Difference between revisions of "1992 AIME Problems/Problem 9"
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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>. | 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | The area of the trapezoid is <math>\frac{(19+92)h}{2}</math>, where <math>h</math> is the height of the trapezoid. | ||
+ | |||
+ | Draw lines CP and BP. We can now find the area of the trapezoid as the sum of the three triangles BPC, CPD, and PBA. | ||
+ | |||
+ | [BPC] = <math>\frac{1}{2} * 50 * r</math> (where <math>r</math> is the radius of the tangent circle.) | ||
+ | |||
+ | [CPD] = <math>\frac{1}{2} * 19 * h</math> | ||
+ | |||
+ | [PBA] = <math>\frac{1}{2} * 70 * r</math> | ||
+ | |||
+ | [BPC] + [CPD] + [PBA] = <math>60r + \frac{19h}{2}</math> = Trapezoid area = <math>\frac{(19+92)h}{2}</math> | ||
+ | |||
+ | <math>60r = 46h</math> | ||
+ | |||
+ | <math>r = \frac{23h}{30}</math> | ||
+ | |||
+ | From Solution 1 above, <math>\frac{h}{70} = \frac{r}{x}</math> | ||
+ | |||
+ | Substituting <math>r = \frac{23h}{30}</math>, we get <math>x = \frac{161}{3}</math> --> <math>\boxed{164}</math>. | ||
== See also == | == See also == |
Revision as of 12:09, 3 March 2014
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, and
this implies that so
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 , we discover that circle 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, .
Solution 4
The area of the trapezoid is , where is the height of the trapezoid.
Draw lines CP and BP. We can now find the area of the trapezoid as the sum of the three triangles BPC, CPD, and PBA.
[BPC] = (where is the radius of the tangent circle.)
[CPD] =
[PBA] =
[BPC] + [CPD] + [PBA] = = Trapezoid area =
From Solution 1 above,
Substituting , we get --> .
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 |
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