Difference between revisions of "1992 AIME Problems/Problem 9"
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this implies that 70/x =50/(92-x) so x = 161/3 | this implies that 70/x =50/(92-x) so x = 161/3 | ||
+ | == Solution 3 == | ||
+ | Extend <math>AD</math> and <math>BC</math> to meet at a point <math>X</math>. Since <math>AB</math> and <math>CD</math> are parallel, <math>\triangle XCD ~ \triangle XAB</math>. If <math>AX</math> is further extended to a point <math>A'</math> and <math>XB</math> is extended to a point <math>B'</math> such that <math>A'B'</math> is tangent to <math>\circle P</math>, we discover that <math>P</math> is the incircle of triangle <math>XA'B'</math>. Then line <math>XP</math> is the angle bisector of <math>\angle AXB</math>. By homothety, <math>P</math> is the intersection of the angle bisector of <math>\triangle XAB</math> with <math>AB</math>. By the angle bisector theorem, | ||
+ | |||
+ | <math>\begin{align*} | ||
+ | \frac{AX}{AP} &= \frac{XB}{BP}\ | ||
+ | \frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\ | ||
+ | \frac{AD}{AP} &= \frac{BD}{PB}\ | ||
+ | &=\frac{7}{5} | ||
+ | \end{align*}</math> | ||
+ | |||
+ | Let <math>7a = AP</math>, then <math>AB = 7a + 5a = 12a</math>. <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1992|num-b=8|num-a=10}} | {{AIME box|year=1992|num-b=8|num-a=10}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 01:15, 19 July 2008
Contents
[hide]Problem
Trapezoid has sides , , , and , with parallel to . A circle with center on is drawn tangent to and . Given that , where and are relatively prime positive integers, find .
Solution 1
Let be the base of the trapezoid and consider angles and . Let and let equal the height of the trapezoid. Let equal the radius of the circle.
Then
and
Let be the distance along from to where the perp from meets .
Then and so now substitute this into to get and .
you don;t have to use trig nor angles A and B ..From similar triangles,
h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
Solution 2
From above, and . Adding these equations yields . Thus, , and .
from solution 1 we get from 1 that h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
Solution 3
Extend and to meet at a point . Since and are parallel, . If is further extended to a point and is extended to a point such that is tangent to $\circle P$ (Error compiling LaTeX. Unknown error_msg), we discover that is the incircle of triangle . Then line is the angle bisector of . By homothety, is the intersection of the angle bisector of with . By the angle bisector theorem,
$
Let , then . . Thus, .
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |