Difference between revisions of "1992 AIME Problems/Problem 9"
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Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. | Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. | ||
− | == Solution == | + | == Solution 1== |
− | + | Let <math>AB</math> be the base of the trapezoid and consider angles <math>A</math> and <math>B</math>. Let <math>x=AP</math> and let <math>h</math> equal the height of the trapezoid. Let <math>r</math> equal the radius of the circle. | |
− | + | Then | |
− | |||
− | + | <math>(1) \sin{A}= \frac{r}{x} = \frac{h}{70}</math> and <math>\sin{B}= \frac{r}{92-x} = \frac{h}{50}</math> | |
− | |||
− | + | Let <math>z</math> be the distance along <math>AB</math> from <math>A</math> to where the perp from <math>D</math> meets <math>AB</math>. | |
− | + | Then <math>h^2 +z^2 =70^2</math> and <math>(73-z)^2 + h^2 =50^2</math> so <math>h =\frac{\sqrt{44710959}}{146}</math> | |
+ | now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>. | ||
− | + | == Solution 2 == | |
+ | From <math>(1)</math> above, <math>x = \frac{70r}{h}</math> and <math>92-x = \frac{50r}{h}</math>. Adding these equations yields <math>92 = \frac{120r}{h}</math>. Thus, <math>x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}</math>, and <math>m+n = \boxed{164}</math>. | ||
− | + | == See also == | |
− | + | {{AIME box|year=1992|num-b=8|num-a=10}} | |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 13:50, 23 June 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 .
Solution 2
From above, and . Adding these equations yields . Thus, , and .
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 |