IMO Proof Problem

Given a triangle \(ABC\) let \(I\) be the centre of its inscribed circle. The internal bisectors of the angles \(A\), \(B\) and \(C\) meet the opposite sides at \(A'\), \(B'\) and \(C'\) respectively. Prove that

14<AI⋅BI⋅CIAA′⋅BB′⋅CC′≤827\dfrac {1}{4} < \dfrac {AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leq \dfrac {8}{27}

#Geometry #IMO #Sharky

Note by Sharky Kesa
6 years, 10 months ago

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Comments

Let (a,b,c)=(BC,CA,AB)(a,b,c)=(BC,CA,AB). Some not-so-tedious geometry leads to

AI⋅BI⋅CIAA′⋅BB′⋅CC′=(a+b)(b+c)(c+a)(a+b+c)3.\dfrac{AI\cdot BI\cdot CI}{AA'\cdot BB'\cdot CC'}=\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}. So we need to prove

(a+b)(b+c)(c+a)(a+b+c)3>14,   (a+b)(b+c)(c+a)(a+b+c)3≤827.\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3} > \dfrac 1 4, ~~~ \dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\le \dfrac 8 {27}. For the first one, clearing denominators, expanding and cancelling gives

(a2b+b2c+c2a+ab2+bc2+ca2)−(a3+b3+c3)−2abc>−4abc(a^2b+b^2 c+c^2a+ab^2+bc^2+ca^2)-(a^3+b^3+c^3)-2abc> -4abc
which factors into (a+b−c)(b+c−a)(c+a−b)>−4abc.(a+b-c)(b+c-a)(c+a-b)>-4abc. This is true because by the triangle inequalities, (a+b−c)(b+c−a)(c+a−b)>0(a+b-c)(b+c-a)(c+a-b)>0.

For the second part, directly apply AM-GM inequality

(a+b)(b+c)(c+a)(a+b+c)3≤(2(a+b+c)/3a+b+c)3=(23)3=827.\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\le \left(\dfrac{2(a+b+c)/3}{a+b+c}\right)^3=\left(\dfrac{2}{3}\right)^3=\dfrac{8}{27}.

Jubayer Nirjhor - 6 years, 10 months ago

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A slightly easier way of doing the LHS is to use the (Ravi) substitution of AB=x+y,BC=y+z,CA=z+xAB = x+y, BC = y+z, CA = z + x , and then we want to show that

(1+xx+y+z)(1+yx+y+z)(1+zx+y+z)>2 ( 1 + \frac{x}{ x+y+z} ) ( 1 + \frac{y}{x+y+z} ) ( 1 + \frac{z}{ x+y+z} ) > 2

This follows immediately by expanding out the LHS and looking at 4 terms.

Calvin Lin Staff - 6 years, 10 months ago

When did you turn 13?

Bogdan Simeonov - 6 years, 10 months ago

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I haven't.

Sharky Kesa - 6 years, 10 months ago
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