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 A,B,C,D are collinear and if they satisfy
\frac{AC}{CB}=\frac {AD}{DB}, then the four points ACBD are said to form a harmonic range.
Now let us move onto some theorems.
Theorem 1.
If (ACBD) is a harmonic range and if O is the midpoint of AB then OB^2=OC\cdot OD.
Proof.
After componendo and dividendo on \frac{AC}{CB}=\frac{AD}{DB}; we get \frac{AC+CB}{AC-CB}=\frac{AD+DB}{AD-DB}; leading to
\frac{2OB}{2OC}=\frac{2OD}{2OB}\iff OB^2=OC\cdot OD.
The converse of this theorem is also obviously true.
dot((0,0));dot((-5,0));dot((+5,0));dot((3,0));dot((8,0));draw((-5,0)--(8,0));label("$A$", (-5,0), S);label("$O...
Theorem 2.
If AX, BY, CZ are three concurrent cevians of a triangle \triangle ABC with X, Y, Z lying on BC, CA, AB respectively, then ZY\cap BC=T\implies (BXCT) is a harmonic division.
Proof
Using Menelaus we have,
\frac{|AZ|}{|BZ|}\cdot\frac{|BT|}{|CT|}\cdot \frac{|CY|}{|AY|}=1;
And using Ceva we obtain
\frac{|AZ|}{|BZ|}\cdot\frac{|BX|}{|CX|}\cdot\frac{|CY|}{|AY|}=1;
Which jointly lead to \frac{|BX|}{|CX|}=\frac{|BT|}{|CT|}\implies (BXCT)=-1; ie the division is harmonic.
size(10cm, 8cm);label("$A$", (0,5), N);label("$B$", (-1,0), S);label("$C$", (5,0), S);label(&qu...
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 OA, OB, OC, OD pass through O and cut by a line d in A,B,C,D respectively. Assume that (ACBD) is harmonic, and therefore let another transversal cut these at A',B',C',D'; it is sufficient to check that the four points A', B', C', D' form a harmonic division.
Note that using the sine rule we get
\frac{AC}{CB}=\frac{OA}{OB}\cdot\frac{\sin\angle AOC}{\sin\angle COB}; and \frac{AD}{BD}=\frac{OA}{OB}\cdot\frac{\sin\angle AOD}{\sin\angle BOD}.
Comparing these two and using the fact that (ACBD) is harmonic, we get \frac{\sin\angle AOC}{\sin\angle COB}=\frac{\sin\angle AOD}{\sin\angle BOD}.
Now again we can use similar arguments to arrive at \frac{A'C'}{C'B'}=\frac{A'D'}{B'D'}; ie (A'C'B'D') is harmonic.
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff =...
Note: Such four concurrent lines are called to form a harmonic range or a harmonic pencil, denoted as O(ACBD) or sometimes as O(AB; CD).
Theorem 4.
Any two of the following statements implore the third statement to be true, too.
Statement 1: The pencil O(AB; CD) is harmonic.
Statement 2: OB bisects \angle AOC internally.
Statement 3: OB and OD are perpendicular to each other.
Proof.
I will prove that (1), (3)\implies (2), and the rest is left for the reader to prove.
Refer to the diagram. Draw EF\parallel AO such that E\in OC, F\in OD. Now,
AO\parallel EF\implies \frac{AC}{CB}=\frac{AO}{EB}; and AO\parallel BF\implies \frac{AD}{BD}=\frac{AO}{BF}.
Using \frac{AC}{CB}=\frac{AD}{BD}, we get EB=BF, ie B is the midpoint of EF.
Now, we already have \angle OBF=\angle BOA=\frac{\pi}{2}\land EB=BF\implies \triangle OBE\cong\triangle OBF. Therefore OB bisects \angle EOF, but since this is perpendicular to OA, therefore OA is the external bisector of \angle EOF.
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff =...
Theorem 5.
If (AB; CD) and (A'B'; C'D') are harmonic, and if AA', CC', BB' concur, then the line DD' is also concurrent with these lines.
Proof.
Let OD' meet BC at some point D'. We obtain that O(A'B'C'D') is harmonic, implying (AB, CD') is also harmonic. So we get
\frac{AC}{CB}=\frac{AD'}{BD'}. But also \frac{AC}{CB}=\frac{AD}{BD}; so that
\frac{AD'}{BD'}=\frac{AD}{BD}\implies \frac{AD'}{AD'-BD'}=\frac{AD}{AD-BD}\implies AD=AD', so that D' and D are coincident.
Image
Theorem 6. (NEW)
Let ABCD be a convex quadrilateral. If K=AD\cap BC, M=AC\cap BD, P=AB\cap KM, Q=DC\cap KM, then we must have K,M,P,Q to form a harmonic division.
Proof.
Let R=BQ\cap AK. Then from Theorem 2 in \triangle KDC with cevians KQ,CA,BD and transversal RQB, we observe that the division K,R,D,A is harmonic.
So the range B(K,R,D,A) is harmonic and using KP as a transversal, we see that (KMPQ)=-1. We are done.
import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwwwff ...
\Box

