Difference between revisions of "2017 AMC 12B Problems/Problem 15"
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First we note that <math>A'B'C'\sim ABC</math> due to symmetry. WLOG, let <math>B = (0, 0)</math> and <math>AB = 1</math> Therefore, <math>C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)</math>. Using the condition that <math>CC' = 3</math>, we get <math>C' = (4, 0)</math> and <math>B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)</math>. It is easy to check that <math>B'C' = \sqrt{37}</math>. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is <math>\boxed{\textbf{(E) } 37 : 1}</math> | First we note that <math>A'B'C'\sim ABC</math> due to symmetry. WLOG, let <math>B = (0, 0)</math> and <math>AB = 1</math> Therefore, <math>C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)</math>. Using the condition that <math>CC' = 3</math>, we get <math>C' = (4, 0)</math> and <math>B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)</math>. It is easy to check that <math>B'C' = \sqrt{37}</math>. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is <math>\boxed{\textbf{(E) } 37 : 1}</math> | ||
− | Solution by | + | Solution by <i>mathwiz0803</i> |
{{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}} | ||
{{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:15, 30 June 2017
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
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 |
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