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Difference between revisions of "1953 AHSME Problems/Problem 42"

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==Solution==
 
==Solution==
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<asy>
 +
size(400);
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draw((0,0)--(41,0));
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draw((0,0)--(45/41,200/41)--(1645/41,-160/41));
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draw((0,0)--(1600/41,-360/41)--(41,0));
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draw(circle((0,0),5));
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draw(circle((41,0),4));
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label("$A$",(0,0),W);
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label("$B$",(41,0),E);
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label("$C$",(45/41,200/41),N);
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label("$D$",(1645/41,-160/41),SE);
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label("$E$",(1600/41,-360/41),E);
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</asy>
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 +
Let <math>A</math> be the center of the circle with radius <math>5</math>, and <math>B</math> be the center of the circle with radius <math>4</math>. Let <math>\overline{CD}</math> be the common internal tangent of circle <math>A</math> and circle <math>B</math>. Extend <math>\overline{BD}</math> past <math>D</math> to point <math>E</math> such that <math>\overline{BE}\perp\overline{AE}</math>. Since <math>\overline{AC}\perp\overline{CD}</math> and <math>\overline{BD}\perp\overline{CD}</math>, <math>ACDE</math> is a rectangle. Therefore, <math>AC=DE</math> and <math>CD=AE</math>.
 +
 +
Since the centers of the two circles are <math>41</math> inches apart, <math>AB=41</math>. Also, <math>BE=4+5=9</math>. Using the [[Pythagorean Theorem]] on right triangle <math>ABE</math>, <math>CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40</math>. The length of the common internal tangent is <math>\boxed{\textbf{(E) } 40\text{ inches}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 00:59, 14 February 2020

Problem

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. The length of the common internal tangent is:

$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Solution

[asy] size(400); draw((0,0)--(41,0)); draw((0,0)--(45/41,200/41)--(1645/41,-160/41)); draw((0,0)--(1600/41,-360/41)--(41,0)); draw(circle((0,0),5)); draw(circle((41,0),4)); label("$A$",(0,0),W); label("$B$",(41,0),E); label("$C$",(45/41,200/41),N); label("$D$",(1645/41,-160/41),SE); label("$E$",(1600/41,-360/41),E); [/asy]

Let $A$ be the center of the circle with radius $5$, and $B$ be the center of the circle with radius $4$. Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$. Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$. Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$, $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$.

Since the centers of the two circles are $41$ inches apart, $AB=41$. Also, $BE=4+5=9$. Using the Pythagorean Theorem on right triangle $ABE$, $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$. The length of the common internal tangent is $\boxed{\textbf{(E) } 40\text{ inches}}$

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 41 		Followed by
Problem 43
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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