Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 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
Prove the inequality
Butterfly   0
6 minutes ago

Prove
$$x^2+y^2+7\ge 3(x+y)+\frac{9}{xy+2}~~(x,y>0).$$
0 replies
Butterfly
6 minutes ago
0 replies
Kaprekar Number
CSJL   5
N 8 minutes ago by Adywastaken
Source: 2025 Taiwan TST Round 1 Independent Study 2-N
Let $k$ be a positive integer. A positive integer $n$ is called a $k$-good number if it satisfies
the following two conditions:

1. $n$ has exactly $2k$ digits in decimal representation (it cannot have leading zeros).

2. If the first $k$ digits and the last $k$ digits of $n$ are considered as two separate $k$-digit
numbers (which may have leading zeros), the square of their sum is equal to $n$.

For example, $2025$ is a $2$-good number because
\[(20 + 25)^2 = 2025.\]Find all $3$-good numbers.
5 replies
CSJL
Mar 6, 2025
Adywastaken
8 minutes ago
Projective geometry
definite_denny   0
9 minutes ago
Source: IDK
Let ABC be a triangle and let DEF be the tangency point of incircirle with sides BC,CA,AB. Points P,Q are chosen on sides AB,AC such that PQ is parallel to BC and PQ is tangent to the incircle. Let M denote the midpoint of PQ. Let EF intersect BC at T. Prove that TM is tangent to the incircle
0 replies
definite_denny
9 minutes ago
0 replies
Problem 7 of RMO 2006 (Regional Mathematical Olympiad-India)
makar   11
N 15 minutes ago by SomeonecoolLovesMaths
Source: Functional Equation
Let $ X$ be the set of all positive integers greater than or equal to $ 8$ and let $ f: X\rightarrow X$ be a function such that $ f(x+y)=f(xy)$ for all $ x\ge 4, y\ge 4 .$ if $ f(8)=9$, determine $ f(9) .$
11 replies
makar
Sep 13, 2009
SomeonecoolLovesMaths
15 minutes ago
Hardest in ARO 2008
discredit   26
N 22 minutes ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
26 replies
discredit
Jun 11, 2008
JARP091
22 minutes ago
Inequality
Kei0923   2
N an hour ago by Kei0923
Source: Own.
Let $k\leq 1$ be a fixed positive real number. Find the minimum possible value $M$ such that for any positive reals $a$, $b$, $c$, $d$, we have
$$\sqrt{\frac{ab}{(a+b)(b+c)}}+\sqrt{\frac{cd}{(c+d)(d+ka)}}\leq M.$$
2 replies
Kei0923
Jul 25, 2023
Kei0923
an hour ago
PAMO 2023 Problem 2
kerryberry   6
N an hour ago by justaguy_69
Source: 2023 Pan African Mathematics Olympiad Problem 2
Find all positive integers $m$ and $n$ with no common divisor greater than 1 such that $m^3 + n^3$ divides $m^2 + 20mn + n^2$. (Professor Yongjin Song)
6 replies
kerryberry
May 17, 2023
justaguy_69
an hour ago
My Unsolved Problem
ZeltaQN2008   0
an hour ago
Source: IDK
Given a positive integer \( m \) and \( a > 1 \). Prove that there always exists a positive integer \( n \) such that \( m \mid (a^n + n) \).

P/s: I can prove the problem if $m$ is a power of a prime number, but for arbitrary $m$ then well.....
0 replies
ZeltaQN2008
an hour ago
0 replies
Computing functions
BBNoDollar   3
N an hour ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
3 replies
BBNoDollar
May 21, 2025
wh0nix
an hour ago
Computing functions
BBNoDollar   8
N an hour ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
8 replies
BBNoDollar
May 18, 2025
wh0nix
an hour ago
Find the remainder
Jackson0423   1
N 2 hours ago by wh0nix

