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

by RoRi
  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.

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  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.

2 comments

  2. 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.

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