AY 2017
Problem of the Week Solution
Week #2
Hexagon ABCDEF is inscribed in a circle, with
AB = BC = CD = 2
DE = EF = F A = 1
Find the radius of the circle.
Figure 1: Picture taken from [1]
This problem and solution is adapted from [1] where we illustrate a few more details of
the solution. The idea of this solution is to introduce symmetry without changing the
answer to the problem. The first inscribed circle above represents the picture for the given
problem and the second circle is found by rearranging the first circle to introduce
rotational symmetry. Notice that we haven’t changed the radius of the circle since all six
triangles are isosceles whose side lengths are all equal to the radius of the circle.
Now we denote the center of the circle by O and note that XY = 1 and Y Z = 2.
Notice that XOY Z is exactly one-third of the entire hexagon and so if we let the radius of
the circle be r then OX = OY = OZ = r. Since XOY Z is exactly one third of the entire
hexagon we know that 6 XOZ = 120◦ . We also notice that all of the interior angles of the
hexagon are equal and so by the fact that the sum of the interior angles of a hexagon is
720◦ we deduce that 6 XY Z = 720
= 120◦ .
6
Now we will finish the problem by applying the law of cosines twice, obtaining the
equations below
(XZ)2 = r2 + r2 − 2r2 cos(120◦ ) = 2r2 (1 + 21 ) = 3r2
(XZ)2 = 12 + 22 − 2 · 2 · 1 · cos(120◦ ) = 5 + 4 · 12 = 7
q
√
where the second equation yields that XZ = 7 and the first equation yields that r = 73 .
References
[1] P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, pg.297 (2007).
Posted: 20 October 2016
Submit NLT: 27 October 2016