Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.8 Shortlist Find the answer

I don't like this solution, but I couldn't find a better one this late at night (or this early in the morning; it's 4:15 AM here :)).

Let $S=KA\cap \Omega$ , and let $T$ be the antipode of $K$ on $\Omega$ . Let $X,Y$ be the touch points between $\Omega$ and $CA,AB$ respectively.

The line $AD$ is parallel to $KT$ and is cut into two equal parts by $KS,KN,KD$ , so $(KT,KN;KS,KD)=-1$ . This means that the quadrilateral $KTSN$ is harmonic, so the tangents to $\Omega$ through $K,S$ meet on $NT$ . On the other hand, the tangents to $\Omega$ through the points $X,Y$ meet on $KS$ , so $KXSY$ is also harmonic, meaning that the tangents to $\Omega$ through $K,S$ meet on $XY$ .

From these it follows that $BC,XY,TN$ are concurrent. If $P=XY\cap BC$ , it's well-known that $(B,C;K,P)=-1$ , and since $\angle KNP=\angle KNT=\frac{\pi}2$ , it means that $N$ lies on an Apollonius circle, so $NK$ is the bisector of $\angle BNC$ .

From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$ , we get $B'C'\|BC$ , so there's a homothety of center $N$ which maps $\Omega$ to the circumcircle of $BNC$ .

Solution

To solve this geometric configuration problem, let's analyze the given setup and deduce the needed relationships.

1. Setup Clarifications:
- Define $S = KA \cap \Omega$ where $\Omega$ is a circle and $K$ and $A$ are points on or outside of it.
- Let $T$ be the antipode of $K$ on $\Omega$ , meaning $KT$ is a diameter of the circle.

2. Special Points and Lines:
- $X$ and $Y$ are the points where the circle $\Omega$ is tangent to lines $CA$ and $AB$ , respectively.
- The line $AD$ is parallel to $KT$ and is divided into two equal segments by points $K, S, N,$ and $D$ .

3. Harmonic Division:
- The given condition $(KT, KN; KS, KD) = -1$ indicates that the points $K, T, S, N$ form a harmonic division, creating unique geometric properties like equal division and angle bisectors.

4. Tangency and Harmonic Conjugates:
- The tangents to $\Omega$ at $K$ and $S$ intersect at line $NT$ , a property of collinear points in a harmonic set.
- Similarly, $KXSY$ is harmonic, implying by extension that the tangents from $X$ and $Y$ to $\Omega$ meet on line $KS$ .

5. Concurrent Lines:
- From these harmonic properties, it follows that lines $BC, XY,$ and $TN$ are concurrent. Designate the point of concurrency as $P = XY \cap BC$ .

6. Apollonius Circle and Angle Bisector:
- The known result $(B, C; K, P) = -1$ helps establish that $N$ , lying on specific geometric loci (Apollonius circle), forces $NK$ to bisect $\angle BNC$ .

7. Homothety and Parallelism:
- If points $B' = NB \cap \Omega$ and $C' = NC \cap \Omega$ , the parallelism $B'C' \parallel BC$ indicates the possibility of a homothety centered at $N$ transforming $\Omega$ onto the circumcircle of triangle $BNC$ .

Through this derivation, we can conclude by the harmonic and homothetic properties that such configurations lead to parallel and bisecting lines, confirming the unique relationships described by the problem.

Final relationships being sought in the problem:
$\boxed{N \text{ is the center of homothety, bisecting }\angle BNC \text{ and mapping } \Omega \rightarrow \Gamma_{\triangle BNC}}$

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Note: Additional diagrams and constructs may enhance the geometric intuition and verification of these analytic results for thorough understanding.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.