Put three congruent triangles inside a unit square so that they don't overlap one another.
What is the maximum possible area of one of the triangles?
This is the best configuration I could find.
Call the square ABCD. Place the first triangle with one vertex at A and the altitude along AC. Place the other two triangles with their sides on BC and CD respectively and their third vertex at the foot of the altitude of the first triangle.
Then calling the triangles altitude h we have h+h/sqrt(2)=1 or
h = 2-sqrt(2)
From which the area A=h^2/sqrt(3) = (6-4sqrt(2))/sqrt(3)
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Posted by Jer
on 2019-02-17 11:48:40 |
Possible solution
