Difference between revisions of "2017 AMC 12B Problems/Problem 15"
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Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. | Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. | ||
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+ | ==Solution 2 (inspection)== | ||
+ | Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | ||
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+ | By symmetry, <math>\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'</math>. Adding the areas of these three triangles and <math>\triangle ABC</math> for the total area of <math>\triangle A'B'C'</math> gives a ratio of <math>(12 + 12 + 12 + 1) : 1</math>, or <math>\boxed{\textbf{(E) } 37 : 1}</math>. | ||
[[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]] | [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]] |
Revision as of 19:46, 16 February 2017
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
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
.