Locus of the circumcenter of triangle PST

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Locus of the circumcenter of triangle PST

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Source: USA TSTST 2013, Problem 4

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#1 h Aug 13, 2013, 6:05 AM • 4 Y

Y by Amir Hossein, starchan, Adventure10, and 1 other user

Circle $\omega$ , centered at $X$ , is internally tangent to circle $\Omega$ , centered at $Y$ , at $T$ . Let $P$ and $S$ be variable points on $\Omega$ and $\omega$ , respectively, such that line $PS$ is tangent to $\omega$ (at $S$ ). Determine the locus of $O$ -- the circumcenter of triangle $PST$ .

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#2 h Aug 13, 2013, 4:46 PM • 4 Y

Y by Dukejukem, Amir Hossein, Adventure10, Mango247

Let $ST,PS$ meet $\Omega$ at $M,K$ , Hence $YM\parallel XS$ , which means that $YM\perp PK$ $\Rightarrow$ $M$ is the midpoint of arc $PK$ .
Since $\angle POT=2(180-\angle PST)=\angle SXT=\angle MYT$ , hence $\triangle POT\sim \triangle SXT\sim \triangle MYT$ . By their spiral similarity transformation, we obtain $PTSM\sim OTXY$ . Since $PM^2=MS*MT$ ( $\angle MPK=\angle MKP=\angle PTM$ ). Hence from $PTSM\rightarrow OTXY$ (just angle chasing) we have $OY^2=YX*YT=$ constant. Since $Y$ is fixed and the value of $OY$ is constant, hence the locus of $O$ is a circle centered at $Y$ with radius $\sqrt {YX*YT}$ .
Note that $\angle PMT$ varies from $0$ to $180$ depending on which direction $P$ approaches $T$ , this means that $\angle OYT$ also satisfies this, hence $O$ only travels on the half circle with diameter on $TY$ .

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#3 h Aug 23, 2013, 9:59 PM • 2 Y

Y by Adventure10, Mango247

This problem is very similar to problem 5 of Turkey Team Selection Test 2002 at this link http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3006277&sid=52a1bd7941586703ea04862b89c35c9e#p3006277
where it is asked to prove that $PM^2 = MS*MT$ , and with triangles $MPS$ and $YOX$ similar, we have $YO^2 = YX*YT$ .

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#4 h Apr 8, 2016, 8:37 PM • 2 Y

Y by Amir Hossein, Adventure10

Let $P'$ be the second intersection of $PS$ with $\Omega$ . Let $M$ be the second intersection of $TS$ with $\Omega$ . There is a homothety centered at $T$ that sends $S$ to $M$ and $X$ to $Y$ , so $\triangle TXS \sim \triangle TYM$ . Also, $M$ is the midpoint of arc $PP'$ . So angle chasing yields that $\triangle TP'M \sim \triangle TSP$ . Therefore, $\triangle TXS \sim \triangle TYM \sim \triangle TOP$ . Therefore there is a spiral similarity centered at $T$ mapping $\triangle OXY$ to $\triangle PSM$ . Since $\triangle PSM \sim \triangle TPM$ , we have that $\triangle OXY\sim \triangle TOY$ . Therefore $OY^2=YX\cdot YT$ which is fixed. It is easy to verify that the locus is then a circle centered at $Y$ with radius $\sqrt{YX\cdot YT}$ . $\Box$

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#5 h Jun 26, 2018, 7:53 AM • 5 Y

Y by Amir Hossein, v_Enhance, CyclicISLscelesTrapezoid, Adventure10, Mango247

$[asy] unitsize(100); pair T = (0, 0), X = (0, 0.6), Y = (0, 1); pair O = Y + dir(230) * sqrt(abs(Y-X)*abs(Y-T)); pair S = reflect(X, O) * T; pair P = reflect(Y, O) * T; pair H = extension(Y, O, T, P); draw(P--O--T^^S--X--T^^X--Y, gray(0.6)); draw(circle(X, abs(X-T))); draw(circle(Y, abs(Y-T))); draw(H--Y^^O--X); draw(circumcircle(O, X, T), dotted); draw(P--S--T--cycle); dot(T^^X^^Y^^O^^S^^P^^H); label("$T$", T, dir(270)); label("$X$", X, dir(50)); label("$Y$", Y, dir(60)); label("$S$", S, dir(S-X)); label("$P$", P, dir(P-Y)); label("$O$", O, dir(110)); label("$H$", H, dir(H-Y)); [/asy]$
The answer is the circle with center $Y$ and radius $\sqrt{YX \cdot YT}$ , minus the two points on line $XY$ .