Find the remainder when
\[
\frac{5^{2000} - 1}{4}
\]is divided by \(64\).
1 reply
Jackson0423
3 hours ago
wh0nix
2 hours ago
IMO 2018 Problem 1
juckter   170
N 2 hours ago by Adywastaken
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
170 replies
juckter
Jul 9, 2018
Adywastaken
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   2
N 2 hours ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
2 replies
ICE_CNME_4
Yesterday at 7:23 PM
wh0nix
2 hours ago
Minimum of this fuction
persamaankuadrat   1
N 2 hours ago by alexheinis
Source: KTOM January 2020
If $x$ is a positive real number, find the minimum of the following expression

$$\lfloor x \rfloor + \frac{500}{\lceil x\rceil^{2}}$$
1 reply
persamaankuadrat
4 hours ago
alexheinis
2 hours ago
Trapezium with two right-angles: prove < AKB = 90° and more
Leonardo   6
N May 22, 2025 by FrancoGiosefAG
Source: Mexico 2002
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
6 replies
Leonardo
May 8, 2004
FrancoGiosefAG
May 22, 2025
Trapezium with two right-angles: prove < AKB = 90° and more
G H J
G H BBookmark kLocked kLocked NReply
Source: Mexico 2002
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Leonardo
128 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
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darij grinberg
6555 posts
#2 • 2 Y
Y by Adventure10, Mango247
Proving < AKB = 90° is very easy: Since < MAD = 90° and < MKD = 90°, the points A and K lie on the circle with diameter MD, and thus < AKM = < ADM. In the right-angled triangle DAM, we have < ADM = 90° - < DMA. Thus, < AKM = 90° - < DMA. Similarly, < BKM = 90° - < CMB. Hence,

< AKB = < AKM + < BKM = (90° - < DMA) + (90° - < CMB)
= 180° - < DMA - < CMB = < CMD = 90°.

Now we are going to show $\frac{KP}{PA}+\frac{KQ}{QB}=1$. In fact, let U be the orthogonal projection of the point K on the line AB, and let the line AC meet the line KU at T. Also let the line AK meet the line BC at Q.

Since < AKB = 90°, the point K lies on the circle with diameter AB. The center of this circle is the midpoint M of the segment AB. Hence, MK = MA = MB. Now, since MK = MB, the triangle KMB is isosceles, so that < MKB = < MBK. Thus, < CKB = 90° - < MKB = 90° - < MBK = < CBK, and it follows that the triangle KCB is isosceles, so that CK = CB. Similarly, DK = DA.

The lines AD, BC and KU are parallel to each other, since all of them are perpendicular to the line AB. Thus, from AD || BC, we have by Thales CQ : DA = CK : DK. But DK = DA; thus, CQ = CK. Together with CK = CB, this yields CQ = CB, and thus the point C is the midpoint of the segment BQ.

Since KU || BC, Thales yields TK : TU = CQ : CB. Since CQ = CB, we thus have TK = TU, and hence the point T is the midpoint of the segment KU.

Now, we have defined the point T as the point of intersection of the lines AC and KU and proved that this point T is the midpoint of the segment KU. Instead, we could have defined the point T as the midpoint of the segment KU and would be able to conclude that this point T lies on the line AC. Similarly, the same point T lies on the line BD. Altogether, the point T lies on the three lines AC, BD and KU.

Now, the points P and Q join the scene. Applying the van Aubel theorem to triangle AKB, whose cevians AQ, BP and KU concur at the point T, we get $\frac{KT}{TU} = \frac{KP}{PA}+\frac{KQ}{QB}$. Since the point T is the midpoint of the segment KU, we have $\frac{KT}{TU}=1$; thus, $\frac{KP}{PA}+\frac{KQ}{QB}=1$, and we are done.

Darij
This post has been edited 1 time. Last edited by darij grinberg, Jan 28, 2005, 5:18 PM
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Huynh Anh Hao
31 posts
#3 • 2 Y
Y by Adventure10, Mango247
Leonardo wrote:
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.
We can easily prove $\angle AKB = 90^{\circ}$
We call $I$ is the intersection of $AC$ and $BD$.
=> $\frac{IB}{IA}= \frac{AB}{CD}= \frac{BK}{CK}$
=> $IK$ // $CD$ // $AB$
=> $\frac{KP}{PA}= \frac{IK}{AD}= \frac{KC}{DC}$
$\frac{KQ}{QB}= \frac{IK}{CB}= \frac{KD}{DC}$
=> The proof is completed.
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Virgil Nicula
7054 posts
#4 • 1 Y
Y by Adventure10
Prove easily that $KA\perp KB\ .$ Denote $B'\in BC\cap AK$ and $A'\in AD\cap BK\ .$ Then $CB=CB'$ and $DA=DA'\ .$

