IMO 1996 SL G7

Origin: GBR | Category: Geometry

Problem

Let $\triangle ABC$ be an acute-angled triangle with circumcenter $O$ and circumradius $R$. Let $AO$ meet the circle $BOC$ again in $A'$, let $BO$ meet the circle $COA$ again in $B'$, and let $CO$ meet the circle $AOB$ again in $C'$. Prove that

$$OA' \cdot OB' \cdot OC' \geq 8R^3.$$

When does equality hold?

Solution

Let $A_1$ be the point of intersection of $OA'$ and $BC$; similarly define $B_1$ and $C_1$. From the similarity of triangles $\triangle OBA_1$ and $\triangle OA'B$ we obtain

$$OA_1 \cdot OA' = R^2.$$

Now it is enough to show that

$$8OA_1 \cdot OB_1 \cdot OC_1 \leq R^3.$$

Thus we must prove that

$$\lambda\mu\nu \leq \frac{1}{8},$$

where

$$\frac{OA_1}{OA} = \lambda,\qquad \frac{OB_1}{OB} = \mu,\qquad \frac{OC_1}{OC} = \nu. \tag{1}$$

On the other hand, we have

$$\frac{\lambda}{1+\lambda} + \frac{\mu}{1+\mu} + \frac{\nu}{1+\nu} = \frac{S_{OBC}}{S_{ABC}} + \frac{S_{AOC}}{S_{ABC}} + \frac{S_{ABO}}{S_{ABC}} =1.$$

Simplifying this relation, we get

$$1=\lambda\mu+\mu\nu+\nu\lambda+2\lambda\mu\nu \geq 3(\lambda\mu\nu)^{2/3}+2\lambda\mu\nu,$$

which cannot hold if

$$\lambda\mu\nu>\frac{1}{8}.$$

Hence

$$\lambda\mu\nu\leq\frac{1}{8},$$

with equality if and only if

$$\lambda=\mu=\nu=\frac{1}{2}.$$

This implies that $O$ is the centroid of $\triangle ABC$, and consequently, that the triangle is equilateral.

Second solution

In the official solution, the inequality to be proved is transformed into

$$\cos(A-B)\cos(B-C)\cos(C-A) \geq 8\cos A\cos B\cos C.$$

Since

$$\frac{\cos(B-C)}{\cos A} = -\frac{\cos(B-C)}{\cos(B+C)} = \frac{\tan B\tan C+1}{\tan B\tan C-1},$$

the last inequality becomes

$$(xy+1)(yz+1)(zx+1) \geq 8(xy-1)(yz-1)(zx-1),$$

where we write

$$x=\tan A,\qquad y=\tan B,\qquad z=\tan C.$$

Using the relation

$$x+y+z=xyz,$$

we can reduce this inequality to

$$(2x+y+z)(x+2y+z)(x+y+2z) \geq 8(x+y)(y+z)(z+x).$$

This follows from the AM–GM inequality:

$$2x+y+z=(x+y)+(x+z)\geq 2\sqrt{(x+y)(x+z)},$$

etc.