Fixed point in a small configuration

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Fixed point in a small configuration

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Source: Balkan MO Shortlist 2024 G3

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1324 posts

#1 h Apr 27, 2025, 10:23 PM

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Let $A, B, C, D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$ . Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$ . Prove that $ZT$ passes through a fixed point independent of $\omega$ .

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mathuz

#2 h Apr 28, 2025, 6:06 AM

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Let $M$ be the midpoint of $CD$ , and $N$ be the intersection of $ZT$ and $CD$ .

By DIT, there exists an involution so that $(A,B)$ , $(C,D)$ , $(N,M)$ are reciprocal pairs. In other words, there exists an inversion on the line $\ell$ passing through the points $A,B,C,D$ that maps the following points to each other: $A \leftrightarrow B, \quad C\leftrightarrow D, \quad N \leftrightarrow M.$ Note that defining an inversion on a line only requires two pairs of points. So $N$ is a fixed point w.r.t. $A,B,C,D$ .

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Tamam

#3 h Apr 28, 2025, 6:45 AM • 1 Y

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Let $ZT$ intersect $CD$ at $K$ and let $M$ be the midpoint of $CD$ . Since $(A,K;C,D)=(X,T;C,D)=(M,B;C,D)$ $\frac{KC}{KD}$ is constant so $K$ is a constant point.

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#4 h Apr 28, 2025, 7:49 PM

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Actually good problem, especially surprising given it's a balkan shortlist problem.

Let $ZY \cap CD = F$ , $XT \cap CD = G$

I first claim that $F$ is fixed. Consider two circles $X_1Z_1T_1Y_1$ and $X_2Z_2T_2Y_2$ . By PoP on $A$ , $Z_1Z_2X_2X_1$ is cyclic, and a quick angle chase gives that $Z_1Z_2Y_2Y_1$ is also cyclic.
Now by radical axis theorem, $Z_1Y_1$ , $Z_2Y_2$ and $CD$ concur, so $F$ is fixed. Similar argument proves $G$ is fixed.

Now another easy angle chase gives that $BTFZ$ and $AZGT$ are cyclic. Let $TZ \cap CD = P$ . Power of a point gives us that $BP\times PF = AP\times PG$ , so viewing the position of $P$ as a variable, this gives us a linear equation for $P$ which has exactly one solution.

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