Difference between revisions of "1992 AHSME Problems/Problem 30"
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== Solution == | == Solution == | ||
− | <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=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: | + | [[Category: Intermediate Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:00, 1 January 2015
Problem
Let be an isosceles trapezoid with bases and . Suppose and a circle with center on is tangent to segments and . If is the smallest possible value of , then =
Solution
Note that the center of the circle is the midpoint of , call it . When we decrease , the limiting condition is that the circle will eventually be tangent to segment at and segment at . That is, and .
From here, we drop the altitude from to ; call the base . Since , we have Thus, . Furthermore,
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 |
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