over.)

MaryM 2021-03-12 22:01:05
And what's the equation for

$b$ ?

Zhaom 2021-03-12 22:02:06

$b=\frac{a+1}{4}+\frac{b+1}{4}+\frac{1}{2}$

functionalmath 2021-03-12 22:02:06
(a+1)/4+(b+1)/4+1/2

ancientwarrior 2021-03-12 22:02:06
b=1/4(a+1)+1/4(b+1)+1/2

rockyrockrock 2021-03-12 22:02:06
b=1/4(a+1)+1/4(b+1)+1/2

MaryM 2021-03-12 22:02:09
We have

$b = 1 + \dfrac14a + \dfrac14b$ .

MaryM 2021-03-12 22:02:15
So we have a system of three linear equations in three variables:

MaryM 2021-03-12 22:02:38

$\begin{align*} s &= 1 + \dfrac14s + \dfrac34a, \\ a &= 1 + \dfrac18s + \dfrac12a + \dfrac18b, \\ b &= 1 + \dfrac14a + \dfrac14b. \end{align*}$

MaryM 2021-03-12 22:02:44
Now what?

rockyrockrock 2021-03-12 22:03:10
SOLVE!

MaryM 2021-03-12 22:03:14
We need to solve for

$s$ to solve the problem.

MaryM 2021-03-12 22:03:23
Let's clean these up by clearing denominators and moving the variables to one side.

MaryM 2021-03-12 22:03:30

$\begin{align*} 3s - 3a &= 4, \\ -s + 4a - b &= 8, \\ -a + 3b &= 4. \end{align*}$

MaryM 2021-03-12 22:03:40
We can eliminate

$b$ by adding 3 times the second equation to the third.

MaryM 2021-03-12 22:03:48
And then we have:

$\begin{align*} 3s - 3a &= 4, \\ -3s + 11a &= 28. \end{align*}$

ancientwarrior 2021-03-12 22:03:49
we need to solve for s in the end

Irving1004 2021-03-12 22:03:49
solve for s!

MaryM 2021-03-12 22:03:54
We eliminate

$a$ by multiplying the top equation by 11 and the bottom one by 3:

MaryM 2021-03-12 22:03:59

$\begin{align*} 33s - 33a &= 44, \\ -9s + 33a &= 84. \end{align*}$

MaryM 2021-03-12 22:04:04
And then adding gives

$24s = 128$ .

MaryM 2021-03-12 22:04:14
What is the answer?

rockyrockrock 2021-03-12 22:04:30
s=16/3

rockyrockrock 2021-03-12 22:04:30
19

functionalmath 2021-03-12 22:04:30
16

MaryM 2021-03-12 22:04:33
So

$s = \dfrac{128}{24} = \dfrac{16}{3}$ .

MaryM 2021-03-12 22:04:37
And our final answer is

$16 + 3 = \boxed{019}$ .

MaryM 2021-03-12 22:05:06
Only three problems left!

happyhari 2021-03-12 22:05:17
yay!

Zhaom 2021-03-12 22:05:22
yay

MaryM 2021-03-12 22:05:31
13. Circles

$\omega_1$ and

$\omega_2$ with radii

$961$ and

$625,$ respectively, intersect at distinct points

$A$ and

$B.$ A third circle

$\omega$ is externally tangent to both

$\omega_1$ and

$\omega_2.$ Suppose line

$AB$ intersects

$\omega$ at two points

$P$ and

$Q$ such that the measure of minor arc

$\stackrel{\displaystyle\frown}{PQ}$ is

$120^\circ\!.$ Find the distance between the centers of

$\omega_1$ and

$\omega_2.$

functionalmath 2021-03-12 22:05:47
daigram!

Zhaom 2021-03-12 22:05:47
diagram

MaryM 2021-03-12 22:05:49
Let's try to sketch a picture.

MaryM 2021-03-12 22:05:54
It's a little bit hard to draw because we don't know if

$\overline{AB}$ crosses the segment connecting the two centers or not.

MaryM 2021-03-12 22:05:57
That is, we could have something like this:

MaryM 2021-03-12 22:06:01

MaryM 2021-03-12 22:06:06
Or we could have something like this:

MaryM 2021-03-12 22:06:11

MaryM 2021-03-12 22:06:25
For now, let's use the first one, and sketch the third circle and

$\overline{BAPQ}$ :

MaryM 2021-03-12 22:06:46

MaryM 2021-03-12 22:06:59
That looks reasonable. I could believe that arc

$PQ$ is

$120^\circ$ in this picture.

MaryM 2021-03-12 22:07:09
Let's also set

$d = O_1O_2$ to be what we're looking for.

MaryM 2021-03-12 22:07:33
Do we know

$OO_1$ and

$OO_2$ ?

functionalmath 2021-03-12 22:08:06
yes

panda2018 2021-03-12 22:08:06
961+r,625+r

Irving1004 2021-03-12 22:08:06
r+961 and r+625

MaryM 2021-03-12 22:08:13
Yes: they connect two tangent circles, so these lengths are the sum of the radii.

MaryM 2021-03-12 22:08:17
That is, we have

$OO_1 = 961+r$ and

$OO_2 = 625+r$ .

MaryM 2021-03-12 22:08:22
And we know that

$\angle POQ = 120^\circ$ .

MaryM 2021-03-12 22:08:27
How can we get that angle to connect to the rest of the picture?

functionalmath 2021-03-12 22:09:04
connect OP and OQ

MaryM 2021-03-12 22:09:10
One idea is to draw line

$OP$ . In particular, we can make the blue triangle shown:

MaryM 2021-03-12 22:09:17

MaryM 2021-03-12 22:09:32
Do you notice anything interesting about that triangle?

ancientwarrior 2021-03-12 22:10:18
looks like a right triangle

panda2018 2021-03-12 22:10:18
<D is right

Irving1004 2021-03-12 22:10:18
Is it right

MaryM 2021-03-12 22:10:26
It is a right triangle, but we know more than that.

Zhaom 2021-03-12 22:10:48
is a 30-60-90 triangle

CircleInvert 2021-03-12 22:10:48

$30$ -

$60$ -

$90$

MaryM 2021-03-12 22:11:27
The blue triangle

$\triangle OCD$ is a 30-60-90 right triangle!

Zhaom 2021-03-12 22:11:39
wait, what's

$D$ ?

MaryM 2021-03-12 22:11:52
Right, I think we didn't define

$D$ .

MaryM 2021-03-12 22:12:14
But it is the intersection of

$AB$ and a perpendicular from

$O$ .

MaryM 2021-03-12 22:12:17
Sorry about that.

Zhaom 2021-03-12 22:13:15
do you mean

$O_1O_2$ , not

$AB$ ?

MaryM 2021-03-12 22:13:21
Yes! Sorry again.

MaryM 2021-03-12 22:13:40
It's definitely not on

$AB$ .

YaoAOPS 2021-03-12 22:14:14
Why is it 30-60-90

MaryM 2021-03-12 22:14:23
That's because

$\angle OPQ=30^\circ$ .

MaryM 2021-03-12 22:14:42
Which follows from the fact that

$\angle POQ= 120^\circ$ .

Irving1004 2021-03-12 22:14:53
Yeah

MaryM 2021-03-12 22:15:26
Well, we know that

$\angle OCO_1 = 120^\circ$ and

$\angle OCO_2 = 60^\circ$ .

MaryM 2021-03-12 22:15:47
I don't have any really great ideas, but I know we eventually need to compute the length

$O_1O_2$ , which splits as

$O_1C + CO_2$ .

