+0 A regular triangle with sides of length 3 is inscribed into a circle. A second circle is inscribed between the triangle and the first circle. Find the radius of the second circle.
Explain the formula please
+15000 From the center M of the circles, draw rays to the corner E and to the center S of the side belonging to the corner. The result is a right-angled triangle M, E, S with sides s/2, r and R.
Then:
\(R^2=(\frac{s}{2})^2+r^2\\r=\sqrt{R^2-(\frac{s}{2})^2}\\ \frac{s}{2}=R\cdot sin(60^{\circ})\)
\(r=\sqrt{R^2-R^2\cdot sin^260^{\circ}}\\ r=R\cdot \sqrt{1-sin^260^{\circ}}\\ r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)
or very simply:
In the right triangle SME, angle SME=60°.
So is:
\(r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)
!
+129895 
Let A = (0,0)
B= (3,0)
C = (1.5 , 1.5 sqrt 3)
D is the center of the circumcircle of ABC and is found by the line with the equation
y = tan (30°) x = (1/sqrt3)x
The x coordinate of D =1.5
So y = (1/sqrt3) (1.5) = (3/2)/sqrt3 = sqrt (3) / 2
AD is the radius of the circumcircle of ABC = sqrt ( 1.5*2 + (sqrt(3) / 2)^2 ) = sqrt [ (3/2)^2 + 3/4] =
sqrt [ 9/4 + 3/4] = sqrt 3
So the equation of the circumcircle is
(x -1.5)^2 + (y - sqrt (3) / 2) = 3
F = (1,5 , 0)
E = (1.5 , sqrt (3)/2 - sqrt (3) ) = (1.5 , -sqrt (3) / 2 )
The radius of the small circle = EF / 2 = ( 0 - -sqrt (3)/2 ) / 2 = [ sqrt (3) / 2] / 2 =
sqrt (3) / 4
The equation of the small circle is (x -1.5)^2 + (y - sqrt (3) / 4 )^2 = 3/16
