3. INCIRCLE AND EXCIRCLES

The circumcircle of a triangle is the circle determined by, and passing through, the three vertices of the triangle. Now in the plane, there is a partial duality: an interchange of points and lines.

Exploring

Given triangle ABC, can you suggest a circle which is determined by the three sides of the triangle? In fact, is there more than one such circle? How many?

The centre of the circumcircle is the point of intersection of the three perpendicular bisectors of the sides of the triangle. How would you determine the centres of these new circles? And are these centres related in any nice way? Draw some diagrams of your own and explore these ideas.

When you have made as much progress as you can, check your results.




                                 
































 The Circles   Extensions 

The Incircle and Excircles

We can associate with any triangle ABC, an incircle (inscribed circle) which lies within the triangle and has the three sides of the triangle as tangents, and three excircles (escribed circles) which also have the three sides of the triangle as tangents, but which lie exterior to the triangle. We take the centres of these circles to be I, I1, I2, I3 respectively as in the diagram.

Now, how do we locate the centres of these circles? Let's start with the incentre.

In this diagram, the two right-angled triangles PIB, RIB have IB in common, and PI = RI. We deduce that these two triangles are congruent, and hence PBI = RBI. That is, the incentre I lies on the (internal) bisector of B, and similarly, on the bisectors of A and C. In other words

The internal bisectors of a triangle are concurrent and meet at the incentre of the triangle.

We can similarly show that each excentre lies at the intersection of one internal bisector and two external bisectors of the angles of the triangle.

Finally, is there any nice relationship between the incentre and the three excentres?

                                   

You might notice we are here using the despised A.S.S. rule which we indicated before as being invalid! In fact this rule is valid for precisely one situation, and that is when the angle being used is a right angle.

In the top figure at right, we see that the criterion allows for two non-congruent possibilities.

In the bottom figure there are again two possibilities, but they are congruent triangles.



































P
 The Circles   Extensions 

We have asked if there is any nice relationship between the incentre and the three excentres.

Look now at the adjacent diagram.

Points I3, A, I2 are collinear since AI3 and AI2 are bisectors of vertically opposite angles. Similarly for the other sides of this triangle.

Points A, I and I1 are collinear, since I and I1 lie on the same bisector of the angle at A. Similarly for the lines through B and C meeting at I.

Now look for example at the underlying (pale) angles at A.
We know that CAB and BAX add to 180°, since CAX is a straight line. Since AI1 and AI3 are angle bisectors, I1AI3 = 90°, so AI1 is an altitude of I1I2I3. In the same way, the white lines through B and C are altitudes, so I is the orthocentre of I1I2I3.

Now there's a nice surprise!

                                 


































 The Circles   Extensions 

Extensions

1. We have shown that I is the orthocentre of I1I2I3. What is the orthocentre of II2I3 say? What about other permutations?

2. Can you find the nine-point circle of I1I2I3? This is easy!

3. Consider ABC with M on AC, and BM the (internal) angle bisector of A. Show that M is the midpoint of AC, if and only if AM AC.

4. Is it possible for the incentre to be the centroid of a triangle? the orthocentre? the circumcentre? (Use your answer to Q3.)


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

http://en.wikipedia.org/wiki/Incircle

http://mathworld.wolfram.com/Incircle.html

                                 





































Hints and Solutions

1. We have shown that I is the orthocentre of I1I2I3. What is the orthocentre of II2I3 say? What about other permutations?

•  The orthocentre is I1. Similarly I2, I3.
     Check the perpendiculars to the sides from the vertices.

2. Can you find the nine-point circle of I1I2I3? This is easy!

•  The circle determined by A, B, C – the circumcircle of ABC.

3. Consider ABC with M on AC, and BM the (internal) angle bisector of B. Show that M is the midpoint of AC, if and only if BM AC.

•  If BM AC, then ABM CBM and AM = MC.
    Prove the converse using a contradiction argument.

4. Is it possible for the incentre to be the centroid of a triangle? the orthocentre? the circumcentre? (Use your answer to Q3.)

•  The answer is 'yes' to all three, and occurs when the triangle is equilateral.