MaryM 2021-03-12 22:15:53
And we know a nice angle at

$C$ .

MaryM 2021-03-12 22:16:00
So maybe we can set up some Law of Cosines computations for some of these triangles to try to get at these lengths.

MaryM 2021-03-12 22:16:07
For example, we can write down the Law of Cosines for

$\triangle OCO_1$ , using angle

$\angle OCO_1 = 120^\circ$ .

MaryM 2021-03-12 22:16:11
What do we get?

Zhaom 2021-03-12 22:16:55

$O_1O^2=CO^2+CO_1^2+CO \cdot CO_1$

MaryM 2021-03-12 22:16:58
We get

$(OO_1)^2 = (OC)^2 + (O_1C)^2 - 2(OC)(O_1C)\cos(120^\circ).$

MaryM 2021-03-12 22:17:10
But

$\cos(120^\circ)=\frac 12$ .

MaryM 2021-03-12 22:17:15
So this is

$(OO_1)^2 = (OC)^2 + (O_1C)^2 + (OC)(O_1C).$

MaryM 2021-03-12 22:17:20
That's not too bad, I guess.

MaryM 2021-03-12 22:17:27
We can do something similar for

$\triangle OCO_2$ .

MaryM 2021-03-12 22:17:32
What do we get?

Zhaom 2021-03-12 22:17:46
Do you mean

$\cos\left(120^\circ\right)=-\frac{1}{2}$ ?

MaryM 2021-03-12 22:18:03
So many typos! Yes, it's

$-\frac 12$ .

MaryM 2021-03-12 22:18:31
But well, what do we get for

$\triangle OCO_2$ using the law of Cosines?

Zhaom 2021-03-12 22:19:55
I meant

$O_1O^2=CO^2+CO_1^2-CO \cdot CO_1$

MaryM 2021-03-12 22:19:58
We get essentially the same thing, except now we're using

$\cos \angle OCO_2 = \cos 60^\circ=\frac 12.$

MaryM 2021-03-12 22:20:09
(I guess typing is getting harder and harder.)

MaryM 2021-03-12 22:20:22
But well, we get

$(OO_2)^2 = (OC)^2 + (O_2C)^2 - (OC)(O_2C).$

MaryM 2021-03-12 22:20:32
Let's put these side-by-side and see if we can do anything:

MaryM 2021-03-12 22:20:37

$\begin{align*} (OO_1)^2 &= (OC)^2 + (O_1C)^2 + (OC)(O_1C), \\ (OO_2)^2 &= (OC)^2 + (O_2C)^2 - (OC)(O_2C). \end{align*}$

MaryM 2021-03-12 22:20:42
What can we do with these equations?

Lamboreghini 2021-03-12 22:21:18
we can get rid of OC^2!

panda2018 2021-03-12 22:21:18
subtract them

happyhari 2021-03-12 22:21:18
subtract and get rid of (OC)^2

Irving1004 2021-03-12 22:21:18
subtract?

Zhaom 2021-03-12 22:21:18
subtract the second from the first

MaryM 2021-03-12 22:21:22
When we subtract them, the

$(OC)^2$ term cancels, and we get

$(OO_1)^2 - (OO_2)^2 = (O_1C)^2 - (O_2C)^2 + (OC)(O_1C + O_2C).$

MaryM 2021-03-12 22:21:29
Hey...what that last term at the end,

$O_1C + O_2C$ ?

panda2018 2021-03-12 22:21:47
O_1O_2

Lamboreghini 2021-03-12 22:21:47
O1O2

MaryM 2021-03-12 22:21:50
That's

$O_1O_2$ . That's what we want to find!

MaryM 2021-03-12 22:21:54
We called it

$d$ , remember?

MaryM 2021-03-12 22:21:59
So let's substitute

$d$ in:

$(OO_1)^2 - (OO_2)^2 = (O_1C)^2 - (O_2C)^2 + (OC)d.$

MaryM 2021-03-12 22:22:05
So I think maybe we're making progress.

MaryM 2021-03-12 22:22:18
But I don't really know what to do with the other terms that involve point

$C$ .

MaryM 2021-03-12 22:22:26
Perhaps there's another point related to the blue angle that we can use?

YaoAOPS 2021-03-12 22:23:11
CPO2

MaryM 2021-03-12 22:23:15
How about point

$P$ ?

MaryM 2021-03-12 22:23:22
We can do the same Law of Cosines calculations using triangles

$\triangle PO_1C$ and

$\triangle PO_2C$ that we just did for

$\triangle OO_1C$ and

$\triangle OO_2C$ .

MaryM 2021-03-12 22:23:54
We will get two equations and then we will be able to subtract them, just like we did before.

MaryM 2021-03-12 22:23:58
And we'll get exactly the same end equation, except

$O$ will be replaced by

$P$ .

MaryM 2021-03-12 22:24:07
That is, we'll get

$(PO_1)^2 - (PO_2)^2 = (O_1C)^2 - (O_2C)^2 + (PC)d.$

MaryM 2021-03-12 22:24:11
Let's put these two equations side-by-side:

MaryM 2021-03-12 22:24:16

$\begin{align*} (OO_1)^2 - (OO_2)^2 &= (O_1C)^2 - (O_2C)^2 + (OC)d, \\ (PO_1)^2 - (PO_2)^2 &= (O_1C)^2 - (O_2C)^2 + (PC)d. \end{align*}$

MaryM 2021-03-12 22:24:21
Any ideas?

rockyrockrock 2021-03-12 22:24:44
subtract?

Z_Lu 2021-03-12 22:24:44
subtract

panda2018 2021-03-12 22:24:44
subtract

MaryM 2021-03-12 22:24:47
We can get rid of a lot of stuff we didn't know how to deal with if we subtract them!

MaryM 2021-03-12 22:24:51
If we subtract them, the

$(O_1C)^2 - (O_2C)^2$ terms will cancel.

MaryM 2021-03-12 22:24:56
What's left is

$(OO_1)^2 - (OO_2)^2 - (PO_1)^2 + (PO_2)^2 = (OC - PC)d.$

MaryM 2021-03-12 22:25:05
But what is

$OC - PC$ ?

Zhaom 2021-03-12 22:25:18

$r$

MaryM 2021-03-12 22:25:20
It's

$OP$ . But that's the radius of the little circle

$\omega$ , which we called

$r$ .

MaryM 2021-03-12 22:25:25
So we've got

$(OO_1)^2 - (OO_2)^2 - (PO_1)^2 + (PO_2)^2 = rd.$

MaryM 2021-03-12 22:25:39
What about those other terms?

MaryM 2021-03-12 22:25:43
Can we rewrite

$OO_1$ as something nicer?

rockyrockrock 2021-03-12 22:25:58
Substitute 961+r,625+r

panda2018 2021-03-12 22:25:58
961+r

Zhaom 2021-03-12 22:25:58

$961+r$

MaryM 2021-03-12 22:26:00
Yes: as we discussed a while earlier,

$OO_1$ is the sum of the radii of

$\omega_1$ and

$\omega$ .

MaryM 2021-03-12 22:26:03
So

$OO_1 = 961 + r$ .

MaryM 2021-03-12 22:26:06
Similarly

$OO_2 = 625 + r$ .

MaryM 2021-03-12 22:26:11
So now we have

$(961+r)^2 - (625+r)^2 - (PO_1)^2 + (PO_2)^2 = rd.$

MaryM 2021-03-12 22:26:19
How can we get at

$(PO_1)^2$ and

$(PO_2)^2$ ?

Irving1004 2021-03-12 22:27:09
Pythagoras?

MaryM 2021-03-12 22:27:12
If we let

$R$ be the foot of the perpendicular from

$P$ to

$\overline{O_1 O_2},$ then we can apply Pythagoras to right triangles

$PRO_1$ and

$PRO_2.$

MaryM 2021-03-12 22:27:16

MaryM 2021-03-12 22:27:25
Pythagoras gives us that

$\begin{align*} (PO_1)^2 - (PO_2)^2 &= (O_1 R)^2 + (PR)^2 - (O_2 R)^2 - (PR)^2 \\ &= (O_1 R)^2 - (O_2 R)^2. \end{align*}$

MaryM 2021-03-12 22:27:32
How can we deal with

$(O_1 R)^2$ and

$(O_2 R)^2?$

MaryM 2021-03-12 22:28:06
Can we use Pythagoras again?

functionalmath 2021-03-12 22:28:57
yes

MaryM 2021-03-12 22:29:00
We can apply Pythagoras to right triangles

$ARO_1$ and

$ARO_2.$

MaryM 2021-03-12 22:29:31
Pythagoras gives us that

$(AO_1)^2 = (O_1 R)^2 + AR^2$ and

$(AO_2)^2 = (O_2 R)^2 + AR^2.$ Subtracting these equations gives us

$(O_1 R)^2 - (O_2 R)^2 = (AO_1)^2 - (AO_2)^2 = 961^2 - 256^2.$

Irving1004 2021-03-12 22:29:45
Ohh I see

MaryM 2021-03-12 22:29:47
Therefore,

$(PO_1)^2 - (PO_2)^2 = 961^2 - 256^2,$ so the equation

$(961+r)^2 - (625+r)^2 - (PO_1)^2 + (PO_2)^2 = rd$ becomes

$(961 + r)^2 - (625 + r)^2 - 961^2 + 256^2 = rd.$

MaryM 2021-03-12 22:30:07
If you're familiar with the radical axis, then it gives us this result quickly: Since

$A$ and

$B$ are the intersection points of circles

$\omega_1$ and

$\omega_2,$ line

$AB$ is the radical axis of the two circles. And since

$P$ lies on this radical axis, it has the same power with respect to both circles. In other words,

$(PO_1)^2 - 961^2 = (PO_2)^2 - 256^2,$ so

$(PO_1)^2 - (PO_2)^2 = 961^2 - 256^2.$

Lamboreghini 2021-03-12 22:30:29
not 256, 625

MaryM 2021-03-12 22:30:33
true!

MaryM 2021-03-12 22:30:51
If you're familiar with the radical axis, then it gives us this result quickly: Since

$A$ and

$B$ are the intersection points of circles

$\omega_1$ and

$\omega_2,$ line

$AB$ is the radical axis of the two circles. And since

$P$ lies on this radical axis, it has the same power with respect to both circles. In other words,

$(PO_1)^2 - 961^2 = (PO_2)^2 - 625^2,$ so

$(PO_1)^2 - (PO_2)^2 = 961^2 - 625^2.$

MaryM 2021-03-12 22:31:05
But well, what happens when we expand the left side?

Irving1004 2021-03-12 22:31:10
Cancellations occurr!

Irving1004 2021-03-12 22:31:10
we get 672d = rd and r = 672!

MaryM 2021-03-12 22:31:19
We get

$(961^2 + 2r \cdot 961 + r^2) - (625^2 + 2r \cdot 625 + r^2) + 625^2 - 961^2 = rd.$

MaryM 2021-03-12 22:31:26
The

$961^2$ and

$625^2$ terms cancel!

MaryM 2021-03-12 22:31:31
And the

$r^2$ terms cancel!

MaryM 2021-03-12 22:31:35
We have

$2r(961-625) = rd$ .

MaryM 2021-03-12 22:31:40
And the

$r$ 's cancel too!

MaryM 2021-03-12 22:31:48
And we're left with

$2(961-625) = d$ .

functionalmath 2021-03-12 22:31:58
d=672

MaryM 2021-03-12 22:32:00
We get

$d = 2(961 - 625) = 2(336) = \boxed{672}$ .

MaryM 2021-03-12 22:32:12
This question was pretty hard!

Irving1004 2021-03-12 22:32:34
Ah I think I somehow guessed that.

happyhari 2021-03-12 22:32:43
agreeable

MaryM 2021-03-12 22:32:46
Wait a second...remember that we had two possible configurations at the beginning.

MaryM 2021-03-12 22:32:51
What if we chose the wrong one? Maybe we got the wrong answer?

Irving1004 2021-03-12 22:33:27
Doesn't matter

rockyrockrock 2021-03-12 22:34:04
will it give u the same answer

MaryM 2021-03-12 22:34:09
I think we're OK. The key idea of our solution used Law of Cosines repeatedly on

$\triangle OO_1O_2$ and the cevian

$\overline{OC}$ that passed through

$P$ .

MaryM 2021-03-12 22:34:14
The other key step of our solution used power of a point at point

$P$ , along the lines

$AB$ ,

$PO_1$ , and

$PO_2$ .

MaryM 2021-03-12 22:34:34
In either configuration, we'll end up doing the exactly the same calculations. None of these depend on whether

$\overline{AB}$ intersects

$\overline{O_1O_2}$ .

MaryM 2021-03-12 22:34:40
So we never actually used the fact whether

$\overline{AB}$ interesects

$\overline{O_1O_2}$ or not.

MaryM 2021-03-12 22:34:45
Hence, our solution is good, regardless of which initial drawing we used.

MaryM 2021-03-12 22:35:04
(And in fact, we used the wrong one. If you draw this carefully to scale, you'll get that

$\overline{AB}$ misses

$\overline{O_1O_2}$ .)

functionalmath 2021-03-12 22:35:13
2 more

MaryM 2021-03-12 22:35:23
14. For any positive integer

$a,\,\sigma(a)$ denotes the sum of the positive integer divisors of

$a.$ Let

$n$ be the least positive integer such that

$\sigma(a^n)-1$ is divisible by

$2021$ for all positive integers

$a.$ Find the sum of the prime factors in the prime factorization of

$n.$

MaryM 2021-03-12 22:35:32
Let's start by experimenting.

MaryM 2021-03-12 22:35:37
What's a good number to experiment on?

Zhaom 2021-03-12 22:35:55

$2$

Mathdreams 2021-03-12 22:35:55
2

MaryM 2021-03-12 22:35:58

$a = 2$ is probably the simplest. (

$a=1$ doesn't really give us any information.)

MaryM 2021-03-12 22:36:06
So let's try to answer a simpler question: what's the smallest value of

$n$ such that

$\sigma(2^n)-1$ is divisible by

$2021$ ?

MaryM 2021-03-12 22:36:15
What exactly is

$\sigma(2^n)$ ?

Zhaom 2021-03-12 22:36:50

$\sum^n_{k=0}2^n$

ancientwarrior 2021-03-12 22:36:50
2^(n+1)-1?

happyhari 2021-03-12 22:36:55
the sum of all powers of 2 from 2 to 2^n

MaryM 2021-03-12 22:36:58
It's the sum of the divisors of

$2^n$ , by definition.

MaryM 2021-03-12 22:37:03
They're

$1$ ,

$2$ ,

$2^2$ , etc. up to

$2^n$ .

MaryM 2021-03-12 22:37:24
Their sum

$1 + 2 + 2^2 + \cdots + 2^n$ is a geometric series.

