**Problem**: (2007 Mock AIME 6 - #2) Draw in the diagonals of a regular octagon. What is the sum of all distinct angle measures, in degrees, formed by the intersections of the diagonals in the interior of the octagon?

**Solution**: Consider the circumscribed circle of the octagon. Each diagonal is a chord of this circle, and we know that the angle between two chords that intercept arcs of measure $ \alpha $ and $ \beta $ is $ \frac{\alpha+\beta}{2} $.

Now, any pair of diagonals can together intercept $ 2 $, $ 3 $, $ 4 $, $ 5 $, or $ 6 $ little arcs (between vertices of the octagon). $ 1 $ is not possible, because then on one side there is no little arc which means the intersection is not in the interior. This also means $ 7 $ is not possible (they come in supplement pairs - except $ 90^{\circ} $ of course). Since each little arc is $ 45^{\circ} $ and we have to account for the division by $ 2 $, the final sum is

$ 45 \cdot \frac{1}{2} \cdot (2+3+4+5+6) = 450 $.

QED.

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Comment: This wasn't an extremely hard problem, but it needed some clever thinking. Most people overcounted, which is reasonable given that there are $ 20 $ diagonals in an octagon. Pretty hard to draw accurately. Using arcs on a circle proved to be much easier and less prone to careless mistakes.

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Practice Problem: Let $R$ be a set of $13$ points in the plane, no three of which lie on the same line. At most how many ordered triples of points $(A,B,C)$ in $R$ exist such that $\angle ABC$ is obtuse?

That's a nice solution... I just guessed it was 90(n-3) based on small cases :P

ReplyDeleteGreat Solution :)

ReplyDeleteP.S. How do you make smilies? :huh: