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Gengar_in_Galar

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Gengar_in_Galar

#1 h Mar 10, 2025, 12:00 PM • 1 Y

Y by kiyoras_2001

Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov

This post has been edited 1 time. Last edited by Gengar_in_Galar, Mar 11, 2025, 9:49 AM

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#2 h Mar 10, 2025, 3:53 PM

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used up a lot of time on this to no avail, this is prolly a troll

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1533 posts

#3 h Mar 10, 2025, 6:58 PM • 1 Y

Y by MS_asdfgzxcvb

Let $M_{IJ}$ be the midpoint of $I$ , $J$ . Let $P = (M_{AD}M_{BD}M_{CD}) \cap (M_{AB}M_{BD}M_{BC})$ so
$\measuredangle M_{AD}E - \measuredangle EM_{AB} = \measuredangle M_{AD}E - \measuredangle EM_{BD} + \measuredangle EM_{BD} - \measuredangle EM_{AB} = \measuredangle M_{AD}M_{CD} - \measuredangle M_{CD}M_{BD} + \measuredangle M_{BD}M_{BC} - \measuredangle M_{AB}M_{BC} = \measuredangle AC - \measuredangle BC + \measuredangle CD - \measuredangle AC = \measuredangle DCB = \measuredangle M_{AD}M_{AC}M_{AB}$ so $P$ lies on $(M_{AD}M_{AC}M_{AB})$ and we are done by symmetry.

This post has been edited 1 time. Last edited by YaoAOPS, Mar 10, 2025, 6:58 PM

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49 posts

ND_

#4 h Mar 12, 2025, 9:25 AM

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Z is simply the Gergonne-Steiner Point.

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FKcosX

#5 h Mar 17, 2025, 1:31 PM

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Suppose the four given points are
$A,\; B,\; C,\; D,$
with no three collinear and not all four concyclic. (Thus, for each vertex the circle through the other three is well–defined.) We shall now define for each vertex $X$ the “vertex–circle”
$\omega_X=(\text{three points other than }X);$
in other words,
$\omega_A=(BCD),\quad \omega_B=(ACD),\quad \omega_C=(ABD),\quad \omega_D=(ABC).$ Next, for each vertex $X$ we perform the homothety (dilation) $h_X$ with center $X$ and ratio $1/2$ . (A homothety with ratio $1/2$ sends every point $Y$ to the midpoint of $XY$ .) Denote
$\omega'_X = h_X(\omega_X).$ Thus, for example,
$\omega'_A \text{ is the circle through } M_{AB},\; M_{AC},\; M_{AD},$ where $M_{AB}$ is the midpoint of $AB$ , etc. (Similarly, $\omega'_B$ passes through $M_{BA},M_{BC},M_{BD}$ ; note that $M_{BA}=M_{AB}$ .)

Observe that if $X$ and $Y$ are two distinct vertices then by construction the circle $\omega'_X$ contains the midpoint $M_{XY}$ (since $M_{XY}=h_X(Y)$ ) and similarly $\omega'_Y$ contains $M_{XY}$ (since $M_{XY}=h_Y(X)$ ). Hence for any two vertices $X$ and $Y$ the circles $\omega'_X$ and $\omega'_Y$ have the point $M_{XY}$ in common.

For each vertex $X$ , note that the homothety $h_X$ with center $X$ and ratio 2 sends $\omega'_X$ back to the circumcircle $\omega_X$ . (Indeed, if a point $P$ lies on $\omega'_X$ then by definition $P=h_X(Q)$ for some $Q\in\omega_X$ ; applying the inverse homothety of ratio 2 shows that $2P-X=Q\in\omega_X$ .) In particular, if a point $Z$ lies on $\omega'_X$ then the point
$X' = 2Z-X$ lies on $\omega_X$ . (This is the same as saying that reflecting $X$ in $Z$ lands on $\omega_X$ .)

Thus, if we can show that there exists a point $Z$ lying simultaneously on all four circles $\omega'_A,\omega'_B,\omega'_C,\omega'_D$ then for each vertex $X$ the reflection of $X$ in $Z$ lies on $\omega_X$ (that is, on the circle through the three vertices other than $X$ ). This is exactly the property desired in the original problem.

We now prove that the four circles $\omega'_A,\omega'_B,\omega'_C,\omega'_D$ have a common point. The proof will use two key facts:

1. Common “side–midpoints”: As we have shown earlier, for any two vertices $X$ and $Y$ the circles $\omega'_X$ and $\omega'_Y$ meet at the midpoint $M_{XY}$ .

2. The Miquel configuration for the complete quadrilateral:
In any quadrilateral
$ABCD$
the four circles
$\omega_A,\omega_B,\omega_C,\omega_D$
(i.e. the circumcircles of the triangles determined by three of the four vertices) have a well‐known associated configuration: The three circles determined by the triangles
$ABC$ , $BCD$ ,
and
$CDA$
have a common “Miquel point”. (In our situation the four points are not concyclic so these circles are distinct and their “Miquel point” is defined by three of them.)

Now, fix two vertices, say $A$ and $B$ . The circles $\omega'_A$ and $\omega'_B$ meet in $M_{AB}=H$ and in a second point, say $Z_{AB}$ . (Because two distinct circles have two intersections unless they are tangent, and in our configuration tangency is avoided by the general‐position hypothesis.) Next, consider the circles $\omega'_A$ and $\omega'_C$ ; they meet in $M_{AC}=I$ and in a second point $Z_{AC}$ . One now shows by using the homothetic relationships
$h_A(\omega'_A)=\omega_A,\quad h_B(\omega'_B)=\omega_B,\quad h_C(\omega'_C)=\omega_C,$ and the standard fact about Miquel points in the complete quadrilateral $ABCD$ (namely, that the circles $\omega_A,\omega_B,\omega_C,\omega_D$ “rotate” about a unique Miquel point when three are taken at a time) that the second intersections $Z_{AB}$ and $Z_{AC}$ must in fact coincide $=Z.$ (One way to see this is to “lift” the configuration via the homothety of ratio 2 from each vertex. For example, applying the homothety with center $A$ sends the pair $\omega'_A$ and $\omega'_B$ to $\omega_A$ and a circle through the image of $M_{AB}=H$ ; the unique intersection point of the corresponding circumcircles coming from the Miquel configuration forces the corresponding “half‐points” to agree.)

