Difference between revisions of "1967 AHSME Problems/Problem 34"

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== See also ==
 
== See also ==
{{AHSME box|year=1967|num-b=33|num-a=35}}   
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[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
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{{MAA Notice}}

Revision as of 00:40, 16 August 2023

Problem

Points $D$, $E$, $F$ are taken respectively on sides $AB$, $BC$, and $CA$ of triangle $ABC$ so that $AD:DB=BE:CE=CF:FA=1:n$. The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is:

$\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad \textbf{(B)}\ \frac{1}{(n+1)^2}\qquad \textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qquad \textbf{(D)}\ \frac{n^2}{(n+1)^2}\qquad \textbf{(E)}\ \frac{n(n-1)}{n+1}$

Solution

WLOG, let's assume that $\triangle ABC$ is equilateral. Therefore, $[ABC]=\frac{(1+n)^2\sqrt3}{4}$ and $[DBE]=[ADF]=[EFC]=n \cdot sin(60)/2$. Then $[DEF]=\frac{(n^2-n+1)\sqrt3}{4}$. Finding the ratio yields $\fbox{A}$. -Dark_Lord

See also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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