IMO 1991 SL 4

Origin: FRA

Problem

Let ABC be a triangle and M an interior point in ABC. Show that at least one of the angles ∡MAB, ∡MBC, and ∡MCA is less than or equal to 30◦.

Solution

Assume the contrary, that \angleMAB, \angleMBC, \angleMCA are all greater than 30◦. By the sine Ceva theorem, it holds that sin \angleMAC sin \angleMBA sin \angleMCB = sin \angleMAB sin \angleMBC sin \angleMCA > sin3 30◦= 1 8 . (∗) On the other hand, since \angleMAC+\angleMBA+\angleMCB < 180◦−3\cdot30◦= 90◦, Jensen’s inequality applied on the concave function ln sin x (x \in[0, \pi]) gives us sin \angleMAC sin \angleMBA sin \angleMCB < sin3 30◦, contradicting (∗).

Second solution. Denote the intersections of PA, PB, PC with BC, CA, AB by A1, B1, C1, respectively. Suppose that each of the angles \anglePAB, \anglePBC, \anglePCA is greater than 30o and denote PA = 2x, PB = 2y, PC = 2z. Then PC1 > x, PA1 > y, PB1 > z. On the other hand, we know that PC1 PC + PC1 + PA1 PA + PA1 + PB1 PB + PB1 = SABP SABC

• SPBC SABC
• SAPC SABC = 1. Since the function t p+t is increasing, we obtain x 2z+x + y 2x+y + z 2y+z < 1. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen’s inequality) yields x 2z + x + y 2x + y + z 2y + z \geq (x + y + z)2 x(2z + x) + y(2x + y) + z(2y + z) = 1.