center of (XYZ) lies on a fixed circle

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center of (XYZ) lies on a fixed circle

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Source: All-Russian 2022 11.3

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1386 posts

#1 h Apr 19, 2022, 5:35 PM • 4 Y

Y by buratinogigle, ImSh95, Rounak_iitr, Funcshun840

An acute-angled triangle $ABC$ is fixed on a plane with largest side $BC$ . Let $PQ$ be an arbitrary diameter of its circumscribed circle, and the point $P$ lies on the smaller arc $AB$ , and the point $Q$ is on the smaller arc $AC$ . Points $X, Y, Z$ are feet of perpendiculars dropped from point $P$ to the line $AB$ , from point $Q$ to the line $AC$ and from point $A$ to line $PQ$ . Prove that the center of the circumscribed circle of triangle $XYZ$ lies on a fixed circle.

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678 posts

#3 h Apr 19, 2022, 9:03 PM • 4 Y

Y by guptaamitu1, ImSh95, SerdarBozdag, Mahdi_Mashayekhi

Beautiful problem.

Let $O, M, N$ be the center of $(XYZ)$ , and the midpoints of $\overline{AB}, \overline{AC}$ , respectively.. Observe that $APXZ$ and $AQCZ$ are cyclic, and the perpendicular bisectors $p_1, p_2$ of $\overline{XZ}, \overline{ZY}$ bisect $\overline{AP}, \overline{AQ}$ , respectively. Moreover $\widehat{PZX} + \widehat{BPZ} = \widehat{PAB} + \widehat{BAQ} = \frac{\pi}{2}\to XZ\perp{BP}\hspace{0.1 cm} (\text{analogously}\hspace{0.1 cm} ZY\perp{CQ}).$ So, $p_1\parallel BP$ and $p_2\parallel CQ\to$ $M\in p_1$ and $N\in p_2 \implies \widehat{MON} = \frac{\pi}{2} - \widehat{A}$ and hence there's a fixed circle passing through $M, N$ and $O$ .

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892 posts

#4 h Apr 20, 2022, 4:12 PM • 1 Y

Y by ImSh95

Other way of proving $M \in p_1$ is reflecting $X$ across $M$ (call it $X'$ ) and writing $AX'/AZ=BX/AZ=PX/PZ$ which implies $XPZ \sim X'AZ \implies \angle X'ZX=90$ .

Nevertheless I could not see the fixed circle. Very hard and nice problem.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h May 15, 2022, 5:39 AM • 1 Y

Y by Mango247

Let $l_1$ and $l_2$ be perpendicular bisectors of $XZ,YZ$ and $O'$ be center of $XYZ$ .
Claim $: l_1$ and $l_2$ bisect $AB,AC$ .
Proof $:$ Note that $AP,AQ$ are diameters of $APXZ$ and $AQYZ$ so $l_1$ and $l_2$ bisect $AP,AQ$ so we need to prove $l_1 || PB$ and $l_2 || QC$ . Note that $\angle PZX = \angle PAX = \angle PAB = \angle PQB \implies XZ || BQ$ and Note that $l_1 \perp XZ$ and $PB || BQ$ so $PB || l_1$ . we prove the other one with same approach.
Now if $M,N$ be midpoints of $AB,AC$ then $\angle MO'N = \frac{\angle XO'Y}{2} = \angle ZXY + \angle ZYX = \angle PZX + \angle QZY = \angle 90 - \angle A$ so $\angle MO'N$ is fixed so $O'$ lies on $MO'N$ which is fixed.

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1162 posts

#6 h Jul 12, 2022, 3:55 AM • 1 Y

Y by Mango247

Let the midpoints of $AP, AQ, AZ, XZ, YZ$ be $M, N, F, D, E$ respectively. In addition, define the circumcenters of $ABC$ and $XYZ$ as $O$ and $O_1$ respectively, and let the circle with diameter $AO$ be $\omega$ .

Because $APXZ$ is cyclic with diameter $AP$ and $AQYZ$ is cyclic with diameter $AQ$ , we know $MD$ and $NE$ are the perpendicular bisectors of $XZ$ and $YZ$ respectively. It follows that $O_1 = MD \cap NE$ .

