Difference between revisions of "1953 AHSME Problems/Problem 14"
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<cmath>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}</cmath> | <cmath>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}</cmath> | ||
If circle <math>Q</math> is inside circle <math>P</math> and it is tangent to circle <math>P</math>, then <math>PQ</math> is <math>p-q</math>. | If circle <math>Q</math> is inside circle <math>P</math> and it is tangent to circle <math>P</math>, then <math>PQ</math> is <math>p-q</math>. | ||
+ | [asy] | ||
+ | pair P = (0,0); | ||
+ | pair Q = (3,0); | ||
+ | draw(Circle(P, 4)); | ||
+ | draw(Circle(Q, 1)); | ||
+ | draw(P -- (4,0)); | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | [/asy] | ||
<cmath>\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}</cmath> | <cmath>\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}</cmath> | ||
If circle <math>Q</math> is outside circle <math>P</math> and it is tangent to circle <math>P</math>, then <math>PQ</math> is <math>p+q</math>. | If circle <math>Q</math> is outside circle <math>P</math> and it is tangent to circle <math>P</math>, then <math>PQ</math> is <math>p+q</math>. |
Revision as of 14:50, 15 July 2018
Problem 14
Given the larger of two circles with center and radius and the smaller with center and radius . Draw . Which of the following statements is false?
Solution
We will test each option to see if it can be true or not. If circle is inside circle and it is tangent to circle , then is . [asy] pair P = (0,0); pair Q = (3,0); draw(Circle(P, 4)); draw(Circle(Q, 1)); draw(P -- (4,0)); dot(P); dot(Q); [/asy] If circle is outside circle and it is tangent to circle , then is . If circle is outside circle and it is not tangent to circle , then is greater than . If circle is inside circle and it is not tangent to circle , then is greater than . Since options A, B, C, and D can be true, the answer must be .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |
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