Start the New Year strong with our problem-based courses! Enroll today!

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k a January Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jan 1, 2025
Happy New Year!!! Did you know, 2025 is the first perfect square year that any AoPS student has experienced? The last perfect square year was 1936 and the next one will be 2116! Let’s make it a perfect year all around by tackling new challenges, connecting with more problem-solvers, and staying curious!

We have some fun new things happening at AoPS in 2025 with new courses, such as self-paced Introduction to Algebra B, more coding, more physics, and, well, more!

There are a number of upcoming events, so be sure to mark your calendars for the following:

[list][*]Accelerated AIME Problem Series classes start on January 6th and 7th. These AIME classes will run three times a week throughout the month of January. With this accelerated track, you can fit three months of contest tips and training into four weeks finishing right in time for the AIME I on February 6th.
[*]Join our Math Jam on January 7th to learn about our Spring course options. We'll work through a few sample problems, discuss how the courses work, and answer your questions.
[*]RSVP for our New Year, New Challenges webinar on January 9th. We’ll discuss how you can meet your goals, useful strategies for your problem solving journey, and what classes and resources are available.
Have questions? Our Academic Success team is only an email away, write to us at success@aops.com.[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Jan 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Diophantine Equation with Primes and Squares
djmathman   13
N 13 minutes ago by MuradSafarli
Source: Indian RMO 2013 Mumbai Region Problem 2
Find all triples $(p,q,r)$ of primes such that $pq=r+1$ and $2(p^2+q^2)=r^2+1$.
13 replies
djmathman
Feb 1, 2014
MuradSafarli
13 minutes ago
Vector units
youochange   2
N 42 minutes ago by EaZ_Shadow
\[
\left( \frac{\hat{\jmath}}{\hat{k}} \right) \times \hat{k} = ?
\]
2 replies
youochange
3 hours ago
EaZ_Shadow
42 minutes ago
Minimum of an expresion with squares and square roots
DensSv   3
N 43 minutes ago by ehuseyinyigit
Source: 2023 Shortlist for NMO in Romania, for 8th grade
For $a$ and $b$ real numbers, define
$$E(a,b)=\sqrt{\bigg(\frac{1}{2}-a\bigg)^{2}+\bigg(\frac{1}{2}-b\bigg)^{2}}+\sqrt{\bigg(\frac{1}{2}+a\bigg)^{2}+\bigg(\frac{1}{2}+b\bigg)^{2}}$$a) Prove that $E(a,b)\geq \sqrt{2}$ for every real number $a,b$ and determine the equality case.
b) Find $min\{E(a,a+2)|a\in \mathbb{R}\}$.
3 replies
DensSv
Yesterday at 9:45 PM
ehuseyinyigit
43 minutes ago
Junior Balkan Mathematical Olympiad 2020- P3
Lukaluce   12
N an hour ago by wizixez
Source: JBMO 2020
Alice and Bob play the following game: Alice picks a set $A = \{1, 2, ..., n \}$ for some natural number $n \ge 2$. Then, starting from Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers and should differ by $1$ from an already chosen number. The game ends when all numbers from the set $A$ are chosen. Alice wins if the sum of all the numbers that she has chosen is composite. Otherwise Bob wins. Decide which player has a winning strategy.

Proposed by Demetres Christofides, Cyprus
12 replies
Lukaluce
Sep 11, 2020
wizixez
an hour ago
thanks u!
Ruji2018252   1
N an hour ago by arqady
$a,b,c>0$ and $a+b+c=3$. Prove:
$$\dfrac{\sqrt{a^3+b^3+8}}{ab+2}+\dfrac{\sqrt{b^3+c^3+8}}{bc+2}+\dfrac{\sqrt{a^3+c^3+8}}{ac+2}\ge 3$$
1 reply
Ruji2018252
3 hours ago
arqady
an hour ago
GCD Functional Equation
pinetree1   57
N an hour ago by D4N13LCarpenter
Source: USA TSTST 2019 Problem 7
Let $f: \mathbb Z\to \{1, 2, \dots, 10^{100}\}$ be a function satisfying
$$\gcd(f(x), f(y)) = \gcd(f(x), x-y)$$for all integers $x$ and $y$. Show that there exist positive integers $m$ and $n$ such that $f(x) = \gcd(m+x, n)$ for all integers $x$.

