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

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you don't have to use trig  nor angles A and B. From similar triangles,  
 
you don't have to use trig  nor angles A and B. From similar triangles,  
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<math>h/r = 70/x  and h/r = 50/ (92-x)</math>
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<math>h/r = 70/x</math> and <math>h/r = 50/ (92-x)</math>
  
 
this implies that <math>70/x =50/(92-x)</math>  so <math>x = 161/3</math>
 
this implies that <math>70/x =50/(92-x)</math>  so <math>x = 161/3</math>

Revision as of 23:12, 13 March 2012

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 $P$, we discover that circle $P$ 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, $m + n = 164$.

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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