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Solutions – sum of the angles

December 27, 2007 am31 10:20 am

comma-three-squares.png(Way cool that this image is fine for illustrating my problem, but is really from a fairly advanced seeming math blog (on game theory?), on a topic that I do not understand: Bosker Blog.

Consider rectangle ADEH with side ABCD and side EFGH such that AB = BC = CD = DE = EF = FG = GH = HA.

Find the sum of ∠FAD, ∠GAD, and ∠HAD ∠EAD, ∠FAD, and ∠GAD.

Answers in the comments section, above.

Questions? Over here.

9 Comments leave one →
  1. Brent permalink
    December 28, 2007 pm31 7:29 pm 7:29 pm

    ∠GAD is clearly pi/4. Some simple trig shows that ∠FAD = arctan(1/2) and ∠EAD = arctan(1/3), so we want to find arctan(1/2) + arctan(1/3). Using the sum formula for tangent:

    tan(arctan(1/2) + arctan(1/3)) = (1/2 + 1/3)(1 – (1/2)(1/3)) = (5/6)/(5/6) = 1.

    So ∠EAD + ∠FAD = arctan(1) = pi/4, and the sum of all three is pi/2. Astounding!

    I’m still not satisfied with this solution, though, since it seems like there ought to be a nice *geometric* explanation of why these three angles add up to a right angle, and I’m not seeing it…

  2. December 29, 2007 am31 12:35 am 12:35 am

    A solution without trig would be nice, especially since the answer looks so simple….

  3. TRMilne permalink
    December 29, 2007 am31 6:59 am 6:59 am

    If you accept Pythagoras as a “geometric explanation” then it’s easy. Label the three angles, largest to smallest, as “a,” “b,” and “c.”

    As noted, a is 45deg.
    b is the smallest angle in a 1 – 2 – rt(5) triangle.
    c is the smallest angle in a 1 – 3 – rt(10) triangle.

    Construct a triangle D(-2,1) E(1,2) F(0,0) and see that D = b+c and that it appears to be a right triangle with right angle F.

    Use slopes to show that the legs DF and EF are perpendicular, i.e., m1=-1/m2
    Distance formula gives you two equal legs.
    An isosceles right triangle has angles of 45-45-90
    and each base angle is b + c = 45

    so a+b+c = 90

  4. TRMilne permalink
    December 29, 2007 am31 6:59 am 6:59 am

  5. December 29, 2007 am31 7:06 am 7:06 am

    Nice diagram. You’ve used up both of my non-trig solutions!

  6. January 17, 2008 pm31 8:33 pm 8:33 pm

    the easiest way I see is to draw the lattice points (0,0), (4,2), and (3,-1). if you call angles EAD, FAD, and GAD X, Y, and Z respectively, you see that you get an isoceles triangle where one of the angles is X+Y and the other angle is 90 – X – Y, so X+Y = 45, and X+Y+Z = 90.

  7. January 17, 2008 pm31 8:33 pm 8:33 pm

    it also looks like I am 3 weeks late for the carnival. =)

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