2013 AIME I Problems/Problem 12
Contents
Problem
Let be a triangle with and . A regular hexagon with side length 1 is drawn inside so that side lies on , side lies on , and one of the remaining vertices lies on . There are positive integers and such that the area of can be expressed in the form , where and are relatively prime, and c is not divisible by the square of any prime. Find .
Solution 1
First, find that . Draw . Now draw around such that is adjacent to and . The height of is , so the length of base is . Let the equation of be . Then, the equation of is . Solving the two equations gives . The area of is .
Solution 2 (Cartesian Variation)
Use coordinates. Call the origin and be on the x-axis. It is easy to see that is the vertex on . After labeling coordinates (noting additionally that is an equilateral triangle), we see that the area is times times the coordinate of . Draw a perpendicular of , call it , and note that after using the trig functions for degrees.
Now, get the lines for and : and , whereupon we get the ordinate of to be , and the area is , so our answer is .
Solution 3 (Trig)
Angle chasing yields that both triangles and are -- triangles. First look at triangle . Using Law of Sines, we find:
Simplifying, we find . Since , WLOG assume triangle is equilateral, so . So .
Apply Law of Sines again,
Simplifying, we find .
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Evaluating and reducing, we get thus the answer is
Solution 4 (Trig)
With some simple angle chasing we can show that and are congruent. This means we have a large equilateral triangle with side length and quadrilateral . We know that . Using Law of Sines and the fact that we know that and the height to that side is so . Using an extremely similar process we can show that which means the height to is . So the area of . This means the area of quadrilateral . So the area of our larger triangle is . Therefore .
Solution 5 (Elementary Geo)
We can find that . This means that the perpendicular from to is perpendicular to as well, so let that perpendicular intersect at , and the perpendicular intersect at . Set . Note that , so and . Also, , so . It's easy to calculate the area now, because the perpendicular from to splits into a (PHQ) and a (PHR). From these triangles' ratios, it should follow that , so the area is . . By Mathscienceclass
Solution 6 (Combination of 1 & 2)
We can observe that (because & are both ). Thus we know that is equivalent to the height of the hexagon, which is . Now we look at triangle and apply the Law of Sines to it. . From here we can solve for and get that . Now we use the Sine formula for the area of a triangle with sides , , and to get the answer. Setting and we get the expression which is . Thus our final answer is . By AwesomeLife_Math
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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