Difference between revisions of "1953 AHSME Problems/Problem 42"
(Created page with "The answer is E just cause it can be") |
m |
||
(4 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
− | The | + | ==Problem== |
+ | |||
+ | The centers of two circles are <math>41</math> inches apart. The smaller circle has a radius of <math>4</math> inches and the larger one has a radius of <math>5</math> inches. | ||
+ | The length of the common internal tangent is: | ||
+ | |||
+ | <math>\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} </math> | ||
+ | |||
+ | ==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== | ||
+ | |||
+ | {{AHSME 50p box|year=1953|num-b=41|num-a=43}} | ||
+ | |||
+ | {{MAA Notice}} |
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
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.