Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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Solution by sp1729 | Solution by sp1729 | ||
+ | |||
+ | ===Solution 5 (Barycentric Bash) === | ||
+ | Set <math>\triangle ABC</math> as the reference triangle, we use barycentric coordinates | ||
+ | <math>A=(1,0,0), B=(0,1,0), C=(0,0,1)</math> | ||
+ | a simple area comparison yields <math>A'=(4,0,-3), B'=(-3,4,0), C'=(0,-3,4)</math> | ||
+ | Now since the coordinates are homogenized (<math>-3+4=1</math>), we can use the area formula to get <math>[A'B'C']=[ABC] \left| | ||
+ | So the answer is <math>\boxed{\textbf{(E) } 37 : 1}</math>, and we're done | ||
+ | |||
+ | Solution by Blast_S1 | ||
==See Also== | ==See Also== |
Revision as of 16:41, 10 July 2017
Contents
[hide]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
Solution 5 (Barycentric Bash)
Set as the reference triangle, we use barycentric coordinates
a simple area comparison yields
Now since the coordinates are homogenized (
), we can use the area formula to get
So the answer is
, and we're done
Solution by Blast_S1
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.