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

(Solution 4)
m
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Then
 
Then
  
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<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>.
Line 44: Line 39:
 
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.)
  
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[CPD] = <math>\frac{1}{2} * 19 * h</math>
+
<math>[CPD] = \frac{1}{2} \cdot 19 \cdot h</math>
  
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[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>
Line 60: Line 55:
 
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
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