A similar argument shows that the second intersection of $\omega'_B$ and $\omega'_C$ agrees with these, and then by symmetry the same point $Z$ lies on all four circles.

Thus, we obtain a unique point $Z$ such that
$Z\in\omega'_A\cap\omega'_B\cap\omega'_C\cap\omega'_D.$

Because the homothety with center $X$ (and ratio 2) sends $\omega'_X$ to $\omega_X$ , the fact that
$Z\in\omega'_X$ implies that the point
$X' = 2Z - X,$ i.e. the reflection of $X$ in $Z$ , lies on $\omega_X$ . Since this holds for every vertex $X\in\{A,B,C,D\}$ , the point $Z$ is exactly the desired point with the property that reflecting any one of $A$ , $B$ , $C$ , $D$ in $Z$ lands on the circle determined by the three remaining vertices.

This completes the proof of the concurrency of the four circles.

Hence, such a unique point $Z$ that satisfies the question conditions surely exists and is constructed as said in the solution.

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130 posts

NicoN9

#6 h Apr 13, 2025, 8:53 AM

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I believe that the claim in here could easily be proved by Miquel point, but I didn't find it so I simply angle chased.

Let $A$ , $B$ , $C$ , $D$ be points with counterclockwise, and let the midpoint of 6 segments $AB$ , $BC$ , $CD$ , $DA$ , $AC$ , $BD$ be $A_1$ , $B_1$ , $C_1$ , $D_1$ , $X$ , $Y$ , respectively. We start by the following claim.

claim. Four circles $A_1XD_1$ , $A_1YB_1$ , $B_1XC_1$ , $C_1YD_1$ are concurrent.
proof.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.2523573454339205, xmax = 11.33924787294384, ymin = -6.9289892229154875, ymax = 6.01258763471356; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pair A = (1.644409529211571,3.883035734543391), B = (-2.147249007373794,-2.3643788996029493), C = (5.712750992626203,-2.3643788996029493), D = (4.3927509926262065,2.3156211003970504), A_1 = (-0.2514197390811115,0.7593284174702208), B_1 = (1.7827509926262046,-2.3643788996029493), C_1 = (5.052750992626205,-0.024378899602949478), D_1 = (3.0185802609188888,3.099328417470221), X = (3.678580260918887,0.7593284174702208), Y = (1.1227509926262063,-0.024378899602949478), K = (1.538042621190102,-0.6133811946579423); draw(A--B--C--D--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw(circle((1.7135802609188875,1.4681745713163756), 2.0889441997866967), linewidth(2.) + dotted); draw(circle((-0.820796891657402,-1.8356359951701189), 2.6566954369565416), linewidth(2.) + dotted); draw(circle((3.4495651782426715,-1.2388371846310886), 2.0112467977635373), linewidth(2.) + dotted); draw(circle((3.0877509926262054,0.9201889478083394), 2.180237009676515), linewidth(2.) + dotted); draw(A--B, linewidth(2.) + zzttqq); draw(B--C, linewidth(2.) + zzttqq); draw(C--D, linewidth(2.) + zzttqq); draw(D--A, linewidth(2.) + zzttqq); draw(A--C, linewidth(2.)); draw(B--D, linewidth(2.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (1.4470187180941552,4.088594441293252), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.4917220646625085,-2.572778218944981), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (5.785078842881449,-2.1916097560975616), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (4.62342257515598,2.34611003970505), NE * labelscalefactor); dot(A_1,linewidth(4.pt) + dotstyle); label("$A_1$", (-0.7855394214407279,1.0210958593306874), NE * labelscalefactor); dot(B_1,linewidth(4.pt) + dotstyle); label("$B_1$", (1.7555836642087326,-2.899494044242769), NE * labelscalefactor); dot(C_1,linewidth(4.pt) + dotstyle); label("$C_1$", (5.240552467385136,-0.01350425411230805), NE * labelscalefactor); dot(D_1,linewidth(4.pt) + dotstyle); label("$D_1$", (3.0624469653998836,3.308106636415203), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (3.842934770277932,0.8758888258650037), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (1.1203028927963674,0.1680045377197964), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (1.1747555303459987,-0.8847464549064095), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $K$ be the intersection of first two circles, different from $A_1$ . We have $\begin{align*} \measuredangle B_1KX &= \measuredangle B_1KA_1+\measuredangle A_1KX \\ &= \measuredangle B_1YA_1+\measuredangle A_1D_1X \\ &= \measuredangle CDA + \measuredangle BCD \\ &= \measuredangle BDA = \measuredangle B_1C_1X \end{align*}$ thus $K$ also lies on circle $B_1XC_1$ . Similarly, for circle $C_1YD_1$ as well. $\blacksquare$

Now, the point $K$ in the claim is the desired point. This is since for example, $Z$ lies on circle $D_1XA_1$ , to reflect $A$ to circle $BCD$ , and same for $B$ , $C$ , $D$ . (or simply take the homothety to $Z$ with ratio $2$ , wrt $A$ then $Z$ must lie on circle $BCD$ .) So we are done.

This post has been edited 3 times. Last edited by NicoN9, Apr 13, 2025, 10:56 AM
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