1984 AIME Problems/Problem 3

From AoPSWiki

Problem

A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ , $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ , $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ .

size(200);pathpen=black;pointpen=black;pair A=(0,0),B=(12,0),C=(4,5);D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B...

Solution

By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ 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 $K = \dfrac{ab\sin C}{2}$ 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 $2x,\ 3x,\ 7x$ . Thus, the corresponding side on the large triangle is $12x$ , and the area of the triangle is $12^2 = \boxed{144}$ .

1984 AIME (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

1984 AIME Problems/Problem 3

From AoPSWiki

Problem

Solution

See also