# Harmonic Divisions – A Powerful & Rarely Used Tool!

Long since I have ever posted anything on my blog, and more importantly, I have not posted anything on geometry before (except for a similarity trivia), and therefore this is my first extensive attempt on geometry.
As the title explains clearly, I want to post some useful results and solve some problems related to Harmonic Divisions, because I have not found a very good book or ebook explaining the concepts clearly.
HARMONIC DIVISIONS
Though all of us know what a harmonic division is, if four points are collinear and if they satisfy
then the four points are said to form a harmonic range.
Now let us move onto some theorems.
Theorem 1.
If is a harmonic range and if is the midpoint of then
Proof.
After componendo and dividendo on we get leading to

The converse of this theorem is also obviously true.

Theorem 2.
If are three concurrent cevians of a triangle with lying on respectively, then is a harmonic division.
Proof
Using Menelaus we have,

And using Ceva we obtain

Which jointly lead to ie the division is harmonic.

Theorem 3.
If four concurrent lines are such that one transversal cuts them harmonically, then every transversal intersects these four lines to form a harmonic division.
Proof
Assume the lines pass through and cut by a line in respectively. Assume that is harmonic, and therefore let another transversal cut these at it is sufficient to check that the four points form a harmonic division.
Note that using the sine rule we get
and
Comparing these two and using the fact that is harmonic, we get
Now again we can use similar arguments to arrive at ie is harmonic.

Note: Such four concurrent lines are called to form a harmonic range or a harmonic pencil, denoted as or sometimes as
Theorem 4.
Any two of the following statements implore the third statement to be true, too.
Statement 1: The pencil is harmonic.
Statement 2: bisects internally.
Statement 3: and are perpendicular to each other.
Proof.
I will prove that and the rest is left for the reader to prove.
Refer to the diagram. Draw such that Now,
and
Using we get ie is the midpoint of
Now, we already have Therefore bisects but since this is perpendicular to therefore is the external bisector of

Theorem 5.
If and are harmonic, and if concur, then the line is also concurrent with these lines.
Proof.
Let meet at some point We obtain that is harmonic, implying is also harmonic. So we get
But also so that
so that and are coincident.

Theorem 6. (NEW)
Let be a convex quadrilateral. If then we must have to form a harmonic division.
Proof.
Let Then from Theorem 2 in with cevians and transversal we observe that the division is harmonic.
So the range is harmonic and using as a transversal, we see that We are done.

The point lies inside triangle so that . The point is such that is a parallelogram. Prove that .
Proof.
Assume that is to the left of Now, complete the parallelogram Then note that is also a parallelogram, leading us to the fact that is also a parallelogram. Therefore, so that is a cyclic quadrilateral. Now, where the first equality comes from and the second one from the cyclic quadrilateral. So, we are done.

Now we have already done some easy theorems which might come in handy here and there, so we need to keep a tab on them. Let’s move onto some simple applications.
Note: From now on I might denote as the cross-ratio and obviously denotes the case when is harmonic.
Note 2: I assume the reader to have some knowledge on Poles and Polars, Inversion, Homothety and Cross ratio. So hope I will not be too ambiguous while using these notations and notions.

If are the altitudes of a triangle and if meet at respectively, then show that and intersect on a point lying on
Solution
Let meet in then we have to show that are collinear.
Using Theorem 2, we have that is harmonic, ie
Similarly we obtain three other relations, and multiplying them out we get

So that using the converse of Menelaus theorem, is a transversal to the triangle and therefore a straight line.
Note: Even if I used directed segments in this proof, I think it is best to avoid using them.

The tangent at a point of a circle cuts a diametre at ; and is the perpendicular to the line joining to the midpoint of cuts at Prove that
Solution
Let us assume Note that we have, from the similarity of and that
So using Theorem 4, we obtain that is harmonic. So the pencil is also harmonic, leading to (suing as a transversal) is harmonic.
But, this will be harmonic if and only if But using we get but Therefore and

Let be a right angled triangle at . is a point on . Let be the midpoint of . intersects the perpendicular bisector of at . Prove that .
Solution
Let We will show that if is a point on extended such that Then we must have is the perpendicular from onto
Now let and meet at then; since is the midpoint of it is forced that is harmonic.
Now this also implies that is a harmonic range, so that using as a transversal to this range it is implored that are harmonic.
Now since we must have Therefore using we have and therefore
Hence is the midpoint of

