3 concurrent diagonals

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3 concurrent diagonals

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Source: Austrian MO 2005 round 1

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#1 h Jun 27, 2005, 5:16 PM • 2 Y

Y by Adventure10, Mango247

We're given two congruent, equilateral triangles $ABC$ and $PQR$ with parallel sides, but one has one vertex pointing up and the other one has the vertex pointing down. One is placed above the other so that the area of intersection is a hexagon $A_1A_2A_3A_4A_5A_6$ (labelled counterclockwise). Prove that $A_1A_4$ , $A_2A_5$ and $A_3A_6$ are concurrent.

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#2 h Jun 28, 2005, 3:58 AM • 2 Y

Y by Adventure10, Mango247

Since the corresponding sides $AB \parallel PQ,\ BC \parallel QR,\ CA \parallel RP$ of the equilateral triangles $\triangle ABC \cong \triangle PQR$ are parallel, these 2 triangles are centrally similar, which means that the lines $AP, BQ, CR$ connecting their corresponding vertices concur at their similarity center $H$ . Since the 2 equilateral triangles are oppositely congruent, their similarity coefficient is -1. Therefore, for the oriented segments, we have $HA = -HP,\ HB = -HQ,\ HC = -HR$ and consequently, $H$ is the common midpoint of the segments $AP, BQ, CR$

The quadrilateral $AA_2PA_5$ is a parallelogram, because the lines $AA_2 \equiv AB \parallel PQ \equiv QA_5$ and $A_5A \equiv CA \parallel RP \equiv A_2P$ are parallel. Hence, its diagonals $AP, A_2A_5$ cut each other in half at the midpoint $H$ of $AP$ .

The quadrilateral $BA_4QA_1$ is a parallelogram, because the lines $BA_4 \equiv BC \parallel QR \equiv QA_1$ and $A_1B \equiv AB \parallel PQ \equiv A_4Q$ are parallel. Hence, its diagonals $BQ, A_1A_4$ cut each other in half at the midpoint $H$ of $BQ$ .

The quadrilateral $CA_6RA_3$ is a parallelogram, because the lines $CA_6 \equiv CA \parallel RP \equiv RA_3$ and $A_3C \equiv BC \parallel QR \equiv A_6R$ are parallel. Hence, its diagonals $CR, A_3A_6$ cut each other in half at the midpoint $H$ of $CR$ .

As a result, $A_1A_4,\ A_2A_5,\ A_3A_6$ all pass through and concur at the similarity center $H$ .

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#3 h Jul 30, 2005, 8:44 PM • 2 Y

Y by Adventure10, Mango247

There is an even simpler way of formulating the solution:

Since the triangles ABC and PQR have parallel sides, they are homothetic, and since they are congruent, their homothetic factor is either 1 or -1 (where 1 would actually mean that there is a translation mapping one to the other). The condition that one of the triangles has one vertex pointing up and the other one has the corresponding vertex pointing down specifies that the factor is -1 rather than 1. Thus, there is a homothety with factor -1, i. e. a reflection in a point, which maps the triangle ABC to the triangle PQR. In other words: There exists a point Z such that the triangle PQR is the reflection of the triangle ABC in the point Z. More precisely, the reflection in the point Z maps the points A, B, C to the points P, Q, R.

The points $A_1$ , $A_2$ , $A_3$ , $A_4$ , $A_5$ , $A_6$ were defined as points of intersections of the sides of triangles ABC and PQR, without specifying which point is the intersection of which sides; let's WLOG assume that $A_1=CA\cap QR$ , $A_2=AB\cap QR$ , $A_3=AB\cap RP$ , $A_4=BC\cap RP$ , $A_5=BC\cap PQ$ and $A_6=CA\cap PQ$ . The reflection in the point Z maps the points A, B, C to the points P, Q, R. Of course, it must then also map the points P, Q, R back to the points A, B, C. Hence, it maps the point $CA\cap QR$ to the point $RP\cap BC$ . In other words, our reflection maps the point $A_1$ to the point $A_4$ . Hence, the line $A_1A_4$ passes through the center Z of our reflection. Similarly, the lines $A_2A_5$ and $A_3A_6$ also pass through this center Z. Hence, the lines $A_1A_4$ , $A_2A_5$ and $A_3A_6$ are concurrent (namely, at the point Z).

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