Difference between revisions of "1967 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
| − | 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 | + | 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 == | ||
Latest revision as of 21:13, 23 April 2024
Problem
Points 
, 
, 
are taken respectively on sides 
, 
, and 
of triangle 
so that 
. The ratio of the area of triangle 
to that of triangle is:
Solution
WLOG, let's assume that 
is equilateral. Therefore, ![$[ABC]=\frac{(1+n)^2\sqrt3}{4}$](http://latex.artofproblemsolving.com/0/e/7/0e705ee830055f3121c2b4d0a8356a95098d95e8.png)
and ![$[DBE]=[ADF]=[EFC]=n \cdot \sin(60)/2$](http://latex.artofproblemsolving.com/1/5/a/15a51c122ec1eb40c874b3696890e343527ec44a.png)
. Then ![$[DEF]=\frac{(n^2-n+1)\sqrt3}{4}$](http://latex.artofproblemsolving.com/c/2/4/c24e7edd2d04ecb15f7bd59b7fc12f6b451d3ead.png)
. Finding the ratio yields 
. -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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