First we show $O$ must lie on this circle. Let $H = \overline{YO} \cap \overline{TP}$ . Then
$\measuredangle HOT = \measuredangle PST = \measuredangle OXT,$ so $\overline{YO}$ is tangent to $(OXT)$ , giving $YO^2 = YX \cdot YT$ .

For the other direction, suppose that $YO^2 = YX \cdot YT$ but $O$ is not on line $XY$ . Then let the circle with center $O$ and radius $OT$ intersect $\Gamma$ and $\omega$ at $P \neq T$ and $S \neq T$ . Then reversing the angle chase above gives $\overline{PS}$ tangent to $\omega$ , as desired. If $O$ is on line $XY$ , then we get $P = S = T$ contradiction.

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#6 h Jul 16, 2018, 7:19 PM • 2 Y

Y by Adventure10, Mango247

v_Enhance wrote:

Circle $\omega$ , centered at $X$ , is internally tangent to circle $\Omega$ , centered at $Y$ , at $T$ . Let $P$ and $S$ be variable points on $\Omega$ and $\omega$ , respectively, such that line $PS$ is tangent to $\omega$ (at $S$ ). Determine the locus of $O$ -- the circumcenter of triangle $PST$ .

We will prove that $O$ lies on a circle centered at $Y$ with radius $\sqrt{YX \cdot YT}$ . The other direction is just reversing the angle-chase.

Extend $\overline{PS}$ to meet $\Omega$ again at $Q \ne P$ . Then $\angle OTP=90^{\circ}-\angle TSQ$ and $\angle YTP=-90^{\circ}-\angle TQP$ hence $\angle YTO=\angle QTS$ . Moreover $\overline{XO} \perp \overline{TS}$ and $\overline{YO} \perp \overline{TP}$ so $\angle XOY=\angle PTS$ . Now since $\overline{TS}$ bisects angle $PTQ$ ; we conclude that $\angle YOX=\angle YTO$ as desired. $\blacksquare$

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#7 h Feb 27, 2021, 3:07 PM • 3 Y

Y by centslordm, Imayormaynotknowcalculus, HamstPan38825

Solution from Twitch Solves ISL:

The answer is a circle centered at $Y$ with radius $\sqrt{YX \cdot YT}$ , minus the two points on line $XY$ itself.
We let $PS$ meet $\Omega$ again at $P'$ , and let $O'$ be the circumcenter of $\triangle TPS'$ . Note that $O'$ , $X$ , $O$ are collinear on the perpendicular bisector of line $\overline{TS}$ Finally, we let $M$ denote the arc midpoint of $PP'$ which lies on line $TS$ (by homothety).
$[asy] import graph; size(9cm); pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffdxqq = rgb(1.,0.84313,0.); pen yqqqqq = rgb(0.50196,0.,0.); pair X = (-2.,0.), Y = (2.,0.), T = (-9.50185,0.), P = (7.49229,10.10580), O = (5.46686,-5.82994), M = (-0.78960,11.15843); pair S = (-3.81946,7.27786); pair Pp = (-7.60181,6.33228); pair Op = (-4.49671,1.94937); fill(T--S--P--cycle, palegreen); fill(T--Op--Y--cycle, palegreen); fill(T--S--Pp--cycle, palecyan); fill(T--O--Y--cycle, palecyan); draw(circle(X, 7.50185), linewidth(0.6) + red); draw(circle(Y, 11.50185), linewidth(0.6) + red); draw(circle(Y, 6.78287), linewidth(0.6) + qqwuqq); draw(T--Op--Y--O--T--Y, grey); draw(T--Pp--M--P--T--M, black); draw(Pp--P, black); draw(Op--O, grey); dot("$X$", X, dir((8.005, 21.422))); dot("$Y$", Y, dir((8.687, 21.422))); dot("$T$", T, dir(T-Y)); dot("$S$", S, dir(135)); dot("$P$", P, dir(P-Y)); dot("$O$", O, dir(-45)); dot("$P'$", Pp, dir(Pp-Y)); dot("$O'$", Op, dir(40)); dot("$M$", M, dir((8.912, 22.013))); [/asy]$
By three applications of Salmon theorem, we have the following spiral similarities all centered at $T$ : $\begin{align*} \triangle TSP &\overset{+}{\sim} \triangle TO'Y \\ \triangle TP'S &\overset{+}{\sim} \triangle TYO \\ \triangle TP'P &\overset{+}{\sim} \triangle TO'O. \end{align*}$ However, the shooting lemma also gives us two similarities: $\begin{align*} \triangle TP'M &\overset{+}{\sim} \triangle TSP \\ \triangle TMP &\overset{+}{\sim} \triangle TP'S. \end{align*}$ Putting everything together, we find that $TP'MP \overset{+}{\sim} TO'YO.$ Then by shooting lemma, $YO' = YX \cdot YT$ , so $O$ indeed lies on the claimed circle.
As the line $\overline{O'O}$ may be any line through $X$ other than line $XY$ (one takes $S$ to be the reflection of $T$ across this line) one concludes the only two non-achievable points are the diametrically opposite ones on line $XY$ of this circle (because this leads to the only degenerate situation where $S=T$ ).

