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Difference between revisions of "1991 AHSME Problems/Problem 10"

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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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<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 $P$ is $9$ units from the center of a circle of radius $15$. How many different chords of the circle contain $P$ and have integer lengths?

(A) 11 (B) 12 (C) 13 (D) 14 (E) 29

Solution

$\fbox{B}$ Let the chord be $AB$, and let the diameter passing through $P$ be $CD$. Then we have $PC = 15-9 = 6$ and $PD = 15+9 = 24$. By power of a point, we now get $6 \times 24 = AP \times PB \implies AP \times PB = 144$. Now AM-GM gives $AB = AP + PB \geq 2 \sqrt{AP \times PB} = 2 \times 12 = 24$. Clearly we also know $AC \leq 30$ as the diameter is the longest chord of a circle. Hence there are 7 possible values of $AC$, and each gives two chords as we can reflect the chord, except that for the $24$ and the $30$ we can't do this as it gives the same chord, so the answer is $7 \times 2 - 2 = 12.$

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

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