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.
In the first case the radius is equal to the side of the triangle (and to the side of the square).
Say AB = BE = 1. Triangle height is sqrt(3)/2. Say the circle of radius R is centered at O and F is the midpoint of ED and G the midpoint of BC. EF = BG =1/2. R= OA = OE. R^2 = OA^2 = OE^2.
Call OG = x. Then
OA^2 = (x + sqrt(3)/2)^2
OE^2 = (1+x)^2 + 1/4
equating and solving the quadratic gives
x = 1-sqrt(3)/2.
R = x + AG
with AG = sqrt(3)/2, R = 1
For the second case, the same result obtains: again, the radius is equal to the side of the triangle (and to the side of the square).
We put F midway between E and D, and place the circle's center at O, x distance above F then:
R = R
OA = OE
OA^2 = OE ^2
(AF + FO )^2 = OF^2 + EF^2
(1-sqrt(3)/2 +x)^2 = x^2 + 1/4
gives
x=sqrt(3)/2 and R=1
Edited on April 13, 2020, 4:07 pm
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