Difference between revisions of "1992 AIME Problems/Problem 9"

Revision as of 02:15, 19 July 2008

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

Solution 3

Extend $AD$ and $BC$ to meet at a point $X$ . Since $AB$ and $CD$ are parallel, $\triangle XCD ~ \triangle XAB$ . If $AX$ is further extended to a point $A'$ and $XB$ is extended to a point $B'$ such that $A'B'$ is tangent to $\circle P$ (Error compiling LaTeX. Unknown error_msg), we discover that $P$ is the incircle of triangle $XA'B'$ . Then line $XP$ is the angle bisector of $\angle AXB$ . By homothety, $P$ is the intersection of the angle bisector of $\triangle XAB$ with $AB$ . By the angle bisector theorem,

$\begin{align*} \frac{AX}{AP} &= \frac{XB}{BP}\\ \frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\ \frac{AD}{AP} &= \frac{BD}{PB}\\ &=\frac{7}{5} \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Let $7a = AP$ , then $AB = 7a + 5a = 12a$ . $AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}$ . Thus, $m + n = 164$ .

@@ Line 29: / Line 29: @@
 this implies that  70/x =50/(92-x)   so x = 161/3
+== Solution 3 ==
+Extend <math>AD</math> and <math>BC</math> to meet at a point <math>X</math>. Since <math>AB</math> and <math>CD</math> are parallel, <math>\triangle XCD ~ \triangle XAB</math>. If <math>AX</math> is further extended to a point <math>A'</math> and <math>XB</math> is extended to a point <math>B'</math> such that <math>A'B'</math> is tangent to <math>\circle P</math>, we discover that <math>P</math> is the incircle of triangle <math>XA'B'</math>. Then line <math>XP</math> is the angle bisector of <math>\angle AXB</math>. By homothety, <math>P</math> is the intersection of the angle bisector of <math>\triangle XAB</math> with <math>AB</math>. By the angle bisector theorem,
+<math>\begin{align*}
+\frac{AX}{AP} &= \frac{XB}{BP}\\
+\frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\
+\frac{AD}{AP} &= \frac{BD}{PB}\\
+&=\frac{7}{5}
+\end{align*}</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>.
 == See also ==
 {{AIME box|year=1992|num-b=8|num-a=10}}
 [[Category:Intermediate Geometry Problems]]

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

Page

Toolbox

Search