1970 AMC 12 Problem 7

Question

Inside square

ABCD

with side

s,

quarter-circle arcs with radii

s

and centers at

A

and

B

are drawn. These arcs intersect at a point

X

inside the square. How far is

X

from side

CD?

\tfrac12s(\sqrt3+4)

\tfrac12s\sqrt3

\tfrac12s(1+\sqrt3)

\tfrac12s(\sqrt3-1)

\tfrac12s(2-\sqrt3)

The two radii to

X

and side

AB

form an equilateral triangle

The altitude from

X

to

AB

is

\frac{s\sqrt3}{2}

; subtract it from the square’s height

Since

AX=BX=AB=s,

triangle

ABX

is equilateral. Thus the perpendicular distance from

X

to

AB

is

\frac{s\sqrt3}{2}.

The distance between the parallel sides

AB

and

CD

is

s,

so the requested distance is

s-\frac{s\sqrt3}{2}=\frac s2(2-\sqrt3).

Therefore, the correct answer is E.

Problem 7 in other years