2020 AMC 12A Problems/Problem 24
- 1 Problem
- 2 Solution 1 (Rotations)
- 3 Solution 2 (Intuition)
- 4 Solution 3 (Answer Choices)
- 5 Solution 4 (Area)
- 6 Solution 5 (Coordinate Bashing)
- 7 Solution 6 (Diagram Nuke)
- 8 Solution 7 (Theorem Nuke)
- 9 Solution 8 (More Coordinate Bashing)
- 10 Video Solutions
- 11 See Also
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 (Rotations)
We begin by rotating counterclockwise by about , such that and . 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 .
Rotate counterclockwise around point to . Then , so is an equilateral triangle. Note that is a -- triangle, hence , and , so and the answer is .
Rotate counterclockwise by around point to . Then , and , so is an equilateral triangle. Note that is a -- triangle, hence , and , so and the answer is .
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
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 ~ edits by aopspandy
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 (which only works for isosceles triangles), it is more flexible and can actually be generalized for any point in a general triangle (although that requires use of Heron's, and potentially Law of Sines and Cosines).
Solution 5 (Coordinate Bashing)
Suppose , , and . So . Since and , we have Solving the equations, we have From (and a fair amount of algebra), we can have . The answer is .
Solution 6 (Diagram Nuke)
Drawing out a rough sketch, it appears that . By Pythagorean, our answer is .
Proof of this fact can be found in the Video Solution by Richard Rusczyk below.
Solution 7 (Theorem Nuke)
We can use the following theorem:
We know that , and . Plugging these into our formula, we get . Let . Then, we have . Solving for , we get . If is equal to , then we have , but this is not possible since is inside of the triangle. This means that , and therefore .
Solution 8 (More Coordinate Bashing)
Set the points . Then, we want to find the intersection of the three circles Subtracting circle 1 from circle 2 yields , which can be rewritten as . Subtracting circle 1 from circle 3 yields , or . Subtracting from this yields , or .
We then substitute our values for and back into , which gives us Solving this equation (the algebra is surprisingly not bad!) gives us . Since must be greater than 1, the answer is .
- curiousmind888 & TGSN
Video Solution by Richard Rusczyk - https://www.youtube.com/watch?v=xnAXGUthO54&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=4 - AMBRIGGS
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