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Problem 1314

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Geometry Difficulty 5.5 Prove it Czech-Polish-Slovak Mathematics Competition · Czech Republic

On a circle of radius $r$ , the distinct points $A, B, C, D$ , and $E$ lie in this order, satisfying $AB = CD = DE > r$ . Show that the triangle with vertices lying in the centroids of the triangles $ABD$ , $BCD$ , and $ADE$ is obtuse.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Denote by $P$ , $Q$ , and $R$ the centroids of the triangles $ABD$ , $BCD$ , and $ADE$ , respectively. Let $K$ and $L$ be the midpoints of the segments $BD$ and $AD$ , respectively. Since $P$ and $Q$ are centroids, they divide in the same ratio the medians $AK$ and $CK$ , respectively. That is, $AP : PK = CQ : QK = 2 : 1$ , and we have $PQ \parallel AC$ . Similarly $PR \parallel BE$ . Hence the angle $QPR$ is of the same measure as the angle $CXE$ determined by the lines $AC$ and $BE$ (here, $X$ is the intersection point of $AC$ and $BE$ , see Fig. 1).

Fig. 1

Denote by $\varphi$ the measure of the inscribed angle determined by the chord $AB$ of the given circle. Since $CD = DE = AB$ , we have $\angle CAE = 2\varphi$ , and therefore from the triangle $AXE$ we conclude
$\angle CXE = 180^\circ - \angle AXE = \varphi + 2\varphi = 3\varphi.$
Since $AB > r$ , we have $\varphi > 30^\circ$ , and so $\angle QPR = 3\varphi > 90^\circ$ .

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