1984 AIME Problems/Problem 3
Contents
Problem
A point is chosen in the interior of such that when lines are drawn through parallel to the sides of , the resulting smaller triangles , , and in the figure, have areas , , and , respectively. Find the area of .
Solution 1
By the transversals that go through , all four triangles are similar to each other by the postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as . Thus, the corresponding side on the large triangle is , and the area of the triangle is .
Solution 2
Alternatively, since the triangles are similar by , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume That means that the base of is 4, the base of is 6, and the base of is 14. Since the quadrilaterals underneath and are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is . Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is .
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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