MaryM 2021-03-12 22:37:30
Its sum is

$\dfrac{2^{n+1}-1}{2-1} = 2^{n+1} - 1$ .

MaryM 2021-03-12 22:37:42
So

$\sigma(2^n)-1 = 2^{n+1} - 2$ .

MaryM 2021-03-12 22:37:55
And the question is: for what value of

$n$ is this a multiple of

$2021$ ?

MaryM 2021-03-12 22:38:17
We can divide it by

$2$ , because the

$2$ is not going to help us be a multiple of

$2021$ .

MaryM 2021-03-12 22:38:25
So the equivalent question is: when is

$2^n - 1$ a multiple of

$2021$ ?

MaryM 2021-03-12 22:38:35
If we wanted to write this using modular arithmetic, the question becomes: when is

$2^n \equiv 1 \pmod{2021}$ ?

Zhaom 2021-03-12 22:39:01

$2021=43 \cdot 47$

Irving1004 2021-03-12 22:39:01
When it is a multiple of 43 and 47

MaryM 2021-03-12 22:39:05
We know that

$2021 = 43 \cdot 47$ . (Hopefully you memorized this before taking the test.)

MaryM 2021-03-12 22:39:11
The Chinese Remainder Theorem tells us that we can separately solve for each prime:

MaryM 2021-03-12 22:39:15

$\begin{align*} 2^n &\equiv 1 \pmod{43}, \\ 2^n &\equiv 1 \pmod{47}. \end{align*}$

MaryM 2021-03-12 22:39:20
And now what? How do we solve these?

JimY 2021-03-12 22:39:36
FLT

Zhaom 2021-03-12 22:39:36
Fermat's Little Theorem

MaryM 2021-03-12 22:39:38
Now we can use Fermat's Little Theorem!

MaryM 2021-03-12 22:39:44
FLT states that

$a^{p-1} \equiv 1 \pmod{p}$ so long as

$a \not\equiv 0 \pmod{p}$ .

MaryM 2021-03-12 22:39:50
When we apply FLT to our system of equations, what does it tell us?

jxwis2010 2021-03-12 22:40:31
n = 42p and n=46q

Mathdreams 2021-03-12 22:40:31
n is a multiple of 42 and 46!

Arrowhead575 2021-03-12 22:40:31
Lcm Of 42 and 46

Zhaom 2021-03-12 22:40:31

$n \equiv 0\pmod{42}$ and

$n \equiv 0\pmod{46}$

MaryM 2021-03-12 22:40:38
It tells us that

$n = 43-1 = 42$ is a solution to the first equation, and

$n = 47-1 = 46$ is a solution to the second equation.

MaryM 2021-03-12 22:40:45
So, we can take

$n = \text{lcm}[42,46]$ as a common solution.

MaryM 2021-03-12 22:40:53
Given that

$42 = 2 \cdot 3 \cdot 7$ and

$46 = 2 \cdot 23$ are prime factorizations, what do we get for a common

$n$ ?

jxwis2010 2021-03-12 22:41:23
2*3*7*23

MaryM 2021-03-12 22:41:26
We get

$n = 2 \cdot 3 \cdot 7 \cdot 23$ . (It makes sense to write it this way -- we don't really care what

$n$ actually is, we just care about its prime factors. Remember what the question is asking!)

MaryM 2021-03-12 22:41:33
Anybody see a problem with all of this?

peace09 2021-03-12 22:42:07
...We only considered

$a=2...$

mathgeek23 2021-03-12 22:42:07
this only works for a=2

Mathdreams 2021-03-12 22:42:10
We need to consider higher cases.

MaryM 2021-03-12 22:42:12
We've only deal with one value of

$a$ .

jxwis2010 2021-03-12 22:42:15
this may not be the smallest n

MaryM 2021-03-12 22:42:22
Also, FLT doesn't guarantee us that this is the smallest such

$n$ . It only guarantees that this

$n$ will work. There might be a smaller

$n$ that also works.

MaryM 2021-03-12 22:42:30
Let's set this annoyance aside for now.

MaryM 2021-03-12 22:42:34
What should we try next?

MaryM 2021-03-12 22:42:54
Hint: pick a more general

$a$ this time.

Zhaom 2021-03-12 22:43:20

$a$ is a prime number.

MaryM 2021-03-12 22:43:23
Let's be bold and try it for a general prime

$p$ .

MaryM 2021-03-12 22:44:14
What is

$\sigma(p^n)$ ?

rockyrockrock 2021-03-12 22:45:01
1+p^1+p^2+...+p^n

MaryM 2021-03-12 22:45:05
The divisors of

$p^n$ are

$1$ ,

$p$ ,

$p^2$ , etc. up to

$p^n$ .

MaryM 2021-03-12 22:45:08
So their sum is

$\sigma(p^n) = 1 + p + p^2 + \cdots + p^n$ .

MaryM 2021-03-12 22:45:12
And what does this work out to?

CircleInvert 2021-03-12 22:45:35

$\frac{p^{n+1}-1}{p-1}$

Lamboreghini 2021-03-12 22:45:35
1+p+p^2+...+p^n=(p^(n+1)-1)/(p-1)

Zhaom 2021-03-12 22:45:35

$\frac{p^{n+1}-1}{p-1}$

MaryM 2021-03-12 22:45:38
This is a geometric series, so we get

$\sigma(p^n) = \dfrac{p^{n+1}-1}{p-1}$ .

MaryM 2021-03-12 22:45:59
So we must have

$\dfrac{p^{n+1}-1}{p-1} - 1$ be a multiple of

$2021$ .

Irving1004 2021-03-12 22:46:15
Yeah

MaryM 2021-03-12 22:46:17
This simplifies to

$\dfrac{p^{n+1}-p}{p-1}$ being a multiple of

$2021$ .

MaryM 2021-03-12 22:46:27
And factoring out a

$p$ , we need

$\dfrac{p(p^n-1)}{p-1}$ to be a multiple of

$2021$ .

MaryM 2021-03-12 22:46:33
Is this the same as

$p(p^n-1)$ being a multiple of

$2021$ ?

Ferocious_Bunny 2021-03-12 22:47:06
Not necessarily

jj_ca888 2021-03-12 22:47:06
unless p-1 has some factors

Zhaom 2021-03-12 22:47:06
no, what if

$p-1$ is a multiple of

$2021$

CircleInvert 2021-03-12 22:47:06
Only if

$p-1$ is relatively prime to

$2021$

JimY 2021-03-12 22:47:10
Almost, only if p-1 isn't a multiple of 2021

MaryM 2021-03-12 22:47:13
Not necessarily. If

$p-1$ is a multiple of

$43$ or

$47$ , we might have a problem.

MaryM 2021-03-12 22:47:21
Let's put that aside for a moment, and assume (for now) that

$p-1$ is not a multiple of

$43$ or

$47$ .

MaryM 2021-03-12 22:47:26
In that case, dividing by

$p-1$ doesn't affect whether the number is a multiple of

$2021$ or not.

MaryM 2021-03-12 22:47:38
So we just need to determine when

$p(p^n-1)$ is a multiple of

$2021$ .

MaryM 2021-03-12 22:47:45
How do we go about it?

MaryM 2021-03-12 22:48:39
Can we get rid of the factor

$p$ ?

ancientwarrior 2021-03-12 22:49:34
it might be 43 or 47?

