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.

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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by 2025 Beijing
sqing   11
N 2 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
1 viewing
sqing
Saturday at 4:56 PM
sqing
2 minutes ago
A functional equation
super1978   1
N 19 minutes ago by CheerfulZebra68
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(x)(y+f(y)))=xf(y)+f(xy) $$for all $x,y \in \mathbb R$
1 reply
super1978
26 minutes ago
CheerfulZebra68
19 minutes ago
Prove that IMO is isosceles
YLG_123   4
N 2 hours ago by Blackbeam999
Source: 2024 Brazil Ibero TST P2
Let \( ABC \) be an acute-angled scalene triangle with circumcenter \( O \). Denote by \( M \), \( N \), and \( P \) the midpoints of sides \( BC \), \( CA \), and \( AB \), respectively. Let \( \omega \) be the circle passing through \( A \) and tangent to \( OM \) at \( O \). The circle \( \omega \) intersects \( AB \) and \( AC \) at points \( E \) and \( F \), respectively (where \( E \) and \( F \) are distinct from \( A \)). Let \( I \) be the midpoint of segment \( EF \), and let \( K \) be the intersection of lines \( EF \) and \( NP \). Prove that \( AO = 2IK \) and that triangle \( IMO \) is isosceles.
4 replies
YLG_123
Oct 12, 2024
Blackbeam999
2 hours ago
Geometric mean of squares a knight's move away
Pompombojam   0
2 hours ago
Source: Problem Solving Tactics Methods of Proof Q27
Each square of an $8 \times 8$ chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.

