Difference between revisions of "1953 AHSME Problems/Problem 42"
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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>. | 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}</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 
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
![[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]](http://latex.artofproblemsolving.com/6/e/8/6e8731daebe13a53605a0de780cb4bc2def89b24.png)
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