Since $MN$ is the $A$ -midline of $APQ$ , we have $F \in MN$ and $AF \perp MN$ . Thus, $DMFZ$ is cyclic with diameter $MZ$ , so $\angle O_1MN = \angle DMF = 180^{\circ} - \angle DZF = 180^{\circ} - \angle XZA = \angle XPA.$ Analogously, we find $\angle O_1NM = \angle YQA$ , implying $\angle MO_1N = 180^{\circ} - \angle O_1MN - \angle O_1NM$ $= 180^{\circ} - \angle XPA - \angle YQA$ $= 180^{\circ} - (90^{\circ} - \angle XAP) - (90^{\circ} - \angle YAQ)$ $= \angle PAQ - \angle XAY = 90^{\circ} - \angle BAC$ which is fixed.

Since $\angle AMO = \angle ANO = 90^{\circ}$ and $\angle MAN = 90^{\circ}$ , we know $MN$ is a diameter of $\omega$ . Now, define $R$ as the projection of $M$ onto $NO_1$ and $S$ as the projection of $N$ onto $MO_1$ . It's clear that $R$ and $S$ both lie on $\omega$ , so $\angle MAS = \angle MNS = 90^{\circ} - \angle O_1MN = 90^{\circ} - \angle XPA = \angle MAX$ which means $S \in AB$ . But $S \in \omega$ gives $\angle ASO = 90^{\circ}$ , so $S$ is actually the midpoint of $AB$ . Similarly, we deduce that $R$ is the midpoint of $AC$ .

Thus, $\angle SO_1R = \angle MO_1N$ is fixed, which finishes since $R$ and $S$ are both fixed points. $\blacksquare$

Remarks: What an amazing problem! It might be too hard for its placement, however. My GGB can be found here.

Discovering $\angle MO_1N = 90^{\circ} - \angle A$ is easy, but noticing that the midpoints of $AB$ and $AC$ lie on $O_1M$ and $O_1N$ respectively requires a lot more insight. In fact, it actually took me an hour to even construct the two midpoints!

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#7 h Jan 2, 2023, 1:04 AM • 1 Y

Y by Mango247

sorry

Let $O_1$ be the circumcenter of $XYZ$ , and let $O'$ be the reflection of $O$ over $\overline{BC}$ . Notice that $AZXP$ and $AZYQ$ are cyclic, so $\angle XZY=180^\circ-\angle PZX-\angle QZY=180^\circ-\angle PAX-\angle QAY=180^\circ-(\angle PAQ-\angle BAC)=\angle BAC+90^\circ.$ Thus, $\angle XO_1Y=180^\circ-2\angle BAC$ . We have $\angle OBO'=\tfrac{360^\circ-2\angle BOC}{2}=180^\circ-2\angle BAC$ , so $\angle XO_1Y=\angle OBO'$ . We also have $O_1X=O_1Y$ and $BO=BO'$ , so $\triangle XYO_1 \sim \triangle OO'B$ .

Let $A=a$ , $B=b$ , and $C=c$ be points on the unit circle, and let $P=p$ , $Q=-p$ . Then, $O'=b+c$ . We have $X=\frac{\left(\frac{1}{a}-\frac{1}{b}\right)p+(a-b)\frac{1}{p}+\frac{b}{a}-\frac{a}{b}}{2\left(\frac{1}{a}-\frac{1}{b}\right)}=\frac{(b-a)p^2+ab(a-b)+(b^2-a^2)p}{2(b-a)p}=\frac{p^2+(a+b)p+ab}{2p},$ and similarly, $Y=\frac{-p^2+(a+b)p-ab}{2p}.$ Since $\tfrac{\overrightarrow{XO_1}}{\overrightarrow{XY}}=\tfrac{\overrightarrow{OB}}{\overrightarrow{OO'}}=\tfrac{b}{b+c}$ , we have
$\begin{align*} O_1 &= X+\frac{b}{b+c}(Y-X) \\ &= \frac{p^2+(a+b)p+ab}{2p}+\frac{b}{b+c} \cdot \frac{-2p^2-a(b+c)+(c-b)p}{2p} \\ &= \frac{p}{2}+\frac{a+b}{2}+\frac{ab}{2p}-\frac{bp}{b+c}-\frac{ab}{2p}+\frac{b(c-b)}{2(b+c)} \\ &= \left(\frac{b}{b+c}+\frac{1}{2}\right)p+\frac{a+b}{2}+\frac{b(c-b)}{2(b+c)}. \end{align*}$ Thus, $O_1$ lies on the fixed circle centered at $\tfrac{a+b}{2}+\tfrac{b(c-b)}{2(b+c)}$ with radius $|\tfrac{b}{b+c}+\tfrac{1}{2}|$ . $\square$

This post has been edited 2 times. Last edited by CyclicISLscelesTrapezoid, Nov 6, 2023, 5:23 AM

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#8 h Jan 2, 2023, 3:46 PM • 1 Y

Y by CyclicISLscelesTrapezoid

Let Mb, Mc be midpoints of AC, AB and R, S = AO \cap (AB), (AC). Then ZMcX ~ ZOR and ZMbY ~ ZOS by spiral similarity, and OZ = OR = OS so ZMb = MbY, ZMc = McX. If P’, Q’ are the midpoints of AP, AQ then ZX || McQ’ and ZY || MbP’, so P’Mc \cap Q’Mb is the circumcenter of XYZ, and P’, Q’, Mb, Mc are on (AO) the angle between these two lines is clearly fixed, meaning the circumcenter travels along a circle through Mb, Mc.

