## Friday, May 1, 2009

### Problem 287: Regular Octagon, Diagonals

Proposed Problem

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
Ajit:ajitathle@gmail.com

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

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
Moratuwa
Sri Lanka

5. It also follows that HME are collinear