2019 AMC 12B Problems/Problem 25
Let be a convex quadrilateral with and Suppose that the centroids of and form the vertices of an equilateral triangle. What is the maximum possible value of the area of ?
Solution 1 (vectors)
Place an origin at , and assign position vectors of and . Since is not parallel to , vectors and are linearly independent, so we can write for some constants and . Now, recall that the centroid of a triangle has position vector .
Thus the centroid of is ; the centroid of is ; and the centroid of is .
Hence , , and . For to be equilateral, we need . Further, . Hence we have , so is equilateral.
Now let the side length of be , and let . By the Law of Cosines in , we have . Since is equilateral, its area is , while the area of is . Thus the total area of is , where in the last step we used the subtraction formula for . Alternatively, we can use calculus to find the local maximum. Observe that has maximum value when e.g. , which is a valid configuration, so the maximum area is .
Let , , be the centroids of , , and respectively, and let be the midpoint of . , , and are collinear due to well-known properties of the centroid. Likewise, , , and are collinear as well. Because (as is also well-known) and , we have . This implies that is parallel to , and in terms of lengths, . (SAS Similarity)
We can apply the same argument to the pair of triangles and , concluding that is parallel to and . Because (due to the triangle being equilateral), , and the pair of parallel lines preserve the angle, meaning . Therefore is equilateral.
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:
Let , where due to the Triangle Inequality in . By breaking the quadrilateral into and , we can create an expression for the area of . We use the formula for the area of an equilateral triangle given its side length to find the area of and Heron's formula to find the area of .
Substituting , the expression becomes
We can ignore the for now and focus on .
By the Cauchy-Schwarz inequality,
The RHS simplifies to , meaning the maximum value of is .
Thus the maximum possible area of is .
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