Let $\Gamma$ be the circumscribed circle of a triangle $ABC$ and let $D$ be a point at line segment $BC$. The circle passing through $B$ and $D$ tangent to $\Gamma$ and the circle passing through $C $and $D$ tangent to $\Gamma$ intersect at a point $E \ne D$. The line $DE$ intersects $\Gamma$ at two points $X$ and $Y$ . Prove that $|EX| = |EY|$.
Problem
Source: 2021 Dutch IMO TST 2.1
Tags: equal segments, geometry
cadaeibf
29.12.2021 13:45
Let $O$ be the circumcenter and let $Z$ be the intersection of the tangents at $B$ and $C$. By radical center theorem, $DE$ passes trough $Z$. Therefore $ZD\cdot ZE=ZB^2=ZC^2$. Therefore, by inverting in $Z$ with radius $ZB=ZC$, we get that $E$ lies on the inverse of the line $BC$, which is the circle $(ZBC)$. This passes through $O$, because $\angle ZBO=\angle OCZ=\frac{\pi}{2}$, which implies $OZ$ is its diameter. Therefore $OE\perp DE=XY$. So, since $OE$ is perpendicular to the chord $XY$, $E$ must be its midpoint, as wanted.
demmy
02.12.2023 12:40
Let $Z$ be the intersection of the tangents from $B$ and $C$ , $P$ = $\overleftrightarrow{BE} \cap \Gamma$
. By radical center theorem, $Z$ lies on $\overleftrightarrow{DE}$
Since $D$ lies on polar of $Z$ wrt. $\Gamma$
Thus $$-1 = (D,Z;Y,X) \stackrel{C}{=} (B,C;Y,X) $$Notice that $\overline{CP} \parallel \overline{XY}$
Thus
$$-1 = (B,C;Y,X) \stackrel{P}{=} (E,P_{\infty};Y,X) $$Hence $E$ is midpoint of $\overline{XY}$
AshAuktober
10.05.2024 20:04
Simple angle chase?! (Solved with Shreya (IMOTC '24)) Let the centres of the two unlabeled circles (henceforth $\omega_1$, $\omega_2$) be $O_1$ and $O_2$, and that of $\Gamma$ respectively. From some easy angle chasing one can see that $OO_1DO_2$ is a parallelogram; therefore $O_1O_2$ bisects $OD$; i. e. the centres of the circles $\omega_1, \omega_2, (OD)$ are collinear. Since they all pass through $D$, the mentioned circles are coaxial. Therefore, $E$ lies on $(OD)$, which is enough. $\square$