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As much as I hate to ask for a hint on this, I've gotta admit--I'm stuck. None of my previous attempts to solve this were successful, and I can't think of a fresh way to look at the problem.

This is from Apostol's Calculus, Volume 1.

The following problem and the accompanying picture are on page 195.

An isosceles triangle is inscribed in a circle of radius r. If the angle $2\alpha$ at the apex is restricted to lie between $0$ and $\frac {\pi}{2},$ find the largest value and the smallest value of the perimeter of the triangle. Give full details of your reasoning.

Here's a brief description of the picture: An isosceles triangle is inscribed in a circle. The diameter is drawn and bisects the triangle. At one end of the diameter is the base: a chord perpendicular to the diameter. At the other end is the apex of the triangle (again, bisected by the diameter), which touches the circle.

Let $C$ be the center of the circle. Draw a radius from $C$ to either vertex at the base. This radius will form an angle $\theta=2\alpha$ with the diameter. Since the diameter bisects the triangle, it suffices to minimize or maximize either half of the perimeter of the triangle. Half of base is then $r\sin 2\alpha$. Using the law of cosines and the trig identity $\cos (x- \frac{\pi}{2})=\sin(x)$, either leg will have length $$\sqrt {2r^2-2r^2\mathrm {sin}2\alpha},$$ so that the perimeter of half the triangle is $$P(x)=r\sin 2\alpha+\sqrt{2r^2-2r^2 \mathrm {sin}2\alpha}.$$ Differentiating gives you $$P'(x)=2r\cos2\alpha - \frac{2r^2\cos 2\alpha}{\sqrt{2r^2-2r^2 \mathrm {sin}2\alpha}}.$$ At that point, I think, "This can't be right. I supposed to set that equal to $0$ and solve? That can't be right."

Is there a better approach?

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  • $\begingroup$ Why not set that equal to $0$ and solve? :) $\endgroup$
    – Evan
    Commented Aug 11, 2013 at 1:13

1 Answer 1

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Hint: Half the base is indeed $r\sin 2\alpha$. We now find the length of one of the other sides of the triangle. By trigonometry this is $\dfrac{r\sin 2\alpha}{\sin\alpha}$, which is $2r\cos\alpha$.

So we want to maximize/minimize $r\sin 2\alpha +2r\cos\alpha$. We can forget about $2r$, and only worry about $\sin\alpha\cos\alpha +\cos\alpha$.

Examining the derivative of this should not be difficult. We get a quadratic in $\sin\alpha$. It even factors nicely.

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  • $\begingroup$ I can't see how the length of either leg is $\frac {r \mathrm{sin}2\alpha}{\mathrm {sin} \alpha}$. $\endgroup$
    – Ryan
    Commented Aug 11, 2013 at 1:49
  • $\begingroup$ You have drawn the diagram. So you have have split the triangle into two halves, each right-angled. The angle at the top of each is $\alpha$. So if $x$ is the length of the remaining side, then $\frac{r\sin 2\alpha}{x}=\sin \alpha$. Solve for $x$. $\endgroup$ Commented Aug 11, 2013 at 1:53
  • $\begingroup$ Aha! Thanks so much Andre! $\endgroup$
    – Ryan
    Commented Aug 11, 2013 at 2:07
  • $\begingroup$ You are welcome. $\endgroup$ Commented Aug 11, 2013 at 2:10

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