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

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?

• Let the right angle sides be a,b and the hypotenuse be c. Besides $a^2+b^2=c^2$, are you aware of the formulas $r(a+b+c) = ab$ and $a+b-c=2r$? – XYSquared Aug 2 '19 at 22:50

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.

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.

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 $$c_{min}=2(1+\sqrt{2})r$$.

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!

• If you were to formulate the equations in terms of $a$ and $b$ first, and then set $b=ka$ at the end, the symmetry that you mention would be much more apparent. – amd Aug 3 '19 at 1:31
• It seems like there might be a typo in “$c$ is maximized as $2(\sqrt2+1)c$.” You’ve got $c$ defined in terms of itself there. – amd Aug 3 '19 at 1:32