1
$\begingroup$

This question is simple. Since we know that a right-angled triangle inscribed in a circle has the hypotenuse and the diameter being one and the same. What if we inscribed a circle in that triangle. And then we inscribed a triangle in that circle...

It's a never-ending pattern looking like this:

enter image description here

My question is. Considering the largest diameter/hypotenuse to be x, what is the Nth term in the sequence of decreasing diameters. The non-right-angle should be represented by a y

$\endgroup$
  • 1
    $\begingroup$ That definitely ought to depend on which right triangle you inscribe in the circle $\endgroup$ – jgon Apr 7 '18 at 18:10
  • $\begingroup$ I don't understand $\endgroup$ – yolo Apr 7 '18 at 18:14
  • 1
    $\begingroup$ Like you can inscribe an isosceles right triangle in the circle, or you could inscribe a very skewed right triangle with one of the angles almost zero. The diameters for the isosceles right triangle case would decrease much more slowly than for the case where you're inscribing right triangles with one of the angles almost 0. $\endgroup$ – jgon Apr 7 '18 at 18:16
  • $\begingroup$ Ok, editing...now $\endgroup$ – yolo Apr 7 '18 at 18:20
  • 1
    $\begingroup$ The circumradius-to-inradius ratio for the triangle is the scale factor at each stage. Both parts of the ratio are easily determined for a right triangle. $\endgroup$ – Blue Apr 7 '18 at 18:30
1
$\begingroup$

Consider an $a$-$b$-$c$ right triangle with hypotenuse $c$. Since the hypotenuse is the triangle's circumdiameter, the circumradius is $c/2$. For any triangle, $$\frac12\cdot\text{inradius}\cdot\text{perimeter} = \text{area}$$ so our triangle, whose area is $ab/2$, has inradius $ab/(a+b+c)$.

In the sequence of inscribed figures, the inradius of one triangle is the circumradius of the next. Thus, each triangle is similar to the next, with scale factor $$s := \frac{\text{inradius}}{\text{circumradius}} = \frac{2a b}{c(a+b+c)} \tag{$\star$}$$ so that, in general,

$$\text{hypotenuse of $n$-th triangle} = s^n\cdot\text{hypotenuse of initial triangle}$$

where the $n=0$ case is the initial triangle itself.

If one of the acute angles in our triangle is $\theta$, then its legs are $c\cos\theta$ and $c\sin\theta$, and we can rewrite $(\star)$ as $$s = \frac{2\cos\theta\sin\theta}{1+\cos\theta+\sin\theta} = \frac{\sin 2\theta}{1+\cos\theta+\sin\theta} \tag{$\star\star$}$$

$\endgroup$
  • $\begingroup$ Brilliant work! $\endgroup$ – yolo Apr 7 '18 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.