如何计算圆与等边三角形的6个交点的坐标？

Question

I know of an equilateral triangle the center (cx,cy) and the radius (r) of a blue circle which conscripts it.

If I draw a green circle of any radius (radius), assuming the circle is large enough to have this intersection, can I get the coordinates of the 6 intersection points (P1, P2, P3...)?

I'm looking for P5JS/processing but any other clue can help me...

Thank you in advance

Answer 1

Distance from the center to top point is r .
Distance from the center to the lowest triangle side is r/2 (median intersection point is center, they are divided in 1:2 ratio).
Horizontal distance from cx to p4 (and p5) is (Pythagoras' theorem)

dx = sqrt(radius^2 - r^2/4)

So coordinates of p4 and p5 are (relative to center)

p4x = dx
p4y = r/2
p5x = -dx
p5y = r/2

Other points might be calculated using rotation by 120 degrees

p2x = p4x*(-1/2) - p4y*(sqrt(3)/2)
p2y = p4x*(sqrt(3)/2) + p4y*(-1/2)

and so on.

And finally add cx,cy to get absolute coordinates

Answer 2

If you want to test... ;-)

 function setup() { createCanvas(500, 500); const cx = width / 2; const cy = height / 2; const r = 250; // taille du triangle const radius = 180; // externe noFill(); strokeWeight(3); stroke(0, 0, 0); drawTriangle(cx, cy, r); strokeWeight(1); stroke(0, 0, 255); circle(cx, cy, r * 2); strokeWeight(2); stroke(8, 115, 0); circle(cx, cy, radius * 2); noStroke(); fill(215, 0, 0); dx = sqrt(Math.pow(r / 2, 2) - Math.pow(r / 2, 2 / 4)); p4x = dx; p4y = r / 2; circle(cx + p4x, cy + p4y, 20); text("p4", cx + p4x, cy + p4y + 30); p5x = -dx; p5y = r / 2; circle(cx + p5x, cy + p5y, 20); text("p5", cx + p5x - 10, cy + p5y + 30); p6x = p4x * (-1 / 2) - p4y * (sqrt(3) / 2); p6y = p4x * (sqrt(3) / 2) + p4y * (-1 / 2); circle(cx + p6x, cy + p6y, 20); text("p6", cx + p6x - 30, cy + p6y); p2x = p6x * (-1 / 2) - p6y * (sqrt(3) / 2); p2y = p6x * (sqrt(3) / 2) + p6y * (-1 / 2); circle(cx + p2x, cy + p2y, 20); text("p2", cx + p2x + 10, cy + p2y - 10); p1x = p5x * (-1 / 2) - p5y * (sqrt(3) / 2); p1y = p5x * (sqrt(3) / 2) + p5y * (-1 / 2); circle(cx + p1x, cy + p1y, 20); text("p1", cx + p1x - 20, cy + p1y - 10); p3x = p1x * (-1 / 2) - p1y * (sqrt(3) / 2); p3y = p1x * (sqrt(3) / 2) + p1y * (-1 / 2); circle(cx + p3x, cy + p3y, 20); text("p3", cx + p3x + 20, cy + p3y - 10); noFill(); stroke(0, 255, 255); triangle(cx + p2x, cx + p2y, cx + p4x, cx + p4y, cx + p6x, cx + p6y); stroke(255, 0, 255); // prettier-ignore triangle( cx + p1x, cx + p1y, cx + p3x, cx + p3y, cx + p5x, cx + p5y, ) } function drawTriangle(cx, cy, radius) { noFill(); trianglePoints = []; for (var i = 0; i < 3; i++) { var x = cx + radius * cos((i * TWO_PI) / 3.0 - HALF_PI); var y = cy + radius * sin((i * TWO_PI) / 3.0 - HALF_PI); trianglePoints[i] = { x, y, }; } triangle( trianglePoints[0].x, trianglePoints[0].y, trianglePoints[1].x, trianglePoints[1].y, trianglePoints[2].x, trianglePoints[2].y ); }

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