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Problem 96

Geometry Difficulty 1.2 Multiple choice CEMC Cayley · Canada · 2022

Points AA, BB, CC, DD, EE, and FF are evenly spaced around the circle
with centre OO, as shown.

The measure of AOC\angle AOC is

Pick one

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Official solution

Solution 1

Since AA, BB, CC, DD, EE, and FF are equally spaced around the circle,
moving from one point to the next corresponds to moving 16\frac{1}{6} of the way around the
circle.

Therefore, moving from AA to CC corresponds to moving 26\frac{2}{6} (or 13\frac{1}{3}) of the way around the
circle.

Since moving around the whole circle corresponds to moving through 360°360\degree, then moving 13\frac{1}{3} of the way around the circle
corresponds to moving through $13×360°=120°$.\$\frac{1}{3} \times 360\degree = 120\degree\$.

Thus, AOC=120°\angle AOC = 120\degree.

Solution 2

We join AA, BB, CC, DD, EE, and FF to OO.

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Since AA, BB, CC, DD, EE, and FF are equally spaced around the circle,
the angles made at the centre by consecutive points are equal. That is,
AOB=BOC=COD=DOE=EOF=FOA\angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOF = \angle FOA Since these 6 angles form
a complete circle, the sum of their measures is 360°360\degree.

Therefore, AOB=BOC=COD=DOE=EOF=FOA=16×360°=60°\angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOF = \angle FOA = \tfrac{1}{6} \times 360\degree = 60\degree This means that $\$\angle AOC = \angle AOB + \angle BOC = 60°+60°=120°$.60\degree + 60\degree = 120\degree\$.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.