Isogonal Conjugates and Circles

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Isogonal Conjugates and Circles

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Source: ELMO 2014 Shortlist G5, by Sammy Luo

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#1 h Jul 24, 2014, 1:34 PM • 2 Y

Y by narutomath96, Adventure10

Let $P$ be a point in the interior of an acute triangle $ABC$ , and let $Q$ be its isogonal conjugate. Denote by $\omega_P$ and $\omega_Q$ the circumcircles of triangles $BPC$ and $BQC$ , respectively. Suppose the circle with diameter $\overline{AP}$ intersects $\omega_P$ again at $M$ , and line $AM$ intersects $\omega_P$ again at $X$ . Similarly, suppose the circle with diameter $\overline{AQ}$ intersects $\omega_Q$ again at $N$ , and line $AN$ intersects $\omega_Q$ again at $Y$ .

Prove that lines $MN$ and $XY$ are parallel.

(Here, the points $P$ and $Q$ are isogonal conjugates with respect to $\triangle ABC$ if the internal angle bisectors of $\angle BAC$ , $\angle CBA$ , and $\angle ACB$ also bisect the angles $\angle PAQ$ , $\angle PBQ$ , and $\angle PCQ$ , respectively. For example, the orthocenter is the isogonal conjugate of the circumcenter.)

Proposed by Sammy Luo

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#2 h Jul 24, 2014, 2:09 PM • 2 Y

Y by mhq, Adventure10

If you extend $AP$ to meet $w_P$ at $P'$ , and let $l_P, l_Q$ be tangents at $A$ in obvious way, $P'X \parallel l_P \implies PM \parallel l_Q$ . Then invert about $A$ swapping $B, C$ and the result follows as it preserves point at infinity of corresponding lines, fyi $XY \mapsto MN$

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#3 h Jul 26, 2014, 8:30 AM • 5 Y

Y by narutomath96, rkm0959, Hasin_Ahmad, Adventure10, Mango247

It is well-known that for any pair of isogonal conjugate $P,Q$ with respect to $\triangle ABC$ , $\angle BPC+\angle BQC=180^\circ+\angle BAC$ ,etc.

Since $\angle BMC+\angle BNC=\angle BPC+\angle BQC=180^\circ+\angle BAC$ and,without loss of generality,

$\begin{align*}\angle AMC+\angle ANC& =360^\circ-(90^\circ+\angle BMP+\angle BMC)+(90^\circ+\angle CNQ)\\& =360^\circ-(90^\circ+\angle BCP+\angle BPC)+(90^\circ+\angle CBQ)\\& =360^\circ-(90^\circ+180^\circ-\angle CBP)+(90^\circ+\angle ABP)\\& =180^\circ+\angle CBP+\angle ABP\\& =180^\circ+\angle ABC\end{align*}$
we obtain that $M,N$ are isogonal conjugate with respect to $\triangle ABC$ ,which follows that $\triangle APM\sim \triangle AQN$ ,and then

$\begin{align*}\frac{PM}{QN}& =\frac{AP}{AQ}\\& =\frac{\sin \angle ABP\cdot BP}{\sin \angle BAP}\cdot \frac{\sin \angle CAQ}{\sin \angle ACQ\cdot CQ}\\& =\frac{\sin \angle CBQ}{CQ}\cdot \frac{BP}{\sin \angle BCP}\\& =\frac{\sin \angle BQC}{BC}\cdot \frac{BC}{\sin \angle BPC}\\& =\frac{\sin \angle BQC}{\sin \angle BPC}\\& =\frac{PX}{QY}\end{align*}$

so $\triangle XPM\sim \triangle YQN$ ,and hence $\frac{AM}{AN}=\frac{PM}{QN}=\frac{MX}{NY}$ ,therefore $MN\parallel XY$ ,as desired.

$Q.E.D.$

//cdn.artofproblemsolving.com/images/456396f789bc2d03f088ab3fcca2a4b81ce37839.jpg

//cdn.artofproblemsolving.com/images/456396f789bc2d03f088ab3fcca2a4b81ce37839.jpg

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#4 h Oct 24, 2014, 11:32 PM • 2 Y

Y by Adventure10, Mango247

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#5 h Mar 15, 2019, 6:39 AM • 3 Y

Y by Smkh, Adventure10, Mango247

It is easy when ones known $HM$ point
Clearly $N$ is the the $A-HM$ point of $\triangle ABC$ and through angle chasing one can show $M$ is the midpoint of the symmedian chord of $(ABC)$ , $X$ is the intersection of the tangents at $B$ and $C$ and $Y$ is a point such that $ABYC$ is a $\parallel$ gm. Consider an Inversion $\phi$ centered at $A$ with radius $\sqrt{AB.AC}$ followed by the reflection along the $A$ angle bisector. $\phi :(ABC) \longrightarrow BC$ and $\phi : (BHC) \longrightarrow (BOC)$ Hence $M \longrightarrow Y$ and $N \longrightarrow X \implies AM.AY=AN.AX \implies \frac{AM}{AX}=\frac{AN}{AY} \implies MN \parallel XY$ .

