1959 AMC 12 Problem 2

Question

Through a point

P

inside triangle

ABC

a line is drawn parallel to the base

\overline{AB},

dividing the triangle into two equal areas. If the altitude to

\overline{AB}

has length

1,

then the distance from

P

to

\overline{AB}

is:

\dfrac12

\dfrac14

2-\sqrt2

\dfrac{2-\sqrt2}{2}

\dfrac{2+\sqrt2}{8}

Let

x

be the distance from

P

to the base

The small triangle above the parallel line has altitude

1-x

and half the original area

Let

x

be the requested distance. The triangle above the parallel line is similar to the original triangle, with linear ratio

1-x.

Its area is half the original area, so

(1-x)^2=\frac12.

Since

0\lt x\lt1,

we have

1-x=\frac{1}{\sqrt2},

and therefore

x=1-\frac1{\sqrt2}=\frac{2-\sqrt2}{2}.

Therefore, the correct answer is D.

Problem 2 in other years