**I AM BACK!**

Hi guys! Sorry for the long six months hiatus. I have been involved in lots of bridge activities as well as university commitments in this six months. I realise that my brain has became quite rusty after not doing Math Olympiad for half a year. Now that I am back, I will try update as often as possible and get my brain functioning again Anyway I don’t think I will be able to update daily as I did at the start of the year. Back then updating the blog is kinda like my day job. Right now I am having my university vacation break and hopefully I can contribute as much as possible before school term starts again.

Today I want to talk about the techniques involved in proving collinearity in geometry. Proving collinearity is a central problem in MO as evident from the numerous problems which are set regarding this topic. Basically, we would like to show that three or more points in a diagram lies in a common line. Many creative problems can be set on this topic, but there are some common techniques and theorems involved in solving this type of problems. Similar to the series on proving perpendicularity, I will collate these techniques and discuss them over several posts.

**Finding common angle**

The first technique involves identifying a common angle between two lines. Suppose you want to prove that the three points and are collinear. Suppose and are on the same side of and that . If this is the case, we can conclude that , and are collinear. This technique is useful if the conditions given in the question are related to angles. Here’s an illustration:

*(China Second Round Olympiad 2012) In an acute triangle , are distinct points on side such that . Let and be the circumcentres of , respectively. Prove that are collinear.*

I have always hated to draw diagrams in problems involving collinearity because the diagrams are usually very messy. Nevertheless this still has to be done. The diagram is not too hard to construct for this problem. Once constructed, it would seem that there are not clear directions in solving this problem (no similar triangles , no application of Menalaus’ theorem, etc). However, with simple angle chasing, we might be able to prove that .

We start with a fact which is useful to know when it comes ti circumcircles. Given a triangle and its circumcentre , note that . This identity follows immediately upon the construction of the perpendicular bisector from to . We want to prove that . Let us start by writing as . Using the identity mentioned, we note that . We need to find a way to incorporate , and we can do so by writing . Hence, we have

The last equality is achieved upon the application of the aforementioned identity.

**Menalaus’ Theorem**

It will be inappropriate to leave out this theorem in the discussion of collinearity. Menalaus’ theorem is one of the most powerful tools in proving collinearity, especially if triangles are involved. Simply identify the triangle and the three points on the triangle which one would like to prove is collinear, and then check if the corresponding ratios multiply to give us 1.

Here’s a problem from the recent USAMO:

*(USAMO 2012 P5) Let be a point in the plane of , and a line passing through . Let and be the points where the reflections of the line and with respect to intersect lines and respectively. Prove that and are collinear.*

IMO this is a relatively simple problem for USAMO P5 (The diagram is likely to be messy though ). The proof of this problem is very similar to the proof of Ceva’s theorem. Using the converse of Menalaus’ thoerem, the following relationship must hold:

Next, we relate the ratios of lengths to the ratios of areas. Denote the area of as . We have:

We can then express these areas using the sine formula, which simplifies to:

which is true based on the conditions of the question.

There are several other ways to solve the above problem. Coordinate bashing works well for this problem, although it is likely to take up much more time than the above approach. There is also an approach using Simson’s line, which I will discuss below.

**Simson’s Line**

The Simson’s Line approach is related to Menalaus’ theorem because it also involves collinear points on three sides of a triangle. However, this approach works better when circles are involved since the theorems involve the use of the circumcircle of the triangle.

Hmm I think I forgot to include Simson’s theorem in the geometry section above. Anyway, here’s the statement of the theorem:

*Given a point in the plane of . Suppose and are the feet of perpendiculars from to and respectively. Then and are collinear if and only if lies on the circumcircle of .*

Referring to the USAMO problem above, it seems tough to apply Simson’s theorem on the problem above. Yes, we need to prove that the three points and are collinear, but there is no guarantee that the perpendiculars at and will intersect at one point. If the three perpendiculars are not concurrent, we cannot apply Simson’s theorem in this problem.

But who says that we are going to apply Simson’s theorem with regards to ? Clearly, we cannot use since the three perpendiculars are not concurrent. Instead, we are going to create a new triangle such that and lies on three different sides of this triangle. Also, we will guarantee that the perpendiculars from and of this new triangle will intersect at a unique point. To do so, we will need to decide on that special unique point where the perpendiculars will intersect. It comes without surprise that the point that we will choose is . In this new triangle, which I shall call , and will be perpendicular to and respectively. We can easily construct this new triangle as shown by the red lines below:

Now we just have to prove that is on the circumcircle of . We do so by first noting that and are cyclic quadrilaterals, since and . Next, note that . Hence, and are similar. Finally, we have which proves that is a cyclic quadrilateral. Voila!

This is my first time seeing this technique in applying Simson’s theorem. Although the diagram is messy, the approach is elegant and I admit that I doubt I will be able to derive this solution in my first attempt.

I will find some time to collate all geometry problems related to collinearity after describing other techniques involved. Perhaps I will do the same for other kinds of geometry problems such as perpendicularity, parallel lines, etc. It sure feels good to be back on this blog🙂

P/S: Pfff took me two days to write this post… gotta work much harder

Cheers,

ksj