1992 AIME Problems/Problem 9

Problem

Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ , $BC=50^{}_{}$ , $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$ .

Solution 1

Let $AB$ be the base of the trapezoid and consider angles $A$ and $B$ . Let $x=AP$ and let $h$ equal the height of the trapezoid. Let $r$ equal the radius of the circle.

Then

$(1) \sin{A}= \frac{r}{x} = \frac{h}{70}$ and $\sin{B}= \frac{r}{92-x} = \frac{h}{50}$

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 $h =\frac{\sqrt{44710959}}{146}$ now substitute this into $(1)$ to get $x= \frac{11753}{219} = \frac{161}{3}$ and $m+n = 164$ .

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 $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$ . Adding these equations yields $92 = \frac{120r}{h}$ . Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$ , and $m+n = \boxed{164}$ .

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

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
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1992 AIME Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also