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#8 h Sep 20, 2021, 7:52 PM • 1 Y

Y by centslordm

The answer is a circle centered at $A$ with radius $\sqrt{YX \cdot YT}$ . It suffices to show that the length $YO$ , or the ratio $\frac{YO}{YT}$ , is constant. Denote by $R$ the intersection of $\overline{SP}$ with $\Omega$ , while denote $M = \overline{ST} \cap \Omega$ , which is the midpoint of arc $\widehat{RP}$ by the Shooting Lemma.

Claim. A spiral similarity at $T$ sends $\overline{RS} \to \overline{YO}$ .

We will show that $\triangle TSO \sim \triangle TRY$ , which suffices. Notice that both triangles are isosceles by definition; then $\angle SOT = 2\angle RPT = \angle RYT,$ implying the result. $\blacksquare$

Henceforth, $\frac{YO}{YT} = \frac{RS}{RT} = \frac{MS}{MP} = \sqrt{\frac{MS}{MT}}$ by the Shooting Lemma. But $\overline{SX}$ and $\overline{MY}$ are parallel, because they are both perpendicular to $\overline{RS}$ (by definition of tangency and arc midpoints, respectively.) This implies that $\frac{MS}{MT} = \frac{YX}{YT}$ is constant, from where the desired result follows.

This post has been edited 1 time. Last edited by HamstPan38825, Sep 20, 2021, 7:53 PM

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#9 h Oct 3, 2021, 3:00 PM • 1 Y

Y by centslordm

We first fix $S$ and let the tangent from $X$ with respect to $\omega$ hit $\Omega$ at $P_1$ and $P_2$ . Let $O_1$ and $O_2$ be the circumcenters of $(TSP_1)$ and $(TSP_2)$ . Notice that $\measuredangle{TO_2P_2} = 2\measuredangle{TSP_2} = 2\measuredangle{TSP_1} = \measuredangle{TO_1P_1}$ and the fact that $TO_1P_1$ and $TO_2P_2$ are both isosceles gives $T$ to be the Miquel Point of $O_1O_2P_2P_1$ . Additionally, $\measuredangle{TO_2P_2} = 2\measuredangle{TSP_2} = \measuredangle{TXS}$ so $T$ is the Miquel Point of $O_1XSP_1$ and $XO_2P_2S$ as well. Finally, we also have $TYP_1$ and $TO_2S$ both being isosceles and $\measuredangle{TYP_1} = 2\measuredangle{TP_2S} = \measuredangle{TO_2S}$ so $T$ is the Miquel Point of $YO_2SP_1$ and $YO_1SP_2$ . Thus, we have $\measuredangle{O_1TY} = \measuredangle{O_1TP_1} + \measuredangle{P_1TY} = \measuredangle{XTS}+\measuredangle{STO_1} = \measuredangle{YTO_2}$ and $2\measuredangle{TYO_1} = 2\measuredangle{TP_2S} = \measuredangle{TO_2S} = 2\measuredangle{TO_2O_1}$ so $T, O_1, O_2, Y$ are conyclic with $YO_1 = YO_2$ . Finally, notice that $\measuredangle{XO_2Y} = \measuredangle{O_1O_2Y} = \measuredangle{O_1TY} = \measuredangle{YTO_2}$ so $O_2YX \sim TYO_2 \implies YO_2^2 = YX\cdot YT$ . Because $YX$ and $YT$ are fixed, $O_1$ and $O_2$ lie on a fixed circle centered at $Y$ with radius $\sqrt{YX\cdot YT}$ . $\blacksquare$

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#10 h Feb 17, 2022, 1:35 AM • 1 Y

