Construct equilateral triangle ABC and square BCDE with point A outside the square. A circle is drawn containing A, D and E. How does the radius of the circle compare to the side length of the triangle?
Repeat the above but with point A inside the square.
I had a solution like Steven but after seeing the result I decided to try a geometric approach.
Point A outside the square
A
/|\
B-+-C
| F |
|/ \|
E---D
Let the circumcenter of triangle ADE be point F. Point F will be on the bisector of angle EAD, which is also the bisector of angle BAC. Angle BAF = (angle BAC)/2 = 30 degrees.
Triangle ABE is isosceles with AB=BE. Angle ABE = angle ABC + angle CBE = 150 degrees. Then angle BAE = (180 - angle ABE)/2 = 15 degrees.
Triangle AFE is isosceles with AF=FE and angle EAF = angle AEF. Angle AEF = angle BAC - angle BAE = 15 degrees.
Then triangles ABE and AFE have base angles with the same measure of 15 degrees and have the same base AE. Therefore their legs are equal: AB=AE=AF=FE.
AF and FE are radii of the circumcircle thus the circumcircle of ABE has a radius of the same length as the sides of ABC and BCDE.
Point A inside the square
B---C
|\ /|
| A |
E---D
\ /
F
Let the circumcenter of triangle ADE be point F. Angle EFD is a central angle and subtends arc EAD. Triangle EFD is isosceles with vertex angle EFD.
EBA is an isosceles triangle with BE=BA and vertex angle EBA. Angle EBA = angle CBE + angle ABE = 30 degrees. Then angle BAE = (180 - angle ABE)/2 = 75 degrees.
Similarly, angle CAD is also 75 degrees.
Angle EAD = 360 degrees - angle BAE - angle CAD - angle BAC = 150 degrees.
Angle EAD is an inscribed angle of the curcumcircle and subtends arc EA. Arc EA = 2 * angle EAD = 300 degrees. Arc EAD is the supplementary arc to arc EA, then arc EAD = 360 degrees - arc EA = 60 degrees.
Angle EFD = arc EAD = 60 degrees. Then triangle EFD is equilateral which implies ED=EF=DF.
DF and FE are radii of the circumcircle thus the circumcircle of ABE has a radius of the same length as the sides of ABC and BCDE.
Geometric Solution
