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?

1 Answer
Nov 28, 2016

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))#