Difference between revisions of "1991 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
− | <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); | ||
+ | label("<math>6</math>", (12,0), S); | ||
+ | label("<math>15</math>", (O + Q)/2, NW); | ||
+ | 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); | ||
+ | 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); | ||
+ | B2 = 15*dir(-250); | ||
+ | |||
+ | |||
+ | draw(Circle(O,15)); | ||
+ | draw(A--B,red); | ||
+ | draw(A2--B2,red); | ||
+ | draw(O--(15,0)); | ||
+ | |||
+ | dot("<math>O</math>", O, S); | ||
+ | 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 is units from the center of a circle of radius . How many different chords of the circle contain and have integer lengths?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 29
Solution
Let be the center of the circle, and let the chord passing through that is perpendicular to intersect the circle at and . Then and , so by the Pythagorean Theorem, . By symmetry, .
[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, S); dot("", P, NW); dot("", Q, NE); dot("",R,SE);
label("", (-15/2,0), S); label("", (O + P)/2, S); label("", (12,0), S); label("", (O + Q)/2, NW); label("", (P + Q)/2, E); [/asy]
Let be the diameter passing through .
[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, W); dot("", B, E); dot("", O, S); dot("", P, NW); dot("", Q, NE); dot("", R, SE);
label("", (-15/2,0), S); label("", (O + P)/2, S); label("", (12,0), S); label("", (P + Q)/2, E); label("", (P + R)/2, E); [/asy]
Then the longest chord of the circle that passes through is , which has length 30, and the shortest chord is , which has length 24. If we rotate the red chord (while ensuring it passes through ), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer , there are two chords of length passing through , 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, S); dot("", P, N);
[/asy]
Therefore, there are chords of integer length passing through .
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.