1963 AMC 12 Problem 10

Question

Point

P

is taken interior to a square with side-length

a

and such that it is equally distant from two consecutive vertices and from the side opposite these vertices. If

d

represents the common distance, then

d

equals:

\dfrac{3a}{5}

\dfrac{5a}{8}

\dfrac{3a}{8}

\dfrac{a\sqrt2}{2}

\dfrac a2

The equal distances to two consecutive vertices put

P

on their perpendicular bisector

If the distance to the opposite side is

d,

use a right triangle with legs

\frac{a}{2}

and

a-d

Place the opposite side at height

0

and the two vertices at height

a.

Then

P=(\frac{a}{2},d).

Equality of the vertex distance and

d

gives

\left(\frac a2\right)^2+(a-d)^2=d^2,

so

\frac{5a^2}{4}=2ad

and

d=\frac{5a}{8}.

Thus, the correct answer is B.

Problem 10 in other years