Problem on finding the inscribed triangle of maximal area

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Use Lagrange multipliers to find the triangle of largest area that can be inscribed in a circle of radius $r$.

  • Solution

    This problem appears to be a constrained optimization problem: we are to maximize the area of a triangle subject to the constraint that its three points lie on a circle.
    To analyze this problem, we need to choose convenient variables. The variables should be such that the constraint and the formula for area are simple.
    As derived in the next section on Determining the most convenient variables, we introduce the variables $\theta_1, \theta_2, \theta_3$ that are the interior angles between the three points:
    The area of the blue triangle is equal to the sum of the areas of the three isosceles triangles inside. These isosceles triangles have areas $\frac{1}{2} r^2 \sin \theta_1$, $\frac{1}{2} r^2 \sin \theta_2$, $\frac{1}{2} r^2 \sin \theta_3$. The $\theta$ variables must satisfy the constraint $\theta_1 + \theta_2 + \theta_3 = 2 \pi$.
    Our problem can be formulated precisely as: $$ \text{maximize } \frac{1}{2} r^2 \sin \theta_1 + \frac{1}{2} r^2 \sin \theta_2 + \frac{1}{2} r^2 \sin \theta_3 \\\text{ subject to } \theta_1 + \theta_2 + \theta_3 = 2 \pi$$
    Recall that
    To use Lagrange multipliers to solve the problem $$\max f(\theta_1,\theta_2,\theta_3) \text{ subject to } g(\theta_1,\theta_2,\theta_3) = 0,$$

    1. Form the augmented function $$L(\theta_1,\theta_2,\theta_3,\lambda) = f(\theta_1,\theta_2,\theta_3) - \lambda g(\theta_1,\theta_2,\theta_3)$$

    2. Set all partial derivatives of $L$ equal to zero

    3. Solve for $\theta_1,\theta_2,\theta_3$.
    We identify \begin{align}f(\theta_1, \theta_2, \theta_3) &= \frac{1}{2} r^2 \sin \theta_1 + \frac{1}{2} r^2 \sin \theta_2 + \frac{1}{2} r^2 \sin \theta_3 \\ g(\theta_1, \theta_2, \theta_3) &= \theta_1 + \theta_2 + \theta_3 - 2\pi.\end{align}
    Forming the augmented functional, we have \begin{align} L(\theta_1, \theta_2, \theta_3, \lambda) = &\frac{1}{2} r^2 \sin \theta_1 + \frac{1}{2} r^2 \sin \theta_2 + \frac{1}{2} r^2 \sin \theta_3 \\&- \lambda ( \theta_1 + \theta_2 + \theta_3 - 2 \pi)\end{align}
    The critical points of $L$ are given by setting $\partial_{\theta_1} L= \partial_{\theta_2} L= \partial_{\theta_3} L= \partial_\lambda L= 0$:
    \begin{align}
    \partial_{\theta_1}L = \frac{1}{2} r^2 \cos \theta_1 - \lambda &=0 \\
    \partial_{\theta_2}L = \frac{1}{2} r^2 \cos \theta_2 - \lambda &=0 \\
    \partial_{\theta_3}L= \frac{1}{2} r^2 \cos \theta_3 - \lambda &=0 \\
    \partial_{\lambda}L = \theta_1 + \theta_2 + \theta_3 - 2 \pi &= 0
    \end{align}
    The solution to these equations is $\theta_1 = \theta_2 = \theta_3 = 2 \pi / 3$.
    We have thus shows that the inscribed triangle that maximizes area is equilateral.

  • Determining the most convenient variables

    We seek to minimize the area of the triangle subject to the constraint that it is inscribed in the circle.
    We need to find variables in which it is easy to write the constraint and the formula for the triangle's area.

    First attempt: Cartesian coordinates

    We start by using variables corresponding to the $x,y$ coordinates of the three points. Without loss of generality, we can let one of the points be $(r,0)$. The other two points are then $(x_1, y_1)$ and $(x_2, y_2)$.
    The constraint becomes that $x_1^2 + y_1^2 = r^2$ and $x_2^2 + y_2^2 = r^2$.
    Finding a formula for area in terms of $x_1, y_1, x_2, y_2$ is tricky. One strategy would be to use Heron's formula, which relates the area of the triangle to the lengths of all three sides. This formula is relatively messy.
    Another approach to get the area of the triangle would be to use Green's theorem to convert a double integral over the triangle into a line integral over its boundary. This approach is possible, but still requires the two constraints on $(x_1, y_1)$ and $(x_2, y_2)$.

    Second attempt: Angular coordinates

    Because the problem is circular in nature, we consider choosing the variables corresponding to the angles of the three points. Without loss of generality, one of the points can be chosen at $\theta = 0$. We call the other two $\theta_1$ and $\theta_2$.
    The constraint that the three points are on the circle is automatically enforced for any values of $\theta_1$ and $\theta_2$. Writing down the formula for the area in terms of $\theta_1$ and $\theta_2$ is not particularly apparent.

    Third attempt: Interior angle coordinates

    We see that finding a simple formula for area is difficult. This is because our triangle does not have any clear structure. For example, it is not necessarily a right triangle or an isosceles triangle, for which there are nice formulas for area.
    We consider breaking the triangle into three isosceles parts:
    The area of each isosceles triangle is $\frac{1}{2} r^2 \sin \theta$. Hence the area of the blue triangle is $$\frac{1}{2} r^2 \sin \theta_1 + \frac{1}{2} r^2 \sin \theta_2 + \frac{1}{2} r^2 \sin \theta_3.$$
    Not any values of $\theta_1, \theta_2, \theta_3$ are permissible. They must satisfy the constraint that the interior angles adds up to $2\pi$:$$ \theta_1 + \theta_2 + \theta_3 = 2 \pi.$$

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