Ankan Bhattacharya
57 replies
pinetree1
Jun 25, 2019
D4N13LCarpenter
an hour ago
Iran geometry
Dadgarnia   18
N 2 hours ago by cursed_tangent1434
Source: Iranian TST 2020, second exam day 1, problem 3
Given a triangle $ABC$ with circumcircle $\Gamma$. Points $E$ and $F$ are the foot of angle bisectors of $B$ and $C$, $I$ is incenter and $K$ is the intersection of $AI$ and $EF$. Suppose that $T$ be the midpoint of arc $BAC$. Circle $\Gamma$ intersects the $A$-median and circumcircle of $AEF$ for the second time at $X$ and $S$. Let $S'$ be the reflection of $S$ across $AI$ and $J$ be the second intersection of circumcircle of $AS'K$ and $AX$. Prove that quadrilateral $TJIX$ is cyclic.

Proposed by Alireza Dadgarnia and Amir Parsa Hosseini
18 replies
Dadgarnia
Mar 11, 2020
cursed_tangent1434
2 hours ago
JBMO Shortlist 2023 N4
Orestis_Lignos   3
N 2 hours ago by Perelman1000000
Source: JBMO Shortlist 2023, N4
The triangle $ABC$ is sectioned by $AD,BE$ and $CF$ (where $D \in (BC), E \in (CA)$ and $F \in (AB)$) in seven disjoint polygons named regions. In each one of the nine vertices of these regions we write a digit, such that each nonzero digit appears exactly once. We assign to each side of a region the lowest common multiple of the digits at its ends, and to each region the greatest common divisor of the numbers assigned to its sides.

Find the largest possible value of the product of the numbers assigned to the regions.
3 replies
Orestis_Lignos
Jun 28, 2024
Perelman1000000
2 hours ago
circumcircle and altitudes! CMO 2015 P4
aditya21   25
N 2 hours ago by ihatemath123
Source: canadian mathematical olympiad
Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $I$ be a circle with center on the altitude from $A$ in $ABC$, passing through vertex $A$ and points $P$ and $Q$ on sides $AB$ and $AC$. Assume that
\[BP\cdot CQ = AP\cdot AQ.\]Prove that $I$ is tangent to the circumcircle of triangle $BOC$.
25 replies
aditya21
Apr 24, 2015
ihatemath123
2 hours ago
Divisibility
EtacticToe   1
N 3 hours ago by hung9A
Source: Own
Triple natural numbers $a,b,c$ called cold if $6|a+b+c$ and $36|a^2 + b^2 + c^2$ . Find the largest n such that $n|a^3 + b^3 + c^3$ for every triple cold $(a,b,c)$
1 reply
EtacticToe
4 hours ago
hung9A
3 hours ago
D976 : Dominoes 2x1
Dattier   3
N 3 hours ago by Dattier
Source: les dattes à Dattier
Can you, by placing 6 dominoes of size 2x1 on the upper diagonal (the gray part), completely cover the 10x10 grid?

IMAGE
3 replies
Dattier
Yesterday at 9:17 PM
Dattier
3 hours ago
Tangent circles again
AlephG_64   6
N 3 hours ago by Aiden-1089
Source: 1st AGO, problem 6
A triangle $ABC$ with $AB < AC$ has circumcenter $O$. The perpendicular bisector of $BC$ meets line $AB,AC$ at $D,E$ respectively. The circles with diameters $DE,AO$ meet at two points $X,Y$. Let $M$ be the midpoint of $BC$.

