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Difference between revisions of "1959 AHSME Problems/Problem 2"

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[[Category:Introductory Geometry Problems]]

Latest revision as of 10:59, 21 July 2024

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]

Because $[ABDE]=[\triangle EDC]$, we know that $[\triangle EDC]=\frac{1}{2}[\triangle ABC]$. Because the ratio of their areas is $\frac{1}{2}$, we know that the ratio of their side lengths (and thus their altitudes) is $\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$. Thus, the altitude from $C$ to $\overline{DE}$ has length $1*\frac{\sqrt2}{2}=\frac{\sqrt{2}}{2}$. Thus, because $\overline{DE} \parallel \overline{AB}$, the distance from $P$ to $\overline{AB}$ is $1-\frac{\sqrt{2}}{2}=\boxed{\textbf{(D) }\frac{2-\sqrt{2}}{2}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AHSME Problems and Solutions

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