Difference between revisions of "2017 AMC 12B Problems/Problem 15"
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− | Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}absin\theta</math> and calling side <math>AB = 1</math> for ease, we get: <math>4 | + | Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}absin\theta</math> and calling side <math>AB = 1</math> for ease, we get: <math>4\cdot3\cdot\frac{sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}</math> |
Solution by Aadileo | Solution by Aadileo |
Revision as of 13:13, 21 January 2019
Contents
Problem 15
Let be an equilateral triangle. Extend side
beyond
to a point
so that
. Similarly, extend side
beyond
to a point
so that
, and extend side
beyond
to a point
so that
. What is the ratio of the area of
to the area of
?
Solution 1: Law of Cosines
Solution by HydroQuantum
Let .
Recall The Law of Cosines. Letting ,
Since both
and
are both equilateral triangles, they must be similar due to
similarity. This means that
.
Therefore, our answer is .
Solution 2: Inspection
Note that the height and base of are respectively 4 times and 3 times that of
. Therefore the area of
is 12 times that of
.
By symmetry, . Adding the areas of these three triangles and
for the total area of
gives a ratio of
, or
.
Solution 3: Coordinates
First we note that due to symmetry. WLOG, let
and
Therefore,
. Using the condition that
, we get
and
. It is easy to check that
. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is
Solution by mathwiz0803
Solution 4: Computing the Areas
Note that angle is
°, as it is supplementary to the equilateral triangle. Then, using area
and calling side
for ease, we get:
as the area of
. Then, the area of
is
, so the ratio is
Solution by Aadileo
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.