Irving1004 2021-03-12 22:49:34
So we need p to not be 43 or 47

Irving1004 2021-03-12 22:49:34
And we are set

MaryM 2021-03-12 22:49:41
If

$p = 43$ or

$p = 47$ , we get one of the factors of

$2021$ for free!

MaryM 2021-03-12 22:50:20
So, let's suppose that

$p$ is not

$43$ , not

$47$ . In that case, we can ignore the factor

$p$ , since it won't affect divisibility by

$2021$ .

palaashgang 2021-03-12 22:50:27
Yeah

MaryM 2021-03-12 22:50:36
Then, it's just like the

$p=2$ case: we need to solve the system:

MaryM 2021-03-12 22:50:42

$\begin{align*} p^n &\equiv 1 \pmod{43}, \\ p^n &\equiv 1 \pmod{47}. \end{align*}$

MaryM 2021-03-12 22:51:01
What solution do we get if we use the same method we used for

$p=2$ ?

Zhaom 2021-03-12 22:52:09

$n \equiv 0\pmod{42}$ ,

$n \equiv 0\pmod{46}$

YaoAOPS 2021-03-12 22:52:09
same

ancientwarrior 2021-03-12 22:52:09
the same thing?

jxwis2010 2021-03-12 22:52:09
Lcm(42,46)

MaryM 2021-03-12 22:52:12
As before, we want

$n$ to be a multiple of

$43$ and

$47$ .

MaryM 2021-03-12 22:52:18
So we'll definitely need

$n = 2 \cdot 3 \cdot 7 \cdot 23$ (or any multiple of this

$n$ ).

MaryM 2021-03-12 22:52:23
Does this work for

$p=43$ and

$p=47$ too?

functionalmath 2021-03-12 22:52:50
yes

babytiger2010 2021-03-12 22:52:50
yes???

MaryM 2021-03-12 22:52:53
Yes. We get one factor (

$p$ ) for free, and the other factor will be a factor of

$p^n-1$ using the

$n$ that we just found.

MaryM 2021-03-12 22:53:01
Are we done? Is this our

$n$ ?

functionalmath 2021-03-12 22:53:21
no

Zhaom 2021-03-12 22:53:21
no, we still need to consider

$p-1$

MaryM 2021-03-12 22:53:23
Not quite. We had to set a case aside, remember?

MaryM 2021-03-12 22:53:27
We needed

$\dfrac{p(p^n-1)}{p-1}$ to be a multiple of

$2021$ .

MaryM 2021-03-12 22:53:33
We were only allowed to ignore the

$p-1$ denominator if it wasn't a multiple of

$43$ or

$47$ .

MaryM 2021-03-12 22:53:38
So we still have to deal with the case when

$p-1$ is a multiple of

$43$ or

$47$ .

MaryM 2021-03-12 22:53:50
Let's try these cases one at a time.

MaryM 2021-03-12 22:53:54
Suppose

$p-1$ is a multiple of

$43$ .

MaryM 2021-03-12 22:54:08
Then what do we need?

MaryM 2021-03-12 22:54:47
How many factors of

$43$ must divide

$p^n-1$ ?

rockyrockrock 2021-03-12 22:55:01
2 powers of 43?

happyhari 2021-03-12 22:55:01
then we need another 43 multiplied in the numerator

ancientwarrior 2021-03-12 22:55:01
2

jxwis2010 2021-03-12 22:55:01
2

MaryM 2021-03-12 22:55:05
At a minimum, we need

$p^n-1$ to be a multiple of

$43^2$ , because the denominator will cancel out a factor of

$43$ .

MaryM 2021-03-12 22:55:18
More generally, if

$p-1$ is a multiple of

$43^k$ , then we need

$p^n-1$ to be a multiple of

$43^{k+1}$ in order to preserve a power of

$43$ in the numerator after we cancel out a factor of

$43^k$ .

MaryM 2021-03-12 22:55:39
Suppose

$k$ is the maximum power of

$43^k$ in

$p-1$ .

MaryM 2021-03-12 22:55:45
How we we write this statement using modular arithmetic?

MaryM 2021-03-12 22:56:28
Hint: write a congruence

$\mod 43^{k+1}$ .

MaryM 2021-03-12 22:58:16
Further hint: we know that

$p-1=m43^k$ for some

$m$ that is not a multiple of

$43$ . Try to write this as a congruence

$\pmod 43^{k+1}$ .

Zhaom 2021-03-12 22:58:49
typo?

MaryM 2021-03-12 22:59:01
Yes, typo.

MaryM 2021-03-12 22:59:06
I meant:

MaryM 2021-03-12 22:59:17
Further hint: we know that

$p-1=m43^k$ for some

$m$ that is not a multiple of

$43$ . Try to write this as a congruence

$\pmod {43^{k+1}}$ .

Zhaom 2021-03-12 23:01:27

$p-1 \equiv m \cdot 43^k\pmod{43^{k+1}}$ where

$m \cancel{\equiv} 0\pmod{43}$

MaryM 2021-03-12 23:01:35
We can write

$p \equiv (1 + m43^k) \pmod{43^{k+1}}$ for some

$0 \le m < 43$ .

MaryM 2021-03-12 23:01:56
Let's raise it to

$n$ to get

$p^n$ :

MaryM 2021-03-12 23:02:01
We get

$p^n \equiv (1 + m43^k)^n \pmod{43^{k+1}}$ .

MaryM 2021-03-12 23:02:07
And we want this to be equivalent to

$1 \pmod{43^{k+1}}$ .

MaryM 2021-03-12 23:02:46
What do you notice if you expand

$(1+m43^k)^n$ using the Binomial Theorem?

CircleInvert 2021-03-12 23:03:15
Using the Binomial Theorem, this is gives

$p^n\equiv 1+nm43%k \pmod{43^{k+1}}$ (all other terms are 0)

peace09 2021-03-12 23:03:15
All terms except for 1 and

$mn43^k$ are a multiple of

$43^{k+1}$

MaryM 2021-03-12 23:03:18
By the Binomial Theorem, we get

$p^n \equiv (1 + nm43^k + \binom{n}{2}m^243^{2k} + \cdots + m^n43^{nk}) \pmod{43^{k+1}}$ .

MaryM 2021-03-12 23:03:22
All of the higher powers of

$43^k$ are

$0 \pmod{43^{k+1}}$ .

MaryM 2021-03-12 23:03:26
So we're left with

$p^n \equiv (1 + nm43^k) \pmod{43^{k+1}}$ .

MaryM 2021-03-12 23:03:32
How can we guarantee that the second term in the parentheses is

$0 \pmod{43^{k+1}}$ ?

jxwis2010 2021-03-12 23:04:11
n=43u

JimY 2021-03-12 23:04:11
nm is a multiple of 43

happyhari 2021-03-12 23:04:11
not if nm is not divisible by 43

Irving1004 2021-03-12 23:04:11
If n is congruent to 0(mod 43).

MaryM 2021-03-12 23:04:13
We need to take

$n = 43$ . (Or any multiple of

$43$ .) Then that term will be a multiple of

$43^{k+1}$ and go away.

MaryM 2021-03-12 23:04:19
So the conclusion is: if

$p-1$ is a multiple of

$43$ , we need to have

$n$ be a multiple of

$43$ in order to guarantee that

$\sigma(p^n)-1$ is a multiple of

$43$ .

MaryM 2021-03-12 23:04:37
What happens in the case where

$p-1$ is a multiple of

$47$ ?

jxwis2010 2021-03-12 23:04:57
n=47v

happyhari 2021-03-12 23:04:57
same thing, but with 47

Irving1004 2021-03-12 23:04:57
It needs to be a multiple of 47

MaryM 2021-03-12 23:05:00
We can do exactly the same argument for

$47$ instead of

$43$ .

