Archive - Jun 19, 2008

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A circle solution to World's Hardest Easy Geometry Problem

The World's Hardest Easy Geometry Problem ruined my sleep.


Find the measure of x without elementary geometry about triangles.

I tried that problem a few months ago for a few hours and failed, so I cheated and looked up the answer.
But today I dreamed about that problem. In my dream, I saw myself using circle to solve that problem.
I woke up, it's 3 am, great time to work on a geometry problem.

So I did. I succeed!
It is cheating to use circles. Only basic properties about triangles, angles and lines are allowed.
Please don't read this if you want to have your own fun with the problem first.
I also have a link for a triangle only solution.

Note: My graph is different because I changed the name of point E to D, and D to E.
Here is where da solutions at.

First looking show that CE=BE, so it's nice to consider using symmetry to solve the problem.
Construct EI \perp CB
Draw a circle pass through E,D,F
Reflect all points through EI
The graph looks like this:

It's easy to see there are some equations
\begin{eqnarray}
\angle C &=& \frac{\stackrel{\frown}{EF}+\stackrel{\frown}{GF}-\stackrel{\frown}{HD}}{2}\\
\angle ADB &=& \frac{\stackrel{\frown}{FG}}{2}\\
\end{eqnarray}
All the angles listed above can be found directly because the other two angles in the triangle are already known.
After finding all the size of those angles and multiply 2 to each side, it reduce to a few simple linear equation, all we have to do is to solve it.
It's a easy linear system can be computed by hand.
Set

\begin{eqnarray}
a=\stackrel{\frown}{FE}\\
b=\stackrel{\frown}{HD}\\
c=\stackrel{\frown}{FG}\\
\end{eqnarray}


\begin{eqnarray}
40^\circ  &=& a+c-b\\
60^\circ  &=& c\\
\end{eqnarray}
We end up with this

20^\circ  = b-a
c=b because the graph is symmetric.
a = 40^\circ
Thus x = 20

Q.E.D.[Quite Exhausting Demonstration]

I need to start to train my subconscious so I can do math when I'm asleep.

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