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Difference between revisions of "1992 AIME Problems/Problem 9"
(LaTeXed existing solution and added another.)
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Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>.
 
Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>.
  
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== Solution ==
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== Solution 1==
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{{solution}}
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Let <math>AB</math> be the base of the trapezoid and consider angles <math>A</math> and <math>B</math>.  Let <math>x=AP</math> and let <math>h</math> equal the height of the trapezoid.  Let <math>r</math> equal the radius of the circle.
  
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== See also ==
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Then
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{{AIME box|year=1992|num-b=8|num-a=10}}
 
  
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[[Category:Intermediate Geometry Problems]]
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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>
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let AB be the base of the trapezoid and consider angles A and B. Let x equal the distance from A to P and let h equal the height of the trapezoid.  Let r equal the radius of the circle.
 
  
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then
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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>.
  
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  1. sinA= r/x = h/70  and sinB= r/(92-x)  = h/50
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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>
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now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>.  
  
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let z be the distance along AB from A to where the perp from D meets AB
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== Solution 2 ==
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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>.
  
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then h^2 +z^2 =70^2  and (73-z)^2 + h^2 =50^2  so h =(44710959)^.5 /146
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== See also ==
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now substitute this into 1 to get x= 11753/219    kevin raponi
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{{AIME box|year=1992|num-b=8|num-a=10}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 13:50, 23 June 2008

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 $m+n = 164$.

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 $m+n = \boxed{164}$.

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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