Difference between revisions of "1992 AIME Problems/Problem 9"
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| + | let AB be the base of the trapezoid and consider angles A and B. Let x equal the distance from A to P and let h equal the height of the trapezoid. Let r equal the radius of the circle. | ||
| + | |||
| + | then | ||
| + | |||
| + | 1. sinA= r/x = h/70 and sinB= r/(92-x) = h/50 | ||
| + | |||
| + | let z be the distance along AB from A to where the perp from D meets AB | ||
| + | |||
| + | then h^2 +z^2 =70^2 and (73-z)^2 + h^2 =50^2 so h =(44710959)^.5 /146 | ||
| + | now substitute this into 1 to get x= 11753/219 kevin raponi | ||
Revision as of 10:47, 23 June 2008
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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 | ||
let AB be the base of the trapezoid and consider angles A and B. Let x equal the distance from A to P and let h equal the height of the trapezoid. Let r equal the radius of the circle.
then
1. sinA= r/x = h/70 and sinB= r/(92-x) = h/50
let z be the distance along AB from A to where the perp from D meets AB
then h^2 +z^2 =70^2 and (73-z)^2 + h^2 =50^2 so h =(44710959)^.5 /146 now substitute this into 1 to get x= 11753/219 kevin raponi
