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Difference between revisions of "1992 AIME Problems/Problem 9"
 
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Let <math>AP=x</math> so that <math>PB=92-x.</math> Extend <math>AD, BC</math> to meet at <math>X,</math> and note that <math>XP</math> bisects <math>\angle AXB;</math> let it meet <math>CD</math> at <math>E.</math> Using the angle bisector theorem, we let <math>XB=y(92-x), XA=xy</math> for some <math>y.</math>  
 
Let <math>AP=x</math> so that <math>PB=92-x.</math> Extend <math>AD, BC</math> to meet at <math>X,</math> and note that <math>XP</math> bisects <math>\angle AXB;</math> let it meet <math>CD</math> at <math>E.</math> Using the angle bisector theorem, we let <math>XB=y(92-x), XA=xy</math> for some <math>y.</math>  
  
−
Then <math>XD=xy-70, XC=y(92-x)-50,</math> thus <cmath>\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},</cmath> which we can rearrange, expand and cancel to get <math>120x=70\cdot 92,</math> hence <math>AP=x=\frac{161}{3}.</math>
+
Then <math>XD=xy-70, XC=y(92-x)-50,</math> thus <cmath>\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},</cmath> which we can rearrange, expand and cancel to get <math>120x=70\cdot 92,</math> hence <math>AP=x=\frac{161}{3}</math>. This gives us a final answer of <math>161+3=\boxed{164}</math>
  
 
== Solution 2==
 
== Solution 2==
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Substituting <math>r = \frac{23h}{30}</math>, we find <math>x = \frac{161}{3}</math>, hence the answer is <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>.
 +
==Solution 6==
 +
As the problem tells, the circle is tangent to both sides <math>AD,BC</math>, we can make it up to a triangle <math>QAB</math> and <math>P</math> must lie on its angular bisector. Then we know that <math>AP:BP=7:5</math>, which makes <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 09:27, 22 August 2021

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 $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$

Then $XD=xy-70, XC=y(92-x)-50,$ thus \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},\] which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$. This gives us a final answer of $161+3=\boxed{164}$

Solution 2

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 3

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 4

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{BC}{BP}\\ &=\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, $m + n = 164$.

Note: this solution shows that the length of $CD$ is irrelevant as long as there still exists a circle as described in the problem.

Solution 5

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 $\boxed{164}$.

Solution 6

As the problem tells, the circle is tangent to both sides $AD,BC$, we can make it up to a triangle $QAB$ and $P$ must lie on its angular bisector. Then we know that $AP:BP=7:5$, which makes $AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}$. Thus, $m + n = 164$.

See also

1992 AIME (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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