Is it true that all of the numbers must be equal?
0 replies
Pompombojam
2 hours ago
0 replies
No more topics!
easy geometry
andria   10
N Nov 18, 2024 by mcmp
Source: Iran team selection test third exam p1
Point $A$ is outside of a given circle $\omega$. Let the tangents from $A$ to $\omega$ meet $\omega$ at $S, T$ points $X, Y$ are midpoints of $AT, AS$ let the tangent from $X$ to $\omega$ meet $\omega$ at $R\neq T$. points $P, Q$ are midpoints of $XT, XR$ let $XY\cap PQ=K, SX\cap TK=L$ prove that quadrilateral $KRLQ$ is cyclic.
10 replies
andria
Jun 4, 2015
mcmp
Nov 18, 2024
easy geometry
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G H BBookmark kLocked kLocked NReply
Source: Iran team selection test third exam p1
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andria
824 posts
#1 • 6 Y
Y by Rmasters, Davi-8191, itslumi, Adventure10, Mango247, Rounak_iitr
Point $A$ is outside of a given circle $\omega$. Let the tangents from $A$ to $\omega$ meet $\omega$ at $S, T$ points $X, Y$ are midpoints of $AT, AS$ let the tangent from $X$ to $\omega$ meet $\omega$ at $R\neq T$. points $P, Q$ are midpoints of $XT, XR$ let $XY\cap PQ=K, SX\cap TK=L$ prove that quadrilateral $KRLQ$ is cyclic.
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tranquanghuy7198
253 posts
#2 • 4 Y
Y by aleksam, nguyendangkhoa17112003, Adventure10, Mango247
My solution:
$XS\cap{\omega} = \{S, I\}$
Notice that: $XR, XT$ is tangent to $\omega \Rightarrow (TRIS) = -1$
$\Rightarrow T(XRIS) = -1$ (1)
$TR\cap{XY} = W$
$XP = PT, XQ = QR \Rightarrow XK = KW$
$\Rightarrow T(XWKS) = -1$
$\Rightarrow T(XRKS) = -1$ (2)
(1), (2) $\Rightarrow TI \equiv TK$
$\Rightarrow I = TK\cap{SX} = L$
$\Rightarrow L\in{\omega}$
$\Rightarrow \angle{KLR} = \angle{RST} = \angle{XRT} = \angle{XQP} = \angle{KQR}$
$\Rightarrow K, L, R, Q$ are concyclic
Q.E.D
Attachments:
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andria
824 posts
#3 • 5 Y
Y by Tafhim, rashah76, Eliot, Adventure10, solidgreen
Thank you for your solution.
Here's my solution:
Let $ KT\cap \omega=L'$ note that $XY$ is radical axis of point $A,\omega$ and $PQ$ is radical axis of $X,\omega$ so $K$ is radical center of $A,X,\omega$ so $KA^2=KX^2=KL'.KT$ so $KX$ is tangent to $\odot (\triangle XL'T)\longrightarrow \angle XL'K=\angle KXT=\angle 180-\angle STA$ so $X, L', S$ are collinear and $L\equiv L'$ now observe that $\angle KQR=\angle XQP=\angle XRT=\angle KLR$ so $KQLR$ is cyclic.
DONE
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applepi2000
2226 posts
#4 • 4 Y
Y by anantmudgal09, reveryu, Adventure10, Mango247
Let $TR\cap XY=M$ and let the midpoints of $XS, XY$ be $B, N$. Note that $\angle YMR=\angle RTS=\angle YSR$, so $MYRS$ is cyclic. Due to a homothety centered at $X$, this implies $KNQB$ is cyclic. Because the radical axes of $A, X, \omega$ concur, we get $KA=KX\implies \triangle AXK\sim\triangle TAS$. Therefore $XK=\frac{(AS)(AX)}{ST}\implies (XK)(XN)=\frac{XR^2}{2}=(XR)(XQ)$, so $KNQR$ is cyclic. Finally, $(XL)(XB)=\left(\frac{XS}{1+\frac{ST}{XK}}\right)\left(\frac{XS}{2}\right)=(XK)(XN)$, and so $KNBL$ is cyclic. These three combines give $KNQLBR$ is cyclic, as required.
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tenplusten
1000 posts
#5 • 1 Y
Y by Adventure10
What does "radical axes of point and circle " mean? ı think radical axes exist between circle and circle
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den_thewhitelion
262 posts
#6 • 1 Y
Y by Adventure10
A point is a circle with radius 0
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GeoMetrix
924 posts
#7 • 3 Y
Y by amar_04, Eliot, Adventure10
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(30cm); 
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 = -25.983372532706174, xmax = 74.06437578769733, ymin = -40.10491409386843, ymax = 25.965579439801473;  /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); 
 /* draw figures */
draw(circle((8.26,-3.26), 9), linewidth(2) + wrwrwr); 
draw((xmin, -0.34995352372502203*xmin-9.904574766277786)--(xmax, -0.34995352372502203*xmax-9.904574766277786), linewidth(2) + wrwrwr); /* line */
draw((xmin, 0.2690234650970968*xmin + 3.8378588610526587)--(xmax, 0.2690234650970968*xmax + 3.8378588610526587), linewidth(2) + wrwrwr); /* line */
draw((xmin, -1.1255014513011363*xmin-7.513525349498235)--(xmax, -1.1255014513011363*xmax-7.513525349498235), linewidth(2) + wrwrwr); /* line */
