1990 AIME Problem 12

Question

A regular

12

-gon is inscribed in a circle of radius

12.

The sum of the lengths of all sides and diagonals of the

12

-gon can be written in the form

a+b\sqrt2+c\sqrt3+d\sqrt6,

where

a,

b,

c,

and

d

are positive integers. Find

a+b+c+d.

Answer

Group the chords by the number of vertex steps

k,

where

k

is

1,

2,

\ldots,

or

6

For

k\lt6

there are

12

chords of length

24\sin(\frac{k\pi}{12}),

while there are

6

diameters

For

k

equal to

1,

2,

\ldots,

and

5,

there are

12

chords of length

24\sin(\frac{k\pi}{12}),

and there are

6

diameters of length

24.

The five chord lengths are

\begin{gathered}6(\sqrt6-\sqrt2),\quad12,\quad12\sqrt2,\\12\sqrt3,\quad6(\sqrt6+\sqrt2).\end{gathered}

Their sum

U

is

\begin{aligned}U&=12+12\sqrt2\\&\quad+12\sqrt3+12\sqrt6.\end{aligned}

Therefore the total is

\begin{aligned}12U+6(24)&=288+144\sqrt2\\&\quad+144\sqrt3\\&\quad+144\sqrt6.\end{aligned}

Hence

a=288

and

b=c=d=144,

so

a+b+c+d=720.

Problem 12 in other years