Question

Two corners of a triangle have angles of `(3 pi ) / 8` and `pi / 8`. If one side of the triangle has a length of `3`, what is the longest possible perimeter of the triangle?

Answer 1

First, we note that if two angles are `alpha=pi/8` and `beta=(3pi)/8`, as the sum of the internal angles of a triangle is always `pi` the third angle is: `gamma=pi-pi/8-(3pi)/8 = pi/2`, so this is a right triangle.

To maximize the perimeter the known side must be the shorter cathetus, so it is going to be opposite the smallest angle, which is `alpha`.

The hypotenuse of the triangle will then be:

`c=a/sin alpha = 3/sin (pi/8)`

where `sin(pi/8) = sin (1/2pi/4) = sqrt((1-cos(pi/4))/2) =sqrt((1-sqrt(2)/2)/2)`

`c= (3sqrt(2))/sqrt(1-sqrt(2)/2)`

while the other cathetus is:

`b = a/tan(pi/8)`

where `tan(pi/8) = sqrt((1-sqrt(2)/2)/(1+sqrt(2)/2))`

`b=3sqrt((1+sqrt(2)/2)/(1-sqrt(2)/2))`

Finally:

`a+b+c = 3+ (3sqrt(2))/sqrt(1-sqrt(2)/2)+3sqrt((1+sqrt(2)/2)/(1-sqrt(2)/2))`