Track / Stage 5 / 100 of 400 #700 of 1964

Problem 700

AIME late

Geometry Difficulty 5.3 Find the answer

5. As shown in the figure, the area of $\triangle \mathrm{ABC}$ is $60, \mathrm{E}$ and $\mathrm{F}$ are points on $\mathrm{AB}$ and $\mathrm{AC}$ respectively, satisfying $\mathrm{AB}=3 \mathrm{AE}, \mathrm{AC}=3 \mathrm{AF}$ , point $D$ is a moving point on segment $B C$ , let the area of $\triangle F B D$ be $S_{1}$ , the area of $\triangle E D C$ be $S_{2}$ , then the maximum value of $S_{1} \times S_{2}$ is $\qquad$

Your answer

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

Official solution

【Analysis】Since $\frac{A E}{A B}=\frac{A F}{A C}=\frac{1}{3}$ , therefore $\mathrm{EF} / / \mathrm{BC}$
Therefore $\mathrm{S}_{\mathrm{EBD}}=\mathrm{S}_{\mathrm{FBD}}=\mathrm{S}_{1} \rightarrow \mathrm{S}_{1}+\mathrm{S}_{2}=\mathrm{S}_{\mathrm{EBC}}=\frac{2}{3} \mathrm{~S}_{\mathrm{ABC}}=40$
When the sum is constant, the smaller the difference, the greater the product, so when $S_{1}=S_{2}$ , that is, when $D$ is the midpoint, $S_{1} \times S_{2}$ is maximized at $20 \times 20=400$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.