Difference between revisions of "1992 AIME Problems/Problem 9"
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Kevinr9828 (talk | contribs) (→Solution 1) |
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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> | 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>. | + | now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>. |
+ | |||
+ | 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 == | == Solution 2 == |
Revision as of 14:53, 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 .
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
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 |