2013 AMC 10B Problems/Problem 16
Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution 1
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
Solution 2
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
Solution 3
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
Solution 4
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.
Solution 5
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, .
~CoolJupiter
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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