2013 AMC 10B Problems/Problem 16
In triangle , medians and intersect at , , , and . What is the area of ?
Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so AP must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . Since the diagonals of quadrilaterals , and , are perpendicular, the area of is
Note that triangle is a right triangle, and that the four angles (angles and ) that have point are all right angles. Using the fact that the centroid () divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is
From the solution above, we can find that the lengths of the diagonals are and . Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
We know that , and using median properties. So Now we try to find . Since , then the side lengths of are twice as long as since and are midpoints. Therefore, . It suffices to compute . Notice that is a Pythagorean Triple, so . This implies , and then . Finally, .
As from Solution 4, we find the area of to be . Because , the altitude perpendicular to . Also, because , is similar to with side length ratio , so and the altitude perpendicular to . The altitude of trapezoid is then and the bases are and . So, we use the formula for the area of a trapezoid to find the area of
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