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#9 h Mar 18, 2023, 3:30 AM

Y by

Let $E,F$ denote the midpoints of $AB,AC$ and let $O'$ denote the circumcenter of $XYZ$ . The main claim is that $E,F$ lie on the perpendicular bisectors of $XY,XZ$ respectively. Note that $AXYP$ and $AXZQ$ are cyclic, so the midpoint of $AX$ lies on the perpendicular bisector of $XY$ . Hence, by midpoint theorem, we only need to show $BP\perp XY$ . We use complex numbers :

$x-y = \frac 12 (a+ \frac {p^2}a)- \frac 12 (a+b+p- \frac {ab}p)= \frac {(p^2-ab)(p-a)}{2ap}$
Then we have :

$\overline {\left( \frac {x-y}{b-p} \right) } = \overline {\left( \frac {(p^2-ab)(p-a)}{2ap(b-p)}\right)} = \frac {(ab-p^2)(a-p) \frac 1{ap} \frac 1{abp^2}}{\frac 1{ap} \frac 1{bp} (b-p)}= - \left( \frac {x-y}{b-p} \right)$
This proves $XY\perp BP$ , so the main claim is done. Finally note we have :

$\angle EO'F = \pi - \angle XYZ = \angle PXY+ \angle QXZ = \angle PAQ - A = \frac {\pi}2 -A$
which is fixed, so we're done.

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#10 h Mar 25, 2023, 6:17 PM

Y by

here's a hybrid solution containing both synthetic and (debatable) bash elements, found with mxlcv
solution

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mcmp

#11 h Nov 8, 2024, 8:14 AM

Y by

Alright here we go. I can’t find similar triangles as I am predominantly an angle/length/cross-ratio chase nerd, so here we go.

Russian geo :love:

:love:

once again. Let $O’$ be the circumcentre of $(XYZ)$ .

So first let’s try to eliminate $Z$ because it’s not too well-controlled w.r.t. $X$ and $Y$ . Construct $X’=\overline{AB}\cap\overline{QY}$ , and $Y’=\overline{AC}\cap\overline{PX}$ . Note $XYX’Y’$ is clearly cyclic because of the bunch of perpendicular lines. I claim now to eliminate $Z$ that $Z\in(XYX’Y’)$ .
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We angle-chase:
$\begin{align*} \measuredangle XZY&=\measuredangle XZA+\measuredangle AZY\\ &=\measuredangle XPA+\measuredangle AQY\\ &=90^\circ-\measuredangle PAX+90^\circ-\measuredangle YAQ\\ &=\measuredangle BAP+\measuredangle QAC\\ &=\measuredangle BAC-\measuredangle PAQ\\ &=90^\circ+\measuredangle BAC\\ \measuredangle XX’Y&=\measuredangle AX’Y\\ &=90^\circ-\measuredangle YAX’\\ &=90^\circ+\measuredangle BAC=\measuredangle XZY \end{align*}$ (see I told you I was an angle-chasing nerd) So that eliminates $Z$ . Note now that $O’$ is actually the midpoint of $X’Y’$ due to the right angles so we can essentially delete $X$ and $Y$ from the diagram; I choose not to do so however because it’s annoying.

