Find the dimensions and orientation of the largest area (a) square and (b) rectangle you can draw inside an equilateral triangle of side 1.
Source: Past puzzle experience.
(Excuse the bad ascii art - obviously view in fixed-width font!)
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/__\ P
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Assuming the base of the rectangle lies against the base of the triangle...
P is the distance along one edge (0 < P < 1) where rectangle meets triangle.
Height of rectangle is P cos60 = P sqrt(3)/2
Base of rectangle is 1 - 2 P sin60 = 1 - P
Area of rectangle = sqrt(3)/2 (P - P^2)
Differentiating, we get dA/dP = sqrt(3)/2 (1 - 2P)
This has a zero at P=0.5. (simple checking shows this is indeed a maximum)
This gives us a RECTANGLE of area = sqrt(3)/8. (approx 0.2165)
(dimensions sqrt(3)/4 x 1/2) =========
For a SQUARE, base=height, so P sqrt(3)/2 = 1-P
This reduces to P = 1 / (1 + sqrt(3)/2)
Edge length reduces to 1 / (1 + 2/sqrt(3))
for an area of 1/(4/sqrt(3) + 7/3). (approx 0.21539)
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Now check square rotated by 45 degrees...
(Apex of square is at midpoint of triangle base.)
We have a corner triangle formed, with angles 60,45,75. 0.5 (half-base)
length is opposite 75 deg angle.
Using sine rule to calculate side (L) of the square. (a/sinA = b/sinB =
c/sinC)
0.5/sin75 = L/sin60 -> L = sin60/2.sin75 = approx. 0.44828, for an area of
0.20096. Less than by aligning with the base. Therefore, as expected, the
original orientation stands.