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("$O$", 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 <math>2 + 2(29 - 25 + 1) = \boxed{12}</math> chords of integer length passing through <math>P</math>. | ||
== See also == | == See also == | ||
| Line 13: | Line 107: | ||
[[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 
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("$O$", 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]](http://latex.artofproblemsolving.com/b/7/7/b77639fd9005d5d50a1931e1444762dd51ae0f7d.png)
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. 
(This problem was also on 2001 State Target Round!)
