Maths Olympiad Prep

Library / /98 of 348

Geometry Difficulty 4.8 AIME Find the answer

Find the area of triangle EFCEFC given that [EFC]=(56)[AEC]=(56)(45)[ADC]=(56)(45)(23)[ABC][EFC]=\left(\frac{5}{6}\right)[AEC]=\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)[ADC]=\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)\left(\frac{2}{3}\right)[ABC] and [ABC]=203[ABC]=20\sqrt{3}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

By shared bases, we know that [EFC]=(56)[AEC]=(56)(45)[ADC]=(56)(45)(23)[ABC][EFC]=\left(\frac{5}{6}\right)[AEC]=\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)[ADC]=\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)\left(\frac{2}{3}\right)[ABC] By Heron's formula, we find that [ABC]=(15)(8)(2)(5)=203[ABC]=\sqrt{(15)(8)(2)(5)}=20\sqrt{3}, so [AEC]=8039[AEC]=\frac{80\sqrt{3}}{9}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.