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This is a slight variation of my previous question.
There I questioned regarding the area of a polygon, here I am wondering about the perimeter of any polygon.

So my hypothesis is as follows:

There is no such polygon whose perimeter is equal to the difference between the circumferences of its circumscribed and inscribed circle.
The inscribed circle must touch all edges and the circumscribed circle must pass through all vertices.

Can anyone verify my hypothesis or find a counter-proof? I have tried some regular polygons and some common triangles and it has held so far.
I have mainly put this question to check that it holds for regular polygons but I believe that it should also hold for irregular polygons (just gut feeling, could be false). I would appreciate any help that I can receive. Thanks in advance!


Edit:
As it has been proven to not hold for all polygons, I want to ask the following questions:
Are there finitely many irregular polygons that obey this hypothesis?
Are there finitely many irregular polygons that disobey this hypothesis?
Does this hold for regular polygons?
I would appreciate answers to these questions and I would like to thank Alex for his answer. All the best!

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  • $\begingroup$ Could anyone help me with this? $\endgroup$ Dec 29, 2017 at 10:23
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    $\begingroup$ I hope what you're finding from these questions is that gut feeling is not enough reason to believe something in mathematics! $\endgroup$ Dec 29, 2017 at 20:56
  • $\begingroup$ True that! But sometimes, as in the case of Einstein's Cosmological constant, it may just work out! $\endgroup$ Dec 30, 2017 at 14:47

2 Answers 2

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There is such a triangle satisfying the given condition.

Denote $\displaystyle c = \left(\frac{π + 1}{2π - 1}\right)^2$, then it can be verified that $\displaystyle \frac{1}{2} < c <1$. Also denote $\displaystyle x_0 = \frac{c + \sqrt{2c - 1}}{1 - c}$, then $(1 - c)x_0^2 - 2cx_0 + (1 - c) = 0$, which implies $x_0^2 + 1 = c(x_0 + 1)^2$.

Now, take a right triangle $\triangle ABC$ with $BC = 1$, $CA = x_0$, $AB = \sqrt{1 + x_0^2} = \sqrt{c} (x_0 + 1)$. The inradius and the circumradius of $\triangle ABC$ are\begin{gather*} r = \frac{1}{2} (BC + CA - AB) = \frac{1}{2} (1 + x_0 - \sqrt{1 + x_0^2}) = \frac{1 - \sqrt{c}}{2}(x_0 + 1),\\ R = \frac{1}{2} AB = \frac{1}{2} \sqrt{1 + x_0^2} = \frac{\sqrt{c}}{2}(x_0 + 1), \end{gather*} respectively. Note that $(2\sqrt{c} - 1) π = \sqrt{c} + 1$, therefore,\begin{align*} 2πR - 2πr &= π(2 \sqrt{c} - 1)(x_0 + 1) = (\sqrt{c} + 1)(x_0 + 1) = BC + CA + AB. \end{align*}

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A numeric example of suitable isosceles triangle:

enter image description here

\begin{align} R&=1 ,\\ \alpha&\approx 18.5528612^\circ ,\quad \beta=\gamma \approx 80.7235694^\circ ,\\ a&\approx 0.6363589 ,\quad b=c\approx 1.9738442 ,\\ r&\approx 0.2704262 ,\\ a+b+b&\approx 4.5840473 ,\\ 2\pi(R-r)&\approx 4.5840473 . \end{align}

Edit:

Another example is an obtuse isosceles triangle:

enter image description here

\begin{align} R&=1 ,\\ \alpha&\approx 106.8724079^\circ ,\quad \beta=\gamma \approx 36.56379605^\circ ,\\ a&\approx 1.9139069 ,\quad b=c\approx 1.1914349 ,\\ r&\approx 0.3161467 ,\\ a+b+b&\approx 4.2967768 ,\\ 2\pi(R-r)&\approx 4.2967768 . \end{align}

Edit:

And these are the only two suitable shapes of isosceles triangles. This can be shown as follows.

Using known identities for triangles

\begin{align} \sin\alpha+\sin\beta+\sin\gamma&=\frac{a+b+c}{2R} ,\\ \cos\alpha+\cos\beta+\cos\gamma&=\frac{r}{R}+1 . \end{align}

For isosceles triangle we have

\begin{align} \sin2\beta+2\sin\beta&=\frac{a+b+c}{2R} ,\\ -\cos2\beta+2\cos\beta&=\frac{r}{R}+1 ,\\ \end{align}

and condition $a+b+c=2\pi(R-r)$ can be expressed in terms of $\beta$ as follows:

\begin{align} 2\sin\beta\cos\beta+2\sin\beta - \pi(1+2\cos^2\beta-2\cos\beta)=0 , \end{align}

which can be simplified to

\begin{align} 5\pi t^4-2\pi t^2-8t+\pi&=0 ,\qquad t=\tan\tfrac\beta2 . \end{align}

For $\beta\in(0,\tfrac\pi2)$, this equation results in two real values $t_1\approx 0.3303679220$, $t_2\approx 0.8499172732$ with corresponding values of $\beta$, shown above.


Edit: Using that the sides $a,b,c$ of a triangle with semiperimeter $\rho=\tfrac12(a+b+c)$, incircle $r$ and circumcircle $R$ are the roots of cubic polynomial

\begin{align} x^3&-2\rho x^2+(\rho^2+4rR+r^2)x-4\rho rR \tag{g1}\label{g1} \end{align}

and scaling the triangle such that $R=1$, the condition $2\rho=2\pi(R-r)$ converts \eqref{g1} to

\begin{align} x^3-2\pi\,(1-r)\,x^2+((\pi^2+1)\,r^2+(4-2\pi^2)\,r+\pi^2)\,x-4\pi\,r(1-r) \tag{g2}\label{g2} . \end{align}

Recall that the cubic polynomial $ax^3+bx^2+cx+d$ has discriminant \begin{align} \Delta&=b^2c^2-27 a^2 d^2-4(a c^3+b^3 d)+18 abcd ,\\ \text{and in case of \eqref{g2}}\quad \Delta= &-(\pi^2+1)^2\,r^4 \\ &+4\,(\pi^4+6\,\pi^2-3)\,r^3 \\ &-2\,(19\,\pi^2+3\,\pi^4+24)\,r^2 \\ &+4\,(\pi^4+3\,\pi^2-16)\, r \\ &-\pi^2(\pi^2-4) \tag{g3}\label{g3} ,\\ \Delta&>0\quad \forall\ r\in(0.2704262, 0.3161467) , \end{align}

that means, there are infinitely many scalene triangles, which inradius is in this range, as well as there also are infinitely many triangles with $r\in(0,0.2704262)\cup(0.3161467,0.5)$ for which the condition $2\rho=2\pi(R-r)$ does not hold.

For example, let's take $r=0.29$. Solutions to \begin{align} \\ \text{gives }\quad& a\approx 0.7586625863,\quad b\approx 1.722967032,\quad c\approx 1.979431950 ,\\ a+b+c&\approx 4.461061568 ,\\ 2\pi(R-r)&\approx 4.461061568 . \end{align}

enter image description here

Also note that the two endpoints of the $r$-interval correspond to the pair of isosceles triangles shown above.

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