1993 AMC 12 Problem 13

Question

A square of perimeter

20

is inscribed in a square of perimeter

28.

What is the greatest distance between a vertex of the inner square and a vertex of the outer square?

\sqrt{58}

\frac{7\sqrt5}{2}

8

\sqrt{65}

5\sqrt3

A vertex of the inner square divides an outer side into lengths

x

and

y

Use

x+y=7

and

x^2+y^2=25

The outer and inner side lengths are

7

and

5.

At any inner vertex, the two pieces of the outer side have lengths

x

and

y

with

x+y=7,\qquad x^2+y^2=25.

Hence

2xy=49-25=24,

so

\{x,y\}=\{3,4\}.

From an inner vertex on one side, the farther opposite outer vertex has horizontal displacement

4

and vertical displacement

7.

The greatest distance is therefore

\sqrt{4^2+7^2}=\sqrt{65}.

Thus the correct answer is D.

Problem 13 in other years