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1985 AIME Problems/Problem 6

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Problem

As shown in the figure, triangle ABC is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle ABC.

Image:AIME 1985 Problem 6.png

Solution

Let the interior point be P, let the points on \overline{BC}, \overline{CA} and \overline{AB} be D, E and F, respectively. Let x be the area of \triangle APE and y be the area of \triangle CPD. Note that \triangle APF and \triangle BPF share the same altitude from P, so the ratio of their areas is the same as the ratio of their bases. Similarly, \triangle ACF and \triangle BCF share the same altitude from C, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: \frac{40}{30} = \frac{124 + x}{65 + y} or equivalently 372 + 3x = 260 + 4y and so 4y = 3x+ 112.

Applying identical reasoning to the triangles with bases \overline{CD} and \overline{BD}, we get \frac{y}{35} = \frac{x+y+84}{105} so that 3y = x + y + 84 and 2y = x + 84. Substituting from this equation into the previous one gives x = 56, from which we get y = 70 and so the area of \triangle ABC is 56 + 40 + 30 + 35 + 70 + 84 = 315.

See also

1985 AIME (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
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