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

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== Solution ==
 
== Solution ==
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WLOG, let's assume that <math>\triangle ABC</math> is equilateral. Therefore, <math>[ABC]=\frac{(1+n)^2\sqrt3}{4}</math> and <math>[DBE]=[ADF]=[EFC]=n \cdot sin(60)/2</math>. Then <math>[DEF]=\frac{(n^2-n+1)\sqrt3}{4}</math>. Finding the ratio yields <math>\fbox{A}</math>
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WLOG, let's assume that <math>\triangle ABC</math> is equilateral. Therefore, <math>[ABC]=\frac{(1+n)^2\sqrt3}{4}</math> and <math>[DBE]=[ADF]=[EFC]=n \cdot sin(60)/2</math>. Then <math>[DEF]=\frac{(n^2-n+1)\sqrt3}{4}</math>. Finding the ratio yields <math>\fbox{A}</math>.  -Dark_Lord
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=33|num-a=35}}   
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{{AHSME 40p box|year=1967|num-b=33|num-a=35}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:40, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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
All AHSME Problems and Solutions

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