3
$\begingroup$

I would like to see a proof of the following

Claim. Let $A_1,A_2,A_3,A_4,A_5$ be vertices of bicentric pentagon. Let $B_1$ be the intersection point of $A_1A_3$ and $A_2A_5$, $B_2$ the intersection point of $A_1A_3$ and $A_2A_4$ , $B_3$ the intersection point of $A_2A_4$ and $A_3A_5$, $B_4$ the intersection point of $A_1A_4$ and $A_3A_5$, $B_5$ the intersection point of $A_1A_4$ and $A_2A_5$ . Denote $A_1B_1$ by $m$, $A_3B_2$ by $n$, $A_3B_3$ by $o$, $A_5B_4$ by $p$, $A_5B_5$ by $q$, $A_2B_1$ by $r$, $A_2B_2$ by $s$, $A_4B_3$ by $t$, $A_4B_4$ by $u$ and $A_1B_5$ by $v$. Then $$m+o+q+s+u=n+p+r+t+v.$$

enter image description here

The GeoGebra applet that demonstrates this claim can be found here.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

This follows immediately from the fact that the polygon $B_1B_2B_3B_4B_5$ is circumscribed (about a third circle). Let $C_{12},\ldots, C_{51}$ be the five points that touch the corresponding circle. Then, clearly, $|A_1C_{45}|=|A_1C_{12}|$ and so on cyclically. In other words, if we extend these blue and red segments up to the points where they touch this third circle, we will have five blue/red couples of equal segments. But now, the total sum of lengths added to blue segments is $|B_1C_{12}|+\ldots +|B_5C_{51}|$ and the total sum of lengths added to red segments is $|B_1C_{51}|+\ldots$. Finally notice that $|B_1C_{12}|=|B_1C_{51}|$ and so on cyclically.

So one just need to check that $B_1B_2B_3B_4B_5$ is circumscribed. Here is the applet that shows this among other things: https://www.geogebra.org/m/tdwbytj2

Improve this answer
$\endgroup$
3
  • $\begingroup$ (1) Well, if the complex coordinates of the vertices (on the init circle) $A_i$ are $a_i^2$ and the midpoints of "smaller" arcs $A_iA_{i+1}$ are $-a_ia_{i+1}$, we should prove that the lines joining $a_i^2$ and $-a_{i+2}a_{i-2}$ are concurrent (their common point is equidistant from the sides of $B_1B_2B_3B_4B_5$). And we are given that the lines joining $a_i^2$ and $-a_{i-1}a_{i+1}$ are concurrent. The line joining $a$ and $b$ on the unit circle has equation $\bar{z}ab+z=a+b$.... $\endgroup$
    – Fedor Petrov
    Commented May 4, 2021 at 19:23
  • $\begingroup$ (2) So, given that the vectors $(1,a_i^2-a_{i-1}a_{i+1},a_{i-1}a_i^2a_{i+1})$ lie in the 2-dimensional plane, we should prove the same for the vectors $(1,a_i^2-a_{i-2}a_{i+2},a_{i-2}a_i^2a_{i+2})$. Is there some algebraic mumbo-jumbo solving such questions? $\endgroup$
    – Fedor Petrov
    Commented May 4, 2021 at 19:25
  • $\begingroup$ There are both synthetic and algebraic arguments for the circumscribability in this paper by Zaslavsky and Chelnokov m.mathnet.ru/links/d8ff8e98f93c9c98cb43422f95ca69f5/mo426.pdf (pitifully, in Tuussian). $\endgroup$
    – Ilya Bogdanov
    Commented Jun 23 at 14:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .