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

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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
 
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
 
<cmath>\frac{DM}{19/2}=\frac{46}{DM}.</cmath>
 
<cmath>\frac{DM}{19/2}=\frac{46}{DM}.</cmath>
Thus, <math>DM=\sqrt{19\cdot 23}</math>. Furthermore, <math>x^2=DN^2+AN^2=(DM^2-NM^2)+AN^2=19\cdot 23-\left(\frac{19}{2}\right)^2+\left(\frac{73}{2}\right)^2=1679. </math>\boxed{B}$.
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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 ==

Latest revision as of 16:00, 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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