Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Argentina

Consider a semicircle with diameter $AB$ , center $O$ and radius $r$ . Let $C$ be the point on segment $AB$ such that $AC = \frac{2r}{3}$ . Line $l$ is perpendicular to $AB$ at $C$ and $D$ is the common point of $l$ and the semicircle. Let $H$ be the foot of the perpendicular from $O$ to $AD$ and $E$ the intersection of lines $CD$ and $OH$ .

a) Express $AD$ as a function of $r$ .

b) If $M$ and $N$ are the midpoints of $AE$ and $OD$ respectively, find the measure of angle $MHN$ .

Solution

a) Since $CO = \frac{1}{3}r$ and $OD = r$ , Pythagoras' theorem in triangles $COD$ and $ACD$ gives $CD = \sqrt{OD^2 - OC^2} = \frac{2\sqrt{2}}{3}r$ , $AD = \sqrt{AC^2 + CD^2} = \frac{2\sqrt{3}}{3}r$ .

b) As $AD$ is a chord in the semicircle and $O$ its center, $OH \perp AD$ implies that $H$ is the midpoint of $AD$ . So $HM$ and $HN$ are median lines in triangles $ADE$ and $DAO$ , hence $HM \parallel DE$ and $HN \parallel AO$ . On the other hand $DE \perp AO$ , so $HM \perp HN$ . Therefore $\angle MHN = 90^\circ$ .

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