IMO 2012 Shortlist G7

Category: Geometry

Problem

Let ABCD be a convex quadrilateral with non-parallel sides BC and AD. Assume that there is a point E on the side BC such that the quadrilaterals ABED and AECD are circumscribed. Prove that there is a point F on the side AD such that the quadrilaterals ABCF and BCDF are circumscribed if and only if AB is parallel to CD. Solution. Let ω1 and ω2 be the incircles and O1 and O2 the incenters of the quadrilater- als ABED and AECD respectively. A point F with the stated property exists only if ω1 and ω2 are also the incircles of the quadrilaterals ABCF and BCDF. D C E B O1 O2 A F1 F2 O Let the tangents from B to ω2 and from C to ω1 (other than BC) meet AD at F1 and F2 respectively. We need to prove that F1 = F2 if and only if AB k CD. Lemma. The circles ω1 and ω2 with centers O1 and O2 are inscribed in an angle with vertex O. The points P,S on one side of the angle and Q,R on the other side are such that ω1 is the incircle of the triangle PQO, and ω2 is the excircle of the triangle RSO opposite to O. Denote p = OO1 · OO2. Then exactly one of the following relations holds: OP · OR < p < OQ · OS, OP · OR > p > OQ · OS, OP · OR = p = OQ · OS. Proof. Denote ∠OPO1 = u, ∠OQO1 = v, ∠OO2R = x, ∠OO2S = y, ∠POQ = 2ϕ. Because PO1,QO1,RO2,SO2 are internal or external bisectors in the triangles PQO and RSO, we have u + v = x + y (= 90◦ − ϕ). (1) R S O1 O2 x y P u v Q O ϕ ϕ By the law of sines OP OO1

sin(u + ϕ) sinu and OO2 OR

sin(x + ϕ) sinx . Therefore, since x,u and ϕ are acute, OP ·OR ≥ p ⇔ OP OO1 ≥ OO2 OR ⇔ sinxsin(u+ϕ) ≥ sinusin(x+ ϕ) ⇔ sin(x−u) ≥ 0 ⇔ x ≥ u. Thus OP · OR ≥ p is equivalent to x ≥ u, with OP · OR = p if and only if x = u. Analogously, p ≥ OQ · OS is equivalent to v ≥ y, with p = OQ · OS if and only if v = y. On the other hand x ≥ u and v ≥ y are equivalent by (1), with x = u if and only if v = y. The conclusion of the lemma follows from here. 37 Going back to the problem, apply the lemma to the quadruples {B,E,D,F1}, {A,B,C,D} and {A,E,C,F2}. Assuming OE · OF1 > p, we obtain OE · OF1 > p ⇒ OB · OD < p ⇒ OA · OC > p ⇒ OE · OF2 < p. In other words, OE · OF1 > p implies OB · OD < p < OA · OC and OE · OF1 > p > OE · OF2. Similarly, OE · OF1 < p implies OB · OD > p > OA · OC and OE · OF1 < p < OE · OF2. In these cases F1 6= F2 and OB · OD 6= OA · OC, so the lines AB and CD are not parallel. There remains the case OE · OF1 = p. Here the lemma leads to OB · OD = p = OA · OC and OE · OF1 = p = OE · OF2. Therefore F1 = F2 and AB k CD. Comment. The conclusion is also true if BC and AD are parallel. One can prove a limit case of the lemma for the configuration shown in the figure below, where r1 and r2 are parallel rays starting at O′ and O′′, with O′O′′ ⊥ r1,r2 and O the midpoint of O′O′′. Two circles with centers O1 and O2 are inscribed in the strip between r1 and r2. The lines PQ and RS are tangent to the circles, with P,S on r1, and Q,R on r2, so that O,O1 are on the same side of PQ and O,O2 are on different sides of RS. Denote s = OO1 + OO2. Then exactly one of the following relations holds: O′ P + O′′ R < s < O′′ Q + O′ S, O′ P + O′′ R > s > O′′ Q + O′ S, O′ P + O′′ R = s = O′′ Q + O′ S. O2 R S O1 Q P r1 r2 O O′ O′′ Once this is established, the proof of the original statement for BC k AD is analogous to the one in the intersecting case. One replaces products by sums of relevant segments.38