Difference between revisions of "1991 IMO Problems/Problem 1"

Revision as of 17:26, 30 August 2008

Given a triangle $ABC$ let $I$ be the center of its inscribed circle. The internal bisectors of the angles $A,B,C$ meet the opposite sides in $A^\prime,B^\prime,C^\prime$ respectively. Prove that

$\frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}$

We have $\prod\frac{AI}{AA^\prime}=\prod\frac{1}{1+\frac{IA^\prime}{IA}}$ . From Van Aubel's Theorem, we have $\frac{IA}{IA^\prime}=\frac{AB^\prime}{B^\prime C}+\frac{AC^\prime}{C^\prime B}$ which from the Angle Bisector Theorem reduces to $\frac{b+c}{a}$ . We find similar expressions for the other terms in the product so that the product simplifies to $\prod\frac{1}{1+\frac{a}{b+c}}=\prod\frac{b+c}{a+b+c}$ . Letting $a=x+y,b=y+z,c=z+x$ for positive reals $x,y,z$ , the product becomes $\frac{1}{8}\prod\frac{x+2y+z}{x+y+z}=\frac{1}{8}\prod\left(1+\frac{y}{x+y+z}\right)$ . To prove the right side of the inequality, we simply apply AM-GM to the product to get