1983 AIME Problem 14

Question

In the adjoining figure, two circles of radii

6

and

8

are drawn with their centers

12

units apart. At

P,

one of the points of intersection, a line is drawn in such a way that the chords

QP

and

PR

have equal length. Find the square of the length of

QP.

Answer

Put the centers at

(-6,0)

and

(6,0)

and find the coordinates of

P

If

Q=P-\ell u

and

R=P+\ell u,

subtract the two equal-radius equations

Put the center of the radius-

8

circle at

O_1=(-6,0)

and the other center at

O_2=(6,0).

Subtracting the two circle equations gives

P=\left(\frac76,\frac{\sqrt{455}}6\right),

where the positive

y

-coordinate selects the pictured intersection.

Let the common chord length be

\ell,

and let

u

be the unit vector from

Q

toward

R.

Then

Q=P-\ell u

and

R=P+\ell u.

Comparing

Q

with

P

in the first circle and

R

with

P

in the second gives

\begin{aligned} u\mathbin{\cdot}(P-O_1)&=\frac{\ell}{2},\\ u\mathbin{\cdot}(P-O_2)&=-\frac{\ell}{2}. \end{aligned}

Adding shows that

u\perp P,

so

u=\frac{(\sqrt{455},-7)}{\sqrt{504}}.

Subtracting the two dot-product equations gives

\ell=u\mathbin{\cdot}(O_2-O_1) =\frac{12\sqrt{455}}{\sqrt{504}}.

Therefore

\ell^2=\frac{144\cdot455}{504}=130.

Problem 14 in other years