Difference between revisions of "1991 AHSME Problems/Problem 10"
m (→Solution) |
(→See also) |
||
(One intermediate revision by one other user not shown) | |||
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> | |
import graph; | import graph; | ||
Line 25: | Line 25: | ||
draw(Q--R,red); | draw(Q--R,red); | ||
− | dot(" | + | dot("$O$", O, S); |
− | dot(" | + | dot("$P$", P, NW); |
− | dot(" | + | dot("$Q$", Q, NE); |
− | dot(" | + | dot("$R$",R,SE); |
− | label(" | + | label("$15$", (-15/2,0), S); |
− | label(" | + | label("$9$", (O + P)/2, S); |
− | label(" | + | label("$6$", (12,0), S); |
− | label(" | + | label("$15$", (O + Q)/2, NW); |
− | label(" | + | label("$12$", (P + Q)/2, E); |
[/asy] | [/asy] | ||
− | Let | + | Let $AB$ be the diameter passing through $P$. |
[asy] | [asy] | ||
Line 57: | Line 57: | ||
draw(Q--R); | draw(Q--R); | ||
− | dot(" | + | dot("$A$", A, W); |
− | dot(" | + | dot("$B$", B, E); |
− | dot(" | + | dot("$O$", O, S); |
− | dot(" | + | dot("$P$", P, NW); |
− | dot(" | + | dot("$Q$", Q, NE); |
− | dot(" | + | dot("$R$", R, SE); |
− | label(" | + | label("$15$", (-15/2,0), S); |
− | label(" | + | label("$9$", (O + P)/2, S); |
− | label(" | + | label("$6$", (12,0), S); |
− | label(" | + | label("$12$", (P + Q)/2, E); |
− | label(" | + | label("$12$", (P + R)/2, E); |
[/asy] | [/asy] | ||
− | Then the longest chord of the circle that passes through | + | 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(" | + | dot("$O$", O, S); |
− | dot(" | + | dot("$P$", P, N); |
− | + | </asy> | |
Line 107: | 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 14:27, 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,
.
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!)