1992 AIME Problems/Problem 9

Revision as of 13:09, 3 March 2014 by Sanqiang3 (talk | contribs) (Solution 4)

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. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

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

Solution 4

The area of the trapezoid is $\frac{(19+92)h}{2}$, where $h$ 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] = $\frac{1}{2} * 50 * r$ (where $r$ is the radius of the tangent circle.)

[CPD] = $\frac{1}{2} * 19 * h$

[PBA] = $\frac{1}{2} * 70 * r$

[BPC] + [CPD] + [PBA] = $60r + \frac{19h}{2}$ = Trapezoid area = $\frac{(19+92)h}{2}$

$60r = 46h$

$r = \frac{23h}{30}$

From Solution 1 above, $\frac{h}{70} = \frac{r}{x}$

Substituting $r = \frac{23h}{30}$, we get $x = \frac{161}{3}$ --> $\boxed{164}$.

See also

1992 AIME (ProblemsAnswer KeyResources)
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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