Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.9 National olympiad Prove it Greece

Let $AB\Gamma$ an acute angled scalene triangle with $AB < A\Gamma < B\Gamma$ . Let $\Delta, E, Z$ be the midpoints of the sides $B\Gamma, A\Gamma, AB$ , respectively, and let $BK, G\Lambda$ be altitudes. At the extension of $\Delta Z$ , to the part of $Z$ , we consider a point $M$ , such that the parallel from $M$ to $K\Lambda$ intersect the extensions of $\Gamma A, BA$ and $\Delta E$ at points $\Sigma, T$ and $N$ , respectively. If the circumcircle of the triangle $MB\Delta$ , say $(c_1)$ , intersects the line $\Delta N$ at point $P$ and the circumcircle of the triangle $N\Gamma\Delta$ , say $(c_2)$ , intersects the line $\Delta M$ at point $\Pi$ , prove that: $\Sigma T \parallel \Pi P$ . (E. Psychas)

Solution

Since $\Delta, E, Z$ are the midpoints of the sides $B\Gamma, A\Gamma, AB$ , respectively, the quadrilaterals $AE\Delta Z, ZE\Delta B$ and $ZE\Gamma\Delta$ are parallelograms.
From $\Delta M \parallel \Gamma \Sigma$ and $MN \parallel K\Lambda$ we conclude that: $\hat{M} = \hat{\Sigma}_1$ and $\hat{\Sigma}_1 = \hat{K}$ .
From the cyclic quadrilateral $B\Lambda K\Gamma$ ( $B\Lambda\Gamma = B\hat{K}\Gamma = 90^\circ$ ) we get: $\hat{K} = \hat{B}$ .

From the last three equalities of angles we arrive at the equality $\hat{M} = \hat{B}$ , and from this we conclude that the points $M, B, \Delta, T$ , are cocyclic.

In a similar fashion, since $\Delta N // BT$ and $MN // K\Lambda$ we have the equalities $\hat{N} = \hat{T}_1$ and $\hat{T}_1 = \hat{\Lambda}$ . Also, from the cyclic quadrilateral $B\Lambda K\Gamma$ (since $B\hat{\Lambda}\Gamma = B\hat{K}\Gamma = 90^\circ$ ) we have that: $\hat{\Lambda} = \hat{\Gamma}$ .
From the last three equalities of angles we have $\hat{N} = \hat{\Gamma}$ , and from this we conclude that the points $N, \Gamma, \Delta, \Sigma$ , are cocyclic.

Since the points $B, \Delta, P, T, M$ belong to the circle $(c_1)$ , from the cyclic quadrilateral $MTP\Delta$ we have: $\hat{T}_2 + \hat{\Delta} = 180^\circ$ .
Since the points $\Gamma, \Delta, \Pi, \Sigma, N$ belong to the circle $(c_2)$ , from the cyclic quadrilateral $\Delta\Pi\Sigma N$ we have: $\hat{\Sigma}_2 + \hat{\Delta} = 180^\circ$ .
From the last two equalities we get:
$\hat{T}_2 = \hat{\Sigma}_2 = 180^\circ - \hat{\Delta} = 180^\circ - \hat{A}.$

The quadrilateral $MTP\Delta$ is inscribed to the circle $(c_1)$ , and hence:
$\hat{TPE} = \hat{P}\Delta = 180^\circ - \hat{M}\Delta = 180^\circ - \hat{B}\Gamma = 180^\circ - \hat{B}.$

The angle $\hat{\Sigma}_2 = 180^\circ - \hat{A}$ is an external angle of the triangle $\Sigma\Pi M$ , and so:
$\Sigma\hat{\Pi}Z = \Sigma\hat{\Pi}M = \hat{\Sigma}_2 - \hat{M} = 180^\circ - \hat{A} - \hat{B} = \hat{\Gamma} = \Sigma\hat{E}Z.$
The quadrilateral $\Sigma\Pi\Gamma\Pi$ is inscribed into the circle $(c)$ and since $\hat{T}_2 = \hat{\Sigma}_2$ it is isosceles trapezium and so $\Sigma T \parallel \Pi P$ .

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