I've written this problem myself and I'm wondering if there's a quicker/more geometrical way to answer it (e.g. one without the use of calculus).

Show that if the height of a right-angled triangle is the diameter of a circle and its base is the reciprocal of the radius, then the hypotenuse of the triangle is shortest when the area of the circle is $2\sqrt2$ times smaller than the circumference.


Let $r$ be the radius of the circle. Then the hypotenuse can be expressed as $$\text{hypotenuse}=f(r)=\sqrt{(2r)^2+\left(\frac1r\right)^2}=(4r^2+r^{-2})^{1/2}$$ Differentiating, we have $$f'(r)=\frac12(4r^2+r^{-2})^{-1/2}(8r-2r^{-3})=\frac{4r^4-1}{r^3\sqrt{4r^2+\dfrac1{r^2}}}=0$$ for stationary points. So $$4r^4-1=0\implies r=\left(\frac14\right)^{1/4}=\frac1{\sqrt2}\implies f(r)=2$$ Now at $r=\dfrac12<\dfrac1{\sqrt2}$ and at $r=1>\dfrac1{\sqrt2}$, $f(r)=\sqrt5>2$ so $$r=\frac1{\sqrt2}$$ is a minimum, meaning that the hypotenuse is shortest at this $r$.

Circumference $=2\pi r=\pi\sqrt2$ and Area $=\pi r^2=\dfrac\pi2=\dfrac1{2\sqrt2}(\pi\sqrt2)$ so the hypotenuse is shortest when the area is $2\sqrt2$ times smaller than the circumference.

  • $\begingroup$ "base" of a triangle is ambiguous, because any side can be a "base" (with a corresponding "height"). Since you're dealing with right triangles, just write something like: "One leg is the diameter of a circle, and the other leg's length is the reciprocal of the radius of that circle." $\endgroup$ – Blue Dec 27 '17 at 20:37

Right Triangle

Let a circle have radius $r$. Now let a right-angled triangle as in the figure have $a=\frac{1}{r}$ and $b=2r$.

Now $\sin\,\alpha=\frac{a}{c}= \frac{1}{rc}$, and $\cos\,\alpha=\frac{b}{c}= \frac{2r}{c}$.

Also $\sin\,\beta=\frac{b}{c}= \frac{2r}{c}$, and $\cos\,\beta=\frac{a}{c}= \frac{1}{rc}$.

Hence $\sin\,\alpha=\cos\,\beta$ and $\cos\,\alpha=\sin\,\beta$, or in other words $\alpha=\beta=45^{\circ}$. Using $\tan\,\alpha=\frac{a}{b}=\frac{1/r}{2r}=\frac{1}{2r^2}=1$, or $\tan\,\beta=\frac{b}{a}= \frac{2r}{1/r}=2r^2=1$ we find $r=\frac{1}{\sqrt{2}}$.

For the ratio of the circumference of the circle to its area we have $$\frac{C_{\circ}}{A_{\circ}}=\frac{2\pi r}{\pi r^2}=\frac{2\pi \frac{1}{\sqrt{2}}}{\pi \frac{1}{2}}=2\sqrt{2}$$ as required.

  • $\begingroup$ A great way to tackle the problem. Thank you! $\endgroup$ – TheSimpliFire Dec 28 '17 at 13:29
  • $\begingroup$ My pleasure, thanks. Note doing it this way gives you $r$ directly. Hence you side step having to worry about whether the hypotenuse is minimum explicitly. The geometry / trig just works out the situation for you due to the constraints on $\alpha$ and $\beta$. $\endgroup$ – Daniel Buck Dec 28 '17 at 14:14

Yes there is an easy way to solve the problem. Let $r$ be the radius of the circle. The hypotenuse is minimized when its square, $4r^2 + (1/r)^2$ is minimized. Notice that $$4r^2 + (1/r)^2 = (2r-1/r)^2 + 4$$ which is minimized when $2r=1/r$. That gives you the desired value of $r= \frac{\sqrt2}{2}$. For this value of $r$, the area of the circle is $\pi/2$ and the circumference $\sqrt{2}\pi.$ That is the area of the circle is $2 \sqrt{2}$ times smaller than the circumference.

  • $\begingroup$ Nice answer. I've edited it to use LaTeX for formatting the math; take a look at it to see how it's done, and you can use it in further answers that you write. $\endgroup$ – John Hughes Dec 27 '17 at 23:06
  • $\begingroup$ Thanks John for your attention. $\endgroup$ – Mohammad Riazi-Kermani Dec 28 '17 at 0:43
  • $\begingroup$ That is a very nice trick to avoid the use of calculus. Thanks! $\endgroup$ – TheSimpliFire Dec 28 '17 at 13:23

You can simply use AM$\ge$GM

$$f^2(r)=\frac{{(2r)^2+\left(\frac1r\right)^2}}{2} \ge{2} $$

$$\implies f(r) \ge 2$$

It is "$=$" when $2r=\frac{1}{r}$ or $r=\frac{1}{\sqrt 2}$

Now, you can easily prove the further part.


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