Find the minimal value the hypotenuse of a right triangle whose radius of inscribed circle is $r$

Question

Find the minimal value the hypotenuse of a right triangle whose radius of inscribed circle is $r$.

I tried to use the radius of the circle to calculate the smallest possible side lengths of the triangle, but I was unable to figure out how to calculate the side lengths with only the radius of the inscribed circle.

Does anyone understand how I am supposed to solve this?

Answer 1

Place the circle in the first quadrant so that it’s tangent to both coordinate axes, i.e., center it on the point $(r,r)$, and let the hypotenuse of the triangle be a line segment with endpoints on the positive $x$- and $y$-axes that’s tangent to the circle. You can then parameterize the hypotenuse length in various ways, such as by its slope. The symmetry of the situation should give you a strong hint as to what the answer must be.

Answer 2

We use the formulas $a^2+b^2=c^2$ and $a+b-c=2r$ for inscribed circle in a right triangle. I will explain my intuitions as we go:

We know that $a$ and $b$ can change freely (think about the triangle changes its shape around the fixed circle), so we set $b = ka$ where $k > 0 \in \mathbb{R}$, and our job is to express $c$ in terms of $r$ and $k$, the two degrees of freedom.

Lengthy algebra: $$ (k+1)a - c = 2r \text{ hence } a = \frac{c+2r}{k+1}\\ a^2 + (ka)^2 = c^2 \text{ and hence } c = a\sqrt{k^2+1} = \sqrt{k^2+1}\frac{c+2r}{k+1} $$ Solve for $c$ and we get $$c = \frac{2r}{\frac{k+1}{\sqrt{k^2+1}}-1}$$ And our only job now is to maximize $\dfrac{k+1}{\sqrt{k^2+1}}$, which from standard calculus gives us $\sqrt{2}$. Therefore, $c$ is maximized as $2(\sqrt{2}+1)r$.

Also, as other people already pointed it out, the symmetry gives some hint as to the desired triangle should be isosceles, which would make the problem much easier. Hopefully this helps!

Answer 3

A clever way to relate the sides of a right triangle, a, b and c, to the radius $r$, is via the area matching, i.e.

$$Area = \frac{1}{2}a b = \frac{1}{2}r(a+b+c)\tag{1}$$

where the last expression is just the sum of three areas making up the triangle. To proceed, let $\theta$ be an angle of the triangle, then $a=c\sin\theta$, $b=c\cos\theta$ and as a result (1) becomes

$$c(\theta)=2r\frac{1+\sin\theta+\cos\theta}{\sin 2\theta}\tag{2}$$

Now, it is straightforward to find the minimum $c$ by solving $c’(\theta)=0$.

Actually, it'd be much cleaner not to even bother with $a$ and $b$, rather just observe that the hypotenuse $c$ is the sum of two segments intersected by the perpendicular radius, i.e.

$$c(\theta)=r\left[ \cot\left( \theta/2\right) + \cot\left( \beta/2\right) \right]$$

where $\theta/2$ and $\beta/2=\pi/4 - \theta/2$ are half of the two acute angles of the right triangle, because the lines from their vertexes to the center of the inscribed circle bisect $\theta$ and $\beta$. Then, $c’(\theta)=0$ simply leads to

$$\csc\left( \theta/2\right) = \csc\left( \pi/4 - \theta/2 \right)$$

which yields $\theta=\pi/4$ (not surprisingly, an isosceles right triangle) and $c_{min}=2(1+\sqrt{2})r$.

Answer 4

In the standard notation by C-S $$r=\frac{a+b-c}{2}\leq\frac{\sqrt{(1^2+1^2)(a^2+b^2)}-c}{2}=\frac{(\sqrt2-1)c}{2}=\frac{c}{2(1+\sqrt2)}.$$ Thus, $$c\geq2(1+\sqrt2)r.$$ The equality occurs for $a=b$, which says that we got a minimal value.