Difference between revisions of "1991 AHSME Problems/Problem 10"
m (→Solution) |
(Added a solution with explanation) |
||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | <math>\fbox{B}</math> Let the chord be <math>AB</math>, and let the diameter passing through <math>P</math> be <math>CD</math>. Then we have <math>PC = 15-9 = 6</math> and <math>PD = 15+9 = 24</math>. By power of a point, we now get <math>6 \times 24 = AP \times PB \implies AP \times PB = 144</math>. Now AM-GM gives <math>AB = AP + PB \geq 2 \sqrt{AP \times PB} = 2 \times 12 = 24</math>. Clearly we also know <math>AC \leq 30</math> as the diameter is the longest chord of a circle. Hence there are 7 possible values of <math>AC</math>, and each gives two chords as we can reflect the chord, except that for the <math>24</math> and the <math>30</math> we can't do this as it gives the same chord, so the answer is <math>7 \times 2 - 2 = 12.</math> |
== See also == | == See also == |
Revision as of 06:17, 24 February 2018
Problem
Point is units from the center of a circle of radius . How many different chords of the circle contain and have integer lengths?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 29
Solution
Let the chord be , and let the diameter passing through be . Then we have and . By power of a point, we now get . Now AM-GM gives . Clearly we also know as the diameter is the longest chord of a circle. Hence there are 7 possible values of , and each gives two chords as we can reflect the chord, except that for the and the we can't do this as it gives the same chord, so the answer is
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.