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

(Added a solution with explanation)
(→‎See also)
 
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== Solution ==
 
== Solution ==
−
<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>
+
Let <math>O</math> be the center of the circle, and let the chord passing through <math>P</math> that is perpendicular to <math>OP</math> intersect the circle at <math>Q</math> and <math>R</math>. Then <math>OP = 9</math> and <math>OQ = 15</math>, so by the Pythagorean Theorem, <math>PQ = 12</math>. By symmetry, <math>PR = 12</math>.
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<asy>
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import graph;
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unitsize(0.15 cm);
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pair O, P, Q, R;
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 +
O = (0,0);
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P = (9,0);
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Q = (9,12);
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R = (9,-12);
 +
 
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draw(Circle(O,15));
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draw((-15,0)--(15,0));
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draw(O--Q);
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draw(Q--R,red);
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dot("$O$", O, S);
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dot("$P$", P, NW);
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dot("$Q$", Q, NE);
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dot("$R$",R,SE);
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label("$15$", (-15/2,0), S);
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label("$9$", (O + P)/2, S);
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label("$6$", (12,0), S);
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label("$15$", (O + Q)/2, NW);
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label("$12$", (P + Q)/2, E);
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[/asy]
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Let $AB$ be the diameter passing through $P$.
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[asy]
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import graph;
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unitsize(0.15 cm);
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pair A, B, O, P, Q, R;
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A = (-15,0);
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B = (15,0);
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O = (0,0);
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P = (9,0);
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Q = (9,12);
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R = (9,-12);
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draw(Circle(O,15));
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draw(A--B,red);
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draw(Q--R);
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dot("$A$", A, W);
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dot("$B$", B, E);
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dot("$O$", O, S);
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dot("$P$", P, NW);
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dot("$Q$", Q, NE);
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dot("$R$", R, SE);
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label("$15$", (-15/2,0), S);
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label("$9$", (O + P)/2, S);
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label("$6$", (12,0), S);
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label("$12$", (P + Q)/2, E);
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label("$12$", (P + R)/2, E);
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[/asy]
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 +
Then the longest chord of the circle that passes through $P$ is $AB$, which has length 30, and the shortest chord is $QR$, which has length 24. If we rotate the red chord (while ensuring it passes through $P$), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer $n=25,26,27,28,29$, there are two chords of length $n$ passing through $P$, as seen in this picture:
 +
 
 +
[asy]
 +
import graph;
 +
 
 +
unitsize(0.15 cm);
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pair O, P,A,B,A2,B2;
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O = (0,0);
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P = (9,0);
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A = 15*dir(20);
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B = 15*dir(250);
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A2 = 15*dir(-20);
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B2 = 15*dir(-250);
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draw(Circle(O,15));
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draw(A--B,red);
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draw(A2--B2,red);
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draw(O--(15,0));
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dot("$O$", O, S);
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dot("$P$", P, N);
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</asy>
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 +
 
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Therefore, there are <math>2 + 2(29 - 25 + 1) = \boxed{12}</math> chords of integer length passing through <math>P</math>.
  
 
== See also ==
 
== See also ==
Line 13: Line 107:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
 +
(This problem was also on 2001 State Target Round!)

Latest revision as of 14:27, 23 June 2021

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

Let $O$ be the center of the circle, and let the chord passing through $P$ that is perpendicular to $OP$ intersect the circle at $Q$ and $R$. Then $OP = 9$ and $OQ = 15$, so by the Pythagorean Theorem, $PQ = 12$. By symmetry, $PR = 12$.

[asy] import graph; unitsize(0.15 cm); pair O, P, Q, R; O = (0,0); P = (9,0); Q = (9,12); R = (9,-12); draw(Circle(O,15)); draw((-15,0)--(15,0)); draw(O--Q); draw(Q--R,red); dot("$O$", O, S); dot("$P$", P, NW); dot("$Q$", Q, NE); dot("$R$",R,SE); label("$15$", (-15/2,0), S); label("$9$", (O + P)/2, S); label("$6$", (12,0), S); label("$15$", (O + Q)/2, NW); label("$12$", (P + Q)/2, E); [/asy] Let $AB$ be the diameter passing through $P$. [asy] import graph; unitsize(0.15 cm); pair A, B, O, P, Q, R; A = (-15,0); B = (15,0); O = (0,0); P = (9,0); Q = (9,12); R = (9,-12); draw(Circle(O,15)); draw(A--B,red); draw(Q--R); dot("$A$", A, W); dot("$B$", B, E); dot("$O$", O, S); dot("$P$", P, NW); dot("$Q$", Q, NE); dot("$R$", R, SE); label("$15$", (-15/2,0), S); label("$9$", (O + P)/2, S); label("$6$", (12,0), S); label("$12$", (P + Q)/2, E); label("$12$", (P + R)/2, E); [/asy] Then the longest chord of the circle that passes through $P$ is $AB$, which has length 30, and the shortest chord is $QR$, which has length 24. If we rotate the red chord (while ensuring it passes through $P$), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer $n=25,26,27,28,29$, there are two chords of length $n$ passing through $P$, as seen in this picture: [asy] import graph; unitsize(0.15 cm); pair O, P,A,B,A2,B2; O = (0,0); P = (9,0); A = 15*dir(20); B = 15*dir(250); A2 = 15*dir(-20); B2 = 15*dir(-250); draw(Circle(O,15)); draw(A--B,red); draw(A2--B2,red); draw(O--(15,0)); dot("$O$", O, S); dot("$P$", P, N); [/asy]


Therefore, there are $2 + 2(29 - 25 + 1) = \boxed{12}$ chords of integer length passing through $P$.

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. AMC logo.png


(This problem was also on 2001 State Target Round!)

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