Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.2 National olympiad, round 2 Prove it Greece

Let $AB\Gamma$ be an acute angled triangle inscribed in the circle $c(O, R)$ (with $AB < A\Gamma < B\Gamma$ ) and let $\Delta, E, Z$ be the touching points of the incircle of the triangle with the sides $B\Gamma$ , $A\Gamma$ , $AB$ , respectively. The circumcircle of the triangle $AEZ$ (say, $(c_1)$ ) intersects the circle $(c)$ at point $A'$ . The circumcircle of the triangle $B\Delta Z$ (say, $(c_2)$ ) intersects the circle $(c)$ at point $B'$ . The circumcircle of the triangle $\Gamma\Delta E$ (say, $(c_3)$ ) intersects the circle $(c)$ at point $\Gamma'$ . Prove that:

(α) The quadrilateral $\Delta EA'B'$ is cyclic.

(β) The lines $\Delta A'$ , $EB'$ and $Z\Gamma'$ are concurrent.

Solution

From the inscribed in the circle $c_1$ quadrilateral $AA'IZ$ we have:
$A\hat{A}'I = A\hat{Z}I = 90^\circ = T A\hat{A}'I \quad . (\alpha)$
From the inscribed quadrilateral $\Gamma\Delta IE$ (since $\Gamma I$ bisector), we have
$\hat{\Delta}_1 = \frac{\hat{\Gamma}}{2} \qquad (1)$
From the inscribed quadrilateral $B\Delta IZ$ (since $BI$ bisector), we have:
$\hat{\Delta}_2 = \frac{\hat{B}}{2} \qquad (2)$
From the inscribed quadrilateral $B\Delta IB'$ we have: $\hat{\Delta}_3 = \hat{B}_1$ .
From the inscribed quadrilateral $BB'A'A$ we have $\hat{B}_1 = \hat{A}'_1 = 90^\circ - \hat{A}'_2$ .
Hence: $\hat{\Delta}_3 + \hat{A}'_2 = 90^\circ$ .
From the inscribed quadrilateral $AEIA'$ we have: $\hat{A}'_3 = \frac{\hat{A}}{2}$
Hence:
$\hat{\Delta}_1 + \hat{\Delta}_2 + \hat{\Delta}_3 + \hat{\Delta}'_2 + \hat{\Delta}'_3 = 180^\circ$
Therefore the quadrilateral $A'Ε\Delta B'$ is cyclic. Similarly we prove that the quadrilaterals $\Delta ZA'\Gamma'$ and $ΖΕ\Gamma'B'$ are cyclic.

Finally, we conclude that the lines $\Delta A'$ , $EB'$ and $Z\Gamma'$ are concurrent at the radical point of three circles.

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