Y by centslordm

Let $Q=\Omega\cap\overline{PS}$ ; we claim $Q$ lies on $\gamma,$ the circle with center $Y$ and radius $\sqrt{YX\cdot YT}.$ Let $Z$ be the center of $(PST)$ and note $\angle YZT=\tfrac{1}{2}\angle PZT=\angle QST=\tfrac{1}{2}\angle SXT=180-\angle TXZ$ as $\overline{XZ}$ is the perpendicular bisector of $\overline{ST}.$ Hence, $\triangle XYZ\sim\triangle ZYT$ and $YZ^2=YX\cdot YT.$

Suppose $Z'$ lies on $\gamma$ on the opposite side of $\overline{XY}$ as $S.$ Let the circle with radius $\overline{ZT}$ and center $Z$ intersect $\omega$ and $\Omega$ at $S'$ and $P'.$ Then, since $YZ^2=YX\cdot YT,$ we know $\triangle XYZ'\sim\triangle Z'YT.$ Thus, $180-\angle TSP=\tfrac{1}{2}\angle P'Z'T=\angle YZ'T=\angle Z'XY=\tfrac{1}{2}\angle S'XT$ and $S=S'.$ $\square$

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#11 h May 24, 2023, 12:32 AM

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Note that $XO\perp ST$ and $YO\perp PT$ since $XS=XT$ and $YP=YT$ .

Let $\angle PST=\alpha$ . Then, $\angle TSX=90-\alpha$ , so $\angle SXO=\angle TXO=\alpha,$ so $\angle XTS=90-\alpha$ . Then, $\angle PTO=90-\alpha.$ Thus, $TS$ and $TO$ are isogonal in $\triangle TFX$ , so $\angle FTS=\angle XTO$ . However, if we let $E$ and $F$ be the feet from $O$ to $PT$ and $ST$ , then from $FOET$ cyclic, $\angle YOX=\angle FTS$ . Thus, $\angle XTO=\angle YOX$ , so $(OTX)$ is tangent to $OY$ , thus $OY=\sqrt{YX\cdot YT}$ so the locus of $O$ is the circle centered at $Y$ with radius $\sqrt{YX\cdot YT}.$

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#12 h Sep 25, 2023, 6:29 PM • 1 Y

Y by asdf334

We present a motivated solution (what's spiral similarity?). Let $\measuredangle$ denote directed angles modulo $180^\circ$ .

For a point $O$ , let $\omega_1$ be the circle through $T$ centered at $O$ , and redefine $P$ and $S$ as the second intersections of $\omega_1$ with $\Omega$ and $\omega$ , respectively. Why is this helpful? We want to find a necessary and sufficient condition for $O$ that implies $\overline{PS}$ is tangent to $\omega$ . This is equivalent to $\measuredangle TXS=\measuredangle TOP$ , which is equivalent to $\measuredangle TXO=\measuredangle TOY$ . Notice that this is exactly the condition for $\overline{OY}$ to be tangent to the circumcircle of $OTX$ , so $YO^2=YT \cdot YX$ . Thus, it is equivalent for $O$ to lie on the circle centered at $Y$ with radius $\sqrt{YT \cdot YX}$ . Other than the points on $\overline{XY}$ , all choices of $O$ on this circle create a nondegenerate configuration, so this is the locus of $O$ .

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#13 h Jan 2, 2024, 3:13 PM

Y by

Let $Q = PS \cap \Omega$ , and $M = TS \cap (ABC)$ , which is evidently the bottom point. Consider the two spiral similarities at $T$ :

$S_T: \triangle TMQ \mapsto \triangle TPS \implies \triangle TOY \sim \triangle TPM$
$S_T: \triangle TOP \mapsto \triangle TXS \implies \triangle TOX \sim \triangle TPS$

These imply $\triangle YOX$ is similar to $\triangle MPS$ , so angle chasing gives
$\angle YOX = \angle SPM = \angle MTP = \angle YTO \implies \triangle YOX \sim \triangle TYO,$
implying the length condition $YO^2 = YX \cdot YT$ , which is fixed. Thus the locus of $O$ is a circle centered at $Y$ with radius $\sqrt{YX \cdot YT}$ , but it can be seen that it must exclude the points on line $XY$ . $\blacksquare$

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#14 h Dec 20, 2024, 6:01 AM

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Subjective Rating (MOHs) $\text{[math]}$

I thought the hint meant to look at the two tangents from $P$ ;-;

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