Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.8 National Olympiad, round 2 Prove it Hong Kong

Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$ . Diagonals $AC$ and $BD$ meet at $E$ . $F$ and $G$ are points on segments $AB$ and $CD$ respectively. Suppose $AF = 15$ , $FB = 13$ , $BE = 30$ , $ED = 13$ , $DG = 7.5$ and $GC = 6.5$ . Let $P$ be a point such that $PF \perp AB$ and $PG \perp CD$ . Find $\frac{PE}{PO}$ .

Solution

The answer is $\frac{PE}{PO} = 5$ .
Note that $\frac{AF}{FB} = \frac{DG}{GC}$ . Now if we let $M$ and $N$ be the perpendicular feet from $O$ to $AB$ and $CD$ respectively, we have $\frac{AM}{MB} = \frac{DN}{NC}$ . Also if we let $H$ and $K$ be the perpendicular feet from $E$ to $AB$ and $CD$ respectively, then by the fact

$\triangle EAB \sim \triangle EDC$ , we have $\frac{AH}{HB} = \frac{DK}{KC}$ . Therefore, $P$ , $O$ , $E$ are collinear, and $\frac{PE}{PO} = \frac{FH}{FM} = \frac{GK}{GN}$ .

Now $AF = 15$ and $AM = \frac{AF + FB}{2} = 14$ . Since $\triangle EAB \sim \triangle EDC$ with $\frac{AB}{DC} = 2$ , we have $EA = 2ED = 26$ . By the cosine formula, we have
$AH = AE \cos A = 26 \times \frac{26^2 + 28^2 - 30^2}{2 \times 26 \times 28} = 10.$
Hence we conclude $\frac{PE}{PO} = \frac{FH}{FM} = \frac{AF - AH}{AF - AM} = \frac{15 - 10}{15 - 14} = 5$ .

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