MaryM 2021-03-12 23:05:03
And we get the same conclusion: if

$p-1$ is a multiple of

$47$ , we need to have

$n$ be a multiple of

$47$ in order to guarantee that

$\sigma(p^n)-1$ is a multiple of

$47$ .

MaryM 2021-03-12 23:05:17
So, we've shown that in order to ensure that

$\sigma(p^n) - 1$ is a multiple of

$2021$ for all primes

$p$ , we must have that

$n$ is a multiple of

$2 \cdot 3 \cdot 7 \cdot 23 \cdot 43 \cdot 47$ .

MaryM 2021-03-12 23:05:25
That takes care of the primes. But what if

$a$ is composite?

jxwis2010 2021-03-12 23:06:37
Decomposes into primes

peace09 2021-03-12 23:06:37
Somehow use what we know about primes

peace09 2021-03-12 23:06:37
Prime factorization of

$a?$

Irving1004 2021-03-12 23:06:37
Same thing

Zhaom 2021-03-12 23:06:37
then, the sum of the divisors of

$a^n$ is the product of the sum of the divisors of

$p_k^n$ some primes

$p_k$ so it is still

$1\pmod{2021}$

CircleInvert 2021-03-12 23:06:37
Use the fact that

$\sigma$ is multiplicative

MaryM 2021-03-12 23:07:15
If we let

$a=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ , then

$\begin{align*}\sigma(a^n)&=\sigma(p_1^{n\alpha_1})\sigma(p_2^{n\alpha_2})\cdots \sigma(p_k^{n\alpha_k}) \end{align*}$

YaoAOPS 2021-03-12 23:08:05
why

MaryM 2021-03-12 23:08:07
The product on the right hand side spams all divisors of

$a^n$ .

MaryM 2021-03-12 23:08:34
For example

$\sigma(p^b)\sigma(q^c) = (1+p+\cdots+p^b)(1+q+\cdots+q^c).$

MaryM 2021-03-12 23:08:53
If we expand the right hand side, we get all the terms of the form

$p^iq^j$ with

$0 \le i \le b$ and

$0 \le j \le c$ .

MaryM 2021-03-12 23:09:01
But those are all the divisors of

$p^bq^c$ !

YaoAOPS 2021-03-12 23:09:13
oh

MaryM 2021-03-12 23:09:16
The same thing happens if we add more primes.

MaryM 2021-03-12 23:09:39
So,

$\begin{align*}\sigma(a^n)&=\sigma(p_1^{n\alpha_1})\sigma(p_2^{n\alpha_2})\cdots \sigma(p_k^{n\alpha_k})\\ &\equiv 1\cdot 1\cdots 1\pmod{2021}. \end{align*}$ This means that it's enough to just handle the prime powers!

MaryM 2021-03-12 23:09:55
I think we're done!

happyhari 2021-03-12 23:10:10
yay!

Zhaom 2021-03-12 23:10:12
but we didn't get the answer yet

MaryM 2021-03-12 23:10:18
And our final answer is the sum of these primes.

MaryM 2021-03-12 23:10:22
That's

$2+3+7+23+43+47 = \boxed{125}$ .

MaryM 2021-03-12 23:10:39
Last problem!

Irving1004 2021-03-12 23:10:48
shoot

happyhari 2021-03-12 23:10:48
yay!!

MaryM 2021-03-12 23:10:51
15. Let

$S$ be the set of positive integers

$k$ such that the two parabolas

$y = x^2-k \qquad \text{and} \qquad x = 2(y-20)^2 - k$ intersect in four distinct points, and these four points lie on a circle with radius at most

$21.$ Find the sum of the least element of

$S$ and the greatest element of

$S.$

MaryM 2021-03-12 23:11:17
What does the graph of

$y=x^2-k$ look like?

Lamboreghini 2021-03-12 23:11:38
parabola, upward facing

functionalmath 2021-03-12 23:11:38
parabola

Zhaom 2021-03-12 23:11:38
a parabola with vertex

$(0,-k)$

JimY 2021-03-12 23:11:38
parabola with vertex at 0, -k

CircleInvert 2021-03-12 23:11:38

$y=x^2$ translated

$k$ down

MaryM 2021-03-12 23:11:42
It's an upwards-opening parabola with vertex at

$(0,-k)$ that's symmetric across the

$y$ -axis.

MaryM 2021-03-12 23:11:46

MaryM 2021-03-12 23:11:50
How about the other parabola?

Lamboreghini 2021-03-12 23:12:12
sideways facing

jxwis2010 2021-03-12 23:12:12
Horizontal

happyhari 2021-03-12 23:12:12
a sideways parabola

rockyrockrock 2021-03-12 23:12:16
opens up to the side

MaryM 2021-03-12 23:12:18
It's a right-opening parabola.

MaryM 2021-03-12 23:12:23
Where is its vertex?

Irving1004 2021-03-12 23:12:49
horizontal with vertex (-k, 20)

Rubikscube3.1415 2021-03-12 23:12:49
(-k, 20)

ancientwarrior 2021-03-12 23:12:49
(-k, 20)

rockyrockrock 2021-03-12 23:12:52
-k, 20

JimY 2021-03-12 23:12:55
(-k, 20)

sugar_rush 2021-03-12 23:12:55

$(-k, 20)$

MaryM 2021-03-12 23:12:57
The minimum value of

$x$ occurs when

$y-20 = 0$ , or

$y = 20$ .

MaryM 2021-03-12 23:13:01
And at that point,

$x=-k$ .

MaryM 2021-03-12 23:13:04
So the vertex is at

$(-k,20)$ .

MaryM 2021-03-12 23:13:10

MaryM 2021-03-12 23:13:15
The example I've drawn shows 4 intersection points.

MaryM 2021-03-12 23:13:20
But does that always happen?

Zhaom 2021-03-12 23:13:44
no

Lamboreghini 2021-03-12 23:13:44
no

MaryM 2021-03-12 23:13:48
The vertex might be to the right of the left blue branch, like so:

MaryM 2021-03-12 23:13:54

MaryM 2021-03-12 23:14:08
When does this happen?

Lamboreghini 2021-03-12 23:15:17
when k is too small

MaryM 2021-03-12 23:15:19
This happens when

$-k$ is larger than the value in the right parabola that makes

$y=20$ .

MaryM 2021-03-12 23:15:24
What value of the left branch of the blue curve gives a

$y$ value of

$20$ ?

ancientwarrior 2021-03-12 23:15:54

$-\sqrt{20+k}$

MaryM 2021-03-12 23:15:57
We must have

$20 = x^2 - k$ .

MaryM 2021-03-12 23:16:03
So

$x^2 = 20+k$ , and

$x = -\sqrt{20+k}$ .

MaryM 2021-03-12 23:16:08
And what do we conclude?

Zhaom 2021-03-12 23:16:40
when

$\sqrt{k+20}>k$

CircleInvert 2021-03-12 23:16:40
When

$k^2-k<20$

ancientwarrior 2021-03-12 23:16:40
sqrt(20+k)>k

ancientwarrior 2021-03-12 23:16:40
k<sqrt(20+k)

MaryM 2021-03-12 23:16:44
The red parabola's vertex is too far to the right if

$-\sqrt{20+k} < -k$ .

MaryM 2021-03-12 23:16:47
That is,

$\sqrt{20+k} > k$ .

