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

$1+\sqrt2$

$2\sqrt2$

$2+\sqrt2$

$3\sqrt2$

$3+\sqrt2$

Answer: C

Concepts:coordinate geometry circle distance formula optimization

Difficulty rating: 2310

Small Hint:

Place $D=(0,0),$ $A=(1,0),$ $B=(1,1),$ and $C=(0,1)$

Big Hint:

Substitute $P=(x,y)$ into the distance equation and complete the square

Solution:

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.

1983 AMC 12 Problem 29

Answer: C

Difficulty rating: 2310

Small Hint:

Big Hint:

Solution:

Problem 29 in Other Years