1005 – A geometry problem
I was fumbling through my archives and found this problem: Find the angle marked with “?”.
I remember quite clearly the stories around it. It was back in 2001, when I was in the last year of my middle school (or something like the 9th grade). Someone in my class brought this question up but none of us could solve it analytically. (And yes, if you draw the graph carefully with a protractor, you can actually measure the “?” angle.) So the mystery lingered, and lingered, until one day our math teacher came up with a piece of paper written with the solution. It was a full piece of notebook paper, perhaps B5, with a huge amount of trignometry equations.
Over the years, the solution is lost but the problem is still there, interestingly. I just tried another 10 minutes but could not get it right. Are there easy ways to solve it analytically?
I remember the answer is 20 degrees. Too neat to be true, no?
Update on January 12, 2015: Ioanid Roşu provided a stunning quick solution to the problem. The idea is to make use of the angle sizes and create an extra piece of information that facilitates the anlaysis. Below is the idea (and his original illustration):
Start with two isosceles triangles, MBA and ACN, and one equilateral triangle, AMN. Note that the points B, A, and C are on the same line by construction. Interestingly, we then get back the original quadralateral, but with a new angle size known: angle BMN is of 80 degrees. Then there are many ways to proceed. The rest is left to the readers. :-)