Difference between revisions of "1958 AHSME Problems/Problem 47"

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== Problem ==
 
== Problem ==
<math> ABCD</math> is a rectangle (see the accompanying diagram) with <math> P</math> any point on <math> \overline{AB}</math>. <math> \overline{PS} \perp \overline{BD}</math> and <math> \overline{PR} \perp \overline{AC}</math>. <math> \overline{AF} \perp \overline{BD}</math> and <math> \overline{PQ} \perp \overline{AF}</math>. Then <math> PR \plus{} PS</math> is equal to:
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<math> ABCD</math> is a rectangle (see the accompanying diagram) with <math> P</math> any point on <math> \overline{AB}</math>. <math> \overline{PS} \perp \overline{BD}</math> and <math> \overline{PR} \perp \overline{AC}</math>. <math> \overline{AF} \perp \overline{BD}</math> and <math> \overline{PQ} \perp \overline{AF}</math>. Then <math> PR + PS</math> is equal to:
  
 
<asy>
 
<asy>
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<math> \textbf{(A)}\ PQ\qquad  
 
<math> \textbf{(A)}\ PQ\qquad  
 
\textbf{(B)}\ AE\qquad  
 
\textbf{(B)}\ AE\qquad  
\textbf{(C)}\ PT \plus{} AT\qquad  
+
\textbf{(C)}\ PT + AT\qquad  
 
\textbf{(D)}\ AF\qquad  
 
\textbf{(D)}\ AF\qquad  
 
\textbf{(E)}\ EF</math>
 
\textbf{(E)}\ EF</math>
 
  
 
== Solution ==
 
== Solution ==

Revision as of 08:38, 18 May 2016

Problem

$ABCD$ 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]

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

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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