Concurrent Chords in a Circle, Equally Inclined

What Might This Be About?

Problem

Concurrent Chords in a Circle, Equally Inclined, problem

Solution

Vladimir Dubrovsky
30 March, 2017

We'll make use of the Lagrange formula for the moment of inertia, also known as Steiner's and Huygens-Steiner theorem:

For unit masses placed at points $A_1,A_2,\ldots,A_n,\,$ the moment of inertia with respect to a point $X\,$ is defined as $\displaystyle I_X=\sum_{k=1}^nXA_k^2,\,$ and the Lagrange formula takes the form $\displaystyle I_X=nXG^2+I_G,\,$ where $G\,$ is the centroid (center of mass) of these material points.

In particular, $I_X,\,$ as a function of $X,\,$ depends only on the distance $XG.$

The centroid of the six endpoints of the given chords coincides with the centroid of the midpoints of the chords, which are the projections of $O\,$ - the center of the given circle - on the chords, and hence form a triangle inscribed in the circle $\mathbb{c}\,$ with diameter $BO.\,$ By the Inscribed Angle Theorem, the triangle is equilateral.

Concurrent Chords in a Circle, Equally Inclined

Therefore, its centroid $G\,$ is the center of circle $\mathbb{c},\,$ or the midpoint of $BO.\,$ It follows that $BG=GO,\,$ so, by the Lagrange formula,

$\displaystyle BN^2+BM^2+BP^2+BQ^2+BR^2+BS^2=I_B=I_G=6R^2$

and we are done.

This proof readily generalizes to $n\ge 3\,$ chords drawn through the same point at equal angles to each other: the sum of the squares of their pieces is equal to $nR^2.$

Acknowledgment

The problem has been discussed elsewhere as one of the properties (viz., #3) of the configuration of six concurrent chords. Vladimir Dubrovsky has commented with his solution, reproduced above.

 

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  20. Concurrent Chords in a Circle, Equally Inclined

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