GL(s,R)

August 6, 2010

A geometry question

Filed under: Uncategorized — Adam Glesser @ 8:59 am
Tags: , ,

One of the things most pleasing to me about this whole inter-web-o-blog-o-sphere is that I can ask so many people for help in such a short amount of time. I am going to put the over-under on an answer at three hours (I would put it lower if I was on the West Coast) [Update: It turns out the winning time is 33 minutes. Thanks, Justin!].

In a 2008 MAA Mathematics Magazine “Proof Without Words” article, Sidney H. Kung gave the following proof of the tangent of the sum formula:  The proof is transparent to me in all but one step. Why does \angle BDF =\beta? I guess it must have something to do with the line segments being chords in the circle, since I haven’t used the circle anywhere in the proof, but I’m just missing something obvious or forgetting something from geometry. Help!

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4 Comments »

  1. Angles BAF and BDF are both inscribed angles that cut off the same arc, and are therefore equal. Is that what you were looking for?

    That’s a cute diagram! I especially like how (at least I saw) that EC is equal to tan(b). (With information flowing from BC to DC to EC).

    Comment by Justin Lanier — August 6, 2010 @ 9:30 am | Reply

    • That’s perfect, Justin; thanks for showing me. Indeed, BC to DC to EC is how I got tan(b), as well. It is quite remarkable how the picture will lead most people to consider the same path. Incidentally, in the original publication, the author is apparently not confident that his picture is sufficiently clear and so adds six lines of equations.

      Comment by Adam Glesser — August 6, 2010 @ 9:39 am | Reply

  2. Justin Lanier is clutch with the plane geometry, I’m tellin ya!

    Okay so let me try to work thru the whole thing here:

    Begin with angle \alpha+\beta at point A. Cut off 1 unit on the outside ray of angle \alpha and label the point C. Construct a perpendicular to AC at C and mark the point (B) where the ray on the other side of angle \alpha intersects this perpendicular. Construct a perpendicular to AB and mark the point (F) where this perpendicular intersects the last original ray out of A. Since ABF is a right angle, the circle with diameter AF contains B. Label the point D where this circle intersects AC. Find the point E that is collinear with B and C and makes DE parallel to BF.

    Since D is on the circle with AF as a diameter, ADF is a right angle. Thus DF is parallel to BC and EDFB is a parallelogram.

    By construction, the length of BC is \tan(\alpha). By Justin’s argument, the measure of angle BDF is \beta, and since DF and BC are parallel, this means angle CBD is \beta too. Thus length CD is \tan(\alpha)\tan(\beta).

    Meanwhile, the measure of angle DFB is \alpha by Justin’s same argument: DFB and DAB are inscribed angles that cut off the same arc (DB), so congruent. Because of the parallelogram, this means angle DEB is \alpha too, which means that the length of EC is \tan(\beta).

    Thus EB is \tan(\alpha)+\tan(\beta) and because of the parallelogram, so is DF. Since AC is 1 and CD is \tan(\alpha)\tan(\beta), AD must be 1-\tan(\alpha)\tan(\beta). And the conclusion comes from considering what we now know about the triangle FAD.

    Wow.

    Yeah that’s awesome.

    Comment by benblumsmith — August 9, 2010 @ 7:06 pm | Reply

    • Thanks for the detailed explanation, Ben. I like how you get that DFB is \alpha; I had compared triangles BFA and DFA to reach the same conclusion, but I prefer your way of using Justin’s argument again. These proof by pictures articles are a lot of fun and I’m currently enjoying the first collection of them put out by Roger Nelsen: http://bit.ly/cjDLWb

      Comment by Adam Glesser — August 9, 2010 @ 9:46 pm | Reply


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