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

$[asy] import geometry; point A=(0,0); point B=(10,0); point C=(4,6); point P=(3,6(-sqrt(2)/2+1)); point D,E; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NW); // Point P and triangle CDE dot(P); label("P",P,N); pair[] d=intersectionpoints(parallel(P,line(A,B)),(B--C)); D=d[0]; dot(D); label("D",D,NE); pair[] e=intersectionpoints(parallel(P,line(A,B)),(A--C)); E=e[0]; dot(E); label("E",E,NW); draw(D--E); [/asy]$

$\boxed{\textbf{(D)}}$

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1959 AHSME Problems/Problem 2

Problem 2

Solution

See also