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Monday, August 31, 2015

Problem of the Week

We haven't done a geometry problem in a while, so here's one for this week's Problem of the Week. 

Check out my written-up solution after the break to see if you got it right!

Did you find a different way to reach this value? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week!


  1. I solved it via similar triangles, using the fact that the hypotenuse of the smaller right triangle is 1/cos(theta) from SOHCAHTOA. It gets the same answer, of course.

    Your proof does not address what happens when theta is a right angle. Of course your formula gets the right result, but to be complete the proof should mention it.

  2. i solved it with direction and orthogonal vectors, linked with the equation of l.
    same answer !

  3. If you look at the congruent right triangles each with the short side of length 1 which are generated by the radius of circle B along the x-axis and the radius along line m, you can write it as tan(1/2*(\pi - \theta)).