Prove area of triangle with a given base and a given vertex angle is maximum when triangle is isoceles

Question

Consider the set of triangles having a given base and a given vertex angle. Show that the triangle having the maximum area will be isosceles.

I have taken $a$ and $\alpha$ to be the given base and vertex angle respectively. Area of triangle in terms of only sides is $\frac{1}{4}\sqrt{4b^2c^2-b^2-c^2+a^2}$ - got this from cosine formula.

Now since $b$ and $c$ are variables (while $a$ is a constant) I have taken them to be $x$ and $y$ respectively, and tried to maximise this expression $4x^2y^2-x^2-y^2+a^2$. But upon taking partial derivatives and equating them to 0, I am getting specific values of $x$ and $y$, instead of a relation as asked in the question. Why?

Answer 1

Consider a triangle with the law of sines setup/notation:

${\displaystyle {\frac {a}{\sin \alpha}}\,=\,{\frac {b}{\sin \beta}}\,=\,{\frac {c}{\sin \gamma}}}$

Recall that Angle-Angle-Side describes a unique triangle with area given by

$\tag 1 a^2 \frac{sin(\beta) sin(\gamma)}{2 sin(\alpha)}$

If both $a$ and $\alpha$ are set, we can let $\beta$ be a variable, and can write

$\tag 2 \gamma = \pi - (\beta + \alpha) $

A useful trigonometric identity allows us to write

$\tag 3 sin(\gamma) = sin(\beta + \alpha)$

To maximize (1), we can ignore the constant multiplicative factors, finding that the function

$f(\beta) = sin(\beta) sin(\beta + \alpha)$

is of interest. Using the sine angle addition identity and the sine double angle identity, you can write

$f(\beta) = cos(\alpha) sin^2(\beta) + sin(\alpha) (.5) sin(2\beta)$

Using the double angle identity again, you will find that the derivative of $f$ with respect to the variable $\beta$ is equal to

$\tag 4 cos(\alpha) sin(2 \beta) + sin(\alpha) cos(2\beta)$

If you set the expression (4) to $0$, after some algebra you can write

$\tag 5 tan(2\beta) = - tan(\alpha)$

But using the useful identity,

$tan(\pi - \theta) = -tan(\theta)$, we find that the derivative is zero when

$\tag 6 \beta = \frac{\pi - \alpha}{2}$

When $\beta$ takes this value so does $\gamma$, so we are looking at an isosceles triangle when the derivative is $0$. Examining (1) it is easy to see that this is the maximum possible area for these constrained triangles.

Answer 2

$b$ and $c$ are variables, indeed, but they're not independent. That's why setting the partials WRT $b$ and $c$ to zero doesn't work.

You might instead put two vertices at $(\pm \frac{a}{2}, 0)$, and then the top vertex at $(0, q)$, where $q$ is chosen to make the vertex angle be $\alpha$. (Hint: you're going to need an arctangent somewhere). That's the supposedly optimizing triangle, but what are the other possible locations for that third vertex?

Answer: Take the circle containing the three points I just described. (There's only one such circle). Then the "top" point can be at any point on the arc of the circle that's above the $x$-axis, because of a theorem about angles subtended by an arc of a circle.

NOW you've got a single variable --- the position of that third point on the upper arc of the circle --- and can use some calculus to optimize. Or you can do some nice geometric arguments and avoid the calculus altogether. :)