inradius of P A_j A_{j+k} independent of P, if inradii of P A_iA_{i+1} are equal

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inradius of P A_j A_{j+k} independent of P, if inradii of P A_iA_{i+1} are equal

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Source: 15th -b QEDMO problem 10 (19. - 22. 10. 2017) https://artofproblemsolving.com/community/c1512515_qedmo_2005

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#1 h May 30, 2021, 1:19 AM • 2 Y

Y by centslordm, Mango247

Let $\ell$ be a straight line and $P \notin \ell$ be a point in the plane. On $\ell$ are, in this arrangement, points $A_1, A_2,...$ such that the radii of the incircles of all triangles $P A_iA_{i + 1}$ are equal. Let $k \in N$ . Show that the radius of the incircle of the triangle $P A_j A_{j + k}$ does not depend on the choice of $j \in N$ .

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pi_quadrat_sechstel

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pi_quadrat_sechstel

#2 h Dec 28, 2023, 10:35 PM

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parmenides51 wrote:

Let $\ell$ be a straight line and $P \notin \ell$ be a point in the plane. On $\ell$ are, in this arrangement, points $A_1, A_2,...$ such that the radii of the incircles of all triangles $P A_iA_{i + 1}$ are equal. Let $k \in N$ . Show that the radius of the incircle of the triangle $P A_j A_{j + k}$ does not depend on the choice of $j \in N$ .

W.l.o.g. assume that $l$ is the $x$ -axis and $P=(0,1)$ .

For $t<s$ let $T:=(t,0),S:=(s,0)$ . Let $r$ be the radius of the incircle of $PST$ . We claim that
$-\ln(1-2r)=\sinh^{-1}(s)-\sinh^{-1}(t).$ To check this note that the area of $PST$ is $\frac{s-t}{2}$ and the perimeter $(s-t)+\sqrt{1+s^2}+\sqrt{1+t^2}$ . So we have
$\begin{align*} r=\frac{s-t}{s-t+\sqrt{1+s^2}+\sqrt{1+t^2}}\\ 1-2r=\frac{-s+t+\sqrt{1+s^2}+\sqrt{1+t^2}}{s-t+\sqrt{1+s^2}+\sqrt{1+t^2}}\\ =\frac{\frac{1}{s+\sqrt{1+s^2}}+(t+\sqrt{1+t^2})}{(s+\sqrt{1+s^2})+\frac{1}{t+\sqrt{1+t^2}}}=\frac{t+\sqrt{1+t^2}}{s+\sqrt{1+s^2}}>0 \end{align*}$ In particular $\ln(1-2r)$ is well-defined and we have
$\begin{align*} -\ln(1-2r)=\ln(s+\sqrt{1+s^2})-\ln(t+\sqrt{1+t^2})=\sinh^{-1}(s)-\sinh^{-1}(t) \end{align*}$ proving the claim.

Now let $A_i=(a_i,0)$ for all $i$ and let $r_1$ be the common radius of the incircle of $PA_iA_{i+1}$ . The incircle radius $r_k$ of $PA_jA_{j+k}$ satisfies
$\begin{align*} -\ln(1-2r_k)=\sinh^{-1}(a_{j+k})-\sinh^{-1}(a_j)\\ =\sum_{i=1}^k\sinh^{-1}(a_{j+i})-\sinh^{-1}(a_{j+i-1})=-k\ln(1-2r_1)\\ 1-2r_k=(1-2r_1)^k\\ r_k=\frac{1}{2}\left(1-(1-2r_1)^k\right). \end{align*}$ But this does not depend on $j$ .

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