Friday, May 1, 2009

Problem 287: Regular Octagon, Diagonals

Proposed Problem

 Problem 287: Regular Octagon, Diagonals.

See also:
Complete Problem 287: Regular Octagon, Diagonals
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry


  1. Let G:(0,0) be the origin. Hence, F:(a/V2,a/V2), B:(-a-a/V2,a+a/V2) and A:(-a -a/V2,a/V2) where V=square root. Slope of line BF = (aV2-a-a/V2)/(a/V2+a+a/V2)= 1-V2. Hence, BF:y=(1-V2)x+c and passes thru F:(a/V2,a/V2). On substitution, we obtain c=a. Thus, BF: y =(1-V2)x + a while DG is x=0 or M:(0,a) and MA^2=(a+a/V2)^2+(a-a/V2)^2 = 3a^2. MA = x in the diagram; hence, x=a*V3

  2. Since B and F are opposite vertices, BF bisects angle ABC. angle ABM = 67.5.
    join AG observe that AGMB is a parallelogram, angle AGM = 67.5 and GM = AB = a.
    In triangle AGH apply cosine rule to get AG.
    Now AG will be known GM will be Known and also angle AGM is known.
    AM can be found in terms of a using cosine rule.

  3. Easily M is the incenter of tr. DHF, so MG=GH; since GH and GD are perpendicular, HM=a.sqrt{2} and <AHM=90 degs, so from tr. AHM, right-angled at H, AM=a.sqrt{3}.

    1. To anonymous
      Please give the reason for your statement MG=GH, thanks

  4. The regular octagon is also cyclic and each side subtends an angle of 22.5 degrees at the other 6 points (135/6). So < MGF = 45 and < MFG = 67.5 hence MG = GF = a

    Hence HGM is isoceles and right and < MHG = 45. So < AHM = 135 - 45 = 90. Now applying Pythagoras to Tr. AHM,

    x^2 = 2a^2 + a^2 = 3a^2 and the result follows

    Sumith Peiris
    Sri Lanka

  5. It also follows that HME are collinear

  6. Construct BE meeting DG at X. BXE is congruent to MXE, and both are isosceles right. So DX = XM = asqrt2/2. Now Construct HY such that GHY is 45 and Y is on DG. Because DGFE is an isosceles trapezoid, angle HGM is a right angle, so triangle GHY is isosceles right and GY = a.

    But GD must equal a + a sqrt2, as this is the altitude of the octagon. As DM = asqrt2, YG = a, and D,M,Y and G are collinear by assumption, Y must be concurrent with M. So now it's just a matter of solving right triangles. HM = a sqrt2, so AM = asqrt3. (Once angle GHM was found to be 45, AHM had to be 90.

  7. Alternatively upon realization that Tr.s AMB & CMB are congruent the answer is straightforward

  8. It can be observed that MGF is an isosceles with MG=GF=a
    =>MGH is an isosceles right triangle with MH=Sqrt(2).a
    =>AHM is a right triangle with x^2=AH^2+MH^2