Theorem 7. (Newer) (BMO 2013)

The point P lies inside triangle ABC so that \angle ABP = \angle P CA. The point Q is such that P BQC is a parallelogram. Prove that \angle QAB =\angle CAP.
Proof.
Assume that P is to the left of Q. Now, complete the parallelogram BASP. Then note that BPQC is also a parallelogram, leading us to the fact that ASCQ is also a parallelogram. Therefore, \angle ABP=\angle ASP=\angle ACP, so that ASCP is a cyclic quadrilateral. Now, \angle QAC=\angle ACS=\angle APS, where the first equality comes from AQ\parallel SC and the second one from the cyclic quadrilateral. So, we are done.

DiagBMO.jpg

\Box
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 (ACBD) as the cross-ratio and (ACBD)=-1 obviously denotes the case when (AB; CD) 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. :lol:

\boxed{E1}. If AD,BE,CF are the altitudes of a triangle \triangle ABC, and if DE, EF meet AB, BC at F', D' respectively, then show that FD and F'D' intersect on a point lying on AC.
Solution
Let FD meet AC in E', then we have to show that D',E',F' are collinear.
Using Theorem 2, we have that (BDCD') is harmonic, ie \frac{BD}{DC}=\frac{BD'}{CD'}.
Similarly we obtain three other relations, and multiplying them out we get
\frac{BD'}{D'C}\cdot\frac{CE'}{E'A}\cdot\frac{AF'}{F'C}=\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}\stackrel{\text{Ceva...
So that using the converse of Menelaus theorem, D'E'F' is a transversal to the triangle \triangle ABC, and therefore a straight line.
Note: Even if I used directed segments in this proof, I think it is best to avoid using them.
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff =...

\boxed{E2}. The tangent at a point P of a circle cuts a diametre AB at T; and PN is the perpendicular to AB, the line joining B to the midpoint of TP cuts PN at Q. Prove that AQ\parallel TP.
Solution
Let us assume QA\cap TP=X. Note that we have, from the similarity of \triangle PBN and \triangle PBA; that
\angle BPT=\angle BAP=\angle NPB. So using Theorem 4, we obtain that (ABTN) is harmonic. So the pencil Q(ABTN) is also harmonic, leading to (suing TX as a transversal) (XMTP) is harmonic.
But, this will be harmonic if and only if \frac{XT}{MT}=\frac{XP}{PM}=\frac{XD}{TM}. But using PM=TM, we get XT=XP, but T\neq P. Therefore X=\infty, and AQ\parallel TP.
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff =...

