MISSION · Geometry
The Four Centres of a Triangle
A triangle has no single centre, and each of the four is useful for something different
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The idea in one sentence
“Centre of a triangle” has no single meaning, so there are four of them, each defined by a different set of three lines that — surprisingly — always meet at a point.
Why three lines meeting is surprising
Two lines in a plane cross. Three lines drawn independently do not: they normally make a small triangle of near-misses. When three lines built by the same rule always pass through one point, something is going on, and the word for it is concurrent.
Each of the four centres is a concurrency that has to be proved. They are results, not definitions.
Centroid — the three medians
A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid.
It is the triangle’s balance point. Cut the triangle out of card and it will balance on a pin at the centroid, which is why it is also called the centre of mass.
The centroid cuts every median in the ratio \(2 : 1\), with the longer part next to the vertex. And in coordinates it is the plain average of the three vertices:
\[G = \left( \frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3} \right)\]It is always inside the triangle.
Circumcentre — the three perpendicular bisectors
Take the perpendicular bisector of each side. They meet at the circumcentre, which is equidistant from all three vertices — so a circle centred there passes through all three corners.
The reason is the one fact bisectors carry: a point on the bisector of \(AB\) is equally far from \(A\) and \(B\). A point on two of them is equally far from all three vertices, and therefore lies on the third bisector as well. That is the whole proof.
This one can escape the triangle. It sits inside for an acute triangle, exactly on the midpoint of the hypotenuse for a right triangle, and outside for an obtuse one.
Incentre — the three angle bisectors
Bisect each angle. The three bisectors meet at the incentre, which is equidistant from all three sides — so a circle centred there touches all three sides from within.
Same argument, different distance: a point on the bisector of an angle is equally far from the two arms. Vertices for the circumcentre, sides for the incentre.
The incentre is always inside, because an angle bisector always runs into the triangle.
Orthocentre — the three altitudes
An altitude drops from a vertex perpendicular to the opposite side. The three meet at the orthocentre.
It has no tidy distance property, and it is the one that wanders furthest: inside for acute triangles, exactly at the right-angled vertex for a right triangle, well outside for obtuse ones, where two of the altitudes only meet the opposite sides once those sides are extended.
The Euler line
Three of the four — centroid, circumcentre, orthocentre — are always collinear, on what is called the Euler line. The centroid sits between the other two and always twice as close to the circumcentre as to the orthocentre.
The incentre can also lie on it whenever the triangle is isosceles, because all four centres then lie on the symmetry axis. Only in the equilateral case do all four centres collapse onto the same point, which is presumably why the word “centre” felt unambiguous for so long.
Common mix-ups
- Median against altitude. A median goes to a midpoint; an altitude goes in perpendicular. They coincide only in an isosceles triangle, and then only from the apex.
- Perpendicular bisector against angle bisector. One bisects a side, one bisects an angle. They give different centres and different circles.
- Assuming every centre is inside. Only the centroid and incentre always are.
- Getting the 2:1 ratio backwards. The long piece is on the vertex side.
Check yourself
Name the point where the medians meet, the ratio in which it cuts each median, and the centre of the circle through all three vertices.
The medians meet at the centroid, which cuts each median \(2 : 1\) from the vertex. The circle through all three vertices is centred at the circumcentre, where the perpendicular bisectors meet.
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