draw((xmin, 3.341464458272747*xmin + 7.2452672799432785)--(xmax, 3.341464458272747*xmax + 7.2452672799432785), linewidth(2) + wrwrwr); /* line */
draw((xmin, 27.076223935693086*xmin + 222.04753627358411)--(xmax, 27.076223935693086*xmax + 222.04753627358411), linewidth(2) + wrwrwr); /* line */
draw((-8.139965143017417,1.6480172325085285)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); 
draw((-9.05011357700883,-22.995365580963366)--(5.921918693006509,5.430993947868541), linewidth(2) + wrwrwr); 
draw((-0.22944998216551502,-6.248183227365741)--(1.5319932317881242,-9.237785955259287), linewidth(2) + wrwrwr); 
draw((-9.05011357700883,-22.995365580963366)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); 
draw((5.921918693006509,5.430993947868541)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); 
draw((-3.303985955614646,-3.7948843613753795)--(-0.22944998216551502,-6.248183227365741), linewidth(2) + wrwrwr); 
draw((1.5319932317881242,-9.237785955259287)--(-9.05011357700883,-22.995365580963366), linewidth(2) + wrwrwr); 
draw(circle((-8.010519033264748,-12.846422661194547), 10.202048774834777), linewidth(2) + wrwrwr); 
draw((1.5319932317881242,-9.237785955259287)--(5.921918693006509,5.430993947868541), linewidth(2) + wrwrwr); 
draw((xmin, -0.670286327440097*xmin-8.210911838260941)--(xmax, -0.670286327440097*xmax-8.210911838260941), linewidth(2) + wrwrwr); /* line */
 /* dots and labels */
dot((8.26,-3.26),dotstyle); 
label("$O$", (8.513610816731658,-2.619491508885604), NE * labelscalefactor); 
dot((-22.201848979041344,-2.134959482851484),dotstyle); 
label("$A$", (-18.12247802181733,-1.255204031789175), NE * labelscalefactor); 
dot((5.921918693006509,5.430993947868541),linewidth(4pt) + dotstyle); 
label("T", (6.1748322845663814,5.956029775720519), NE * labelscalefactor); 
dot((5.28719790285137,-11.754848303012167),linewidth(4pt) + dotstyle); 
label("S", (5.525171581187138,-11.259978863829652), NE * labelscalefactor); 
dot((-8.139965143017417,1.6480172325085285),linewidth(4pt) + dotstyle); 
label("$X$", (-7.857838908425279,2.187997696120859), NE * labelscalefactor); 
dot((-8.457325538094988,-6.944903892931825),linewidth(4pt) + dotstyle); 
label("$Y$", (-8.182669260114901,-6.452489658823189), NE * labelscalefactor); 
dot((1.5319932317881242,-9.237785955259287),linewidth(4pt) + dotstyle); 
label("$R$", (1.8221055719254495,-8.72630212065057), NE * labelscalefactor); 
dot((-3.303985955614646,-3.7948843613753795),linewidth(4pt) + dotstyle); 
label("$Q$", (-3.050349703418877,-3.2691522122648555), NE * labelscalefactor); 
dot((-1.109023225005454,3.5395055901885346),linewidth(4pt) + dotstyle); 
label("$P$", (-0.8415033119294489,4.072013735920689), NE * labelscalefactor); 
dot((-9.05011357700883,-22.995365580963366),linewidth(4pt) + dotstyle); 
label("$K$", (-8.76736389315622,-22.499109032290708), NE * labelscalefactor); 
dot((-0.22944998216551502,-6.248183227365741),linewidth(4pt) + dotstyle); 
label("$L$", (0.0030556024635676683,-5.737862885106012), NE * labelscalefactor); 
dot((-8.298645340556202,-2.64844333021165),linewidth(4pt) + dotstyle); 
label("$B$", (-8.052737119439053,-2.0997629461822025), NE * labelscalefactor); 
dot((-1.4263836200830151,-5.053415535251828),linewidth(4pt) + dotstyle); 
label("$C$", (-1.1663336636190706,-4.503507548685434), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]
Reims FTW!!!
Claim: $L\in \omega$
Proof:We'll use phantom points. Let $L'=XS\cap \omega$ and let $B=\odot(L'RK) \cap RS$. Now let $KL' \cap \omega=T'$ We have by reims theorem that $BK\| ST' \implies T=T'$ as by thales we have $BK \| ST$ .Now we have that $L' \in KT $ as well as $L' \in XS$ and these together imply $L'=L$ and we are done. The rest of the proof follows smoothly from trivial angle chase.
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Gaussian_cyber
162 posts
#8 • 1 Y
Y by Eliot
Storage
This post has been edited 2 times. Last edited by Gaussian_cyber, Jul 27, 2020, 3:55 AM
Reason: latex
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rafaello
1079 posts
#9
Y by
[asy]import olympiad;import geometry;
size(10cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;
pair O,S,T,A,X,Y,R,P,Q,K,L,F,M;
O=(0,0);S=dir(250);T=dir(350);A=intersectionpoint(perpendicular(S,line(S,O)),perpendicular(T,line(T,O)));X=midpoint(A--T);path w=circle(O,1);Y=midpoint(A--S);
R=intersectionpoints(circumcircle(O,T,X),w)[1];P=midpoint(X--T); Q=midpoint(X--R);K=extension(X,Y,P,Q);L=extension(T,K,S,X);M=midpoint(X--Y);