In acute triangle we have points and on sides respectively satisfying Let the angle bisector of meet at and are projections of and to respectively, and is the midpoint of If the incenter of lies on the circumcircle of prove that and the incenter of are collinear.
Solution(motal)
Let , the incenter of and the incenter of . so, considering with transversal , the problem is equivalent to proving that . Easy angle chasing leads to isosceles. Furthermore, concyclic. With similar argument we obtain concyclic. . Therefore we have that is concyclic with diameter . In particular we have that both of these facts hold: and is bisector of . Hence considering pencil we can conclude that .
We are done.
Sorry I could not attach a figure, I will attach one as soon as possible.

(Desargues) Let and be perspective triangles perspective about a point If and then are collinear.
Solution
Let us denote Now we will use a little of cross ratio or projective geometry to prove this fact.
Firstly consider the concurrent points and The corresponding lines are concurrent at Therefore using a simple idea of cross-ratio we obtain that
It is also obvious that and So the points are collinear since lies on the line (Why?)
We are done.

Given that is a diameter of a circle and the lines are tangents to the circle, prove that where is the foot of perpendicular from onto and
Solution
Let and
Since and we see that therefore bisects and also So the range is harmonic. Using as a transversal this leads to the fact that are harmonic. So is harmonic, and again, using as a transversal to this range we have that ie is the midpoint of
These types of midpoint considerations can prove a lot of geometric problems easily, and of course this one is obvious from considering symmedians. Apart from that, harmonic divisions stands out to be a hugely useful and rarely used (in my country, at least) method.

(Proposed in Olympiad Marathon by Fersolve)
Let be the intersection of the diagonals of convex quadrilateral The circumcircles
of and meet at and Line meets the circumcircles of and at and respectively. Prove that is the midpoint of (NEW)

Solution
Let us invert everything about to map points to their primes. By Theorem 6 on the four collinear points we have Also we have the following relations:

And therefore we obtain and so, is the midpoint of as required.

I will keep updating this post from time to time because as time goes on, the number of problems you come across also goes on.
I will add some exercises about a month later (after my secondary exams end), so, till then; Rejoice!!
References
[1] Cosmin Pohoata; Harmonic Divisions; 2007 Mathematical Reflections
[2] C. Stanley Ogilvy; (1990) Excursions in Geometry, Dover.
[3] Application 1
[4] Application 2

Last Updated On: 04:30 am UTC; 11 March, 2013.

## 7 thoughts on “Harmonic Divisions – A Powerful & Rarely Used Tool!”

1. Jonathan Yu says:

hi, sorry but for Harmonic Divisions: E5, i don’t see why showing that AIJK is harmonic shows that P,I, and Q are collinear
could you clarify please? thank you

• Potla says:

Well, because we defined $I$ to be the incentre of $ADE,$ and because $(AQL\infty)=-1,$ we get $(AI'JK)=-1$ where $I'=PQ\cap AK.$ If we can show that $(AIJK)=-1,$ then we obviously have $I\equiv I',$ therefore $I=PQ\cap AK.$
I hope I could clarify. 🙂

2. MarukoM says:

I liked the part with the triangles. More importantly how’s Little Brother?

• Potla says:

I am quite well, and seems you’re doing fine as well. 😀

3. Virus123 says:

No, that was a square. Derp.

• MarukoM says:

Oh please. According to the Hippogriff’s Theorom, a shape that looks like a triangle is a triangle. Besides… The quadrangle you mentioned is nothing more than an abstract in this particular thought ecperiment.

>Typed on iPhone while going 70 mph ftw

• Virus123 says:

Gtfo. If you cut a triangle in half and then cut those triangles in half, you now have 4 triangles. If you put 4 triangles together, the obvious result is a square. Since more often than not a square is a square and NOT a triangle, that means that what you saw was, in fact, a square and NOT a triangle.

Also, since you are nothing more than a figment of my imagination, we can presume that we are not having this argument. It follows that this awesome tl;dr blog is also not real.

In closing, apple is gay and you should throw that phone out the window as you go 70mph.