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

m (Solution)
(See also)
 
(One intermediate revision by one other user not shown)
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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;
  
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draw(Q--R,red);
 
draw(Q--R,red);
  
dot("<math>O</math>", O, S);
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dot("$O$", O, S);
dot("<math>P</math>", P, NW);
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dot("$P$", P, NW);
dot("<math>Q</math>", Q, NE);
+
dot("$Q$", Q, NE);
dot("<math>R</math>",R,SE);
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dot("$R$",R,SE);
  
label("<math>15</math>", (-15/2,0), S);
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label("$15$", (-15/2,0), S);
label("<math>9</math>", (O + P)/2, S);
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label("$9$", (O + P)/2, S);
label("<math>6</math>", (12,0), S);
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label("$6$", (12,0), S);
label("<math>15</math>", (O + Q)/2, NW);
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label("$15$", (O + Q)/2, NW);
label("<math>12</math>", (P + Q)/2, E);
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label("$12$", (P + Q)/2, E);
 
[/asy]
 
[/asy]
  
Let <math>AB</math> be the diameter passing through <math>P</math>.
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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);
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dot("$A$", A, W);
dot("<math>B</math>", B, E);
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dot("$B$", B, E);
dot("<math>O</math>", O, S);
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dot("$O$", O, S);
dot("<math>P</math>", P, NW);
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dot("$P$", P, NW);
dot("<math>Q</math>", Q, NE);
+
dot("$Q$", Q, NE);
dot("<math>R</math>", R, SE);
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dot("$R$", R, SE);
  
label("<math>15</math>", (-15/2,0), S);
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label("$15$", (-15/2,0), S);
label("<math>9</math>", (O + P)/2, S);
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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);
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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]
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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>
  
  
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
 +
(This problem was also on 2001 State Target Round!)

Latest revision as of 13:27, 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 (ProblemsAnswer KeyResources)
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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