Problem

1958 AHSME Problem 47

ABCD is a rectangle (see the accompanying diagram) with P any point on \overline{AB}. \overline{PS} \perp \overline{BD} and \overline{PR} \perp \overline{AC}. \overline{AF} \perp \overline{BD} and \overline{PQ} \perp \overline{AF}. Then PR + PS is equal to:

[asy] draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--cycle,dot); draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); [/asy]

\textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT + AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.


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Category: AHSME
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