2. THE NINE-POINT CIRCLE

We have already discovered one circle associated with the triangle – the circumcircle, passing through the three vertices. It is easily seen that it requires just three non-collinear points to determine a circle.

Exploring

In our triangle ABC we have now discovered two new triplets of points: the midpoints of the sides A', B.', C'; the feet of the altitudes D, E, F. Perhaps there are interesting circles circumscribing these triplets? Let’s try two special cases. You might like to carefully draw your own diagrams.

1. The equilateral triangle Let’s consider the circle through the edge midpoints A', B' , C'. Because this is a very special triangle, a lot of our points coincide! In fact, the circle here becomes the incircle. We look at this later. However, this circle passes through three new easily defined points. Can you describe them?

2. An isosceles triangle See the new diagram. This diagram is more interesting! You might find eight distinct points here which could lie on the circle. What are they? Do the three points found above belong here? Do they lie on the medians or the altitudes? What do you think the nine points might be for a general scalene triangle?

Play with this applet ...

Now check your answers ...

                                 
You might try a simple experiment. Draw one point in the plane. How many circles can you draw through it? Draw two points in the plane. Can you draw more than one circle through both these points? Would a third non-collinear point determine such a circle?

Another idea is to use an algebraic approach. The general equation for a circle is:

x2 + y2 + 2fy + 2gx + c = 0

This has three variable coefficients. Inserting the coordinates of three general points (x, y) gives three equations which completely determine these coefficients.
































 The Circle   Extensions 

The Nine-Point Circle

For a general triangle ABC, as before let A', B', C' denote the midpoints of the sides, D, E, F the feet of the altitudes, and U, V, W the midpoints of the segments joining the orthocentre H to the vertices A, B, C. Then:

Nine Point Circle Theorem: There is a circle which passes through the nine points A', B' C', D, E, F, U, V, W called the nine point circle of ABC. Its radius is half that of the circumcircle of the triangle.

Let's use a simpler diagram.

Let A'U meet OH in N.
Since OA' // AD and GA = 2 GA', we deduce that GHA ~ GOA' and twice the size.
Hence AH = 2 A'O, and UH = OA'.

Since OA' // AD and UH = OA, we deduce that NOA' NHU. Hence NA' = NU.
Hence A' and U lie on a circle with centre N, which also passes through D, as A'DH = 90°.
By symmetry, the points B', C', E, F, V and W also lie on this circle, giving us our result.

Finally, since N and U are midpoints of the sides HO, HA of HOA, we have OA = 2NU. That is, the radius of the circumcircle of ABC is twice the radius of the nine-point circle.

                                        

Two triangles are called similar if they have the same shape. This occurs if the angles of one are the same as the angles of the other. The symbol ‘ ~’ is used to denote similarity.

Two triangles can be shown to be similar by showing that two angles of the first triangle are equal (in size) to two angles of the second triangle. It will then follow that the third pair of angles is equal, since the angles of any triangle sum to 180°.

Two triangles are said to be congruent if they have the same shape and the same size. In this case, corresponding angles will be equal, and corresponding edges will have the same length. The symbol ‘’ is used to denote congruence.

There are various tests to determine whether two triangles are congruent.

• S.S.S. (Side-Side-Side)



• S.A.S. (Side-Angle-Side)



• A.S.A (Angle-Side-Angle)




As the name indicates, A.S.S. is not a valid test for congrunecy.
There is a well-known theorem that in any circle ‘the angle subtended by a chord at the centre is twice the angle subtended at the circumference’.

(Wherever did that word ‘subtend’ come from?!)

The proof is simple, with isosceles triangles and external angles.

In the case where AB is a diameter of the circle, the angle subtended at the circumference becomes a right-angle.









































 The Circle   Extensions 

Extensions

1. We know that G trisects OH. Can you show that G also trisects ON?

2. What would be the nine-point circle of HBC? (Don’t look too hard!) What other triangles in the figure have this nine-point circle?

3. Show that AOA'U is a parallelogram, and hence find the position of the midpoint of OU.

4. If A is a right angle, where does it lie in relation to the nine-point circle?

5. Can you find a condition on ABC which ensures that the nine-point circle touches BC? (Some of the nine points may coincide here.)

6. Do the results of this section still hold when ABC is an equilateral triangle? Investigate this.

                                                                         Hints and Solutions ....
For looking up ...

http://mathworld.wolfram.com/Nine-PointCircle.html

http://en.wikipedia.org/wiki/Nine-point_circle                                                                              

                                    







































Hints and Solutions

1. We know that G trisects OH. Can you show that G also trisects ON?

• HAG and OA'G are similar, and AG = 2GA'.

2. What would be the nine-point circle of HBC? (Don’t look too hard!) What other triangles in the figure have this nine-point circle?

• The same circle (it contains A', D, and the midpoint of HC say).
      Other triangles: HCA, HAB.

3. Show that AOA'U is a parallelogram, and hence find the position of the midpoint of OU.

• OA' = AU, and OA' // AU. Diagonals of a parallelogram bisect, so midpoint lies on median AA'.

4. If A is a right angle, where does it lie in relation to the nine-point circle?

• A lies on the circle. For if A', B', C' are midpoints, B'A'C' = 90°, so B'C' is a diameter of the circle.

5. Can you find a condition on ABC which ensures that the nine-point circle touches BC? (Some of the nine points may coincide here.)

• Make the triangle isosceles by taking AB = AC.

6. Do the results of this section still hold when ABC is an equilateral triangle? Investigate this.

• Here O = H = G, and the feet of the perpendiculars coincide with the midpoints of the edges.