Track / Stage 4 / 313 of 340 #573 of 2000

Problem 573

AMC 12 late, AIME early

Geometry Difficulty 5.0 Find the answer

18.55 $ABCD$ is a rectangle (as shown in the figure), $P$ is any point on $AB$ , $PS \perp BD$ , $PR \perp AC$ , $AF \perp BD$ , $PQ \perp AF$ , then $PR + PS$ equals
(A) $PQ$ .
(B) $AE$ .
(C) $PT + AT$ .
(D) $AF$ .
(E) $EF$ .

Your answer

Official solution

[Solution] Connect $P E$ . Given that $A B C D$ is a rectangle, then $A E=B E=D E$ , let it be $a$ .
$\text { Also } \begin{aligned} S_{\triangle A E B} & =\frac{1}{2} \cdot A E \cdot P R+\frac{1}{2} \cdot B E \cdot P S \\ & =\frac{1}{2} a(P R+P S), \end{aligned}$

and $S_{\triangle A E D}=\frac{1}{2} \cdot D E \cdot A F=\frac{1}{2} a A F$ ,
$\begin{array}{ll} \because & S_{\triangle A E B}=S_{\triangle A E D}, \\ \therefore & P R+P S=A F . \end{array}$

Therefore, the answer is $(D)$ .

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