1984 AIME Problem 3

Attempt Problem 3 of the 1984 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1984 AIME solutions, or check the answer key.

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3.

A point PP is chosen in the interior of △ABC\triangle ABC so that when lines are drawn through PP parallel to the sides of △ABC,\triangle ABC, the resulting smaller triangles, t1,t_1, t2,t_2, and t3t_3 in the figure, have areas 4,4, 9,9, and 49,49, respectively. Find the area of △ABC.\triangle ABC.
Answer: 144
Concepts:similarityarea ratio
Small hint:
Each of the three smaller triangles is similar to △ABC\triangle ABC
Big hint:
Convert each area ratio into a linear ratio and add the three linear ratios
Solution:
Let the area of △ABC\triangle ABC be K.K. The three small triangles are similar to △ABC,\triangle ABC, so their corresponding linear ratios are 2K,3K,7K. \frac{2}{\sqrt K},\qquad \frac{3}{\sqrt K},\qquad \frac{7}{\sqrt K}. Each such scale factor is also the perpendicular distance from PP to one side divided by the altitude to that side. These are the three barycentric area ratios [PBC]K,\frac{[PBC]}{K}, [PCA]K,\frac{[PCA]}{K}, and [PAB]K,\frac{[PAB]}{K}, which add to 1.1. Hence 2+3+7K=1. \frac{2+3+7}{\sqrt K}=1. Therefore K=12\sqrt K=12 and K=144.K=144.
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