Apply the Menelaus' theorem to the transversal $\overline{AQC}$ for the triangle $BB'K$, where $CB'=CB$ :

$\frac{AK}{AB'}\cdot \frac{CB'}{CB}\cdot\frac{QB}{QK}=1$ $\Longrightarrow$ $\frac{QK}{QB}=\frac{AK}{AB'}$ $\Longrightarrow$ $\boxed{\ \frac{QK}{QB}=\frac{AA'}{AA'+BB'}\ }\ \ (1)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BPD}$ for the triangle $AA'K$, where $DA'=DA$ :

$\frac{BK}{BA'}\cdot\frac{DA'}{DA}\cdot\frac{PA}{PK}=1$ $\Longrightarrow$ $\frac{PK}{PA}=\frac{BK}{BA'}$ $\Longrightarrow$ $\boxed{\ \frac{PK}{PA}=\frac{BB'}{AA'+BB'}\ }\ \ (2)\ .$

Therefore, from the sum of the relations $(1)$ and $(2)$ we find the required relation $\boxed{\ \frac{PK}{PA}+\frac{QK}{QB}=1\ }\ \ (*)\ .$

Remark. Denote the intersections $R\in MK\cap AC$ and $S\in MK\cap BD\ .$

Apply the Menelaus' theorem to the transversal $\overline{ARQ}$ for the triangle $MBK$ : $\frac{AM}{AB}\cdot\frac{QB}{QK}\cdot\frac{RK}{RM}=1$ $\Longrightarrow$ $\boxed{\ \frac{RK}{RM}=2\cdot\frac{QK}{QB}\ }\ \ (3)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BSP}$ for the triangle $MAK$ : $\frac{BM}{BA}\cdot\frac{PA}{PK}\cdot\frac{SK}{SM}=1$ $\Longrightarrow$ $\boxed{\ \frac{SK}{SM}=2\cdot\frac{PK}{PA}\ }\ \ (4)\ .$

Therefore, from the sum of the relations $(3)$ , $(4)$ and using the relation $(*)$ obtain a new and interesting relation $\boxed{\ \frac{SK}{SM}+\frac{RK}{RM}=2\ }\ \ (**)\ .$
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parmenides51
30653 posts
#5
Y by
Let $ABCD$ be a quadrilateral with $\angle DAB=\angle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\angle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.
This post has been edited 1 time. Last edited by parmenides51, Jan 8, 2023, 12:00 AM
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UI_MathZ_25
116 posts
#6
Y by
Since $AMKD$ and $MBCK$ are cyclic \[\angle AKM = \angle ADM = 90^{\circ} - \angle AMD = \angle BMC = \angle BKC = 90^{\circ} - \angle BKM \square \]By Pappus' theorem on $A-M-B$ and $D-K-C,$ \[E = AK \cap MD, F = AC \cap BD, G = MC \cap BK \]are collinear. Hence, by angle chasing $DM$ and $MC$ are perpendicular bisector of $AK$ and $BK$, respectively. Thereby $EG \parallel AB$.
We say that the parallel to $AB$ that passes through $K$ cut to $BD$ and $AC$ at $J$ and $L$, respectively. Therefore \[\frac{KP}{PA} + \frac{KQ}{QB} = \frac{KJ}{AB} + \frac{KL}{AB} = \frac{JL}{AB} = 1\]where the last equation is given because \[\frac{JL}{AB} = \frac{FL}{AF} = \frac{EK}{AE} = 1 \blacksquare\]
[asy]
import graph; size(13cm); 
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 = -10.366231833421102, xmax = 11.055501321739651, ymin = -4.584365118522112, ymax = 26.38971231092906;  /* image dimensions */
pen zzttff = rgb(0.6,0.2,1); pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); 

