The centres of a triangle

A triangle has three corners, connected by three edges; as it turns out, it also has three natural geometric constructions that give, in their respective senses, the circle a centre – even though one of them may actually lie outside it. In each case, there is a line one can construct in relation to a vertex, an edge or the pair of a vertex and the opposite edge; this gives us three lines; and the interesting fact is that they meet in a single point, rather than each missing the point where the other two meet (which would lead to them forming a triangle), as is more usual for three lines in a plane.

Centre of a circle through the three corners

Given a triangle, construct the (grey) perpendicular bisectors of its three edges. Each point on the perpendicular bisector of an edge is exactly as far from one end of the edge as from the other; so, where two of these perpendicular bisectors meet, the common vertex in which the two bisected edges meet is the same distance (green lines) from this meeting point as each of the other two vertices of the triangle, so these two others are in fact equal distances from the meeting point, which thus lies on the third edge's perpendicular bisector. Thus, the perpendicular bisectors of the three edges all meet at one point.

This point is known as the circumcentre of the triangle. As the circumcentre is the same distance from each of the vertices, there's a circle – known as the circumcircle – centred on the circumcentre, that passes through all three vertices. Notice, by considering a triangle whose vertices are three points on the same side of some diameter of a circle, that the circumcentre needn't actually lie within the triangle. Even in that case, though, I trust it is clear that the triangle won't fit inside any smaller circle than its circumcircle.

Centre of a circle touching each edge

Next imagine we bisect the angles at the three vertices and look at where two of these (grey) angle-bisectors meet, dropping a (green) perpendicular from this point to each of the sides of the triangle. Between each of the bisected angles and the meeting point we have a segment of the angle's bisector as the hypotenuse of two right-angle triangles, one on each side, with an edge into the vertex as one side of each triangle and a perpendicular as the remaining side. Since the two right-angle triangles have the same angle at the vertex, as their shared hypotenuse bisects the original triangle's angle there, and each has right angles facing the bisector, their third angles, at the meeting point, must in fact be equal. As they share their hypotenuse, they are thus necessarily each other's mirror images in it, making the two perpendiculars' lengths equal. By considering the matching pair of right-angle triangles for the other vertex whose bisector met, we find one of these two perpendiculars equal to the third, hence all three are equal and the place where two angle bisectors meet is necessarily the same distance from (the nearest point to it on) each of the three sides.

If we now draw a straight line from the third vertex to the given meeting-point of the first two's bisectors, it forms two right-angle triangles with the perpendiculars and the edges into this third vertex; those triangles share their hypotenuse and have a same-length side opposite the third vertex; that (e.g. via Pythagoras) makes their third sides equal, so the two triangles are again mirror images of one another and we can infer that our line from the third vertex to where the other two bisectors met is in fact the bisector of the angle at the third vertex. So the three angle bisectors all meet in a single point; and it's the same distance from each of the sides of the triangle. This point is known as the incentre of the triangle; it's the centre of the largest circle that lies wholly inside the triangle (with the perpendiculars we dropped as its radii). This circle might even be called the incircle; but the name doesn't ring any bells in my memory.

Balance point

We have one more centre to go. Connect each vertex by a straight (grey) line to the mid-point of the opposite edge. Consider where two of these lines meet: construct, through this point, a (green) line parallel to each edge of the triangle.

Opposite each vertex, the lines parallel to those that meet at the vertex form a small triangle with the middle part of the opposite edge; as the edges of this triangle are one from the original and two parallel to the original's others, it is similar to the original. Between each line through the meeting point and the vertex opposite the side it's parallel to, we get an intermediate-sized triangle, again similar to the original, due to sharing two edges and having the third parallel.

For either of the lines we intersected, between a vertex and an opposite edge's mid-point, scaling down the full triangle towards the vertex at one end pulls the mid-point opposite along the line while scaling the opposite edge to a line parallel to it, with its ends on the other two edges and its centre-point on the line; so, when the scaling brings this edge to pass through the meeting-point, we get an intermediate-sized triangle, with the meeting point as the centre of its edge that passes through that; the two halves of this edge are edges of small triangles, telling us that those two small triangles are the same size as each other (since they're similar and have a corresponding edge equal) and half the size of the intermediate-sized triangle. Doing this for the other intersected line, we discover its vertex's intermediate triangle is twice the two small triangles on either side of it, one of which we already know to be equal to the remaining small triangle; so the three small triangles are all the same size and two of the intermediate triangles are twice as big as them.

But now the third intermediate triangle's edge through this meeting point is made of two corresponding edges of equal small triangles, so is bisected by its edge through the meeting point, which is makes the third intermediate triangle the same size as the other two. It also makes the result of scaling down the third line, from the third vertex to its opposite's edge's middle, scaled towards the vertex at one end of it to make the triangle's opposite edge pass through the meeting point, have the meeting-point as its image of the mid-point that was one of its ends; and this was a scaling towards a point on the line, so merely moves that point along the line, so the meeting point of the other two lines is in fact on this third line. This meeting point is called the centroid of the triangle. (In particular, balancing the point masses at the corners implies that the centroid is the average of the position vectors of the three vertices.)

