Difference between revisions of "2020 AMC 12A Problems/Problem 24"
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− | We begin by rotating <math>\triangle{ABC}</math> by <math>60^{\circ}</math> about <math>A</math>, such that in <math>\triangle{A'B'C'}</math>, <math>B' = C</math>. We see that | + | We begin by rotating <math>\triangle{ABC}</math> by <math>60^{\circ}</math> counterclockwise about <math>A</math>, such that in <math>\triangle{A'B'C'}</math>, <math>B' = C</math>. We see that |
<math>\triangle{APP'}</math> is equilateral with side length <math>1</math>, meaning that <math>\angle APP' = 60^{\circ}</math>. We also see that <math>\triangle{CPP'}</math> is a <math>30-60-90</math> right triangle, meaning that <math>\angle CPP'= 60^{\circ}</math>. Thus, by adding the two together, we see that <math>\angle APC = 120^{\circ}</math>. We can now use the law of cosines as following: | <math>\triangle{APP'}</math> is equilateral with side length <math>1</math>, meaning that <math>\angle APP' = 60^{\circ}</math>. We also see that <math>\triangle{CPP'}</math> is a <math>30-60-90</math> right triangle, meaning that <math>\angle CPP'= 60^{\circ}</math>. Thus, by adding the two together, we see that <math>\angle APC = 120^{\circ}</math>. We can now use the law of cosines as following: | ||
Revision as of 23:41, 20 April 2021
Contents
Problem
Suppose that is an equilateral triangle of side length
, with the property that there is a unique point
inside the triangle such that
,
, and
. What is
?
Solution 1
We begin by rotating by
counterclockwise about
, such that in
,
. We see that
is equilateral with side length
, meaning that
. We also see that
is a
right triangle, meaning that
. Thus, by adding the two together, we see that
. We can now use the law of cosines as following:
giving us that . ~ciceronii
Solution 2 (Intuition)
Suppose that triangle had three segments of length
, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside
(as pictured in the diagram above). Clearly
and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).
Take this equilateral triangle to have side length . The portions of each segment outside this triangle (in red) have length
. Take
to be the intersection of the segments emanating from
and
. By Law of Cosines,
So,
actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines
~ hnkevin42
Solution 3 (Answer Choices)
We begin by dropping altitudes from point down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles
,
, and
. Let
be the side of the equilateral triangle, we use the Heron's formula:
Similarly, we obtain:
By Viviani's theorem,
Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with the equation in this form. Testing , We obtain
on both sides, revealing that our answer is in fact
~ siluweston
Solution 4 (Area)
Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length.
Begin by reflecting over each of the sides. Label these reflected points
. Connect these points to the vertices of the equilateral triangle, as well as to each other.
Observe that the area of the equilateral triangle is half that of the hexagon
.
Note that . The same goes for the other vertices. This means that
is isosceles. Using either the Law of Cosines or simply observing that
is comprised of two 30-60-90 triangles, we find that
. Similarly (pun intended),
and
. Using the previous observation that
is two 30-60-90 triangles (as are the others) we find the areas of
to be
. Again, using similarity we find the area of
to be
and the area of
to be
.
Next, observe that is a 30-60-90 right triangle. This right triangle therefore has an area of
.
Adding these areas together, we get the area of the hexagon as . This means that the area of
is
.
The formula for the area of an equilateral triangle with side length is
(if you don't have this memorized it's not hard to derive). Comparing this formula to the area of
, we can easily find that
, which means that the side length of
is
.
While this approach feels rather convoluted in comparison to Solution 1, it is more flexible and can actually be generalized for any point in an equilateral triangle (although that requires use of Heron's).
~IAmTheHazard
Solution 5
Suppose ,
,
and
. So
. Since
and
, we have
Solving the equations, we have
From
, we can have
. The answer is
.
~Linty Huang
Video Solution
https://www.youtube.com/watch?v=mUW4zcrRL54
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.