A Look at an Octadecagon-1

Geometry Level 3

In an 18 sided regular polygon with vertices labelled $A_1,A_2,\ldots,A_{18}$ in order,
the diagonals $A_1A_{13}$ and $A_6A_{15}$ intersect at a point $M$ ,
the diagonals $A_1A_{13}$ and $A_3A_{14}$ intersect at a point $N$ ,
the diagonals $A_3A_{14}$ and $A_6A_{15}$ intersect at a point $P$ .

Compute the angle $MA_6A_{14}$ in degrees.

The answer is 10.

1 solution

A Former Brilliant Member
May 20, 2016

Let $\angle MA_{16}A_{14}=\alpha$

Note that $\alpha=\angle A_{15}A_{16}A_{14}$

Also, if $O$ denotes the circumcentre of the polygon, then by the incribed angle theorem and the fact that angle subtended by a side at the circumcentre of a regular $n$ sided polygon is $\dfrac{360^\circ}{n}$ , we have, $2\alpha = 2 \angle A_{15}A_{16}A_{14}=\angle A_{15}OA_{14}=\frac{360^{\circ}}{18}=20^{\circ} \\\implies \boxed{\alpha=10^\circ}$

Oh,i didn't know about this incribed angle theorem.good solution..+1

Ayush G Rai - 5 years ago

Thanks! Well, you too gained something :)

I'd like to know your solution as well :)

A Former Brilliant Member - 5 years ago

I cannot write the solution without the diagram.sorry!!

Ayush G Rai - 5 years ago

A Look at an Octadecagon-1

The answer is 10.

1 solution

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