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Difference between revisions of "1998 IMO Problems/Problem 5"

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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.
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Denote the length 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 that 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.
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1998|num-b=4|num-a=6}}
 
{{IMO box|year=1998|num-b=4|num-a=6}}

Latest revision as of 08:33, 27 July 2024

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

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Denote the length 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 that 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.

See Also

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