MaryM 2021-03-12 23:16:58
we can square it (since both sides are positive) to get

$20 + k > k^2$ .

MaryM 2021-03-12 23:17:05
This is

$0 > k^2 - k - 20$ .

MaryM 2021-03-12 23:17:11
And this factors as

$0 > (k-5)(k+4)$ .

MaryM 2021-03-12 23:17:20
So, what bound do we get for

$k$ ?

sugar_rush 2021-03-12 23:17:39

$-4\leq k\leq 5$

ancientwarrior 2021-03-12 23:17:39
so any k in (-4, 5) is not going to intersect at 4 points

rockyrockrock 2021-03-12 23:17:39
-4<k<5?

MaryM 2021-03-12 23:17:45
So since

$k$ is positive, we must have

$k < 5$ .

MaryM 2021-03-12 23:17:56
When

$k < 5$ , we'll definitely not get 4 intersection points.

MaryM 2021-03-12 23:18:11
When

$k>5$ , the red parabola's vertex is to the left of the blue parabola, so we'll get 4 intersection points, as in the first picture...

MaryM 2021-03-12 23:18:50
...so long as the red parabola stays above

$(0,-k)$ at

$x=0$ . (That is, we don't want the red parabola to pass under

$(0,-k)$ , since then we'd only get 2 points from the top red branch.)

MaryM 2021-03-12 23:18:59
Let's check this. What inequality do we need?

Zhaom 2021-03-12 23:19:45
an upper bound for

$k$

MaryM 2021-03-12 23:19:49
When

$x=0$ , we need the smaller

$y$ value of the red parabola to be greater than

$-k$ .

MaryM 2021-03-12 23:19:58
So we have to solve for

$y$ in the second equation when

$x=0$ .

MaryM 2021-03-12 23:20:02
What do we get?

Zhaom 2021-03-12 23:21:05
y=-sqrt(k/2)+20

ancientwarrior 2021-03-12 23:21:05

$y=\sqrt{\frac{k}{2}}+20$ ? and that has to be greater than -k

Rubikscube3.1415 2021-03-12 23:21:05

$-\sqrt{\frac{k}{2}}+20<-k$ , or

$\sqrt{\frac{k}{2}}+20>k$

MaryM 2021-03-12 23:21:08
We get

$(y-20)^2 = \dfrac{k}{2}$ .

MaryM 2021-03-12 23:21:12
So

$y = \pm\sqrt{\dfrac{k}{2}} + 20$ .

MaryM 2021-03-12 23:21:17
Since we want the smaller value, we take the minus sign.

MaryM 2021-03-12 23:21:21
Therefore, the inequality we need to ensure 4 points is

$-\sqrt{\dfrac{k}{2}} + 20 > -k$ .

functionalmath 2021-03-12 23:21:35
solve

MaryM 2021-03-12 23:21:37
We can rearrange this as

$-\sqrt{\dfrac{k}{2}} > -(k+20)$ .

MaryM 2021-03-12 23:21:43
And then multiplying by

$-1$ gives

$\sqrt{\dfrac{k}{2}} < k+20$ .

MaryM 2021-03-12 23:21:48
And then squaring gives

$\dfrac{k}{2} < (k+20)^2$ .

MaryM 2021-03-12 23:21:55
This simplifies to

$0 < 2k^2 + 79k + 800$ .

MaryM 2021-03-12 23:21:58
For what values of

$k$ does this hold?

MaryM 2021-03-12 23:22:47
Hint: We are already assuming that

$k>5$ , so in particular

$k$ is positive.

Rubikscube3.1415 2021-03-12 23:22:54
all k I think

Zhaom 2021-03-12 23:22:54
every value

functionalmath 2021-03-12 23:22:54
always true

MaryM 2021-03-12 23:23:00
This holds for all positive

$k$ , since every term on the right side is positive!

MaryM 2021-03-12 23:23:05
So it'll hold for all the

$k$ 's we care about.

MaryM 2021-03-12 23:23:08
Thus, when

$k > 5$ we definitely get 4 intersection points.

MaryM 2021-03-12 23:23:12
What value(s) of

$k$ are we missing?

MaryM 2021-03-12 23:23:54
We showed that

$k<5$ don't work. But does

$k=5$ work?

palaashgang 2021-03-12 23:24:40
no

Zhaom 2021-03-12 23:24:40
maybe

happyhari 2021-03-12 23:24:40
yes

Rubikscube3.1415 2021-03-12 23:24:40
no, we get exactly 3 intersection points

sugar_rush 2021-03-12 23:24:40
yes

MaryM 2021-03-12 23:24:44
Let's see!

MaryM 2021-03-12 23:24:51
In this case the vertex of the red parabola lies on the blue parabola:

MaryM 2021-03-12 23:24:55

MaryM 2021-03-12 23:24:58
This looks like 3 intersection points, but is it really?

MaryM 2021-03-12 23:25:08
Since I have a fancy computer graphics package, I could zoom in.

MaryM 2021-03-12 23:25:12
But you didn't have a fancy computer graphics package available to you when you took the test.

MaryM 2021-03-12 23:25:16
So let's do it the old-fashioned way.

Zhaom 2021-03-12 23:25:23
solve for

$x$ and

$y$ if

$k=5$

MaryM 2021-03-12 23:25:26
We plug in

$k=5$ to our equations.

MaryM 2021-03-12 23:25:40
Let's see how many times do

$y = x^2 - 5$ and

$x = 2(y-20)^2 - 5$ intersect.

MaryM 2021-03-12 23:25:50
We can substitute

$y$ into the 2nd equation to get

$x = 2((x^2-5)-20)^2 - 5$ .

MaryM 2021-03-12 23:26:00
This is

$x = 2(x^2 - 25)^2 - 5$ .

MaryM 2021-03-12 23:26:14
So, as a quartic in

$x$ , this is

$2x^4 - 100x^2 - x + 1245 = 0$ .

MaryM 2021-03-12 23:26:21
Do we know a root of this equation?

ancientwarrior 2021-03-12 23:27:02
-5

JimY 2021-03-12 23:27:02
x=-5

sugar_rush 2021-03-12 23:27:06

$x=-5$

Rubikscube3.1415 2021-03-12 23:27:06
-5

MaryM 2021-03-12 23:27:08
We know that

$x=-5$ is a root, so this must be divisible by

$x+5$ .

MaryM 2021-03-12 23:27:14
We could do the long division to get a cubic, but maybe we don't have to do all that work?

MaryM 2021-03-12 23:27:18
What will the constant term of the resulting cubic be?

sugar_rush 2021-03-12 23:27:30

$249$

JimY 2021-03-12 23:27:30
249

Lamboreghini 2021-03-12 23:27:34
249

functionalmath 2021-03-12 23:27:35
249

MaryM 2021-03-12 23:27:38
It'll be

$1245/5 = 249$ .

MaryM 2021-03-12 23:27:41
In particular, it's not a multiple of

$5$ .

MaryM 2021-03-12 23:27:46
So we can't have a second root at

$x=-5$ , by the Rational Root Theorem.

MaryM 2021-03-12 23:27:55
What's the conclusion?

functionalmath 2021-03-12 23:28:16
4 roots

MaryM 2021-03-12 23:28:24
We know that

$x=-5$ is not a double root.

MaryM 2021-03-12 23:29:00
Otherwise,

$-5$ would be a root of the resulting polynomial, but it isn't.

MaryM 2021-03-12 23:29:11
So since the quartic must have a second negative root, that root must be different than

$x=-5$ .

