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 == | ||
− | {{AHSME box|year=1967|num-b=33|num-a=35}} | + | {{AHSME 40p box|year=1967|num-b=33|num-a=35}} |
[[Category: Intermediate Geometry Problems]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:13, 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,
and
. Then
. 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 |
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