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#6 h Mar 25, 2019, 5:43 AM • 2 Y

Y by Adventure10, Mango247

We claim that under a root bc inversion, $M$ swaps with $Y$ and $N$ swaps with $X$ . This solves the problem by angle chase under inversion.

It is well known that given isogonal conjugates $P, Q$ in $\triangle ABC$ , the circumcircles $(BPC)$ and $(BQC)$ swap under root bc inversion. Thus, the image of $M$ under root bc inversion (let this be $M^{*}$ ) must lie on $(BQC)$ . Similarly, $N^{*}$ must lie on $(BPC)$ . Extend $AP$ to meet $(BPC)$ again at $Q^{*}$ , the image of $Q$ under root bc and extend $AQ$ to meet $(BQC)$ again at $P^{*}$ , the image of $P$ under root bc.

Angle chasing, we see that $90 = \angle AMP = \angle AP^{*}M^{*} = \angle QP^{*}M^{*}$ and since $M^{*} \in (BQC)$ , we conclude that $M^{*}$ is the antipode of $Q$ on $(BQC)$ . Similarly, $N^{*}$ is the antipode of $P$ on $(BPC)$ .

Notice that $X$ and $Y$ are defined as the antipodes of $P$ in $(BPC)$ and $Q$ in $(BQC)$ respectively, so we have proven the claim. Thus, $\angle ANM = \angle AYX$ , finishing the problem.

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#7 h May 7, 2022, 6:10 PM • 1 Y

Y by centslordm

Claim: $M$ and $N$ are isogonal conjugates.
Proof. Redefine $N$ as the isogonal conjugate of $M.$ It suffices to prove $N=(BQC)\cap(AQ).$ First, we claim $CBQN$ is cyclic. Indeed, $\measuredangle BPC+\measuredangle BQC=\measuredangle BAC+180=\measuredangle BMC+\measuredangle BNC$ and $\measuredangle BPC=\measuredangle BMC.$ Then, $\begin{align*}&\measuredangle MCP=\measuredangle QCN=\measuredangle QBN\\&\implies \measuredangle CBA-\measuredangle NBA-\measuredangle CBQ+\measuredangle MAC=\measuredangle ACB-\measuredangle PCB-\measuredangle ACM+\measuredangle BAN\\&\implies \measuredangle CBA+\measuredangle MAC+\measuredangle ACM-\measuredangle CBQ=\measuredangle ACB+\measuredangle BAN+\measuredangle NBA-\measuredangle PCB\\&\implies \measuredangle CBA+180-\measuredangle CMA-\measuredangle CNQ=\measuredangle ACB+180-\measuredangle ANB-\measuredangle PMB\\&\implies \measuredangle QNA=\measuredangle AMP=90.\end{align*}$ $\blacksquare$

Take the $\sqrt{bc}$ inversion, denoting $Z^*$ as the inverse of $Z.$

Claim: $\omega_P^*=\omega_Q.$
Proof. Let $P'=\overline{AQ}\cap\omega_Q$ and note $\measuredangle QP'B=\measuredangle QCB=\measuredangle ACP$ so $\triangle AP'B\sim\triangle ACP$ or $P^*=P'.$ $\blacksquare$

Hence, $P$ and $Q$ lie on $\omega_Q^*$ and $\omega_P^*,$ respectively. Also, $\measuredangle N^*Q^*A=\measuredangle QNA=90$ so $\overline{PN^*}$ is the diameter of $\omega_Q^*.$ Hence, $Y^*$ is the foot from $P$ to $\overline{AN^*}=\overline{AM}$ so $Y^*=M.$ Similarly, $X^*=N.$ Thus, $AX^*\cdot AN^*=AB\cdot AC=AY^*\cdot AM^*$ and $\triangle AX^*Y^*\sim\triangle AM^*N^*.$ $\square$

This post has been edited 1 time. Last edited by Mogmog8, May 7, 2022, 7:08 PM

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