IMO 1996 SL G4

Origin: USA | Category: Geometry

Problem

Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ in the points $A_1$, $B_1$, $C_1$ respectively. Prove that

$$A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \geq A_1B \cdot B_1C \cdot C_1A.$$

Solution

By the law of cosines applied to $\triangle CA_1B_1$, we obtain

$$A_1B_1^2 = A_1C^2+B_1C^2-A_1C\cdot B_1C \geq A_1C\cdot B_1C.$$

Analogously,

$$B_1C_1^2 \geq B_1A \cdot C_1A$$

and

$$C_1A_1^2 \geq C_1B \cdot A_1B.$$

Multiplying these inequalities yields

$$A_1B_1^2 \cdot B_1C_1^2 \cdot C_1A_1^2 \geq A_1B \cdot A_1C \cdot B_1A \cdot B_1C \cdot C_1A \cdot C_1B. \tag{1}$$

Now, the lines $AA_1$, $BB_1$, $CC_1$ concur, so by Ceva’s theorem,

$$A_1B \cdot B_1C \cdot C_1A = AB_1 \cdot BC_1 \cdot CA_1,$$

which together with $(1)$ gives the desired inequality.

Equality holds if and only if $CA_1=CB_1$, etc.