F=extension(T,R,X,Y);

draw(X--F,deep);draw(P--K,deep);
draw(S--X,deep);
draw(w,deep);draw(circumcircle(K,R,L),med+dashed);
draw(S--Y--A--X,med, StickIntervalMarker(3,2,med));
draw(R--Q--X--P--T,light, StickIntervalMarker(4,1,light));
draw(T--F--A,deep);draw(circumcircle(R,M,A),deep);

clip((1,0.2)--(1.1,-1.4)--(-0.4,-1.4)--(-0.4,0.2)--cycle);

dot("$S$",S,dir(S));
dot("$T$",T,dir(T));
dot("$A$",A,dir(A));
dot("$O$",O,N);
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
dot("$R$",R,dir(R));
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$K$",K,dir(K));
dot("$L$",L,N);
dot("$M$",M,M);
dot("$F$",F,F);
[/asy]


Let $M$ be the midpoint of $XY$. Note that $OTXMR$ is cyclic. Let $F=TR\cap XY$, then $FRMA$ is cyclic as $\measuredangle FMA=90^\circ=\measuredangle TRA=\measuredangle FRA$.

Now, \begin{align*}
\measuredangle XAM=90^\circ-\measuredangle MXA=90^\circ-\measuredangle MXT=90^\circ-\measuredangle MRT=90^\circ-\measuredangle MRF=90^\circ-\measuredangle MAF=\measuredangle AFM,
\end{align*}therefore $XA^2=XM\cdot XF$ and as $K$ is the midpoint of $XF$ and $M$ the midpoint of $XY$, we get that $XT^2=XA^2=XK\cdot XY$. Now, we claim that $L$ lies on $\omega$.
\begin{align*}
\measuredangle LTX=\measuredangle KTX=\measuredangle XYT=\measuredangle  XST=\measuredangle  LST,
\end{align*}hence $L$ lies on $\omega$.
Now, \begin{align*}
\measuredangle QRL=\measuredangle RTL=\measuredangle RTK=\measuredangle  QKL,
\end{align*}we are done.
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Mahdi_Mashayekhi
696 posts
#10
Y by
Note that K lies on radical axis of A and circle $\omega$ and also lies on radical axis of X and $\omega$. So if TK meets $\omega$ at L' we have :
KA^2 = KX^2 = KL'.KT. We will prove X,L',S are collinear so L' is L. we want to prove ∠XL'T = ∠L'ST + L'TS. ∠L'ST + ∠L'TS = ∠STX = ∠KXA and we have
KX^2 = KL'.KT so L'TX is tangent to KX so ∠XL'T = ∠KXA so L' is L. ∠RLK = ∠RST = ∠XRT = ∠XQP = ∠RQK so RLQK is cyclic.
we're Done.
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mcmp
53 posts
#12
Y by
YES I FINALLY HAVE IT I CAN ACTUALLY IDENTIFY SIMILAR TRIANGLES AND POINT CIRCLES LETSSS GOOOOO

Let $Z$ be the midpoint of $XY$ and $U$ the midpoint of $ST$; I claim that $KZLR$ and $KZQR$ are concyclic which clearly finishes.

Let’s start with the first one. First I claim $L\in\omega$, from which the conclusion that $KZLR$ will follow by Reims. Let $L’$ be the second intersection of $\overline{KT}\cap\omega\neq T$. I claim that $X-L-S$. However notice that $KYLS$ is cyclic because $\measuredangle YKL=\measuredangle STK=\measuredangle STL=\measuredangle YSL$ as desired.

We now show that $\overline{XT}$ tangent to $(TYK)$ before we continue. Notice by radax on point circles $(A)$, $(X)$ and $\omega$ it’s clear $AK=KX$. Hence $\measuredangle AXK=\measuredangle KAX=\measuredangle XYA$ and so $\triangle AKX\stackrel{-}{\sim}\triangle YAX$, and so $TX^2=AX^2=XY\cdot XK$ as desired. Hence $\measuredangle XSY=\measuredangle XTY=\measuredangle TKX=\measuredangle LKY=\measuredangle LSY$ as desired. So we have that $L\in\omega$.

For the final argument it suffices to show that $XZ\cdot XK=XQ\cdot XR$ however multiplying both sides by two it STP that $XR^2=XT^2=XY\cdot XK$, which is what we just showed.
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