Familiar configuration

Geometry Level 3

Let A B C ABC be a triangle with area equal to 100 100 , and D , E D, E are interior points of the sides A B , A C AB, AC , respectively.

Let F F be the intersecting point of C D CD and B E BE .

Find the maximum value of the area of triangle D E F DEF .


The answer is 9.017.

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1 solution

Ivo Zerkov
Mar 19, 2018

If BD AB = a \frac{\text{BD}}{\text{AB}}=a and CE AC = b \frac{\text{CE}}{\text{AC}}=b , it’s fairly straightforward using Menelaus’ theorem to show DEF \text{DEF} ’s area to be a b ( 1 a b + a b ) a + b a b \frac{a\cdot b \cdot(1-a-b+a\cdot b)}{a+b-a\cdot b} of ABC \text{ABC} ’s. Maximizing this numerically yields a = b 0.38197 a=b\approx0.38197 , making the answer 9.017 \approx9.017 .

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