MaryM 2021-03-12 23:29:27
That is, there is indeed a fourth intersection point hiding over there on the left side.

MaryM 2021-03-12 23:29:52
That means that

$k=5$ works!

palaashgang 2021-03-12 23:30:01
oh yeah i found 4 now

MaryM 2021-03-12 23:30:07
So we've established that if

$k \ge 5$ , then there are 4 distinct intersection points.

ancientwarrior 2021-03-12 23:30:15
cool

sugar_rush 2021-03-12 23:30:19
But we still have to find the largest value of

$k$ .

MaryM 2021-03-12 23:30:29
Now we have to determine when they lie on a circle with radius at most

$21$ .

MaryM 2021-03-12 23:30:36
Can we determine what the equation of that circle might be?

MaryM 2021-03-12 23:32:17
Hint: We need an equation with an

$x^2$ term and

$y^2$ term in it. Can we find one by combining the equations of the two parabolas?

happyhari 2021-03-12 23:33:17
yes, add them together and then complete the square

MaryM 2021-03-12 23:33:20
Let's add the two given equations together! Any point that lies on both equations will also lie on the equation given by their sum.

MaryM 2021-03-12 23:33:25
We get

$x+y = x^2 - k + 2(y-20)^2 - k$ .

MaryM 2021-03-12 23:33:43
Unfortunately, this ends up not being a circle.

Zhaom 2021-03-12 23:33:50
but we get a

$2y^2$ term, not a

$y^2$ term

Lamboreghini 2021-03-12 23:33:50
wait, that's a 2y^2, which makes it an ellipse

MaryM 2021-03-12 23:33:57
It's an ellipse, because the

$x^2$ term has a different coefficient than the

$2y^2$ term.

MaryM 2021-03-12 23:34:08
Let's try again. What can we do instead?

Zhaom 2021-03-12 23:34:32
divide the second equation by

$2$ and add the

$2$ equations together

Lamboreghini 2021-03-12 23:34:32
double the y=x^2-k equation first

Irving1004 2021-03-12 23:34:32
We can multiply the right one by 1/2

ancientwarrior 2021-03-12 23:34:32
divide both sides by 2 for the second equation

Zhaom 2021-03-12 23:34:32
divide the second equation by

$2$ and then add

MaryM 2021-03-12 23:34:36
Add the first equation to half of the second equation.

MaryM 2021-03-12 23:34:40
Now it's

$y + \dfrac12x = x^2 - k + (y-20)^2 - \dfrac{k}{2}$ .

MaryM 2021-03-12 23:34:43
Is that a circle?

Lamboreghini 2021-03-12 23:34:55
yep!

peace09 2021-03-12 23:34:55
I Think S- Yes?

MaryM 2021-03-12 23:34:57
Yes! It's got an

$x^2$ and a

$y^2$ term.

MaryM 2021-03-12 23:35:01
What's its radius?

MaryM 2021-03-12 23:35:52
A better question would be "how can we find it's radius?"

MaryM 2021-03-12 23:35:59
So, how can we find the radius of this circle?

Zhaom 2021-03-12 23:36:15
complete the square

MaryM 2021-03-12 23:36:18
We need to complete the square in

$x$ and in

$y$ .

MaryM 2021-03-12 23:36:23
Let's expand and get all the constants on one side.

MaryM 2021-03-12 23:36:28
We get

$\dfrac{3}{2}k - 400 = x^2 - \dfrac12x + y^2 - 41y$ .

MaryM 2021-03-12 23:36:50
To complete the square in

$x$ we need to add

$\left(\dfrac14\right)^2 = \dfrac{1}{16}$ .

MaryM 2021-03-12 23:37:02
And for

$y$ we need to add

$\left(\dfrac{41}{2}\right)^2 = \dfrac{1681}{4}$ .

MaryM 2021-03-12 23:37:09
So we get

$\dfrac{3}{2}k - 400 + \dfrac{1}{16} + \dfrac{1681}{4} = \left(x - \dfrac14\right)^2 + \left(y - \dfrac{41}{2}\right)^2.$

MaryM 2021-03-12 23:37:21
So our equation is

$\dfrac{3}{2}k + \dfrac{325}{16} = \left(x-\dfrac14\right)^2 + \left(y-\dfrac{41}{2}\right)^2$ .

MaryM 2021-03-12 23:37:29
What the radius then?

Zhaom 2021-03-12 23:38:02

$\sqrt{\frac{3k}{2}+\frac{325}{16}}$

happyhari 2021-03-12 23:38:05
sqrt((3/2)k + (325/16))

MaryM 2021-03-12 23:38:10
This is a circle with center

$\left(\dfrac14,\dfrac{41}{2}\right)$ and radius

$\sqrt{\dfrac{3}{2}k + \dfrac{325}{16}}$ .

MaryM 2021-03-12 23:38:24
We need the radius to be at most

$21$ .

functionalmath 2021-03-12 23:38:27
sqrt(3k/2+325/16)<=21

MaryM 2021-03-12 23:38:30
So we need

$\sqrt{\dfrac{3}{2}k + \dfrac{325}{16}} \le 21$ .

MaryM 2021-03-12 23:38:35
Square both sides to get

$\dfrac{3}{2}k + \dfrac{325}{16} \le 441$ .

MaryM 2021-03-12 23:38:40
Then multiply through by

$16$ to get

$24k + 325 \le 7056$ .

MaryM 2021-03-12 23:38:45
So

$24k \le 6731$ and

$k \le \dfrac{6731}{24}$ .

MaryM 2021-03-12 23:38:55
The largest allowed value of

$k$ is

$\left\lfloor \dfrac{6731}{24} \right\rfloor$ .

Lamboreghini 2021-03-12 23:39:05
we have our upper bound for k!

MaryM 2021-03-12 23:39:07
A little long division gives that this is

$280$ .

MaryM 2021-03-12 23:39:13
So we must have

$k \le 280$ for the circle to have an allowed radius.

MaryM 2021-03-12 23:39:18
Combining this with the fact that

$k \ge 5$ that we found earlier, we see that we must have

$5 \le k \le 280$ .

MaryM 2021-03-12 23:39:21
So what's the final answer?

functionalmath 2021-03-12 23:39:33
so 285 is our answer

happyhari 2021-03-12 23:39:33
285\

Zhaom 2021-03-12 23:39:33

$280+5=\boxed{285}$

functionalmath 2021-03-12 23:39:33
285

palaashgang 2021-03-12 23:39:38
5+280 = 285

sugar_rush 2021-03-12 23:39:38

$285$

MaryM 2021-03-12 23:39:41
The sum of the smallest and largest

$k$ is

$5 + 280 = \boxed{285}$ .

MaryM 2021-03-12 23:39:45
We made it through the AIME I !!

MaryM 2021-03-12 23:39:51
If you'd like to review this transcript or transcripts of any of our previous math jams, you can do so by visiting our past Math Jams transcripts page. Note that the transcript for this Math Jam may take a little while to show up there.

Lamboreghini 2021-03-12 23:40:03
Wow!

sugar_rush 2021-03-12 23:40:03
Yay!!

palaashgang 2021-03-12 23:40:03
YAY

happyhari 2021-03-12 23:40:03
YAY!!

MaryM 2021-03-12 23:40:10
Please join us again for the AIME II Math Jam on Saturday, March 20th, at 6:30 PM Eastern / 3:30 PM Pacific.

MaryM 2021-03-12 23:40:14
See you there! Bye!

2021 AIME I Discussion

Facilitator: AoPS Staff