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Difference between revisions of "1996 AHSME Problems/Problem 18"

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==Problem==
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A circle of radius <math>2</math> has center at <math>(2,0)</math>. A circle of radius <math>1</math> has center at <math>(5,0)</math>. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the <math>y</math>-intercept of the line?
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<math> \text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3 </math>
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==Solution==
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The two circles are tangent to eaech other at the point <math>(4,0)</math>, since it is both <math>2</math> units from <math>(2,0)</math> and <math>1</math> unit from <math>(5,0)</math>.
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Label the x-intercept of the common tangent line <math>A</math>, and label the y-intercept of the common tangent <math>B</math>.  Triangle <math>\triangle OAB</math> is a right triangle at the origin.
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Label <math>D</math> the point of tangency to the first, big circle, and label <math>E</math> the point of tangency to the small circle.  <math>\angle D</math> and <math>\angle E</math> are both right angles as well.
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Label the center of the big circle <math>F</math>, and the small circle <math>G</math>.
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<math>\triangle DAF  \sim \triangle EAG \sim \triangle OAB</math> because they each have one right angle, and also ahve a common angle <math>A</math>.
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The y-intercept is the length <math>OB</math>.
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We know that <math>DF = 2</math> and <math>EG = 1</math> because they are radii of circles.  From similarity, <math>\frac{DF}{EG} = \frac{FA}{GA}</math>.  Thus, <math>FA = 2GA</math>.
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Looking at line <math>FGA</math>, we know that <math>FG + GA = FA</math>.
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Thus, <math>FG + GA = 2GA</math>, meaning <math>FG = GA</math>.  Since <math>FG = 3</math> because it's the distance of the centers of the circles, so <math>GA = 3</math> as well.  This gives point <math>A</math> as <math>(8,0)</math>.
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Since <math>\triangle EAG</math> is a right triangle with <math>GA = 3</math> and <math>GE = 1</math>, we know that <math>AE = \sqrt{GA^2 - GE^2} = \sqrt{8}</math>
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Thus, all triangles are <math>1:\sqrt{8}:3</math> triangles.
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<math>\triangle OAC</math> has as its middle side <math>OA</math>, which is <math>8</math>.  Thus, all sides are scaled up by a factor of <math>\sqrt{8}</math>, and <math>OB</math>, the shortest side, is <math>\sqrt{8}</math>.  This means the y-intercept is <math>\sqrt{8} = 2\sqrt{2}</math>, and the answer is <math>\boxed{D}</math>.
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=17|num-a=19}}
 
{{AHSME box|year=1996|num-b=17|num-a=19}}

Revision as of 18:50, 19 August 2011

Problem

A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?

$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3$

Solution

The two circles are tangent to eaech other at the point $(4,0)$, since it is both $2$ units from $(2,0)$ and $1$ unit from $(5,0)$.

Label the x-intercept of the common tangent line $A$, and label the y-intercept of the common tangent $B$. Triangle $\triangle OAB$ is a right triangle at the origin.

Label $D$ the point of tangency to the first, big circle, and label $E$ the point of tangency to the small circle. $\angle D$ and $\angle E$ are both right angles as well.

Label the center of the big circle $F$, and the small circle $G$.

$\triangle DAF \sim \triangle EAG \sim \triangle OAB$ because they each have one right angle, and also ahve a common angle $A$.

The y-intercept is the length $OB$.

We know that $DF = 2$ and $EG = 1$ because they are radii of circles. From similarity, $\frac{DF}{EG} = \frac{FA}{GA}$. Thus, $FA = 2GA$.

Looking at line $FGA$, we know that $FG + GA = FA$.

Thus, $FG + GA = 2GA$, meaning $FG = GA$. Since $FG = 3$ because it's the distance of the centers of the circles, so $GA = 3$ as well. This gives point $A$ as $(8,0)$.

Since $\triangle EAG$ is a right triangle with $GA = 3$ and $GE = 1$, we know that $AE = \sqrt{GA^2 - GE^2} = \sqrt{8}$

Thus, all triangles are $1:\sqrt{8}:3$ triangles.

$\triangle OAC$ has as its middle side $OA$, which is $8$. Thus, all sides are scaled up by a factor of $\sqrt{8}$, and $OB$, the shortest side, is $\sqrt{8}$. This means the y-intercept is $\sqrt{8} = 2\sqrt{2}$, and the answer is $\boxed{D}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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