During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

m (Solution)
m (Solution)
Line 8: Line 8:
 
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]
Line 57: Line 57:
 
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 (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

Invalid username
Login to AoPS