Difference between revisions of "2013 AIME I Problems/Problem 12"
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<math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot sin 75^\circ</math>. | <math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot sin 75^\circ</math>. | ||
− | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4} | + | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, thus the answer is \boxed{021}</math> |
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=11|num-a=13}} | {{AIME box|year=2013|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:49, 16 June 2018
Problem 12
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 .
Cartesian Variation Solution
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 ordinate 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 2 (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
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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