Right Triangle Image

Say the sidelengths of the triangle are $a,b,c$ , with $a^2+b^2=c^2$ . Then its area can be written in two ways: $[\Delta ABC]=\frac12 ab=\frac12 ch$

so $h=\frac{ab}{c}$ .

In a right-angled triangle, the inradius is $r=\frac{a+b-c}{2}$ . So $f(\Delta ABC)=\frac{2ab}{(a+b-c)c}$

Since we're looking at a ratio of two lengths, the size of the triangle doesn't matter, only its "shape" (ie its angles); so we can take the sidelengths to be $\cos\theta,\sin\theta,1$ for some $\theta \in \left(0,\frac{\pi}{2}\right)$ .

Plugging these in, $\begin{aligned} f(\Delta ABC)&=\frac{2\cos \theta \sin \theta}{\cos\theta+\sin\theta-1} \\ &=\frac{2\cos \theta \sin \theta}{\cos\theta+\sin\theta-1} \cdot \frac{\cos\theta+\sin\theta+1}{\cos\theta+\sin\theta+1} \\ &=\frac{2\cos \theta \sin \theta}{(\cos\theta+\sin\theta)^2-1} \cdot (\cos\theta+\sin\theta+1) \\ &=\frac{2\cos \theta \sin \theta}{\cos^2 \theta + 2\cos\theta \sin \theta + \sin^2 \theta - 1} \cdot (\cos\theta+\sin\theta+1) \\ &=\cos\theta+\sin\theta+1 \\ &=\sqrt2 \sin \left(\theta+\frac{\pi}{4}\right)+1 \end{aligned}$

In this form, it's easy to see that $f$ is maximised at $\theta=\frac{\pi}{4}$ (and it attains this value) and minimised at $\theta=0$ or $\theta=\frac{\pi}{2}$ (and $f$ doesn't attain this value); so $2<f(\Delta ABC) \le 1+\sqrt2$

Choose two of the sides of the triangle to coincide with the x- and y-axes and choose our circle of radius $r$ to be centered at $(r,r)$ . A third line touches the circle at $(r+r \cos φ, r+r \sin φ)$ .

Together with the axes it forms a triangle containing the circle when φ is in the range $(0,π/2)$ . It does not require calculus to see that the distance h of the line to the origin would be $2r$ for $φ=0$ , (this value for φ is just outside the allowed range!) and reaches a maximum of $r+r\sqrt{2}$ for $φ=\frac{π}{4}$ . Divide by r to get $\boxed{\frac{h}{r} \in (2, 1+\sqrt{2}]}$