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1992 AIME Problems/Problem 9: Difference between revisions

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Solution 4
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Then
Then


<math>(1) \sin{A}= \frac{r}{x} = \frac{h}{70}</math>  and <math>\sin{B}= \frac{r}{92-x}  =  \frac{h}{50}</math>  
<cmath>\sin{A}= \frac{r}{x} = \frac{h}{70}\qquad\text{ and }\qquad\sin{B}= \frac{r}{92-x}  =  \frac{h}{50}.\tag{1}</cmath>


Let <math>z</math> be the distance along <math>AB</math> from <math>A</math> to where the perp from <math>D</math> meets <math>AB</math>.  
Let <math>z</math> be the distance along <math>AB</math> from <math>A</math> to where the perp from <math>D</math> meets <math>AB</math>.  


Then <math>h^2 +z^2 =70^2</math>  and <math>(73-z)^2 + h^2 =50^2</math>  so <math>h =\frac{\sqrt{44710959}}{146}</math>  
Then <math>h^2 +z^2 =70^2</math>  and <math>(73-z)^2 + h^2 =50^2</math>  so <math>h =\frac{\sqrt{44710959}}{146}</math>.
now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>.
We can substitute this into <math>(1)</math> to find that <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>.


you don't have to use trig nor angles A and B. From similar triangles,  
<b>Remark:</b> One can come up with the equations in <math>(1)</math> without directly resorting to trig. From similar triangles,  
<math>h/r = 70/x</math>  and <math>h/r = 50/ (92-x)</math>
<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>


== Solution 2 ==
== Solution 2 ==
From <math>(1)</math> above, <math>x = \frac{70r}{h}</math> and <math>92-x = \frac{50r}{h}</math>. Adding these equations yields <math>92 = \frac{120r}{h}</math>. Thus, <math>x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}</math>, and <math>m+n = \boxed{164}</math>.
From <math>(1)</math> above, <math>x = \frac{70r}{h}</math> and <math>92-x = \frac{50r}{h}</math>. Adding these equations yields <math>92 = \frac{120r}{h}</math>. Thus, <math>x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}</math>, and <math>m+n = \boxed{164}</math>.


We can use <math>(1)</math> from Solution 1 to find that <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>
 
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 ==
== 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 circle <math>P</math>, we discover that circle <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,  
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 circle <math>P</math>, we discover that circle <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*}
<cmath>\begin{align*}
\frac{AX}{AP} &= \frac{XB}{BP}\\
\frac{AX}{AP} &= \frac{XB}{BP}\\
\frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\
\frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\
\frac{AD}{AP} &= \frac{BD}{PB}\\
\frac{AD}{AP} &= \frac{BD}{PB}\\
&=\frac{7}{5}
&=\frac{7}{5}
\end{align*}</math>
\end{align*}</cmath>


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>.
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The area of the trapezoid is <math>\frac{(19+92)h}{2}</math>, where <math>h</math> is the height of the trapezoid.
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.
Draw lines <math>CP</math> and <math>BP</math>. We can now find the area of the trapezoid as the sum of the areas of the three triangles <math>BPC</math>, <math>CPD</math>, and <math>PBA</math>.


[BPC] = <math>\frac{1}{2} * 50 * r</math> (where <math>r</math> is the radius of the tangent circle.)
<math>[BPC] = \frac{1}{2} \cdot 50 \cdot r</math> (where <math>r</math> is the radius of the tangent circle.)


[CPD] = <math>\frac{1}{2} * 19 * h</math>
<math>[CPD] = \frac{1}{2} \cdot 19 \cdot h</math>


[PBA] = <math>\frac{1}{2} * 70 * r</math>
<math>[PBA] = \frac{1}{2} \cdot 70 \cdot r</math>


[BPC] + [CPD] + [PBA] = <math>60r + \frac{19h}{2}</math> = Trapezoid area = <math>\frac{(19+92)h}{2}</math>
<math>[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}</math>


<math>60r = 46h</math>
<math>60r = 46h</math>
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From Solution 1 above, <math>\frac{h}{70} = \frac{r}{x}</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>.
Substituting <math>r = \frac{23h}{30}</math>, we find <math>x = \frac{161}{3}</math>, hence the answer is <math>\boxed{164}</math>.


== See also ==
== See also ==

Revision as of 15:25, 13 March 2015

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

\[\sin{A}= \frac{r}{x} = \frac{h}{70}\qquad\text{ and }\qquad\sin{B}= \frac{r}{92-x} = \frac{h}{50}.\tag{1}\]

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

Remark: One can come up with the equations in $(1)$ without directly resorting to trig. 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 $m+n = \boxed{164}$.

We can use $(1)$ from Solution 1 to find 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*}

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 areas of the three triangles $BPC$, $CPD$, and $PBA$.

$[BPC] = \frac{1}{2} \cdot 50 \cdot r$ (where $r$ is the radius of the tangent circle.)

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

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

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

$60r = 46h$

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

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

Substituting $r = \frac{23h}{30}$, we find $x = \frac{161}{3}$, hence the answer is $\boxed{164}$.

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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