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Difference between revisions of "1972 AHSME Problems/Problem 13"

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== Solution ==  
 
== Solution ==  
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Let the line passing through <math>M</math> parallel to <math>AB</math> intersect <math>AD</math> and <math>BC</math> and <math>S</math> and <math>T</math> respectively. Since <math>M</math> is the midpoint of <math>AE</math>, <math>SM=\frac{5}{2}</math> and <math>TM=12-\frac{5}{2}=\frac{19}{2}</math>. Since <math>\triangle PSM\sim \triangle QTM</math>, <math>PM:MQ=SM:MT=5:9</math>, hence our answer is <math>\fbox{C}</math>.
+
Let the line passing through <math>M</math> parallel to <math>AB</math> intersect <math>AD</math> and <math>BC</math> and <math>S</math> and <math>T</math> respectively. Since <math>M</math> is the midpoint of <math>AE</math>, <math>SM=\frac{5}{2}</math> and <math>TM=12-\frac{5}{2}=\frac{19}{2}</math>. Since <math>\triangle PSM\sim \triangle QTM</math>, <math>PM:MQ=SM:MT=5:19</math>, hence our answer is <math>\fbox{C}</math>.

Latest revision as of 20:55, 8 November 2024

Problem 13

[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram [/asy]

Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE, AD$, and $BC$ at points $M, P$, and $Q$ respectively. The ratio of segment $PM$ to $MQ$ is

$\textbf{(A) }5:12\qquad \textbf{(B) }5:13\qquad \textbf{(C) }5:19\qquad \textbf{(D) }1:4\qquad \textbf{(E) }5:21$

Solution

Let the line passing through $M$ parallel to $AB$ intersect $AD$ and $BC$ and $S$ and $T$ respectively. Since $M$ is the midpoint of $AE$, $SM=\frac{5}{2}$ and $TM=12-\frac{5}{2}=\frac{19}{2}$. Since $\triangle PSM\sim \triangle QTM$, $PM:MQ=SM:MT=5:19$, hence our answer is $\fbox{C}$.

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