Track / Stage 8 / 75 of 180 #1775 of 2000

Problem 1775

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.3 Prove it 67th NMO Selection Tests for BMO and IMO · Romania

Two circles, $\omega_1$ and $\omega_2$ , centred at $O_1$ and $O_2$ , respectively, meet at points $A$ and $B$ . A line through $B$ meets $\omega_1$ again at $C$ , and $\omega_2$ again at $D$ . The tangents to $\omega_1$ and $\omega_2$ at $C$ and $D$ , respectively, meet at $E$ , and the line $AE$ meets the circle $\omega$ through $A$ , $O_1$ , $O_2$ again at $F$ . Prove that the length of the segment $EF$ is equal to the diameter of $\omega$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Begin by noticing that the lines $CO_1$ and $DO_2$ meet at a point $P$ on $\omega$ , since $\angle(PO_1, PO_2) = \angle(O_1C, CB) + \angle(BD, DO_2) = \angle(CB, BO_1) + \angle(O_2B, BD) = \angle(O_2B, BO_1) = \angle(O_1A, AO_2)$ . In what follows, we consider the case where $O_1$ and $O_2$ lie on the segments $CP$ and $DP$ , respectively; the other cases are similar.

Since the angles $PCE$ and $PDE$ are both right, and $2\angle ACP = \angle AO_1P = \angle AO_2P = 2\angle ADP$ (the equality in the middle holds on account of $P$ lying on $\omega$ ), the points $A$ , $C$ , $D$ , $E$ , $P$ all lie on the circle on diameter $EP$ , so $FP$ is a diameter of $\omega$ , and it is therefore sufficient to show that $EF = FP$ . Finally, since $\angle AFP = \angle AO_1P = 2\angle ACP = 2\angle AEP$ (the first, respectively third, equality holds on account of $APFO_1$ , respectively $ACEP$ , being cyclic), it follows that the triangle $EFP$ is isosceles with apex at $F$ .

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.