Friday, May 4, 2012

World's Hardest Easy Geometry Problem

I was stuck on the following geometry problem for a long time:


It was kind of nice in a way, because it reminded me of what it is like to be persistent on a math problem.  I spent hours thinking about it and couldn't find a proof.  Then I came back days later and spent hours again, to no avail.  I don't even want to admit how long it took me.

On a whim I brought it to my class and told my students that I had solved a really hard geometry problem.  I said that anyone else who could solve it would get an A for the quarter.  Heck, if it took me that long then any high school student who solves it has proven their worth.

Surprisingly, quite a few students have taken up the challenge.  After two days, no one has solved it but hey, it took me more than two days.  A number of students are coming in during lunch, study hall, after school, and during breaks to discuss what they have found.  It has actually been one of the coolest things I have done all year.  I have a whole wall of my classroom dedicated to different attempts at the proof.  Some of the students are engaged in what may be the most intense problem solving experience of their lives thus far.

In a way it is kind of cruel because it is a very hard problem and a lot of students won't ever get it.  But some of them will persist and there are a few that are quite close already.  This may be the kind of thing that they remember for the rest of their life. 

Hint:  if you want to try it, try drawing some additional lines inside the triangle.  Aside from that all you need to know are the old triangle congruence rules (SAS, ASA, SSS, etc) and the properties of an isosceles triangle (two sides and two base angles are congruent.

5 comments:

  1. i love that you are doing this with them. something about the idea that this hard problem uses only concept they already know appeals to me (and them from the sound of it!). it seems like this problem is a good way to show kids that math is creative and that simple concepts can be used to solve complex problems.

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  2. Actually, you can do the whole problem with no side properties whatsoever, just angles. If you want to see the outline of one solution, let me know.

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  3. But my way has a system of 4 linear equations and 4 unknowns, but no extra lines necessary.

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  4. I would be interested to see what your solution is. I have seen some attempts to use a system of equations, but they weren't independent.

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    1. You are right. I just found about five or six equations and thought surely four of them would be independent. I didn't actually do the problem. This morning I tried it and saw my mistake. So I quickly used the laws of sines and cosines to find the answer to see if that gave me a clue as to a geometric proof that doesn't take so much trig, because your students probably don't have those laws in their toolkits yet.

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