draw((-4.935382086114243,14.666485582905072)--(-4.471987430283072,14.6832425523788)--(-4.4887443997568,15.146637208209972)--(-4.9521390555879705,15.129880238736243)--cycle, linewidth(2) + ccqqqq); 
draw((8.964466288580857,15.633123269262516)--(8.981223258054586,15.169728613431346)--(9.444617913885757,15.186485582905073)--(9.427860944412028,15.649880238736245)--cycle, linewidth(2) + ccqqqq); 
draw((0.5048812728795858,8.885791671830379)--(0.05681575859019139,9.005176335767642)--(-0.06256890534707155,8.557110821478247)--(0.385496608942323,8.437726157540984)--cycle, linewidth(2) + ccqqqq); 
draw((1.8753492906749505,15.100742375453619)--(2.1644871539575745,14.73823072171654)--(2.526998807694653,15.027368584999165)--(2.237860944412029,15.389880238736243)--cycle, linewidth(2) + ccqqqq); 
 /* draw figures */
draw((-4.9521390555879705,15.129880238736243)--(9.427860944412028,15.649880238736245), linewidth(2) + zzttff); 
draw((2.237860944412029,15.389880238736243)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.7597159241415214,9.808640565274834)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(-4.7597159241415214,9.808640565274834), linewidth(2) + zzttqq); 
draw((9.427860944412028,15.649880238736245)--(9.779170059339375,5.934832022091594), linewidth(2) + zzttqq); 
draw((-4.7597159241415214,9.808640565274834)--(2.237860944412029,15.389880238736243), linewidth(2)); 
draw((2.237860944412029,15.389880238736243)--(0.385496608942323,8.437726157540984), linewidth(2)); 
draw((-4.7597159241415214,9.808640565274834)--(9.427860944412028,15.649880238736245), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(0.385496608942323,8.437726157540984), linewidth(2)); 
draw((0.385496608942323,8.437726157540984)--(9.427860944412028,15.649880238736245), linewidth(2)); 
draw((-2.2833212233228233,11.783803198138612)--(4.906678776677175,12.043803198138612), linewidth(2) + linetype("4 4") + zzttff); 
draw((-8.876118698789208,8.113855372426736)--(-4.7597159241415214,9.808640565274834), linewidth(2)); 
draw((-8.876118698789208,8.113855372426736)--(5.483684752146957,8.616005447031823), linewidth(2) + zzttff); ; 
 /* dots and labels */
dot((-4.9521390555879705,15.129880238736243),dotstyle); 
label("$A$", (-4.857786164951195,15.350962062606659), NE * labelscalefactor); 
dot((9.427860944412028,15.649880238736245),dotstyle); 
label("$B$", (9.525377524942455,15.875575935794258), NE * labelscalefactor); 
dot((2.237860944412029,15.389880238736243),linewidth(4pt) + dotstyle); 
label("$M$", (2.33379567999563,15.569551176434825), NE * labelscalefactor); 
dot((9.779170059339375,5.934832022091594),dotstyle); 
label("$C$", (9.875120107067527,6.148360370440855), NE * labelscalefactor); 
dot((-4.7597159241415214,9.808640565274834),linewidth(4pt) + dotstyle); 
label("$D$", (-4.682914873888658,9.973669862433766), NE * labelscalefactor); 
dot((0.385496608942323,8.437726157540984),linewidth(4pt) + dotstyle); 
label("$K$", (0.7380951490499821,7.394318319261403), NE * labelscalefactor); 
dot((-1.7095574770328914,11.064436864858926),linewidth(4pt) + dotstyle); 
label("$P$", (-1.6226672802942643,11.24148672263713), NE * labelscalefactor); 
dot((2.749073801386117,10.322905991297157),linewidth(4pt) + dotstyle); 
label("$Q$", (2.8365506418004234,10.498283735621365), NE * labelscalefactor); 
dot((-2.2833212233228233,11.783803198138612),linewidth(4pt) + dotstyle); 
label("$E$", (-2.1909989762475086,11.96283079827008), NE * labelscalefactor); 
dot((0.2611707105781351,11.875815423842962),linewidth(4pt) + dotstyle); 
label("$F$", (0.3446347441592743,12.050266443801346), NE * labelscalefactor); 
dot((4.906678776677175,12.043803198138612),linewidth(4pt) + dotstyle); 
label("$G$", (5.000582868699316,12.22513773486388), NE * labelscalefactor); 
dot((-8.876118698789208,8.113855372426736),dotstyle); 
label("$J$", (-8.79239021385827,8.334251508722518), NE * labelscalefactor); 
dot((5.483684752146957,8.616005447031823),linewidth(4pt) + dotstyle); 
label("$L$", (5.568914564652561,8.793288647761669), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 [/asy]
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FrancoGiosefAG
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