# Constructing an Equilateral Triangle Inscribed Inside a Circle

How to plot something like this? I need only the triangle and the 3 circles green. Thank you!

• It is expected that you show your own efforts. The minimum that you could have done is coordinates of the points A, B and C as well as the radius of the green circle. Surely you do not expect us to do it for you. Commented Mar 23, 2016 at 10:45
• Without making extensive guesses, it's impossible to determine what you're really asking. What would your input be? Centers and radii of the circles? The points A ... F? Something else? What, if anything, are you assuming about the shape of the triangle or the relative sizes and positions of the circles? Commented Mar 23, 2016 at 14:13

another one liner:

Graphics[{{Yellow, #}, {Green, Circle /@ #[[1]]}}] &@
SSSTriangle[1, 1, 1]

• Nice! Yours has one less variable to set by hand than mine, too, since you get to use unit-radius circles. (+1) Commented Mar 23, 2016 at 17:01

An alternative, perhaps more direct:

pts = CirclePoints[3]
Graphics[{
Thick, Green, Circle[#, Sqrt[3]] & /@ pts,
Yellow, EdgeForm[Black],
FilledCurve@Line@pts
}]


There's probably an easier way to do this, and there is probably a more informative way to do it - more illuminating for those who are new to Mathematica. But this way was more fun for me to come up with, using primarily RegionIntersection to find the points.

circles = {Circle[{-1/2, 0}, 1], Circle[{1/2, 0}, 1]};
AppendTo[circles,
Circle[RegionIntersection[circles] // First // Last, 1]];
triangle = (RegionIntersection[#1, #2, Disk @@ #3] &) @@@
(RotateRight[circles, #] & /@ Range[3]) // Part[#, All, 1] & // Polygon;
Graphics[{Green, circles, Yellow, EdgeForm[Black], triangle}]


Or, if you are using an older version of Mathematica then you have to get the intersection points yourself,

Graphics[{Green, {Circle[{-(1/2), 0}, 1], Circle[{1/2, 0}, 1],
Circle[{0, Sqrt[3]/2}, 1]}, Yellow, EdgeForm[Black],
Polygon[{{1/2, 0}, {-(1/2), 0}, {0, Sqrt[3]/2}}]}]

• Perfect! Thank you <3 Commented Mar 23, 2016 at 12:34
• There is some error in the code? Commented Mar 23, 2016 at 12:44
• youtube.com/watch?v=UtVjRG7PB_4&feature=youtu.be&t=9s Commented Mar 23, 2016 at 12:46
• @AndreaLeo - not that I'm aware, if it doesn't work for you, you need to be more specific, what version are you using, what error message do you get? Commented Mar 23, 2016 at 12:47
• @AndreaLeo - see the edit. Out of curiosity, do you know how you would algebraically find the points for the circle centers that make up the triangle? Commented Mar 23, 2016 at 13:07

The new in M12 function GeometricScene could be useful for you:

scene = GeometricScene[
{a, b, c}, (* points *)
{
Triangle[{a, b, c}],
CircleThrough[{a, b}, c],
CircleThrough[{b, c}, a],
CircleThrough[{c, a}, b]
}
];
scene //RandomInstance


You can then use FindGeometricConjectures to find conjectures that hold for the scene:

FindGeometricConjectures[scene]["Conclusions"]


{GeometricAssertion[Polygon[{b, a, c}], "Regular"], Inactive[PlanarAngle][{a, b, c}] == Inactive[PlanarAngle][{b, a, c}] == Inactive[PlanarAngle][{b, c, a}] == 60 [Degree]}

So, the triangle is an equilateral triangle.

• You can also style @CarlWoll's elements to match the original color specs. This doesn't affect FindGeometricConjectures at all. Note that synthetic geometry in MMA is in an experimental release; a simple scene like this will run without issues, though. scene = GeometricScene[ (*Vertices*) {a, b, c}, (*Hypotheses*) { Style[Triangle[{a, b, c}], Yellow], Style[CircleThrough[{a, b, c}], Red], Style[CircleThrough[{a, b}, c], Green], Style[CircleThrough[{b, c}, a], Green], Style[CircleThrough[{c, a}, b], Green] }]; scene // RandomInstance Commented Dec 15, 2019 at 3:18
• ^^ Sorry for lack of proper code formatting in that last comment. I haven't figured out how to do that in mini-markdown in a comment yet (if it can be done). Commented Dec 15, 2019 at 3:19