Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Singapore

Let $ABC$ be an acute-angled triangle and let $D$ , $E$ and $F$ be the midpoints of $BC$ , $CA$ and $AB$ respectively. Construct a circle, centred at the orthocentre of triangle $ABC$ , such that triangle $ABC$ lies in the interior of the circle. Extend $EF$ to intersect the circle at $P$ , $FD$ to intersect the circle at $Q$ and $DE$ to intersect the circle at $R$ . Show that $AP = BQ = CR$ .

Solution

Let the radius of the circle be $r$ . Let $X$ , $Y$ and $Z$ be the feet of the altitudes from $A$ , $B$ and $C$ respectively. Let $PE$ intersect the altitude from $A$ at $U$ . We have
$AP^2 = AU^2 + PU^2 = AU^2 + r^2 - UH^2 = r^2 + (AU+UH) \cdot (AU-UH) = r^2 + AH \cdot (AU-UH) = r^2 + AH \cdot (UX - UH) = r^2 + AH \cdot HX.$
Similarly, $BQ = r^2 + BH \cdot HY$ , and $CR = r^2 + CH \cdot HZ$ . Since $AH \cdot HX = BH \cdot HY = CH \cdot HZ$ , we have $AP = BQ = CR$ .

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