2013 AIME II Problems/Problem 5
In equilateral let points and trisect . Then can be expressed in the form , where and are relatively prime positive integers, and is an integer that is not divisible by the square of any prime. Find .
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let be the midpoint of . Then is a 30-60-90 triangle with , and . Since the triangle is right, then we can find the length of by pythagorean theorem, . Therefore, since is a right triangle, we can easily find and . So we can use the double angle formula for sine, . Therefore, .
We find that, as before, , and also the area of is 1/3 the area of . Thus, using the area formula, , and . Therefore,
Let A be the origin of the complex plane, B be , and C be . Also, WLOG, let D have a greater imaginary part than E. Then, D is and E is . Then, . Therefore,
Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, or The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, Since Then So . Therefore,
Solution 5 (Vectors)
Setting up a convinient coordinate system, we let be at point , be at point , and be at point . Then and will be at points and . Then . From here, we see that
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