IMO 1991 SL 3

Origin: PRK

Problem

Let S be any point on the circumscribed circle of \trianglePQR. Then the feet of the perpendiculars from S to the three sides of the triangle lie on the same straight line. Denote this line by l(S, PQR). Suppose that the hexagon ABCDEF is inscribed in a circle. Show that the four lines l(A, BDF), l(B, ACE), l(D, ABF), and l(E, ABC) intersect at one point if and only if CDEF is a rectangle.

Solution

Consider the problem with the unit circle on the complex plane. For conve- nience, we use the same letter for a point in the plane and its corresponding complex number. Lemma 1. Line l(S, PQR) contains the point Z = P +Q+R+S .

Proof. Suppose P ′, Q′, R′ are the feet of perpendiculars from S to QR, RP, PQ respectively. It suffices to show that P ′, Q′, R′, Z are on the same line. Let us first represent P ′ by Q, R, S. Since P ′ \inQR, we have P ′−Q R−Q =  P ′−Q R−Q

, that is, (P ′ −Q)(R −Q) = (P ′ −Q)(R −Q). (1) On the other hand, since SP ′ \perpQR, the ratio P ′−S R−Q is purely imagi- nary. Thus (P ′ −S)(R −Q) = −(P ′ −S)(R −Q). (2) Eliminating P ′ from (1) and (2) and using the fact that X = X−1 for X on the unit circle, we obtain P ′ = (Q + R + S −QR/S)/2 and analogously Q′ = (P + R + S −PR/S)/2 and R′ = (P + Q + S − PQ/S)/2. Hence Z −P ′ = (P + QR/S)/2, Z −Q′ = (Q + PR/S)/2 and Z −R′ = (R + PQ/S)/2. Setting P = p2, Q = q2, R = r2, S = s2 we obtain Z −P ′ = pqr 2s  ps qr + qr ps

, Z −Q′ = pqr 2s  qs pr + pr qs

and Z −P ′ = pqr 2s  rs pq + pq rs

. Since x + x−1 = 2Re x is real for all x on the unit circle, it follows that the ratio of every pair of these differences is real, which means that Z, P ′, Q′, R′ belong to the same line. Lemma 2. If P, Q, R, S are four different points on a circle, then the lines l(P, QRS), l(Q, RSP), l(R, SPQ), l(S, PQR) intersect at one point. Proof. By Lemma 1, they all pass through P +Q+R+S . Now we can find the needed conditions for A, B, . . . , F. In fact, the lines l(A, BDF), l(D, ABF) meet at Z1 = A+B+D+F , and l(B, ACE), l(E, ABC) meet at Z2 = A+B+C+E . Hence, Z1 \equivZ2 if and only if D −C = E −F ⇔CDEF is a rectangle. Remark. The line l(S, PQR) is widely known as Simson line; the proof that the feet of perpendiculars are collinear is straightforward. The key claim, Lemma 1, is a known property of Simson lines, and can be shown elementarily: ∗l(S, PQR) passes through the midpoint X of HS, where H is the orthocenter of PQR.