Tuesday, February 24, 2009

Problem 257: Equilateral Triangle, Circumcircle, Point, Vertices, Side, Distances, Squares

Proposed Problem

See complete Problem 257 at:
gogeometry.com/problem/p257_equilateral_triangle_circumcircle_distance_sides.htm

Level: High School, SAT Prep, College geometry

1. By an earlier problem, we know that d = e + f
hence, LHS =(e+f)^2+e^2+f^2 = 2(e^2+f^2+ef)
Using cosine rule in Tr. BDC, we've: a^2 =e^2+f^2-2efcos(BDC)=e^2+f^2-2ef(-1/2)=e^2+f^2+ef
Hence, LHS = 2(e^2+f^2+ef) = 2a^2
Ajit: ajitathle@gmail.com

If we apply the cosine rule in triangl ABD with cos(60)=1/2:

d^2+e^2-2decos(60)=a^2
d^2+e^2-de=a^2

And then we apply the cosine rule in triangle ADC :

d^2+f^2-2dfcos(60)=a^2
d^2+f^2-df=a^2

Adding these two equations we get

2d^2+e^2+f^2-de-df=2a^2
2d^2+e^2+f^2-d(e+f)=2a^2

In problem 256 we see that d=e+f :

2d^2+e^2+f^2-d^2=2a^2
d^2+e^2+f^2=2a^2

3. We can use the extension of ptolemy given here :

http://gogeometry.com/equilic/cyclic_ptolemy_ratio_diagonal.htm

Applying the extension of ptolemy gives

d/a=(a^2+ef)/(ae+af)........ (1.)

and from problem 256 d=e+f

First setting d =e+f in equation 1 gives

d/a=(a^2+ef)/(ae+af)
(e+f)/a=(a^2+ef)/(ae+af)
(e+f)^2=a^2+ef

a^2=e^2+f^2+ef ......... (2.)

We also have ad = af+ea ,from this follows

d/a=(a^2+ef)/(ae+af)
d^2=a^2+ef

and we get

a^2=d-ef .......(3.)