Inversion exercise

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Source: Balkan MO Shortlist 2024 G5

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#1 h Apr 27, 2025, 10:29 PM • 1 Y

Y by compoly2010

Let $ABC$ be an acute scalene triangle $ABC$ , $D$ be the orthogonal projection of $A$ on $BC$ , $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Let $P$ and $Q$ are points on the minor arcs $\widehat{AB}$ and $\widehat{AC}$ of the circumcircle of triangle $ABC$ respectively such that $PQ \parallel BC$ . Show that the circumcircles of triangles $DPQ$ and $DMN$ are tangent if and only if $M$ lies on $PQ$ .

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awesomeming327.

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awesomeming327.

#3 h Apr 27, 2025, 10:56 PM

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Let $P'$ and $Q'$ be reflections of $P$ and $Q$ across $MN$ . Then $(DPQ)$ and $(DMN)$ are tangent if and only if their reflections across $MN$ are tangent: $(AP'Q')$ and $(AMN)$ . By homothety, $(AMN)$ and $(ABC)$ are tangent.

Let the circumcenter of $(ABC)$ be $O$ , the circumcenter of $(AP'Q')$ be $O'$ . Then $O'$ lies on the perpendicular bisector of $BC$ , because $PQCB$ is an isosceles trapezoid, and $PQQ'P'$ is a rectangle. However, $(AP'Q')$ and $(AMN)$ are tangent if and only if $O'$ also lies on $AO$ . $AO$ and the $BC$ perpendicular bisector intersect only at $O$ , so this is true if and only if $O=O'$ , which is true if and only if the two circles are the same circle.

Since $ABC$ is acute, $O$ is not on $MN$ . Therefore, $P'$ and $Q'$ are on $(ABC)$ if and only if $P'=P$ , $Q'=Q$ . We are done.

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Tamam

#4 h Apr 28, 2025, 7:34 AM

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Let $P_1$ and $Q_1$ be the instersection of $MN$ with minor arcs $AB$ and $AC$ respectively. If we invert with radius $\sqrt{\frac{bc}{2}}$ center $A$ and reflect over $A$ angle bisector. $(DMN)$ goes to $(OCB)$ and $(P_1DQ_1)$ goes to $(Q_1OP_1)$ since $P_1$ and $Q_1$ go to each other. Since triangles $P_1OQ_1$ and $BOC$ are iscoceles their circumcircles are tangent. So we are done with one of the ifs. Let $P$ and $Q$ be on the minor arcs $AB$ and $AC$ respectively such that $PQ$ is parallel to $BC$ . Lets invert with radius $\sqrt{\frac{bc}{2}}$ center $A$ and reflect over $A$ angle bisector. Let the point $X$ 's image under the transformation be $X'$ . $Q'=MN\cap AP$ and $P'=MN\cap AQ$ . It is enough to prove $OP'=OQ'$ in order to prove $(BOC)$ and $(P'OQ')$ are tangent. And it is enough to prove that $P_1Q'=Q_1P'$ . And for this it is enough to prove that $P_1'P'=Q_1'Q'$ . Since $P'P_1'=PP_1\frac{\frac{bc}{2}}{AP AP_1}$ and $Q'Q_1'=QQ_1\frac{\frac{bc}{2}}{AQ AQ_1}$ it is enough to prove that $AP AP_1=AQ\dot AQ_1$ . Since $PP_1=QQ_1$ from the areas of $APP_1$ and $AQQ_1$ being equal it is enough to prove that $dis(A,PP_1)=dis(A,QQ_1)$ . If this was true $A$ would be on the perpndicular bisector of $P_1Q_1$ . That would imply $A$ being on the perpendicular bisector of $BC$ . So that would imply $AB=AC$ which is impossible. So we are done.

This post has been edited 1 time. Last edited by Tamam, Apr 28, 2025, 7:35 AM

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#5 h Apr 28, 2025, 11:40 AM

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reflect $(DMN)$ and $(DPQ)$ across $DE$ will result with $(ABC)$ and $(AMN)$ .
known that $(ABC)$ is tangent to $(AMN)$ done. $\square$

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awesomeming327.

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awesomeming327.

#6 h Apr 28, 2025, 3:28 PM

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Who is E!!!

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Haris1

#7 h Apr 29, 2025, 8:19 AM

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Invert at A with $\sqrt(\frac{bc}{2})$ , now we have to prove that the only points $X,Y$ on $MN$ such that $(OBC)$ and $(OXY)$ are tangent and also $(ABC)$ and $(AXY)$ are tangent, are the ones that also lie on $(ABC)$ . Let $MN$ intersect $(ABC)$ at $K,L$ . Tangent from $A$ intersects $MN$ at T.
We know that $T$ swaps $(X,Y),(K,L),(M,N)$ but if we let tangent at $O$ of circle $(OBC)$ intersects $MN$ at $S$ clearly $S$ =Point At infinity swaps $(X,Y),(K,L)$ so $S=T$ , This forces $AM=AN$ which contradicts scalene condition.

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