Prove that the circumcircle of triangle $MXY$ is tangent to the circumcircle of triangle $ABC$.
6 replies
AlephG_64
Jun 10, 2024
Aiden-1089
3 hours ago
Rooks on marked cells with cherry and strawberry
NO_SQUARES   1
N 3 hours ago by NO_SQUARES
Source: Russian winter TST to IMO 2023 P5
On the checkered board $n\times n$, several cells are marked, with two opposite corners marked. There is a cherry in one of the other two corners, and a strawberry in the fourth corner (corners with berries can be marked or not). Let $A$ be the number of ways to arrange $n$ non-beating rooks on marked squares, $B$ be the number of ways to arrange $n$ non-beating rooks so that one of them stands on the cherry square and the rest on marked ones, and $C$ be the number of ways to arrange $n$ non-beating rooks so that one of them was on a strawberry cage, and the rest were on the marked ones. Prove that $A^2 \geqslant 4BC$.
1 reply
NO_SQUARES
Jan 3, 2025
NO_SQUARES
3 hours ago
(KLB) is tangent to AC; a lot of points
NO_SQUARES   2
N 3 hours ago by NO_SQUARES
Source: Russian summer TST before IMO 2023; 1nd day; groups A and B P4 = group NG P3
On the minor arc $BC$ of circle $\omega$ described about an acute triangle $ABC$, the point $P$ is chosen. The external bisector of the angle $ABC$ intersects the ray $CA$ in point $Q$. Points $M$ and $N$ are chosen on tangent line from $B$ to $\omega$ and on (internal) bisector of angle $ABC$ respectively such that $M$ lies on line $PQ$ and $MN \parallel AC$. Line $PN$ meets circle $\omega$ second time in point $L$ and with line $AC$ in point $K$. Prove that the circumscribed circle of the triangle $KLB$ is tangent to line $AC$.
2 replies
NO_SQUARES
Jan 3, 2025
NO_SQUARES
3 hours ago
concurrent wanted, start with 2 circles, each external to the other
parmenides51   2
N Sep 10, 2021 by WolfusA
Source: 2018 Centroamerican Shortlist G7
Two circles $S_1$ and $S_2$ have centers $O_1$ and $O_2$ respectively, they do not intersect and neither is inside the other. Line $\ell_1$ is tangent to $S_1$ at $A$ and $S_2$ at $ B$, and does not intersect at segment $O_1O_2$. Line $\ell_2$ is tangent to $S_1$ at $C$ and $S_2$ at $D$, and intersects segment $O_1O_2$. Prove that lines $AC$, $BD$, and $O_1O_2$ concur.
2 replies
parmenides51
Sep 10, 2021
WolfusA
Sep 10, 2021
concurrent wanted, start with 2 circles, each external to the other
G H J
G H BBookmark kLocked kLocked NReply
Source: 2018 Centroamerican Shortlist G7
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parmenides51
30514 posts
#1 • 1 Y
Y by centslordm
Two circles $S_1$ and $S_2$ have centers $O_1$ and $O_2$ respectively, they do not intersect and neither is inside the other. Line $\ell_1$ is tangent to $S_1$ at $A$ and $S_2$ at $ B$, and does not intersect at segment $O_1O_2$. Line $\ell_2$ is tangent to $S_1$ at $C$ and $S_2$ at $D$, and intersects segment $O_1O_2$. Prove that lines $AC$, $BD$, and $O_1O_2$ concur.
This post has been edited 1 time. Last edited by parmenides51, Sep 10, 2021, 2:16 AM
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naruto_uzumaki_06
6 posts
#2 • 1 Y
Y by centslordm
An old but funny problem :)