Balancing on it

Now, I described the centroid in its section heading as a balance point; so now to elaborate on that. Briefly: if your triangle is a sheet of some stiff material with uniform mass per unit area and you place its centroid on a pivot-point, the triangle should balance (in theory; of course, your physical reality shall always be a little off centre, so you might need a rounded point to balance it on in practice); likewise, if you put three equal masses on the corners of such a triangular sheet, it'll still balance on a pivot under the centroid. In contrast, a wire-frame of the triangle, with uniform mass per unit length along the edges, has a different balance point. For the corner or edges cases, of course, you'll need some massless rigid sheet to fill the space between, so as to mechanically be able to support the mass at any interior balance point; but the fact that corners and sheet agree means you can combine a uniform sheet with masses at each corner to get a result that'll still balance on the centroid.

For the plain uniform rigid sheet to tip, part of it must go up while the opposite part goes down and some axis stays still. Since the three lines in the construction of the centroid meet, they're not parallel; so they can't all be parallel to the axis that's staying still; indeed, at least two of them aren't. For the triangle to tilt, then, the net torque about at least two of the centroid's construction lines must be non-zero. So let's consider any one of those lines: it joins a vertex to the centre of the opposite edge. When considering torque about a line, we can move any of the triangle's mass parallel to the line without changing the torque; this doesn't change how far off the axis any mass is, or on which side. So sheer the triangle until the bisected edge is perpendicular to our pivot line; this doesn't change areas or the torque. Now, in the sheered triangle, the pivot line still connects the vertex to the opposite edge's mid-point, meeting it in a right angle and thus cutting the sheered triangle into two right-angle triangles, with equal halves of the bisected edge perpendicular to their shared edge, the pivot-line; this makes the two triangles mirror images of each other. Now, reflecting in the pivot-line will reverse the torque; but it won't change the figure, so can't change the torque; so the torque is eequal to its own reverse, making it zero. So the triangle's net torque about each of the construction lines is zero and there can't be any net torque at all; the triangle must balance. This argument applies to both corner weights and the rigid sheet of uniform mass density.


The case of a wire-frame triangle is trickier: sheering does change the lengths of edges, so would change the balance-point in this case. Now, to work out whether it'll balance on a pivot, we must work out the net torque between the pivot's support and the weight of the triangle. That weight acts on each part of the weight, so we have to accumulate (by integrating or summing) the torque's contributions from the weight in various places. By using the pivot point as origin, we can eliminate the pivot's supporting force (its position is the zero, so the cross-product with force is then zero, too). The contribution from each element of the triangle's mass will then be the cross product of the mass's position, relative to the pivot, and the weight of the mass.

Since the weight is just the mass times the gravitational field strength, which I'll assume uniform, we can accumulate the mass of each element times its position, taking the cross product only at the end when we have accumulated its contributions from the whole triangle. We can, further, assume that we start out by placing the triangle in a horizontal plane, so that the gravitational field is perpendicular to all displacements within it; this just makes taking the cross product with it rotate our accumulated position times mass and scale it by a constant; that won't affect whether the result is zero, so we'll only get zero torque if the accumulated mass times position is zero.

Now, in accumulating position times mass, each edge has equal amounts either side of its mid-point at each distance from the mid-point, along a straight line. At each distance from the centre, the average of positions (times equal mass) for parts (of equal length) on opposite sides of the centre is just the centre; so the accumulated position times mass of the whole edge is just the mass of the edge times its mid-position. In effect, then, we reduce the wire-frame of a triangle to point masses at the mid-points of its edges: however, these masses are not equal, as in the cases considered above – each corner's mass is proportional to the length of the edge it was on. As a result, the average of mass times position for these masses is slightly off the centroid in the case illustrated (see the small blue dot near the middle). Indeed, in the case of a triangle with one edge very short compared to the other two, the short edge's mass is tiny compared to the masses of the long edges, whose centre-points are close together; so the balance point of the wire-frame is quite close to the line connecting the mid-points of the two long edges, roughly half-way along the line connecting the short edge to its opposite vertex, rather than roughly a third of the way along it, from the short edge, where the centroid lies.

Notice, however, that the triangle whose vertices are the mid-points of the original is necessarily just a half-scale version of the original (each edge of it is an edge of a triangle in a corner that's the result of scaling the original down by a factor of two towards that corner), half-turned; and each line from a corner of the original to the opposite edge's centre bisects an edge of this inner triangle on its way to its opposite vertex, so the lines that define the outer triangle's centroid are the lines that define the inner's, so they have the same centroid. Of course, we can repeat this construction indefinitely: at each step, we form a triangle whose vertices are the mid-points of the previous triangle's edges; every two steps shrink the triangle towards its centroid by a factor of four. (Indeed, if we interpret a half turn as a scaling by a factor of −1, each step just scales inwards by a factor of −2.)


