What is the radius of smallest circular disc large enough to cover every asute angle isoceles triangle of perimeter L

Dear Student:

To solve the question, first draw an isosceles triangle, base AB, vertex X, so AX = BX and AXB<90°.
Let 2θ = angle AXB
Let S = AX = BX
Now draw the circumscribed circle passing through A, B and X. This is the perimeter of the smallest disc able to cover the triangle.
Mark the circle's centre O.
The triangle is symmetric about the line OX. From a suitable diagram it should be clear that the total length of the sides is:
L = 2(S+Ssin(θ))
L = 2S(1+sin(θ)) (equation 1)
It the radius of the circle is R, note that OX is a radius and from th ediagram that OXcos(θ) = S/2
Rcos(θ) = S/2
S = 2Rcos(θ)
From equation 1:
L = 2(2Rcos(θ))(1+sin(θ))
L = 4R[cos(θ)(1+sin(θ))]
R = (L/4)/[cos(θ)(1+sin(θ))]
Let y = cos(θ)(1+sin(θ))
R = (L/4)/y
To get the smallest possible R, we want the largest possible value
of y.
We find the maximum of y in the usual way using calculus:
y = cos(θ)(1+sin(θ))
dy/dθ = 2cos²(θ) - sin(θ) - 1
For max/min dy/dθ=0:
2cos²(θ) - sin(θ) - 1 = 0
Since cos²(θ) = 1 - sin²(θ):
2(1-sin²(θ)) - sin(θ) - 1 = 0
For neatness, let x = sin(θ)
2(1-x²) - x - 1 = 0
2 - 2x² - x - 1 = 0
2x² + x - 1 = 0
(2x - 1)(x + 1) = 0
x = 1/2 or x = -1
Find if x=1/2 ro x=-1 gives the maximum of y:
y = cos(θ)(1+sin(θ))
= (√(1-x²))(1+x)
When x = 1/2, y = (√(1-½²))(1+½) = (√¾)(3/2) = ¾√3
When x = -1. y = (√(1-1²))(1+1) = 0
So the maximum vaue of y is ¾√3
So the smallest value of R = (L/4)/ (¾√3) = L/(3√3) = L√3/9

Regards

Sumit