Find the area of triangle ABC, whose medians have measures 39, 42, and 45.

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Let's glue an equal triangle to the original one but flipped to obtain a parallelogram (see attached image). Denote the medians `a, b, c .`

It is known that the point of medians intersection divides them 1 : 2; it is also known that each of the sub-triangles (say, green...

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Let's glue an equal triangle to the original one but flipped to obtain a parallelogram (see attached image). Denote the medians `a, b, c .`

It is known that the point of medians intersection divides them 1 : 2; it is also known that each of the sub-triangles (say, green or yellow) have the same area.

This way, the area of the triangle formed by both green and yellow triangles is one-third of the area of the original triangle.

And this composite triangle is formed by sides of lengths `2 / 3 a , ` `2 / 3 b , ` `2 / 3 c . ` By Heron's formula, its area is `A_1 = ( 2 / 3 )^2 sqrt ( s ( s - a ) ( s - b ) ( s - c ) ) ,` where `s = 1 / 2 ( a + b + c ) . ` Here `a = 3 * 13 , ` `b = 3*14, ` `c = 3*15, ` so `s = 3*21 ` and we obtain `A_1 = 4 sqrt (21*8*7*6) = 4*7*3*4.`

Finally, the area in question is `3A_1 = 1008 .`

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