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

(Added a solution with explanation)
m (→‎Solution)
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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>.
 +
 
 +
[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("<math>O</math>", O, S);
 +
dot("<math>P</math>", P, NW);
 +
dot("<math>Q</math>", Q, NE);
 +
dot("<math>R</math>",R,SE);
 +
 
 +
label("<math>15</math>", (-15/2,0), S);
 +
label("<math>9</math>", (O + P)/2, S);
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label("<math>6</math>", (12,0), S);
 +
label("<math>15</math>", (O + Q)/2, NW);
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label("<math>12</math>", (P + Q)/2, E);
 +
[/asy]
 +
 
 +
Let <math>AB</math> be the diameter passing through <math>P</math>.
 +
 
 +
[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("<math>A</math>", A, W);
 +
dot("<math>B</math>", B, E);
 +
dot("<math>O</math>", O, S);
 +
dot("<math>P</math>", P, NW);
 +
dot("<math>Q</math>", Q, NE);
 +
dot("<math>R</math>", R, SE);
 +
 
 +
label("<math>15</math>", (-15/2,0), S);
 +
label("<math>9</math>", (O + P)/2, S);
 +
label("<math>6</math>", (12,0), S);
 +
label("<math>12</math>", (P + Q)/2, E);
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label("<math>12</math>", (P + R)/2, E);
 +
[/asy]
 +
 
 +
Then the longest chord of the circle that passes through <math>P</math> is <math>AB</math>, which has length 30, and the shortest chord is <math>QR</math>, which has length 24. If we rotate the red chord (while ensuring it passes through <math>P</math>), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer <math>n=25,26,27,28,29</math>, there are two chords of length <math>n</math> passing through <math>P</math>, 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);
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B2 = 15*dir(-250);
 +
 
 +
 
 +
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));
 +
 
 +
dot("<math>O</math>", O, S);
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dot("<math>P</math>", P, N);
 +
 
 +
[/asy]
 +
 
 +
 
 +
 
 +
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 ==

Revision as of 02:12, 23 April 2020

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

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