Let $S= AC \cap BD; E = AB \cap CD; M, N \text{be the midpoint of } AB, CD.$
It is easy to show that $O_1E \perp O_2E$, so $AC \perp BD$.
As $M$ is the midpoint of the hypotenuse of the right triangle $ASB$, it follows that $MA^2 = MB^2=MS^2 \Rightarrow Pow_{M/S_1} =Pow_{M/S_2}=Pow_{M/(S, 0)} $.
Similarity, $Pow_{N/S_1} =Pow_{N/S_2}=Pow_{N/(S, 0)} $. Therefore, $S_1,S_2,(S,0)$ is a coaxial system, so $O_1, O_2, S$ are collinear.
Attachments:
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WolfusA
1900 posts
#3 • 1 Y
Y by centslordm
$\definecolor{A}{RGB}{255,0,205}\color{A}\fbox{Solution}$
$\definecolor{A}{RGB}{79,208,220}\color{A}\fbox{Case 1.}\ S_1,S_2\text{ have unequal radii.}$
Assume wlog that $S_1$ has smaller radius than $S_2$. Then there exists intersection $X$ of two common external tangents. Let $Y$ be the intersection of lines $CD, AB$ and $Z$ be the intersection of line $CD$ with second external tangent. Define $F$ as the intersection of $AC, O_1O_2$. Obviously points $X,O_1,O_2$ are collinear. Since $BY=DY$ and $O_1Y$ is $Y$-angle internal bisector we have $$\definecolor{A}{RGB}{127,0,255}\color{A}BD\parallel O_1Y.$$If $x,y,z$ are lengths of sides $YZ,ZX,XY$ respectively, then
$$2\cos^2\frac{\angle X}{2}=\cos{\angle X}+1=\frac{(x+y+z)(y+z-x)}{2yz}.$$Knowing lengths of tangent segments we conclude
$$\frac{y+z-x}{2XO_1}=\cos\frac{\angle X}2=\frac{XF}{y}\implies \frac{XF}{XO_1}=\frac{2y\cos^2\frac{\angle X}2}{y+z-x}=\frac{x+y+z}{2z}=\frac{BX}{XY}\implies \definecolor{A}{RGB}{127,0,255}\color{A}BF\parallel O_1Y.$$Finally we arrived at$$\definecolor{A}{RGB}{127,0,255}\color{A}BF\parallel BD$$which concludes this case.
[asy]
import graph; size(8cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -9.258815757030918, xmax = 3.966319828854776, ymin = 1.0298292725696054, ymax = 11.235293807498413;  /* image dimensions */
pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen xfqqff = rgb(0.4980392156862745,0,1); 
 /* draw figures */
draw((-8.90055,1.51093)--(xmax, 0.9972829188910803*xmax + 10.387296483736005), linewidth(0.7) + uququq); /* ray */
draw((-8.90055,1.51093)--(xmax, -0.001475476684154335*xmax + 1.4977974459988501), linewidth(0.7) + uququq); /* ray */
draw((-1.96042,1.50069)--(-1.36673,9.02428), linewidth(0.7) + uququq); 
draw(circle((-3.880926270770868,3.5817924647647517), 2.078266540315883), linewidth(0.7) + uququq); 
draw(circle((3.6706565153044086,6.69722305972859), 5.204835916290968), linewidth(0.7) + uququq); 
draw((-5.34848210870255,5.053346634732405)--(-1.8091002073029998,3.4183037026684335), linewidth(0.7) + uququq); 
draw((-1.5180497926970027,7.1066662973315635)--(-0.00470811490722763,10.382601161158851), linewidth(0.7) + xfqqff); 
draw((-8.90055,1.51093)--(3.6706565153044086,6.69722305972859), linewidth(0.7) + uququq); 
draw((-2.9734401118048894,3.9561788979456285)--(-1.5180497926970027,7.1066662973315635), linewidth(0.7) + xfqqff); 
draw((-3.880926270770868,3.5817924647647517)--(-1.36673,9.02428), linewidth(0.7) + xfqqff); 
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dot((-8.90055,1.51093),linewidth(3pt) + dotstyle); 
label("$X$", (-9.3,1.6849443480314348), NE * labelscalefactor); 
dot((-1.36673,9.02428),linewidth(3pt) + dotstyle); 
label("$Y$", (-1.5407412742462328,9.085697466139287), NE * labelscalefactor); 
dot((-1.96042,1.50069),linewidth(3pt) + dotstyle); 
label("$Z$", (-1.8478264658689656,1.6030549635987061), NE * labelscalefactor); 
dot((-3.880926270770868,3.5817924647647517),linewidth(3pt) + dotstyle); 
label("$O_{1}$", (-3.85411638447082,3.), NE * labelscalefactor); 
dot((3.6706565153044086,6.69722305972859),linewidth(3pt) + dotstyle); 
label("$O_{2}$", (3.35214944560931,6.8849202595097045), NE * labelscalefactor); 
dot((-1.8091002073029998,3.4183037026684335),linewidth(3pt) + dotstyle); 
label("$C$", (-1.7249923892198724,3.322732036686008), NE * labelscalefactor); 
dot((-0.00470811490722763,10.382601161158851),linewidth(3pt) + dotstyle); 
label("$B$", (-0.6,10.406163790117036), NE * labelscalefactor); 