To obtain the area of a triangle, cut it along the perpendicular, from its vertex with the widest internal angle, to the opposite edge; this decomposes the triangle into a pair of right angle triangles, back-to-back, with a common edge (that isn't the hypotenuse of either). Now cut along the perpendicular bisector of the first cut; on the side of this towards the original vertex, we get a pair of half-scale versions of the two right-angle triangles from the previous cut; half-turn each of these about its corner that wasn't on the original perpendicular, where its perpendicular bisector emerges from the original triangle. This rearranges the original triangle's four pieces into a rectangle that shares one side with the original triangle and whose other sides are half the height of the triangle perpendicular to that edge. So the area of the triangle is half the length of an edge times the height, by which is meant the distance from it to the opposite vertex.

I've only shown this for the edge opposite the widest angle; but it sufficed that the line perpendicular to that edge tha passes through the vertex does so within the edge (i.e. not on its extension outside the triangle); but we can do the same for any edge. Just shear parallel to that edge (a shear doesn't change area, or distances between lines parallel to the shear's direction) to put the opposite vertex where the perpendicular from it to the chosen edge does indeed lie within the edge. This hasn't changed the vertex's distance, along a line perpendicular to the edge, from the line of which the edge forms a segment, or the length of the edge; so hasn't changed either the area or the result of multiplying half base by height.

Now, still with the perpendicular from a vertex hitting the interior of the opposite side, let it sub-divide that side into two piece of lengths a and c, with the perpendicular being of length b. The perimeter of the triangle then consists of a, c and the hypotenuses of the right triangles with a or c as one side and b as the other; let's name the three whole sides A, B, C with:

and refer to half their sum, the semi-perimeter, as d, with 2.d = A +B +C, so

The triangle's area is b.(a +c)/2 = b.B/2. So now let's look at the product of d/2 with the results of subtracting each side from it:

d.(d −A).(d −B).(d −C)
= (B +A +C).(A +C −B).(C +B −A).(B +A −C)/16
= −(A +C +B).(A +C −B).(C −A +B).(C −A −B)/16
= −((A +C).(A +C) −B.B).((C −A).(C −A) −B.B)/16
= −(A.A +2.A.C +C.C −B.B).(C.C −2.A.C +A.A −B.B)/16
= −(A.A −B.B +C.C +2.A.C).(A.A −B.B +C.C −2.A.C)/16
= (4.A.A.C.C −(A.A −B.B +C.C).(A.A −B.B +C.C))/16
= (4.A.A.C.C −(A.A.A.A +B.B.B.B +C.C.C.C +2.A.A.C.C −2.B.B.A.A −2.B.B.C.C))/16
= (2.A.A.B.B +2.C.C.A.A +2.B.B.C.C −A.A.A.A −B.B.B.B −C.C.C.C))/16

in which A.A = c.c +b.b, C.C = a.a +b.b and B.B = a.a +2.a.c +c.c, so

d.(d −A).(d −B).(d −C)
= (2.(c.c +b.b).(a.a +2.a.c +c.c) +2.(a.a +b.b).(c.c +b.b) +2.(a.a +2.a.c +c.c).(a.a +b.b) −(c.c +b.b).(c.c +b.b) −(a.a +2.a.c +c.c).(a.a +2.a.c +c.c) −(a.a +b.b).(a.a +b.b)))/16
= (2.a.a.c.c +2.a.a.b.b +4.a.c.c.c +4.a.b.b.c +2.c.c.c.c +2.b.b.c.c +2.a.a.c.c +2.b.b.c.c +2.a.a.b.b +2.b.b.b.b +2.a.a.a.a +2.a.a.b.b +4.a.a.a.c +4.a.b.b.c +2.a.a.c.c +2.b.b.c.c −c.c.c.c −2.b.b.c.c −b.b.b.b −a.a.a.a −4.a.a.a.c −6.a.a.c.c −4.a.c.c.c −c.c.c.c −a.a.a.a −2.a.a.b.b −b.b.b.b)/16
= (2.a.a.a.a −a.a.a.a −a.a.a.a +4.a.a.a.c −4.a.a.a.c +2.a.a.b.b +2.a.a.b.b +2.a.a.b.b −2.a.a.b.b +4.a.b.b.c +4.a.b.b.c +2.b.b.b.b −b.b.b.b −b.b.b.b +2.a.a.c.c +2.a.a.c.c +2.a.a.c.c −6.a.a.c.c +4.a.c.c.c −4.a.c.c.c +2.b.b.c.c +2.b.b.c.c +2.b.b.c.c −2.b.b.c.c +2.c.c.c.c −c.c.c.c −c.c.c.c)/16
= b.b.(a.a +2.a.c +c.c)/4

which is the square of the area, b.(a +c)/2. This gives rise to Hero's formula (Hero was a not uncommon name in classical Greece), which gives the area of a triangle as the square root of the product of the semi-circumference and the three results of subtracting one edge's length from it.

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