2017 AMC 8 Problems/Problem 16
In the figure below, choose point on so that and have equal perimeters. What is the area of ?
Because we can see that when we draw a line from point to imaginary point that line applies to both triangles. Let us say that is that line. Perimeter of would be , while the perimeter of would be . Notice that we can find from these two equations by setting them equal and then canceling . We find that , and because the height of the triangles is the same, the ratio of the areas is , so that means that the area of .
We know that the perimeters of the two small triangles are and . Setting both equal and using , we have and . Now, we simply have to find the area of . Since , we must have . Combining this with the fact that , we get .
Since point is on line , it will split it into and . Let and . Triangle has side lengths and triangle has side lengths . Since both perimeters are equal, we have the equation . Eliminating and solving the resulting linear equation gives . Draw a perpendicular from point to . Call the point of intersection . Because angle is common to both triangles and , and both are right triangles, both are similar. The hypotenuse of triangle is 2, so the altitude must be Because and share the same altitude, the height of therefore must be . The base of is 4, so .
Using any preferred method, realize . Since we are given a 3-4-5 right triangle, we know the value of . Since we are given , apply the Sine Area Formula to get .
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