Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎See also)
(→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
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>.
 +
 
 +
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>
 +
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}$.
  
 
== See also ==
 
== See also ==

Revision as of 15:33, 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, $x^2=DN^2+AN^2=(DM^2-NM^2)+AN^2=19\cdot 23-\left(\frac{19}{2}\right)^2+\left(\frac{73}{2}\right)^2=1679.$\boxed{B}$.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_30&oldid=66734"