Two points on a circle of radius $1$ are chosen at random. Find the probability that the distance between the two points is at most $3/2.$

Answer 1

Let $O$ be the center of the circle. Suppose the first point we choose is $A.$ Let $B$ be the second point we choose.

[asy]
unitsize(2 cm);

pair A, B, O;

O = (0,0);
A = dir(110);
B = dir(60);

draw(Circle(O,1));
draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));

draw(O--A);
draw(O--B);

label("$O$", O, SW);
label("$A$", A, NW);
label("$B$", B, N);
[/asy]

Then the distance $AB$ is at most $\frac{3}{2}$ if and only if $B$ lies inside the shaded annulus centered at $O.$

[asy]
unitsize(2 cm);

pair A, B, O;

O = (0,0);
A = dir(110);
B = (1.5,0);

draw(Circle(O,1));
draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));
draw(arc(O,3/2,0,180),red);

label("$O$", O, SW);
filldraw(arc(O,1,0,180)--(0,0)--cycle,gray(0.7));
label("$A$", A, N);
label("$B$", B, N);
[/asy]

The probability that the second point we choose lies inside this annulus is the ratio of the areas of the annulus and the whole circle, which is
\[\frac{\pi \left( \left( \frac{3}{2} \right)^2 - 1^2 \right)}{\pi (1)^2} = \boxed{\frac{5}{4}}.\]