Let $ABC$ be a triangle. How to find a bound of ratio of circumradius and inradius. One may think that the edge case is in a triangle that is somehow special. It’s trivial, that when one is dealing with equlateral case, since its incircle and circumcircle are cocentric. To find the ratio, a handful observation comes to mind - the common center of those circles is the centroid of $ABC$ and since centroid divides every median into two pieces with one of them the double of the other, the ratio is exactly $2$.

This gives an impression, that $2$ is an extremal value. One can easily check that if it actually is an extreme, it is the minimal value. Quick look on a isosceles right triangle assures this is a valid constraint.

What happens in a general case? Of course, there are a lot of proofs, some of them applying to $n$-dimensional space, however I’d like to present a method that uses algorithmic, approximation techniques.

Let $\Gamma$ be a constant circle. Every triangle can be generated from three points of this circle, with a help of homothety. Therefore, if we find the triangle with maximal inradius, and it turns out to be equilateral, $\frac{R}{r} \geq 2$.

Take a chord $c$ of $\Gamma$ as constant. Point $C$ varies on the circle. Angle $\angle ACB$ is therefore constant. Since $I$ is the incenter, it lies on the angle bisectors of $\angle A$ and $\angle B$. From the relation $\angle A +\angle B = 180^{\circ} - \angle C$ follows that $\angle A + \angle B$ is constant. Therefore, $\frac{\angle A + \angle B}{2}$ is constant too. $IAB$ is a triangle with to of angles equal to $\frac{\angle A}{2}$ and $\frac{\angle B}{2}$ respectively. This proves that $\angle AIB$ is constant too. This means, that with arbitrary $C$, $AIB$ lie on a circle that is not dependent on $C$. The inradius of $ABC$ is the distance from $I$ to $AB$ and since $A$, $B$, $I$ share the same circle, inradius is greatest when $AI=IB$, i.e. $\angle A = \angle B$. In other words, $ACB$ is isosceles with $AB$ as base. If we can prove that applying this method of leaving one of angles intact and making the other two equal cyclically to triangles $ABC$, $BCA$, $CAB$, $ABC$, $\ldots$ makes the triangle more and more equilateral, it would mean that none of triangles has greater ratio. In other words, triples $(\angle A, \angle B, \angle C)$ converge to $(60^{\circ}, 60^{\circ}, 60^{\circ})$ and inradius is strictly increasing.

I will facilitate my thought with the AM-GM inequality which states that for any sequence of non-negative numbers $a_1, a_2, \ldots, a_n$, the relation holds:

What I really need is the proof for $n = 2$ or $3$. That’s why I won’t provide full proof for all cases. My version is a slightly modified proof by Augustin-Louis Cauchy. If $n=1$, proof is obvious. Let’s prove it works when $n=2$ then.

Squareing both sides, we get

This holds for any real numbers $a_1, a_2$.

Now, let’s see how it works, when $n=4$. For $k=1$ proof is complete. Assuming thesis is correct for $2$, I’ll show why it works for $4$.

Applying it again, for $a_1 = \sqrt{a_1a_2}$ and $a_2=\sqrt{a_3a_4}$, we get

First, I need to show a little lemma:

Multiplying both sides by $12$ finishes the lemma’s proof:

Now, let’s try to do some mangling with symbols:

We can exponentiate it sidewise:

Divide it by the content of the parenthesis to get

i.e. AM-GM for $n=3$.

Coming back to proof of original problem - I’ve already shown that substituting $\angle A$ and $\angle B$ by their arithmetic mean increases the inradius. From AM-GM follows, that

If the growth of geometric mean of $A, B, C$ is associated with increase of inradius, this will be the end of the proof, because the mean is bounded from the ceiling by a constant ($60^{\circ}$ in this case). So, equivalent inequality follows:

Dividing both sides by $\sqrt[3]{C}$, we get

or, in other words

Square root it:

which holds from AM-GM. Therefore, as angles of $ABC$ are nearing to $60$ degrees, inradius strictly increases. Since when they are equal to each other, the ratio is equal to $2$, and every other case has this value smaller, it is the greatest bound, that was to find.

How Brick problems accidentally made me prove a geometric progression property

Back in the times of my primary school, mathematical contests were consisting of problems from a “basket” - what I mean is that you could almost always be sure that if you know certain kinds of problems, you would solve everything with no doubt.

One of these kinds was something which I describe as “brick” since it was “brick” which was often recalled then. Let’s formulate it:

Brick weights 1 kg and a half of the brick. How heavy is the brick?

Solution is pretty straightforward. Let $b$ be the weight of brick in kilograms. Then following equation is equivalent to the problem:

Let’s multiply both sides twofold:

Now substract $b$ from both sides:

This yields the solution.

Generalization

What if brick weights $y$ and $x$-th part of brick? Let’s rewrite the equation:

If $x=1$, there is no solution, except $y$ being $0$ (since it would mean that a brick weights the same as a brick plus something which is obviously false). Therefore the equation can be divided sidewise by $1-x$. This gives the solution:

Another view

Last formula seems extremely familiar. What it remainds me of? Ah, sure, sum of infinite geometric progression! How to express the weight of brick in as a sequence? Let’s write again the original expression:

We could expand it, yeah?

This is a geometric progression! Combining it with the previous formula, somehow opaque proof is complete and I can claim without any doubt that $\sum_{i=0}^{\infty} y \cdot x^i$ is indeed equal to $\frac{y}{1-x}$.

How this formula is proved with classical tools?

Later, I discovered a proof of sum of first $n$ elements of geometric progression within What is Mathematics?. Not that surprising is fact that the flow of proof is similar, based on substracting an element of the sequence from the preceding one. The final formula tells that

For $n \rightarrow \infty$ and $q \in (0; 1)$, these formulas are equivalent.

Thanks to

I’m really thankful to my thorough friend Agnieszka Huć for making me think about bricks and their halves once again, when I know something more than basics about algebra.