Difference between revisions of "1953 AHSME Problems/Problem 42"
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==Solution== | ==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 <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 01:59, 14 February 2020
Problem
The centers of two circles are inches apart. The smaller circle has a radius of inches and the larger one has a radius of inches. The length of the common internal tangent is:
Solution
Let be the center of the circle with radius , and be the center of the circle with radius . Let be the common internal tangent of circle and circle . Extend past to point such that . Since and , is a rectangle. Therefore, and .
Since the centers of the two circles are inches apart, . Also, . Using the Pythagorean Theorem on right triangle , . The length of the common internal tangent is
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 41 |
Followed by Problem 43 | |
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All AHSME Problems and Solutions |
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