Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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== Solution 7: Easiest Solution using Scaling (Can be done in your head, and destroys the problem)== | == Solution 7: Easiest Solution using Scaling (Can be done in your head, and destroys the problem)== | ||
Since equilateral triangles and <math>120^\circ-x^\circ-(60-x)^\circ</math> triangles have the same form for area formulas (and the answer to this problem is simply found by adding <math>\text{(former+3*latter)/former}</math> ), a lot can cancel out when considering the ratio. Suppose a net factor of <math>f</math> is common to the two formulas. Now scale the diagram so that <math>f=1</math>. The area of <math>\Delta ABC</math> is simply then of the form <math>k*k=k^2</math>, whereas the area of <math>\Delta A'AB'</math>, etc... is of the form <math>3k*4k=12k^2</math>. We can now scale down by another factor of <math>k^2</math> to get <math>1</math> and <math>12</math>, respectively. Therefore, the overall area of<math>\Delta A'B'C'</math> is <math>12*3+1=37</math>, and the area of <math>\Delta ABC</math> is simply <math>1</math>, which gives us <math>\boxed{37:1,E}</math>. | Since equilateral triangles and <math>120^\circ-x^\circ-(60-x)^\circ</math> triangles have the same form for area formulas (and the answer to this problem is simply found by adding <math>\text{(former+3*latter)/former}</math> ), a lot can cancel out when considering the ratio. Suppose a net factor of <math>f</math> is common to the two formulas. Now scale the diagram so that <math>f=1</math>. The area of <math>\Delta ABC</math> is simply then of the form <math>k*k=k^2</math>, whereas the area of <math>\Delta A'AB'</math>, etc... is of the form <math>3k*4k=12k^2</math>. We can now scale down by another factor of <math>k^2</math> to get <math>1</math> and <math>12</math>, respectively. Therefore, the overall area of<math>\Delta A'B'C'</math> is <math>12*3+1=37</math>, and the area of <math>\Delta ABC</math> is simply <math>1</math>, which gives us <math>\boxed{37:1,E}</math>. | ||
+ | |||
+ | == Solution 8: Quick Proportionality (essentially a quicker version of solution 7) == | ||
+ | By the proportionality, it becomes clear that the answer is <math>3*4*3+1*1=37</math>. | ||
==See Also== | ==See Also== |
Revision as of 20:40, 5 August 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 Coordinates)
We use barycentric coordinates wrt
It's quite obvious that
now, since the coordinates are homogenized (), we can directly apply the area formula:
So the answer is
Solution 6 (Area Comparison)
First, comparing bases yields that
by congruent triangles,
this yields that
Solutions 5 and 6 by Blast_S1
Solution 7: Easiest Solution using Scaling (Can be done in your head, and destroys the problem)
Since equilateral triangles and triangles have the same form for area formulas (and the answer to this problem is simply found by adding
), a lot can cancel out when considering the ratio. Suppose a net factor of
is common to the two formulas. Now scale the diagram so that
. The area of
is simply then of the form
, whereas the area of
, etc... is of the form
. We can now scale down by another factor of
to get
and
, respectively. Therefore, the overall area of
is
, and the area of
is simply
, which gives us
.
Solution 8: Quick Proportionality (essentially a quicker version of solution 7)
By the proportionality, it becomes clear that the answer is .
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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