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"

(→‎Solution)
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 14: Line 14:
 
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}$.
+
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 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 (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=66736"