dot((-5.34848210870255,5.053346634732405),linewidth(3pt) + dotstyle); 
label("$A$", (-5.6,5.15), NE * labelscalefactor); 
dot((-1.5180497926970027,7.1066662973315635),linewidth(3pt) + dotstyle); 
label("$D$", (-1.3257816401103197,6.94633729783425), NE * labelscalefactor); 
dot((-2.9734401118048894,3.9561788979456285),linewidth(3pt) + dotstyle); 
label("$F$", (-3.2,3.3), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]
$\definecolor{A}{RGB}{79,208,220}\color{A}\fbox{Case 2.}\ S_1,S_2\text{ have equal radii.}$
From assumption quadrilateral $O_1O_2BA$ is a rectangle. Let $M,E,F$ be intersections of $O_1O_2$ with lines $CD,AC,BD$ respectively. Then $$EM\parallel AB\implies\angle ECM=\angle CAM=\angle CEM\implies CM=EM$$and
$$MF\parallel AB\implies\angle MFD=\angle DBA=\angle FDM\implies DM=FM.$$$\triangle O_1CM\equiv\triangle O_2DM\implies CM=DM\implies EM=FM.$
Since points $E,F$ lie on the same half-line $MO_1^{\rightarrow}$ or $MO_2^{\rightarrow}$ we have $$E=F.\blacksquare$$[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(5cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -8.98285127060889, xmax = 5.7312772970898775, ymin = -6.304638685752549, ymax = 5.049839783284332;  /* image dimensions */
pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); 
 /* draw figures */
draw((xmin, 0*xmin + 2.350731059952596)--(xmax, 0*xmax + 2.350731059952596), linewidth(0.7) + uququq); /* line */
draw(circle((-8.132280786612714,-2.3871041935695487), 4.737835253522144), linewidth(0.7) + uququq); 
draw(circle((4.19864012334485,-2.3871041935695487), 4.737835253522144), linewidth(0.7) + uququq); 
draw((xmin, 1.2008643630146694*xmin-0.02521974885766618)--(xmax, 1.2008643630146694*xmax-0.02521974885766618), linewidth(0.7) + uququq); /* line */
draw((xmin, 0.4685908521757041*xmin + 0.38328670657532904)--(xmax, 0.4685908521757041*xmax + 0.38328670657532904), linewidth(0.7) + uququq); /* line */
draw(circle((-1.966820331633932,-2.3871041935695487), 3.9453541960627443), linewidth(0.7) + linetype("4 4") + uququq); 
draw((-8.132280786612714,2.350731059952596)--(-8.132280786612714,-2.3871041935695487), linewidth(0.7) + uququq); 
draw((-8.132280786612714,-2.3871041935695487)--(4.19864012334485,-2.3871041935695487), linewidth(0.7) + uququq); 
draw((4.19864012334485,-2.3871041935695487)--(4.19864012334485,2.350731059952596), linewidth(0.7) + uququq); 
draw((-5.912174527696676,-2.387104193569548)--(-4.491501147617446,-5.418903413470947), linewidth(0.7) + uququq); 
draw((-8.132280786612714,-2.3871041935695487)--(-4.491501147617446,-5.418903413470947), linewidth(0.7) + uququq); 
draw((4.19864012334485,-2.3871041935695487)--(0.5578604843495838,0.6446950263318516), linewidth(0.7) + uququq); 
draw((-8.132280786612714,2.350731059952596)--(-5.912174527696676,-2.387104193569548), linewidth(0.7) + uququq); 
 /* dots and labels */
dot((-8.132280786612714,-2.3871041935695487),linewidth(3pt) + dotstyle); 
label("$O_{1}$", (-8,-2), NE * labelscalefactor); 
dot((4.19864012334485,-2.3871041935695487),linewidth(3pt) + dotstyle); 
label("$O_{2}$", (4.2,-2.7399929878502984), NE * labelscalefactor); 
dot((-8.132280786612714,2.350731059952596),linewidth(3pt) + dotstyle); 
label("$A$", (-8.083148362831643,2.4190629263660575), NE * labelscalefactor); 
dot((4.19864012334485,2.350731059952596),linewidth(3pt) + dotstyle); 
label("$B$", (4,2.464617503975032), NE * labelscalefactor); 
dot((-1.966820331633932,-2.3871041935695487),linewidth(3pt) + dotstyle); 
label("$M$", (-1.9560576744245717,-3.5), NE * labelscalefactor); 
dot((-4.491501147617446,-5.418903413470947),linewidth(3pt) + dotstyle); 
label("$C$", (-4.2,-5.6), NE * labelscalefactor); 
dot((0.5578604843495838,0.6446950263318516),linewidth(3pt) + dotstyle); 
label("$D$", (0.2,1.2), NE * labelscalefactor); 
dot((-5.912174527696676,-2.387104193569548),linewidth(3pt) + dotstyle); 
label("$E/F$", (-5.759864904773944,-2.1363948345313872), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]
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