MathDB
Distance from line parallel to base

Source: 1959 AMC 12 Problem 2

August 13, 2013
AMCaltitudeareaAMC 12

Problem Statement

Through a point PP inside the triangle ABCABC a line is drawn parallel to the base ABAB, dividing the triangle into two equal areas. If the altitude to ABAB has a length of 11, then the distance from PP to ABAB is: <spanclass=latexbold>(A)</span> 12<spanclass=latexbold>(B)</span> 14<spanclass=latexbold>(C)</span> 22<spanclass=latexbold>(D)</span> 222<spanclass=latexbold>(E)</span> 2+28 <span class='latex-bold'>(A)</span>\ \frac12 \qquad<span class='latex-bold'>(B)</span>\ \frac14\qquad<span class='latex-bold'>(C)</span>\ 2-\sqrt2\qquad<span class='latex-bold'>(D)</span>\ \frac{2-\sqrt2}{2}\qquad<span class='latex-bold'>(E)</span>\ \frac{2+\sqrt2}{8}