Bariz - Problem

Source: 2021 Greek Junior MO p4 (served as Greek JBMO TST p4 since the latter didn't take place)

Tags: geometry, equal circles

parmenides51

03.07.2021 20:53

Given a triangle$ABC$ with $AB<BC<AC$ inscribed in circle $(c)$. The circle $c(A,AB)$ (with center $A$ and radius $AB$) interects the line $BC$ at point $D$ and the circle $(c)$ at point $H$. The circle $c(A,AC)$ (with center $A$ and radius $AC$) interects the line $BC$ at point $Z$ and the circle $(c)$ at point $E$. Lines $ZH$ and $ED$ intersect at point $T$. Prove that the circumscribed circles of triangles $TDZ$ and $TEH$ are equal.

Let $O$ be the circumcircle of $ABC$. Since $AB = AH$ and $AC = AE$, the perpendicular bisectors of $BH$ and $CE$ pass through both $A$ and $O$. Thus, $BH \parallel CE$ and $BECH$ is an isosceles trapezoid, giving us $EH = BC$. Since $\angle AZB = \angle AZC = \angle ACZ = \angle ACD$ and $\angle ZBA = 180^{\circ} - \angle DBA = 180^{\circ} - \angle BDA = \angle CDA$, we have $\angle BAZ = \angle DAC$. Together with $AB = AD$ and $AC = AZ$, we have $\triangle ABZ \cong \triangle ADC$ and $BZ = CD$. From those, we get $EH = BC = BD + DC = BZ + BD = ZD$. Since $T = DE \cap ZH$, $sin\angle ZTD = sin\angle ETH$. Thus, from the extended sine law, the circumradii of triangles $TDZ$ and $TEH$ are equal.