1959 AHSME Problems/Problem 2

Problem 2

Through a point $P$ inside the $\triangle ABC$ a line is drawn parallel to the base $AB$ , dividing the triangle into two equal areas. If the altitude to $AB$ has a length of $1$ , then the distance from $P$ to $AB$ is:

$\textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8}$

Solution

$\boxed{\textbf{D}}$

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1959 AHSME Problems/Problem 2

Problem 2

Solution

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