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

m (→‎Solution)
m (→‎Solution)
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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>.
 
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]
+
<asy>
 
import graph;
 
import graph;
  
Line 25: Line 25:
 
draw(Q--R,red);
 
draw(Q--R,red);
  
−
dot("<math>O</math>", O, S);
+
dot("$O$", O, S);
−
dot("<math>P</math>", P, NW);
+
dot("$P$", P, NW);
−
dot("<math>Q</math>", Q, NE);
+
dot("$Q$", Q, NE);
−
dot("<math>R</math>",R,SE);
+
dot("$R$",R,SE);
  
−
label("<math>15</math>", (-15/2,0), S);
+
label("$15$", (-15/2,0), S);
−
label("<math>9</math>", (O + P)/2, S);
+
label("$9$", (O + P)/2, S);
−
label("<math>6</math>", (12,0), S);
+
label("$6$", (12,0), S);
−
label("<math>15</math>", (O + Q)/2, NW);
+
label("$15$", (O + Q)/2, NW);
−
label("<math>12</math>", (P + Q)/2, E);
+
label("$12$", (P + Q)/2, E);
 
[/asy]
 
[/asy]
  
−
Let <math>AB</math> be the diameter passing through <math>P</math>.
+
Let $AB$ be the diameter passing through $P$.
  
 
[asy]
 
[asy]
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draw(Q--R);
 
draw(Q--R);
  
−
dot("<math>A</math>", A, W);
+
dot("$A$", A, W);
−
dot("<math>B</math>", B, E);
+
dot("$B$", B, E);
−
dot("<math>O</math>", O, S);
+
dot("$O$", O, S);
−
dot("<math>P</math>", P, NW);
+
dot("$P$", P, NW);
−
dot("<math>Q</math>", Q, NE);
+
dot("$Q$", Q, NE);
−
dot("<math>R</math>", R, SE);
+
dot("$R$", R, SE);
  
−
label("<math>15</math>", (-15/2,0), S);
+
label("$15$", (-15/2,0), S);
−
label("<math>9</math>", (O + P)/2, S);
+
label("$9$", (O + P)/2, S);
−
label("<math>6</math>", (12,0), S);
+
label("$6$", (12,0), S);
−
label("<math>12</math>", (P + Q)/2, E);
+
label("$12$", (P + Q)/2, E);
−
label("<math>12</math>", (P + R)/2, E);
+
label("$12$", (P + R)/2, E);
 
[/asy]
 
[/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:
+
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]
 
[asy]
Line 93: Line 93:
 
draw(O--(15,0));
 
draw(O--(15,0));
  
−
dot("<math>O</math>", O, S);
+
dot("$O$", O, S);
−
dot("<math>P</math>", P, N);
+
dot("$P$", P, N);
  
−
[/asy]
+
</asy>
  
  

Revision as of 02:14, 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

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