1998 IMO Problems/Problem 5

Problem

Let $I$ be the incenter of triangle $ABC$ . Let the incircle of $ABC$ touch the sides $BC$ , $CA$ , and $AB$ at $K$ , $L$ , and $M$ , respectively. The line through $B$ parallel to $MK$ meets the lines $LM$ and $LK$ at $R$ and $S$ , respectively. Prove that angle $RIS$ is acute.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Denote the lenght of the side MB with x. Denote the angle BKM by a and angle BAC by 2b. Using angle chasing and trigonometry, we can derive: RB = x*cos(b)/cos(a-b); BS = x*cos(a-b)/cos(b); RM = x*sin(a)/cos(a-b); SK = x*sin(a)/cos(b). Let's prove that RI^2 + IS^2 > RS^2, because this will imply that angle RIS is acute. Let's calculate RS^2. It is equal to (RB+BS)^2 = RB^2+BS^2 + 2(x^2). RI^2 - RB^2 = IB^2 (since the angle IBK is equal to 90-a, and angle KBS is equal to a, thus angle IBS is equal to 90 degrees). IS^2-BS^2 = IB^2. So we have to prove than 2*(IB^2)>2*(x^2). And this is true because triangle IMB is a right triangle with IB hypotenuse, so IB > x, as desired.

1998 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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1998 IMO Problems/Problem 5

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