1958 AMC 12 Problem 48

Question

Diameter

\overline{AB}

of a circle with center

O

is

10

units.

C

is a point

4

units from

A,

and on

\overline{AB}.

D

is a point

4

units from

B,

and on

\overline{AB}.

P

is any point on the circle. Then the broken-line path from

C

to

P

to

D:

has the same length for all positions of

P

exceeds

10

units for all positions of

P

cannot exceed

10

units

is shortest when

CPD

is a right triangle

is longest when

P

is equidistant from

C

and

D

Answer: E

Concepts:circle distance formula optimization symmetry

Difficulty rating: 1990

Small hint:

Place the center at the origin and the diameter on the

x

-axis, so

C=(-1,0)

and

D=(1,0)

Big hint:

For

P=(u,v)

on the circle, compare

CP^2=26+2u

and

DP^2=26-2u

Solution:

Place

O=(0,0),

C=(-1,0),

D=(1,0),

and

P=(u,v)

with

u^2+v^2=25.

Then

\begin{aligned} CP^2&=26+2u,\\ DP^2&=26-2u. \end{aligned}

Therefore

\begin{aligned} (CP+DP)^2 &=52\\ &\quad+2\sqrt{676-4u^2}. \end{aligned}

This is largest when

u=0,

exactly when

CP=DP.

Therefore, the correct answer is E.