Track / Stage 4 / 85 of 340 #345 of 1964

Problem 345

AMC 12 late, AIME early

Geometry Difficulty 4.7 Multiple choice

5. As shown in Figure 2, in $\triangle A B C$ , $E F$ $/ / B C, S_{\triangle A E F}=S_{\triangle B C E}$ . If $S_{\triangle B B C}$ $=1$ , then $S_{\triangle C E F}$ equals:

Pick one

Official solution

5. (C).

Let $S_{\triangle C E F}=x$ .
Then $S_{\triangle A E F}=S_{\triangle B C E}=\frac{1-x}{2}, S_{\triangle A B C}=\frac{1+x}{2}$ .
Thus, $\frac{A F}{A C}=\frac{S_{\triangle A E F}}{S_{\triangle M B C}}=\frac{1-x}{1+x}$ .
But $E F \parallel B C,\left(\frac{A F}{A C}\right)^{2}=\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\left(\frac{1-x}{1+x}\right)^{2}$ ,
and $\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\frac{1-x}{2}$ , solving for $x$ yields $x=\sqrt{5}-2$ .

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