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

(New page: Given a triangle <math>ABC</math> let <math>I</math> be the center of its inscribed circle. The internal bisectors of the angles <math>A,B,C</math> meet the opposite sides in <math>A^\prim...) |
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<math>\frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}</math> | <math>\frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}</math> | ||

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

## Latest revision as of 16:26, 30 August 2008

Given a triangle let be the center of its inscribed circle. The internal bisectors of the angles meet the opposite sides in respectively. Prove that

## Solution

We have . From Van Aubel's Theorem, we have which from the Angle Bisector Theorem reduces to . We find similar expressions for the other terms in the product so that the product simplifies to . Letting for positive reals , the product becomes . To prove the right side of the inequality, we simply apply AM-GM to the product to get

To prove the left side of the inequality, simply multiply out the product to get

as desired.