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

## 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 $P$.