## Friday, January 13, 2006

### Ratio This. Topic: Geometry/Inequalities. Level: Olympiad.

Problem: (1991 IMO - #1) 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

$\frac{1}{4} < \frac{AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac{8}{27}$. Solution: We begin by changing the expression in the middle into something nicer to work with. First drop perpendiculars from the incenter $I$ to each side, as in the diagram. Note that $ID = IE = IF = r$ because they are all inradii. Also denote the side opposite angle $A$ side $a$ and similarly for $b$ and $c$.

Note that $AI+IA^{\prime} = AA^{\prime} \Rightarrow \frac{AI}{AA^{\prime}} = 1-\frac{IA^{\prime}}{AA^{\prime}}$. However, we then have $\frac{IA^{\prime}}{AA^{\prime}} = \frac{[IBC]}{[ABC]}$ because the two triangles share a base. Using the common formula of $S = rs$, where $S$ is the area of a triangle with inradius $r$ and semiperimeter $s$, we have $\frac{[IBC]}{[ABC]} = \frac{\frac{1}{2}ra}{\frac{1}{2}r(a+b+c)} = \frac{a}{a+b+c}$. So $1-\frac{IA^{\prime}}{AA^{\prime}} = \frac{b+c}{a+b+c}$. Applying the same respective substitutions for $\frac{BI}{BB^{\prime}}$ and $\frac{CI}{CC^{\prime}}$, we have

$\frac{AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} = \frac{(a+b)(b+c)(c+a)}{(a+b+c)^3}$.

Now, to simplify things, set $a = x+y, b = x+z, c = y+z$, where $x = CE = CD, y = BC = BD, z = AF = AE$. This gets rid of the triangle inequality condition. Set $s = x+y+z$. Then

$\frac{(a+b)(b+c)(c+a)}{(a+b+c)^3} = \frac{(s+x)(s+y)(s+z)}{8s^3}$.

We begin with the left inequality. We have

$\frac{(s+x)(s+y)(s+z)}{8s^3} > \frac{s^3+s^2(x+y+z)}{8s^3} = \frac{2s^3}{8s^3} = \frac{1}{4}$.

For the right, we have

$\sqrt{(s+x)(s+y)(s+z)} \le \frac{3s+x+y+z}{3} = \frac{4s}{3}$ by AM-GM, so then

$\frac{(s+x)(s+y)(s+z)}{8s^3} \le \frac{\frac{64s^3}{27}}{8s^3} = \frac{8}{27}$

as desired. QED.

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Comment: This problem really isn't that hard if you see the ratios and make the right changes. As often happens in geometric inequalities involving triangles, substitution of $x,y,z$ above for $a,b,c$ works well in getting rid of the triangle inequality restraint.

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Practice Problem #1: (1984 AIME - #3) A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_1, t_2,$ and$t_3$ in the figure, have areas $4, 9,$ and $49,$ respectively. Find the area of $\triangle ABC$.

Practice Problem #2: Find the area of a triangle in terms of $x,y,z$ defined above.