1983 AMC 12 Problem 29

Question

A point

P

lies in the same plane as a given square of side

1.

Let the vertices of the square, taken counterclockwise, be

A,

B,

C

and

D.

Also, let the distances from

P

to

A,

B

and

C,

respectively, be

u,

v

and

w.

What is the greatest distance that

P

can be from

D

if

u^2+v^2=w^2?

1+\sqrt2

2\sqrt2

2+\sqrt2

3\sqrt2

3+\sqrt2

Place

D=(0,0),

A=(1,0),

B=(1,1),

and

C=(0,1)

Substitute

P=(x,y)

into the distance equation and complete the square

Place

D=(0,0),

A=(1,0),

B=(1,1),

and

C=(0,1),

with

P=(x,y).

Then

\begin{aligned} &(x-1)^2+y^2\\ &\quad+(x-1)^2+(y-1)^2\\ &=x^2+(y-1)^2, \end{aligned}

which simplifies to

(x-2)^2+y^2=2.

Thus

P

lies on a circle centered

2

units from

D

with radius

\sqrt2.

Its greatest distance from

D

is

2+\sqrt2.

Therefore, the correct answer is C.

Problem 29 in other years