Create equilateral triangle in the middle of canvas?

Question

I want to draw an equilateral triangle in the middle of canvas. I tried this:

ctx.moveTo(canvas.width/2, canvas.height/2-50);
ctx.lineTo(canvas.width/2-50, canvas.height/2+50);
ctx.lineTo(canvas.width/2+50, canvas.height/2+50);
ctx.fill();

But the triangle looks a bit too tall.

How can I draw an equilateral triangle in the middle of canvas?

Someone told me you have to find the ratio of the height of an equilateral triangle to the side of an equilateral triangle.

h:s

What are the two numbers?

Cysticercus

Answer 1

You have to do it with the height of the triangle

var h = side * (Math.sqrt(3)/2);

or

var h = side * Math.cos(Math.PI/6);

So the ratio h:s is equal to:

sqrt( 3 ) / 2 : 1 = cos( π / 6 ) : 1 ≈ 0.866025

See : http://jsfiddle.net/rWSKh/2/

Fanya

So is the ratio h : s = Math.sqrt( 3 ) / 2 : 1? – Cysticercus

ya or Math.cos(Math.PI/6) or simply 0.866 – Fanya

When I rotate this triangle, it is no longer centred around the original centre. How do I fix this? – Stlouis

Could someone explain why this is needed, and why the example in the question doesn't work? – Vlad

@Stlouis (I know this account is gone now) I recommend having a look at answer stackoverflow.com/a/8937325 and add an alpha term in both cos and sin: x = r*cos(angle + alpha) and y = r*sin(angle + alpha). You can then change alpha to any value you want, and it will rotate this triangle whilst maintaining its center. – Cysticercus

@Vlad An equilateral triangle has same length for all three sides. Let side of this equilateral triangle be s. If you cut in a straight line from one of the vertices of this triangle to the middle of the side opposite to the vertex you cut, then you get two right angled triangles since the bisection line is perpendicular to the opposite side. Each triangle has side lengths s, s/2 and h is height of triangle we want to find. We can use Pythagoras's theorem to find h > 0. h^2 + (s/2)^2 = s^2 implies h = (sqrt(3)/2)*s, which contradicts my example in my question with s = h = 100. – Cysticercus

Answer 2

The equation for the three corner points is

x = r*cos(angle) + x_center
y = r*sin(angle) + y_center

where for angle = 0, (1./3)*(2*pi), and (2./3)*(2*pi); and where r is the radius of the circle in which the triangle is inscribed.

Susan

So elegant. Wonderful! – Conney

I used this awesome method to create arrows on canvas - https://mcmap.net/q/268230/-draw-arrow-on-canvas-tag - thanks for it! – Mirza

Answer 3

A simple version where X and Y are the points you want to top of the triangle to be:

var height = 100 * (Math.sqrt(3)/2);
context.beginPath();
context.moveTo(X, Y);
context.lineTo(X+50, Y+height);
context.lineTo(X-50, Y+height);
context.lineTo(X, Y);
context.fill();
context.closePath();

This makes an equilateral triange with all sides = 100. Replace 100 with how long you want your side lengths to be.

After you find the midpoint of the canvas, if you want that to be your triangle's midpoint as well you can set X = midpoint's X and Y = midpoint's Y - 50 (for a 100 length triangle).

Waltner

Answer 4

The side lengths will not be equal given those coordinates.

The horizontal line constructed on the bottom has a length of 100, but the other sides are actually the hypotenuse of a 50x100 triangle ( approx. 112).

Diversiform

Answer 5

I can get you started with drawing an equilateral triangle but I don't have the time to get it centered.

jsFiddle

var ax=0;
var ay=0;
var bx=0;
var by=150;

var dx=bx-ax
var dy=by-ay;
var dangle = Math.atan2(dy, dx) - Math.PI / 3;
var sideDist = Math.sqrt(dx * dx + dy * dy);

var cx = Math.cos(dangle) * sideDist + ax;
var cy =  Math.sin(dangle) * sideDist + ay;

var canvas = document.getElementById('equ');
var ctx = canvas.getContext('2d');

ctx.beginPath();  
ctx.moveTo(ax,ay);  
ctx.lineTo(bx,by);  
ctx.lineTo(cx,cy);  

ctx.fill(); 

Anecdotal

Answer 6

my code for drawing triangle also depending on direction (for lines). code is for Raphael lib.

drawTriangle(x2 - x1, y2 - y1, x2, y2);
function drawTriangle(dx, dy, midX, midY) {
        var diff = 0;
        var cos = 0.866; 
        var sin = 0.500; 

        var length = Math.sqrt(dx * dx + dy * dy) * 0.8; 
        dx = 8 * (dx / length); 
        dy = 8 * (dy / length); 
        var pX1 = (midX + diff) - (dx * cos + dy * -sin); 
        var pY1 = midY - (dx * sin + dy * cos); 
        var pX2 = (midX + diff) - (dx * cos + dy * sin); 
        var pY2 = midY - (dx * -sin + dy * cos); 

        return [
                    "M", midX + diff, midY,
                    "L", pX1, pY1,
                    "L", pX2, pY2,
                    "L", midX + diff, midY
                ].join(","); 
    }

Shoreless