Activity 2-13: Triangle Centres www.carom-maths.co.uk Activity 2-13: Triangle Centres

opposite angles of the quadrilateral add to 180o. Take any quadrilateral. Is it possible to draw a circle through the four vertices? Clearly this is not always possible. It turns out that such a circle is possible if and only if opposite angles of the quadrilateral add to 180o. Task: can you prove this?

It IS possible, however, to draw a circle through the corners of any triangle. Proof: given a triangle ABC, consider the locus of points equidistant from A and B. This will be the perpendicular bisector of AB. Now add the perpendicular bisector of AC. The point where they meet is equidistant from A and B, and from A and C, and thus from A, B and C.

This is then the centre of a circle through A,B and C, the circumcircle. We call D the circumcentre of the triangle. Are there any other centres that a triangle can have? There are lots! The Encyclopedia of Triangle Centres website currently lists 5374 different ones. The circumcentre is Number 3 on the list.

Now for Centre Number 2 on the list… Task: in Geogebra, take a triangle, and join each corner to the midpoint of the opposite side. Now drag a corner… A line joining a corner to the opposite mid-point is called a median, and the three medians appear to be concurrent. Can we prove this?

It helps to know a bit about vectors... So now we can come up with our proof:

r = ld+(1-l)c = l(a+b)/2+(1-l)c. The line CD is r = ld+(1-l)c = l(a+b)/2+(1-l)c. Now put l = 2/3, and we have the point (a + b + c)/3, which is on CD. Now by symmetry, this point must also lie on the other two medians, and is thus the centroid of the triangle. The centroid always lies inside the triangle, and is its centre of gravity.

We will now consider Centre 4 on the list… Task: using Geogebra, pick a triangle and draw the perpendicular from each corner to the opposite side. What do you find? It seems the three perpendiculars always meet: we call this point the orthocentre (Centre 4). Can we prove this?

a and b are at right angles. Vectors again prove to be useful… For two vectors a and b, the scalar product is defined to be a.b = |a||b|cos q, where q is the angle between a and b. Three things: clearly a.b = 0 if and only if a and b are at right angles. Secondly, you CAN multiply out a.(b + c) to give a.b + a.c (the distributive law). Thirdly, a.b = b.a (the commutative law).

a.b + c.d – a.c – b.d = 0, a.d + b.c – b.d – a.c = 0. Her we have that (a – d).(b – c)=0. We also have that (c – d).(b – a) = 0. Multiplying out, we get a.b + c.d – a.c – b.d = 0, a.d + b.c – b.d – a.c = 0. Subtracting these gives a.b + c.d – a.d – b.c = 0, and so (a – c)(b – d) = 0, and BD is perpendicular to AC, and so the three perpendiculars meet at a point.

So P (the circumcentre) Q (the centroid) and R (the orthocentre) One more remarkable thing: these three centres for a triangle always lie on a straight line. This is called the Euler line, after the Swiss mathematician who first proved this fact in 1765. So P (the circumcentre) Q (the centroid) and R (the orthocentre) are collinear, AND QR = 2PQ. Tough to prove!

With thanks to: Wikipedia, for another helpful article. Carom is written by Jonny Griffiths, mail@jonny-griffiths.net