Now dilate everything from $A$ ; let $O’\to K$ , and $R=\overline{BP}\cap\overline{CQ}$ , $R’=\overline{BQ}\cap\overline{CP}$ . I actually claim that (surprisingly) $K=R$ . It suffices to show that $AX’RY’$ is a parallelogram. Meanwhile I am going to try show how this finishes. Define $S$ to be the intersection of $\overline{AC}$ with the line perpendicular to $\overline{AB}$ at $B$ , and $T$ with $B\iff C$ . I claim now that $K=R\in(BCST)$ , which finishes. However it’s actually quite clear that $P$ , $Q$ , $R$ , $R’$ form an orthocentric system (USAMO 2024/5 vibes anyone?) from which it’s pretty clear due to the fact that the only circle that is orthogonal to $(ABC)$ is $(BCRR’)=(BCST)$ .
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draw((-6.505325726427008,7.594783551431004)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(-8.538272802918998,-5.20556409459472), linewidth(0.5) + ccqqqq); draw((-8.538272802918998,-5.20556409459472)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ccqqqq); draw((8.586879107882611,-5.124988505997758)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ccqqqq); draw((-11.647867456922755,11.928938787584304)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(-10.81510214329833,-19.541503984292923), linewidth(0.5) + ffvvqq); draw((2.50481026772986,9.681215085033234)--(-11.647867456922755,11.928938787584304), linewidth(0.5) + ffvvqq); draw((-10.81510214329833,-19.541503984292923)--(-2.50481026772986,-9.681215085033234), linewidth(0.5) + ffvvqq); draw((-5.960434633485641,11.025662544556566)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw(circle((0.09108898903745177,-19.35961997709105), 16.577188629599004), linewidth(0.5) + qqwuqq); draw((-8.538272802918998,-5.20556409459472)--(0,0), linewidth(0.5) + ffvvqq); draw((0,0)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-10.81510214329833,-19.541503984292923)--(-12.498926964267309,-30.14362114348741), linewidth(0.5) + ffvvqq); draw((8.586879107882611,-5.124988505997758)--(12.681104942342206,-8.575618810694674), linewidth(0.5) + ffvvqq); draw((-5.960434633485641,11.025662544556566)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ffvvqq); draw((-2.50481026772986,-9.681215085033234)--(4.346919174826464,-1.5515329060246381), linewidth(0.5) + ffvvqq); draw((2.50481026772986,9.681215085033234)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((8.586879107882611,-5.124988505997758)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw(circle((-6.726416803032106,-2.763066831553197), 15.494376581568185), linewidth(0.5) + qqwuqq); draw((-11.647867456922755,11.928938787584304)--(-15.957643873794087,-15.207348748139617), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((-6.505325726427008,7.594783551431004),dotstyle); label("$A$", (-6.220814181678762,8.275524895441126), NE * labelscalefactor); dot((-8.538272802918998,-5.20556409459472),dotstyle); label("$B$", (-8.228389352506392,-4.53142015983871), NE * labelscalefactor); dot((8.586879107882611,-5.124988505997758),dotstyle); label("$C$", (8.870612964542739,-4.46219342981017), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.28649844100390387,0.6605845923017646), NE * labelscalefactor); dot((2.50481026772986,9.681215085033234),dotstyle); label("$P$", (2.7786607220313075,10.352326796297316), NE * labelscalefactor); dot((-2.50481026772986,-9.681215085033234),dotstyle); label("$Q$", (-2.2056638400234996,-8.961930881665248), NE * labelscalefactor); dot((-5.960434633485641,11.025662544556566),dotstyle); label("$X$", (-5.667000341450449,11.73686139686811), NE * labelscalefactor); dot((4.346919174826464,-1.5515329060246381),dotstyle); label("$Y$", (4.64778243280186,-0.8624034683261078), NE * labelscalefactor); dot((-10.81510214329833,-19.541503984292923),dotstyle); label("$X'$", (-10.512871443448178,-18.861353275746417), NE * labelscalefactor); dot((-11.647867456922755,11.928938787584304),dotstyle); label("$Y'$", (-11.343592203790646,12.636808887239125), NE * labelscalefactor); dot((12.681104942342206,-8.575618810694674),dotstyle); label("$S$", (12.954990036226539,-7.854303201208613), NE * labelscalefactor); dot((-12.498926964267309,-30.14362114348741),dotstyle); label("$T$", (-12.243539694161653,-29.453042970112985), NE * labelscalefactor); dot((-15.957643873794087,-15.207348748139617),dotstyle); label("$R$", (-15.704876195588604,-14.50006928394842), NE * labelscalefactor); dot((16.13982185186898,-23.51189120604247),dotstyle); label("$R'$", (16.41632653765349,-22.807276887373177), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
So now that I’ve reduced it to showing that $K=R$ , notice that there is a surprising components in the setup; $PCRY’$ is cyclic. Note that with an angle chase similar to above, we can immediately get $\measuredangle PY’C=90^\circ+\measuredangle BAC=\measuredangle PRC$ , so combining this with the fact that $ABCP$ is cyclic, $\overline{AB}\parallel\overline{RY’}$ by Reim’s so we actually indeed have $AY’RX’$ a parallelogram by similar logic. Thus we are done!

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