\boxed{E4}Let ABC be a right angled triangle at A. D is a point on CB. Let M be the midpoint of AD. CM intersects the perpendicular bisector of AB at E. Prove that BE\parallel DA.
Solution
Let K=AB\cap CE. We will show that if E is a point on CM extended such that BE\parallel DA, Then we must have L is the perpendicular from E onto AB\implies AE=EB.
Now let DA and BE meet at \infty, then; since M is the midpoint of AD it is forced that (DMA\infty) is harmonic.
Now this also implies that B(DMA\infty) is a harmonic range, so that using CE as a transversal to this range it is implored that C,M,K,E are harmonic.
Now since \angle CAK = 90^{\circ} we must have \angle MAK=\angle KAE. Therefore using DA\parallel BE we have \angle DAB=\angle LBE, and therefore \triangle LAE\cong\triangle BEL.
Hence L is the midpoint of AB.\Box
import graph; size(10cm);draw((0,0)--(4,0));draw((4,0)--(0,5));draw((0,0)--(0,5));draw((0,5)--(2,-2.51));draw((0,0)--(2,2.5))...
\boxed{E5}In acute triangle ABC, we have points D and E on sides AC, AB respectively satisfying \angle ADE=\angle ABC. Let the angle bisector of \angle A meet BC at K.\ P and L are projections of K and A to DE, respectively, and Q is the midpoint of AL. If the incenter of \triangle ABC lies on the circumcircle of \triangle ADE, prove that P,\ Q, and the incenter of \triangle ADE are collinear.
Solution(motal)
Let J\equiv AK\cap BD, R the incenter of \triangle ABC and I the incenter of \triangle ADE. (AQL\infty)=-1 so, considering P(AQL\infty) with transversal AK, the problem is equivalent to proving that (AIJK)=-1. Easy angle chasing leads to \triangle ERD isosceles. Furthermore, \angle{ARD}=\angle{AED}=\gamma\Longrightarrow RKCD concyclic. With similar argument we obtain BERK concyclic. \angle{EID}+\angle{EKD}=180^{\circ}-\frac{\beta}{2}-\frac{\gamma}{2}+(\frac{\beta}{2}+\frac{\gamma}{2})=180^{\circ}. Therefore we have that IEKD is concyclic with diameter KI. In particular we have that both of these facts hold: \angle{IDK}=90^{\circ} and DI is bisector of \angle{JDA}. Hence considering pencil D(AIJK) we can conclude that (AIJK)=-1.
We are done. \Box
Sorry I could not attach a figure, I will attach one as soon as possible.

\boxed{E6}.(Desargues) Let \triangle ABC and \triangle A'B'C' be perspective triangles perspective about a point O. If BC\cap B'C'=X, CA\cap C'A'=Y and AB\cap A'B'=Z, then X,Y,Z are collinear.
Solution
Let us denote OA\cap BC=E, OA\cap B'C'=E'. Now we will use a little of cross ratio or projective geometry to prove this fact.
Firstly consider the concurrent points B,E,C,Y and B',E',C',Y. The corresponding lines BB', EE', CC', YY are concurrent at O. Therefore using a simple idea of cross-ratio we obtain that
O(BY, EC)=O(B'Y, E'C). It is also obvious that O(BY,EC)=A(BY, EC)=A(X,Y,Z,O) and O(B'Y,E'C')=A'(X,Y,Z,O). So the points X,Y,Z are collinear since O lies on the line AA'. (Why?)
We are done. :)
Image

\boxed{E7}.Given that AB is a diameter of a circle and the lines CD,CB are tangents to the circle, prove that DE = EF, where F is the foot of perpendicular from D onto AB and E=DF\cap CA.
Solution
Let U=CD\cap BA and S=BD\cap CA.
Since DF\perp AB and BD\perp DA, we see that \angle ADU=\angle ABD=\angle FDA, therefore DA bisects \angle FDU and also AD\perp BD. So the range D(B,F,A,U) is harmonic. Using CA as a transversal this leads to the fact that S,E,A,C are harmonic. So B(S,E,A,C) is harmonic, and again, using DF as a transversal to this range we have that (FED\infty)=-1, ie E is the midpoint of DF.
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.

\boxed{E8}.(Proposed in Olympiad Marathon by Fersolve)
Let O be the intersection of the diagonals of convex quadrilateral ABCD. The circumcircles
of \triangle OAD and \triangle OBC meet at O and M. Line OM meets the circumcircles of \triangle OAB and \triangle OCD at S and T respectively. Prove that M is the midpoint of ST.(NEW)
import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff ...
Solution
Let us invert everything about O to map points to their primes. By Theorem 6 on the four collinear points M',T',S',O; we have \dfrac{M'T'}{T'O}=\dfrac{M'S'}{S'O}. Also we have the following relations:
M'T'=\dfrac{MT\times r^2}{OM\cdot OT}, T'O=\dfrac{r^2}{OT}, M'S'=\dfrac{MS\times r^2}{OM\cdot OS}, S'O=\dfrac{r^2}{OS};
And therefore we obtain MT=MS and so, M is the midpoint of TS as required.\Box

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.

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

  1. 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

    • 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. 🙂

    • 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

      • 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.

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