Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National Olympiad Prove it Ireland

Two circles $C_1$ and $C_2$ , with centres at $D$ and $E$ respectively, touch at $B$ . The circle having $DE$ as diameter intersects the circle $C_1$ at $H$ and the circle $C_2$ at $K$ . The points $H$ and $K$ both lie on the same side of the line $DE$ . $HK$ extended in both directions meets the circle $C_1$ at $L$ and meets the circle $C_2$ at $M$ . Prove that
(a) $|LH| = |KM|$ ;
(b) the line through $B$ perpendicular to $DE$ bisects $HK$ .

Solution

Extend the line $DE$ in both directions to meet the circles again at $A$ and $C$ . Let $F$ be the centre of the circle with diameter $DE$ . From $D$ , $F$ and $E$ draw perpendiculars $DR$ , $FP$ and $ES$ on $LM$ . Then $DR$ , $PF$ and $SE$ are all parallel. Also, since $|DF| = |FE|$ we have $|RP| = |PS|$ . Also $|RL| = |RH|$ , $|PH| = |PK|$ and $|SK| = |SM|$ . Therefore, $|RP| - |PH| = |PS| - |PK|$ and so $|RH| = |KS|$ . It follows that $2|RH| = 2|KS|$ , i.e., $|LH| = |KM|$ . This proves part (a).

Since $|LH| = |KM|$ and $|PH| = |PK|$ we have $|PL| = |PM|$ . This implies $|PH| \cdot |PL| = |PK| \cdot |PM|$ , and therefore $P$ lies on the radical axis of the two circles $C_1$ and $C_2$ . Since $B$ also lies on the radical axis, the radical axis is $BP$ . Therefore $BP$ is perpendicular to $DE$ . Finally, since $P$ is the midpoint of $HK$ , this is the required result.

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