Home » Blog » Find the extreme values for the perimeter of an isosceles triangle inscribed in a circle

Find the extreme values for the perimeter of an isosceles triangle inscribed in a circle

  1. Consider a circle of radius r with an inscribed isosceles triangle. If the angle at the apex of the isosceles triangle is 2 \alpha, and 0 \leq \alpha \leq \frac{\pi}{2}, find the extreme values (maximum and minimum) of the perimeter of the triangle.
  2. Find the minimum value of the radius r such that the circle of radius r will cover every isosceles triangle with fixed perimeter L.

Rendered by QuickLaTeX.com

  1. In the picture we draw in six radial lines, which divide the triangle into six congruent triangles, each with hypotenuse of length r and legs of lengths r \sin \alpha and r \cos \alpha. As shown in the diagram, two of the edges of the inscribed isosceles triangle have lengths 2 r \cos \alpha, and the third leg as length 2 r \sin (2 \alpha). Therefore, the perimeter is,

        \[ P = 4r \cos \alpha + 2r \sin (2 \alpha). \]

    Taking the derivative,

        \[ P' = -4r \sin \alpha + 4r \cos (2 \alpha). \]

    Now, we must note that the extreme values of the function occur when the derivative is 0 or at an end point of the interval we are looking at; in this case, 0 \leq \alpha \leq \frac{\pi}{4} is the interval stipulated in the problem.

    First, let’s set the derivative equal to zero and find the values of \alpha in the interval at which the derivative is 0.

        \begin{align*}  -4r \sin \alpha + 4r \cos (2 \alpha) = 0 && \implies && 4r \cos (2 \alpha) = 4r \sin \alpha \\  && \implies && \cos (2 \alpha) &= \sin \alpha \\  && \implies && 1 - 2 \sin^2 \alpha &= \sin \alpha &(\text{Trig Ident.})\\  && \implies && 2 \sin^2 \alpha + \sin \alpha - 1 &= 0 \\  && \implies && \sin \alpha = \frac{-1 \pm \sqrt{1 + 8}}{4} \\  && \implies && \sin \alpha = \frac{1}{2}, \text{ or } -1. \end{align*}

    Then, the only value of \alpha in the interval \left[0, \frac{\pi}{4} \right] such that \sin \alpha = \frac{1}{2} or \sin \alpha = -1 is

        \[ \alpha = \frac{\pi}{6}. \]

    Plugging this in to our expression for the perimeter of the isosceles triangle, we find,

        \begin{align*}    P_{\max} &= 4r \cos \alpha + 2r \sin (2 \alpha) \\  &= 4r \cos \left( \frac{\pi}{6} \right) + 2r \sin \left( \frac{\pi}{3} \right) \\  &= 2\sqrt{3} r + \sqrt{3} r \\  &= 3 \sqrt{3} r. \end{align*}

    Examining the sign of the derivative of the perimeter function as \alpha changes, we see that this is a maximum. Furthermore, we see that the minimum occurs at the end point, \alpha = 0. The minimum perimeter is then

        \begin{align*}  P_{\min} &= 4r \cos \alpha + 2r \sin (2 \alpha) \\  &= 4r \cos 0 + 2r \sin 0 \\  &= 4r. \end{align*}

  2. First, we identify the value of \alpha at which the radius necessary to cover the isosceles triangle is at a maximum (the worst case scenario for covering the triangle, if you like). From part (a) we know the perimeter L must satisfy the equation

        \[ L = 4r \cos \alpha + 2r \sin (2 \alpha). \]

    Since L is now a fixed constant we can write r as a function of \alpha,

        \[ r = \frac{L}{4 \cos \alpha + 2 \sin (2 \alpha)}. \]

    Now, the value of r is maximal when the denominator is minimal, so we are looking for a minimum of the function

        \[ f( \alpha) = 4 \cos \alpha + 2 \sin (2 \alpha). \]

    Taking the derivative,

        \[ f'(\alpha) = -4\sin \alpha + 4 \cos (2 \alpha). \]

    We set this equal to zero to find the critical points

        \[ -4 \sin \alpha + 4 \cos (2 \alpha) = 0 \quad \implies \quad \sin \alpha = \cos (2 \alpha). \]

    From part (a) we know this implies \alpha = 0 or \alpha = \frac{\pi}{6}. Again, from part (a) we know this is minimal when \alpha = 0. Since we are trying to minimize this we want the value \alpha = 0.

    Finally, there is a bit of a problem with \alpha = 0 since that corresponds to just a line, not really a triangle. We can either ignore that issue, and just plug in \alpha = 0 or (in what will amount to the same thing) we can find the minimal value of r as the limit as \alpha \to 0:

        \begin{align*}  r &= \lim_{\alpha \to 0} \frac{L}{4 \cos \alpha + 2 \sin (2 \alpha)} \\  &= \frac{L}{4}, \end{align*}

    where we can compute the limit by just evaluating at \alpha = 0 since the function is continuous.


  1. Mihajlo says:

    What about the case when alpha = π/2? The perimeter is 0 then, so the problem text should probably constrain alpha to be in the same interval as in the part a) of the problem.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):