Let $ABC$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Prove that $A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A$.
Problem
Source: 1996 IMO Shortlist
Tags: trigonometry, trig identities, Law of Cosines, IMO Shortlist
ThAzN1
28.03.2006 09:29
By the Law of Cosines, $ A_1B_1^2 = A_1C^2 + B_1C^2 - A_1C\cdot B_1C \geq A_1C\cdot B_1C$ etc so multiply together and by Ceva, $ A_1B \cdot B_1C \cdot C_1A = AB_1\cdot BC_1\cdot CA_1$
pinkpig
08.08.2021 23:21
From Ceva's Theorem, we obtain $$\frac{AC_1}{C_1B}\cdot\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}=1\Leftrightarrow AC_1\cdot BA_1\cdot CB_1=C_1B\cdot A_1C \cdot B_1A.$$Now, from the Law of Cosines, $$A_1C_1^2=C_1C^2+A_1B^2-C_1BA_1B.$$From the trivial inequality, $$A_1C_1^2=C_1C^2+A_1B^2-C_1B\cdot A_1B\geq C_1B\cdot A_1B.$$Similarly, we find that $$A_1B_1^2=B_1C^2+A_1C^2-B_1C\cdot A_1C\geq B_1C\cdot A_1C \text{ and }$$$$ B_1C_1^2=B_1A^2+C_1A^2-B_1A\cdot C_1A\geq B_1A\cdot C_1A.$$Multiplying these inequalities together, we obtain $$(A_1B_1 \cdot B_1C_1 \cdot C_1A_1)^2 \geq A_1B \cdot C_1B \cdot B_1C \cdot A_1C\cdot C_1A \cdot B_1A=$$$$(A_1B\cdot B_1C \cdot C_1A)\cdot(C_1B \cdot A_1C \cdot B_1A)=(A_1B\cdot B_1C \cdot C_1A)^2\Leftrightarrow$$$$A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A,$$as desired.
HamstPan38825
11.09.2022 01:16
By Ceva it suffices to show that $$A_1B_1^2 \cdot A_1C_1^2 \cdot B_1C_1^2 \geq B_1C \cdot CA_1 \cdot A_1B \cdot BC_1 \cdot C_1A \cdot AB_1.$$But $$A_1B_1^2 \geq CB_1 \cdot CA_1 \iff \sin \angle CB_1A_1 \cdot \sin \angle CA_1B_1 \leq \frac 34,$$but this is evident as the angles sum to $120^\circ$.