A Realisation

Geometry Level 5

Let us consider a 1126 sided polygon $\Gamma$ (not necessarily regular) with vertices labelled $A_{1},A_{2},...,A_{1126}$ ,such that the vertices $A_{2},A_{564}$ and $A_{1126}$ , when joined,form an equilateral triangle of side length $a$ units.

Given that the polygon $\Gamma$ has a circumcircle, find the length of that side of $\Gamma$ which subtends an angle of $60^{\circ}$ at the circumcentre.

If your answer is of the form $ka$ ,

input $k$

The answer is 0.577.

2 solutions

Rohith M.Athreya
Jun 30, 2017

Let the polygon $\Gamma$ have a circumcircle $S_{1}$

Consider any three vertices $A_{i_{1}},A_{i_{2}} ,A_{i_{3}}$ forming a triangle with circumcircle $S_{2}$

Clearly, the circle $S_{1}$ circumscribes the triangle.

But a triangle has a unique circumcircle and thus, $S_{1}=S_{2}$

Thus, the circumcircle of a polygon ,if it exists, is the circumcircle of the triangle formed by joining any three vertices.

For side length $a$ , the circumradius is $\frac{a}{\sqrt{3}}$ for the equilateral triangle as well as for the polygon $\Gamma$

The rest I leave to the able reader.

Niranjan Khanderia
Jun 30, 2017

$Let~A_2A_{564}A_{1126}~be~renamed~as~XYZ.~So~we~have~an~equilateral~\Delta ~XYZ,~side=a.\\ Let~the~ midpoint~ on~ arc~ over~ XY~ be~ M.\\ \therefore~~XZM=30_o. ~\implies~XY~subtends ~an~ angle~ of~ 60^o~ at~ the~ circumcentre.\\ Clearly~ZM ~is~ the ~diameter.~R=\dfrac 1 {\sqrt3}*a.\\ So~in ~\Delta~XZM, ~ZX=a,~ZM=2*\dfrac 1 {\sqrt3}*a,~\angle~XZM=30^o.\\ So~by~Cos~Rule, \dfrac {XM}{a^2}=\sqrt{1^2+(\dfrac 2 {\sqrt3})^2-2*Cos30*1*\dfrac 2 {\sqrt3}}=\Large~\color{#D61F06}{0.57735}.$

A Realisation

The answer is 0.577.

2 solutions

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