Difference between revisions of "1992 AHSME Problems/Problem 30"

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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Note that the center of the circle is the midpoint of <math>AB</math>, call it <math>M</math>. When we decrease <math>x</math>, the limiting condition is that the circle will eventually be tangent to segment <math>AD</math> at <math>D</math> and segment <math>BC</math> at <math>C</math>. That is, <math>MD\perp AD</math> and <math>MC\perp BC</math>.
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From here, we drop the altitude from <math>D</math> to <math>AM</math>; call the base <math>N</math>. Since <math>\triangle DNM \sim \triangle ADM</math>, we have
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<cmath>\frac{DM}{19/2}=\frac{46}{DM}.</cmath>
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Thus, <math>DM=\sqrt{19\cdot 23}</math>. Furthermore, <math>x^2=AM^2-DM^2=46^2-19\cdot 23=1679, \boxed{B}.</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1992|num-b=29|num-a=30}}   
 
{{AHSME box|year=1992|num-b=29|num-a=30}}   
  
[[Category: IntermediateGeometry Problems]]
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[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:00, 1 January 2015

Problem

Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19$. Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$. If $m$ is the smallest possible value of $x$, then $m^2$=

$\text{(A) } 1369\quad \text{(B) } 1679\quad \text{(C) } 1748\quad \text{(D) } 2109\quad \text{(E) } 8825$

Solution

Note that the center of the circle is the midpoint of $AB$, call it $M$. When we decrease $x$, the limiting condition is that the circle will eventually be tangent to segment $AD$ at $D$ and segment $BC$ at $C$. That is, $MD\perp AD$ and $MC\perp BC$.

From here, we drop the altitude from $D$ to $AM$; call the base $N$. Since $\triangle DNM \sim \triangle ADM$, we have \[\frac{DM}{19/2}=\frac{46}{DM}.\] Thus, $DM=\sqrt{19\cdot 23}$. Furthermore, $x^2=AM^2-DM^2=46^2-19\cdot 23=1679, \boxed{B}.$

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 30
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