Three concentric circles each have radii 8, 15 and 17.

Find the largest possible area of the equilateral triangle with one vertex on each circle.

Construction:

Construct Circle(8), radius 8 on the origin, with centre O.

Point A is at {0,8}.

Point P is at {-7.5,0} Construct a line through P parallel to the y axis. Construct Circle(15), radius 15 on O. By Pythagoras, the line through P intersects Circle(15) at B in the 3rd quadrant at {-7.5,-(15 sqrt(3))/2)}

Point Q is on the line through P, 4 above B. Construct a line through Q parallel to the x axis. Construct Circle(17), radius 17 on O. By Pythagoras, the line through Q intersects Circle(17) at C in the 4th quadrant at {sqrt(17^2-(((-15 sqrt(3))/2)+4)^2),(-(15 sqrt(3))/2)+4)}, simplifying nicely to {(1/2 (15 + 8 sqrt(3)),(-(15 sqrt(3))/2)+4)}, around {-14.42,-8.99}

ABC is the desired equilateral triangle: sqrt((1/2 (15 + 8 sqrt(3)))^2+((((15 sqrt(3))/2)-4)+8)^2) = sqrt(7.5^2 + (8 + 1/2 (15 sqrt(3)))^2) = sqrt(4^2+(7.5+(1/2 (15 + 8 sqrt(3))))^2

From line AB {0,8}{-7.5,-(15 sqrt(3))/2)}, the side length of the triangle is sqrt(7.5^2 + (8 + 1/2 (15 sqrt(3)))^2), so its area is:

sqrt(3)*(sqrt(7.5^2 + (8 + 1/2 (15 sqrt(3)))^2))^2/4, or more simply:

1/4 (360 + 289 sqrt(3))

Update: there is a much neater way of doing this, well worth a more general problem, which I have now added to the queue.

Edited on December 16, 2018, 3:33 am

Posted by broll on 2018-12-15 23:11:11