Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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− | Looking at the answer choices, we see that all but <math>{\textbf{(E)}</math> has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick <math>\boxed{\textbf{(E) } 37 : 1}</math> | + | Looking at the answer choices, we see that all but <math>{\textbf{(E)}}</math> has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick <math>\boxed{\textbf{(E) } 37 : 1}</math> |
Solution by sp1729 | Solution by sp1729 |
Revision as of 02:11, 27 February 2017
Contents
Problem
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 1
Note that by symmetry, is also equilateral. Therefore, we only need to find one of the sides of
to determine the area ratio. WLOG, let
. Therefore,
and
. Also,
, so by the Law of Cosines,
. Therefore, the answer is
Solution 2
As mentioned in the first solution, is equilateral. WLOG, let
. Let
be on the line passing through
such that
is perpendicular to
. Note that
is a 30-60-90 with right angle at
. Since
,
and
. So we know that
. Note that
is a right triangle with right angle at
. So by the Pythagorean theorem, we find
Therefore, the answer is
.
Solution 3
Let . We start by noting that we can just write
as just
.
Similarly
, and
. We can evaluate the area of triangle
by simply using Heron's formula,
.
Next in order to evaluate
we need to evaluate the area of the larger triangles
.
In this solution we shall just compute
of these as the others are trivially equivalent.
In order to compute the area of
we can use the formula
.
Since
is equilateral and
,
,
are collinear, we already know
Similarly from above we know
and
to be
, and
respectively. Thus the area of
is
. Likewise we can find
to also be
.
.
Therefore the ratio of
to
is
Solution 4 (Elimination)
Looking at the answer choices, we see that all but has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick
Solution by sp1729
See Also
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 |
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 |
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