IMO 1993 SL 4

Origin: SPA

Problem

In the triangle ABC, let D, E be points on the side BC such that \angleBAD = \angleCAE. If M, N are, respectively, the points of tangency with BC of the incircles of the triangles ABD and ACE, show that MB + MD = NC + NE .

Solution

We note that p does not exceed one half of the circumradius of \triangleA′B′C′. However, by the theorem on the nine-point circle, this circumradius is equal to 1 2R, and the conclusion follows. Second solution. By a well-known relation we have cos A+cos B+cos C = 1+ r R (= 1+r when R = 1). Next, recalling that the incenter of \triangleA′B′C′ is at the orthocenter of \triangleABC, we easily obtain p = 2 cosA cos B cos C. Cosines of angles of a triangle satisfy the identity cos2 A+cos2 B+cos2 C+ 2 cos A cos B cos C = 1 (the proof is straightforward: see (SL81-11)). Thus p + 1 3(1 + r)2 = 2 cos A cos B cos C + 1 3(cos A + cos B + cos C)2 \leq2 cos A cos B cos C + cos2 A + cos2 B + cos2 C = 1. 3. Let O1 and \rho be the center and radius of kc. It is clear that C, I, O1 are collinear and CI/CO1 = r/\rho. By Stewart’s theorem applied to \triangleOCO1, OI2 = r \rhoOO2 1 +  1 −r \rho  OC2 −CI \cdot IO1. (1) Since OO1 = R −\rho, OC = R and by Euler’s formula OI2 = R2 −2Rr, substituting these values in (1) gives CI \cdot IO1 = r\rho, or equivalently CO1 \cdot IO1 = \rho2 = DO2

1. Hence the triangles CO1D and DO1I are similar, implying \angleDIO1 = 90◦. Since CD = CE and the line CO1 bisects the segment DE, it follows that I is the midpoint of DE. Second solution. Under the inversion with center C and power ab, kc is transformed into the excircle of <A <BC corresponding to C. Thus CD =

ab s , where s is the common semiperimeter of \triangleABC and \triangle<A <BC, and consequently the distance from D to BC is ab s sin C = 2SABC s = 2r. The statement follows immediately. Third solution. We shall prove a stronger statement: Let ABCD be a convex quadrilateral inscribed in a circle k, and k′ the circle that is tangent to segments BO, AO at K, L respectively (where O = BD \capAC), and internally to k at M. Then KL contains the incenters I, J of \triangleABC and \triangleABD. Let K′, K′′, L′, L′′, N denote the midpoints of arcs BC, BD, AC, AD, AB that don’t contain M; X′, X′′ the points on k defined by X′N = NX′′ = K′K′′ = L′L′′ (as oriented arcs); and set S = AK′ \capBL′′, M = NS \capk, K = K′′M \capBO, L = L′M \capAO. It is clear that I = AK′ \capBL′, J = AK′′ \capBL′′. Furthermore, X′M contains I (to see this, use the fact that for A, B, C, D, E, F on k, lines AD, BE, CF are concurrent if and only if AB \cdotCD \cdotEF = BC \cdotDE \cdotFA, and then express AM/MB by applying this rule to AMBK′NL′′ and show that AK′, MX′, BL′ are concurrent). Analogously, X′′M contains J. Now the points B, K, I, S, M lie on a cir- cle (\angleBKM = \angleBIM = \angleBSM), and points A, L, J, S, M do so as well. Lines IK, JL are parallel to K′′L′ (because \angleMKI = \angleMBI = \angleMK′′L′). On the other hand, the quadrilateral ABIJ is cyclic, and simple calculation with an- gles shows that IJ is also paral- lel to K′′L′. Hence K, I, J, L are collinear. A B C D O K′ K′′ L′ L′′ X′′ X′ S I J K L M N Finally, K \equivK, L \equivL, and M \equivM because the homothety centered at M that maps k′ to k sends K to K′′ and L to L′ (thus M, K, K′′, as well as M, L, L′, must be collinear). As is seen now, the deciphered picture yields many other interesting properties. Thus, for example, N, S, M are collinear, i.e., \angleAMS = \angleBMS. Fourth solution. We give an alternative proof of the more general state- ment in the third solution. Let W be the foot of the perpendicular from B to AC. We define q = CW, h = BW, t = OL = OK, x = AL, \theta = ∡WBO (\theta is negative if B(O, W, A), \theta = 0 if W = O), and as usual, a = BC, b = AC, c = AB. Let \alpha = ∡KLC and \beta = ∡ILC (both angles must be acute). Our goal is to prove \alpha = \beta. We note that 90◦−\theta = 2\alpha. One easily gets tan \alpha = cos \theta 1 + sin \theta , tan \beta = 2SABC a+b+c b+c−a −x . (1)

Applying Casey’s theorem to A, B, C, k′, we get AC \cdot BK + AL \cdot BC = AB \cdotCL, i.e., b
h cos \theta −t  +xa = c(b−x). Using that t = b−x−q−h tan\theta we get x = b(b + c −q) −bh
cos \theta + tan \theta  a + b + c . (2) Plugging (2) into the second equation of (1) and using bh = 2SABC and c2 = b2 + a2 −2bq, we obtain tan \alpha = tan \beta, i.e., \alpha = \beta, which completes our proof. 4. Let h be the altitude from A and ϕ = \angleBAD. We have BM = 1 2(BD + AB −AD) and MD = 1 2(BD −AB + AD), so MB + MD = BD MB \cdot MD = 4BD BD2 −AB2 −AD2 + 2AB \cdot AD

4BD 2AB \cdot AD(1 −cos ϕ) = 2BD sin ϕ 2SABD(1 −cos ϕ)

2BD sin ϕ BD \cdot h(1 −cos ϕ) = h tan ϕ . It follows that MB + MD depends only on h and ϕ. Specially, NC + NE = h tan(ϕ/2) as well.