2013 AIME II Problems/Problem 13
In , , and point is on so that . Let be the midpoint of . Given that and , the area of can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
After drawing the figure, we suppose , so that , , and .
Using cosine law for and ,we get
So, , we get
Using cosine law in , we get
Using cosine law in and , we get
, and according to , we can get
Using and , we can solve and .
Finally, we use cosine law for ,
then , so the height of this is .
Then the area of is , so the answer is .
Let be the foot of the altitude from with other points labelled as shown below. Now we proceed using mass points. To balance along the segment , we assign a mass of and a mass of . Therefore, has a mass of . As is the midpoint of , we must assign a mass of as well. This gives a mass of and a mass of .
Now let be the base of the triangle, and let be the height. Then as , and as , we know that Also, as , we know that . Therefore, by the Pythagorean Theorem on , we know that
Also, as , we know that . Furthermore, as , and as , we know that and , so . Therefore, by the Pythagorean Theorem on , we get Solving this system of equations yields and . Therefore, the area of the triangle is , giving us an answer of .
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