Difference between revisions of "1992 AIME Problems/Problem 9"
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== Solution 2 == | == 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>. | + | 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>. |
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+ | from solution 1 we get from 1 that h/r = 70/x and h/r = 50/ (92-x) | ||
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+ | this implies that 70/x =50/(92-x) so x = 161/3 | ||
== See also == | == See also == |
Revision as of 14:20, 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 .
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
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 |