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 .
Note (different ending): When you have the length of the base , you don't need to find the equations of the lines and . Instead, make an altitude from to , and call the foot . is a triangle and is a triangle. And, they both share . So, we can set as , so is . Since is a triangle, . The base can be written as . Solve this equation and and . Multiply this by base and divide by to get the area of triangle which is . So, -hwan
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 .
.
Evaluating and reducing, we get thus the answer is
Solution 4 (Special Triangles)
As we can see, the angle of can be split into a angle and a angle. This allows us to drop an altitude from point for which intersects at point and at point . The main idea of our solution is to obtain enough sides of that we are able to directly figure out its area (specifically by figuring out side and ).
We first begin by figuring out the length of . This can be easily done, since is simply (given in the problem) and because is an equilateral after some simple angle calculations. Now we need to find . This is when we bring in some simple algebra.
PREPARATION: (45-45-90 Right Triangle)
(30-60-90 Right Triangle)
SOLVING:
so
Finally,
Now, we can finally get the length of by adding up , which is simply
To get and , we first work bit by bit.
(30-60-90 Right Triangle)
(same 30-60-90 Right Triangle)
Since because of 45-45-90 right triangles,
too.
Now, we can finally calculate , and it is .
Finally, the area of can be calculated by , which is equal to . So the final answer is .
-by What do Humanitarians Eat?
Solution 5 (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 6 (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 7 (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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