Difference between revisions of "1953 AHSME Problems/Problem 14"
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We will test each option to see if it can be true or not. Links to diagrams are provided. | We will test each option to see if it can be true or not. Links to diagrams are provided. | ||
<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> | ||
− | Let circle <math>Q</math> be inside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR - QR = PQ = p-q</math> | + | Let circle <math>Q</math> be inside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR - QR = PQ = p-q.</math>[https://latex.artofproblemsolving.com/d/5/8/d5896d95c00fde8b69428d09a084959a86e83dfa.png Diagram A] |
<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> | ||
− | Let circle <math>Q</math> be outside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR + QR = PQ = p+q</math> | + | Let circle <math>Q</math> be outside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR + QR = PQ = p+q.</math>[https://latex.artofproblemsolving.com/d/9/d/d9d46af414fdc1eca2b350c8eb991be067717b49.png Diagram B] |
<cmath>\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}</cmath> | <cmath>\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}</cmath> | ||
− | + | Let circle <math>Q</math> be outside circle <math>P</math> and not tangent to circle <math>P</math>, and the intersection of <math>\overline{PQ}</math> with the circles be <math>R</math> and <math>S</math> respectively. <math>PR = p</math> and <math>QS = q</math>, and <math>PR + QS < PQ</math>, so <math>p+q < PQ.</math> [https://latex.artofproblemsolving.com/2/0/d/20d001912b80a7a6ef4c267abdd424da9ca14784.png Diagram C] | |
<cmath>\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}</cmath> | <cmath>\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}</cmath> | ||
If circle <math>Q</math> is inside circle <math>P</math> and it is not tangent to circle <math>P</math>, then <math>PQ</math> is greater than <math>p-q</math>. | If circle <math>Q</math> is inside circle <math>P</math> and it is not tangent to circle <math>P</math>, then <math>PQ</math> is greater than <math>p-q</math>. |
Revision as of 15:04, 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. Links to diagrams are provided.
Let circle
be inside circle
and tangent to circle
, and the point of tangency be
.
, and
, so
Diagram A
Let circle
be outside circle
and tangent to circle
, and the point of tangency be
.
, and
, so
Diagram B
Let circle
be outside circle
and not tangent to circle
, and the intersection of
with the circles be
and
respectively.
and
, and
, so
Diagram C
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 | |
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.