Difference between revisions of "1984 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
− | + | By the transversals that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity <math>K = absinC</math> 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 <math>2x,\ 3x,\ 7x</math>. Thus, the corresponding side on the large triangle is <math>12x</math>, and the area of the triangle is <math>12^2 = \boxed{144}</math>. | |
== See also == | == See also == |
Revision as of 16:57, 21 March 2009
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
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 equation 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 .
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |