Poles and Polars – Another Useful Tool!

As the title suggests, this post is going to deal with the different aspects of poles and polars, which is a really great tool in case of proving problems using cyclic quadrilaterals and a lot of circular reasoning can be exploited by this. It is easier to deal with lines than with circles, so it is better to work with inversion, poles and polars etc – that’s why projective geometry is actually better in many aspects compared to plain euclidean geometry!

Anyways, for the time being let us know the definitions of poles and polars. Before continuing further, you have to note that a pole is a point, and a polar is a line (which is the opposite of Kedlaya, but yeah, notations don’t matter much. :) ) In any case, thanks to Rijul Saini for giving me a hand on this post, which we delayed for more than 8 months. At last, this is getting published on the blog, so Cheers, everyone! :)

Pole of a line
Take the circle \omega with respect to which you’re applying the polar map. Now, drop a perpendicular to the given line from the centre O of \omega to the given line. Name that point P. Now, invert P with respect to the circle. We get a point P' which lies on the same line as O,P. Then, P' is the Pole of the given line.

Polar of a point
Take the circle \omega with respect to which you’re applying the polar map. Let the centre of the circle be O. Now, invert the given point P with respect to the circle \omega and name the image as P'.
Now, Draw the line perpendicular to OP' which passes through the point P'. Then, this line is the Polar of the point P.

Now let us look into some theorems and lemmas that we will be using while solving problems.

Theorem 1:(La Hire’s theorem)
(i) Every point is the pole of its polar, and every line is the polar of its pole.
(ii) If P lies on the polar of Q then Q lies on the polar of P.

Proof:
(i) Again, direct from the definition.
(ii) \ P is a point on the polar of Q. First, extend OQ to the polar of Q and name that point Q'. If Q'= P then we are through. Otherwise, let the inverse of the point P wrt the circle be P'. Now, OQ \cdot OQ'=OP \cdot OP'=R^2. Now, by similar triangles, we have \angle OP'Q= \angle OQ'P=90^{\circ} and so we are through.\Box

Theorem 2:
Three points are collinear if and only if their polars are concurrent.
Proof:
I shall prove only the forward direction. The reverse direction is entirely analogous.
Let the three points be P,Q,R and their polars be p,q,r. Now, let the line through P,Q,R be l, and let the pole of l be L. Since, P,Q,R lie on the polar of L, therefore, L must lie on the polar of P,Q,R i.e. lines p,q,r. Thus, the lines p,q,r are concurrent at L.\Box

Now let us get to know the poles and polars of some points and lines that are considered to be important.

Polar of a point outside a circle with respect to itself.
Take an arbitrary point P outside a circle. Since the inverse of the point actually lies inside the circle, the polar will be a secant of the circle. Now, let us assume that the inverse point of P w.r.t. (O,r) is P'. The line perpendicular to \overline{OP'P} and passing through P' will intersect the circle in two points which will be symmetric w.r.t PP'. So let one of these two intersections be Q, and since PO\times P'O=r^2=PQ^2, we note that QOP'\sim POQ, so this leads to \angle OQP=90^{\circ}, ie:
Theorem 3:
The polar of a point P lying outside the circle is actually the chord of contact of the point P with respect to (O,r).

Polar of a point lying inside the circle does not have any special property that can be exploited alike to the previous one. However, we can take any arbitrary two chords passing through P, find their poles and join the two corresponding points to obtain the polar of that point P.

Now, let us move on to Theorem 4. The first name was coined by me, and the second one by Rijul Saini. Don’t assume that they are the original names of the theorem.
Theorem 4
(The fundamental theorem of poles and polars, or The theorem of two pascals)
Take a quadrilateral ABCD which is cyclic. If K=AC\cap BD then the polar of K with respect to \odot(ABCD) is the line joining AB\cap CD and AD\cap BC.
Proof 1.
The second name almost gives it away. This problem, however, was directly quoted and asked to prove in the Turkish TST 1993.
Anyways, using Pascal’s theorem in ACCBDD, we obtain AC\cap BD, CC\cap DD, BC\cap DA are collinear. Again using the same in CAADBB we obtain that CA\cap DB, AA\cap BB, AD\cap BC are collinear. Therefore it is noted that the points AC\cap BD, AD\cap BC, CC\cap DD, AA\cap BB are collinear. Denote this line as l.
Since l passes through CC\cap DD and AA\cap BB, which is the polar of AC\cap BD=K, we are done.\Box

Proof 2.
This uses harmonic divisions. :lol:
Let us take S=AB\cap CD, F=AD\cap BC, E=AC\cap BD. If the tangent to \odot(ABCD) at S touches it at M,N respectively such that M\in\overhat{DC}, denote X=MN\cap AD and Y=SE\cap BC, Z=MN\cap CD, T=SE\cap AD.
It can be shown easily that (S,X,A,B); (S,Z,C,D) and (S,E,Y,T) are harmonic. Then using (SXAB)=(SZCD) we note that AD,BC,ZX concur; and F,X,Z are collinear. Similarly we obtain that F,X,E and F,Z,E are collinear. Since XY\equiv MN, therefore E,F,M,N are collinear.
So F lies on the polar XZ of S w.r.t \odot(ABCD). By symmetry S lies on the polar YT of F w.r.t \odot(ABCD). So the intersection of these two polars will be E, and the polar of this pole will be FS. We are done! \Box
Image
There are some useful results that are correlated to harmonic divisions and poles and polars, which I explain in the following few theorems.
Theorem 5 (a).
If P is the pole of a line AB w.r.t \omega then any line \ell through P is cut harmonically by P,AB, and \omega.
Proof.
Let O be the centre of \omega, and let OP\cap \overleftrightarrow{AB}=B, \ \ell\cap \overleftrightarrow{AB}=A, \ PA\cap \omega = F.
Let \omega' be the circle passing through P,A,B. Then its centre is the midpoint O' of BP. If \omega\cap \omega' = C,D; then P is the pole of AB. So OP\cdpt OQ=OC^2. Then \omega and \omega' are orthogonal circles, which means that O'D is tangent to \omega. Therefore we obtain O'E\cdot O'F=O'D^2=O'P^2, and since O' is the midpoint of PA, from Theorem 1 – Harmonic Divisions, we are done.
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff =...
\Box

Theorem 5 (b).
If two lines p,q are self-conjugate lines and if p\cap q = T, then p,q are harmonic with the two tangents drawn to the circle from T.
Proof.
Let the poles of p,q be P,Q. if the two tangents from T meet the circle in X,Y.
Since the polar of P passes through T, from La Hire’s theorem we see that the polar of T passes through P, and similarly, Q. Then X,Y,P,Q are collinear. Now, from Theorem 5(a) we are done.
import graph;size(10cm);draw(circle((0,0),3));dot((1.5,4.45));dot((1.5,1.9));dot((6,0));dot((1.5,2.6));dot((1.5,-2.6));draw((...
\Box
Theorem 6.
Finally, if four points A,B,C,D form a harmonic range then their polars a,b,c,d will create a harmonic pencil.
Proof.
Let X be the pole of \overleftrightarrow{ABCD}, and let Y=OX\cap \overleftrightarrow{ABCD}.
If we draw four perpendicular lines XA', XB', XC', XD' perpendicular to OA,OB,OC,OD, respectively, then we have an interesting result. The polar of X passes through A, so from La Hire’s, the polar of A passes through X, and therefore XA' is the polar of A with respect to our aforementioned circle. It is obvious that the pencils X(A',B',C',D') and O(A,B,C,D) are equiangular pencils, and since O(A,B,C,D) are harmonic, therefore X(A',B',C',D') are also harmonic.\Box

Wow, that was enough of theory for a day already! Let us quickly look into some applications of our Dual Pascal and try to find boring solutions to some problems!

\boxed{E1}. Let XY be the diameter of a circle with two points P,Q on its circumference such that P is closer to X than Q. If PX and QY intersect at S outside the circle, and if the tangents at P,Q to the circle meet at R; then show that RS\perp XY.
Solution
Let T=PQ\cap XY. We already know that(from Theorem 4), S lies on the polar of T. The polar of R is \overline{PQT}, we see that R lies on the polar of T. So SR is the polar of T, and therefore we are done. \Box
import graph;unitsize(0.9cm);draw(circle((0,0),3));dot((-2.236,2));dot((3,0));dot((-3,0));dot((1.3, 2.7));draw((3,0)--(-12.33...

\boxed{E2}. Let ABCD be a tangential quadrilateral with incentre I. Let the opposite sides of ABCD meet each other at E,F; ie let E=AB\cap CD, F=AD\cap BC. Let its incircle touch the sides AB,BC,CD,DA at G,H,K,L, respectively. If P=GK\cap HL, then show that OP\perp EF.
Solution
Note that EG,EK are tangents to the incircle of ABCD, so that KG is the polar of E. Similarly the polar of F is HL. Therefore the pole of the line EF is KG\cap HL=P, and we are done. \Box

\boxed{E3}.(China 2006 Western MO) AB is the diameter of a circle with centre O. C is a point on AB, extended. A line through C cuts the circle with centre O at D,E. OF is a diameter of \odot(BOD) which has centre O_1. Let CF\cap \odot(BOD)=G. Prove that O,E,A,G lie on a circle.
Solution
Let P=AE\cap BD. So the polar of P with respect to the circle (O) passes through BA\cap DE=C and AD\cap BE=J. So PO\perp CJ, and it is obvious that G, O, P are collinear.
So by considering the power of P we have PD\cdot PB=PG\cdot PO=PE\cdot PA; and so E,A,O,G are concyclic points. \Box
Image

\boxed{E4}.(IMO 1985) A circle with center O passes through the vertices A and C of triangle ABC and intersects segments AB and BC again at distinct points K and N, respectively. The circumcircles of triangles ABC and KBN intersects at exactly two distinct points B and M. Prove that \angle OMB=90^{\circ}.
Solution
Note that the spiral similarity centered at M which sends A to C and K to N sends the midpoint of AK(say, M_1) to that of CN(say, M_2). So M is the center of unique spiral similarity that sends A to M_2
and C to M_1, and thus it follows that M,M_1,M_2,B are concyclic. Again since \angle OM_2B=\angle OM_1B, so O,M_1,M_2,B are concyclic, and OB is the diameter of the common circle. So, we are done. \Box

\boxed{E5}. Circles \omega_1 and \omega_2 meet at points O and M. Circle \omega, centered at O, meets circles \omega_1 and \omega_2 in four distinct points A,B,C and D, such that ABCD is a convex quadrilateral. Lines AB and CD meet at N_1. Lines AD and BC meet at N_2. Prove that N_1N_2\perp MO.
Solution
Actually this is equivalent to our last problem. Note that if \odot(ADN_1)\cap \odot(BCN_1)=K, then from simple angle chasing we can show that K=\odot(AOCK)\cap \odot(BODK). Also it is obvious that N_1N_2 passes through K\equiv M. All of this implies that N_1N_1\perp MO, as desired.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...

\boxed{E6}. Let ABC be a triangle with incentre I. Reflections of I in BC,CA, AB are X,Y,Z. Prove that AX,BY,CZ are concurrent.
Note: This is a direct consequence of the isogonic theorem. For details, see here.
Solution
I will use a lemma.
Lemma.
If \triangle DEF is the intouch triangle of \triangle ABC, and if I is the incentre of \triangle ABC, and if \triangle A'B'C' is the triangle directly homothetic to ABC such that A'\in IA, B'\in IB, C'\in IC; then A'B'C' is perspective to \triangle DEF.

Now let us come back to our original problem.
Let DEF be, as usual, the intouch triangle of \triangle ABC. Then let \Omega be the incircle of \triangle ABC. Since XD=DI, we see that the polar of X with respect to \Omega is the perpendicular bisector of \overline{DI}. Similarly, the polar of Y w.r.t \Omega is the perpendicular bisector of \overline{IE} and the polar of Z w.r.t \Omega is the perpendicular bisector of \overline{IF}.
Let these perpendicular bisectors intersect each other to form a triangle KLM.
Denote by \omega_1, \omega_2, \omega_3; the circles (D,DI), (E,EI),(F,FI). Then K is the intersection of LK and KM. Since LK is the radical axis of \Omega and \omega_3; and since KM is the radical axis of \Omega and \omega_2, therefore K is the radical centre of \omega_2,\omega_3,\Omega.
Since AE^2-r^2=AF^2-r^2 where r is the inradius of ABC, we note that A has the same power with respect to \omega_2 and \omega_3. So A,K,I all lie on the radical axis of these two circles, and therefore, collinear.
From here it is also obvious that AK=KI,\ BL=LI,\ CM=MI and so \triangle ABC and \triangle KLM are directly similar and perspective around I. So they are homothetic.
Applying the aforementioned lemma, we note that the intersections U=LM\cap EF,\ V=MK\cap FD, \ W=KL\cap DE are collinear.
Since AE and AF are tangents to \Omega, we again note that A is the pole of EF w.r.t \Omega. Also since the polar of X w.r.t \Omega is LM, it follows from the La Hire’s theorem that the pole of AX w.r.t \Omega is U.
Due to symmetry we note that V,\ W are respectively the poles of BY, \ CZ. We already have seen that U,V,W are collinear. Therefore it is obvious that AX,BY,CZ meet at the pole of \overline{VUW} w.r.t \Omega. We are done. \Box
Image

\boxed{E7}.(Polish MO Second Round 2012) Let ABC be a triangle with \angle A=60^{\circ} and AB\neq AC, I-incentre, O-circumcentre. Prove that perpendicular bisector of AI, line OI and line BC have a common point.
Solution
Assume that the incircle touches BC, CA, AB at D,E,F and that X=AI\cap \odot(DEF). Since \angle FIA=60^{\circ},\angle EIA=60^{\circ}, therefore XEI and XFI are equilateral. Also \angle XFA=30^{\circ}=\angle FAX, therefore AX=XE=IX, and the incircle of ABC bisects AI.
Now, we claim that OI is the perpendicular bisector of DX.
Proof of claim.
Let I' be the reflection of I in BC. Then note that we have \angle BI'C=\angle BIC=120^{\circ}, therefore I' lies on \odot(ABC). Now we already have AI=II'=2r, \ I'O= AO, which give OIA\cong OIA', leading to the fact that OI bisects AI'. This readily implies that OI bisects DX.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...
Coming back to our main proof, note that we have IO as the perpendicular bisector of DX, so OI passes through the pole of DX wrt \odot(DEF). Since pole of DX is XX\cap BC, and because XX is the perpendicular bisector of AI, we are done. \Box

\boxed{E8}.(2012 Indonesia Round 2 TST 4: P2) Let \omega be a circle with center O, and let l be a line not intersecting \omega. E is a point on l such that OE is perpendicular with l. Let M be an arbitrary point on M different from E. Let A and B be distinct points on the circle \omega such that MA and MB are tangents to \omega. Let C and D be the foot of perpendiculars from E to MA and MB respectively. Let F be the intersection of CD and OE. As M moves, determine the locus of F.
Solution(hatchguy)
Let E' be the inverse of E wrt \omega. I claim F is the midpoint of EE'. Clearly, since M is in the polar of E' then E' is in the polar of M, and therefore A,E',B are collinear. It is easy to see that M,E,B,O,A lie on a circle with diameter MO. From this, it’s very natural to think of droping a perpendicular from E to AB. Let G be the foot of this perpendicular. By Simson’s line theorem, we have C,D,F,G are collinear. Also, using that DEGB is cyclic we easily get \angle FGE = \angle DBE = \angle MOE = \angle FEG;The last because MO \parallel EG. Hence FG= FE and we get that F is midpoint of EE' since EGE' is a right triangle. We are done.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...
\Box

\boxed{E9}. Let incircle \omega of \triangle ABC be tangent to BC , CA , AB in D, E , F. A line from A that is parallel to DE meets DF in K and a line from A that is parallel to DF meets DE in L. If M,N be midpoint of AB , AC , show that KL is on MN.
Solution
From simple angle chasing, we see that \angle DEN=\angle KAE=\angle KFE. This leads to the fact that A,F,K,I,L,E lie on a circle with AI as diameter. Thus, IK\perp KA; leading to C,I,K being collinear. Let H be the orthocentre of \triangle BIC. Then note that polar of K passes through the pole of DF wrt \omega, which is B. Also since IK'\perp DE, it follows that K'=BH\cap CI. Similarly, L'=CH\cap BI. Therefore, the pole of KL wrt \omega is H. Thus, HI\perp KL, and HI\perp BC, leading to KL\parallel MN.
Let L_1=AL\cap BC, then note that we have BAL_1 is B- isosceles, and because BI\perp AL_1, therefore L is the midpoint of L_1A. Thence ML\parallel BC. Similarly NK\parallel BC. Using all these, we see that M,N,K,L lie on a line parallel to BC.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...
Result:
This problem teaches us that the polar of the orthocentre of BIC wrt \omega bisects AB and AC. This is a very useful fact. Don’t ask me why, because even I don’t know where it may come in handy. :) \Box

\boxed{E10}.(Romania MOM 2012)Let ABC be a triangle and let I and O denote its incentre and circumcentre respectively. Let \omega_A be the circle through B and C which is tangent to the incircle of the triangle ABC; the circles \omega_B and \omega_C are defined similarly. The circles \omega_B and \omega_C meet at a point A' distinct from A; the points B' and C' are defined similarly. Prove that the lines AA',BB' and CC' are concurrent at a point on the line IO.

Solution(r1234)
Firstly, we’ll cite a lemma:
Lemma.
Let \ell be a line and \Gamma be circle.Suppose P is the pole of \ell wrt \Gamma.Now let \triangle ABC be inscribed in \Gamma.Let \triangle A'B'C' be the circumcevian triangle of \triangle ABC wrt P.Then \ell is the perspective axis of \triangle ABC and \triangle A'B'C'.
Proof of lemma.
Let A_1=BC\cap B'C', B_1=AC\cap A'C', C_1=AB\cap A'B'.
Now applying Pascal’s theorem on the hexagon B'C'ABCA' we get A_1,C_1, C'A\cap CA' are collinear.
Again \triangle AC'B and \triangle A'CB' are perspective.So C_1, C'A\cap CA', BC'\cap B'C are collinear.Hence we conclude that A_1,C_1, BC'\cap B'C are collinear.But this line is nothing but the polar of P i.e \ell.Hence we conclude that A_1B_1C_1\equiv \ell.\Box

Coming back to the main proof,
Clearly AA' is the radical axis of \omega _b, \omega_c, BB' is the radical axis of \omega_c, \omega_a and CC' is the radical axis of \omega_a, \omega_b.So by radical axis theorem AA', BB', CC' are concurrent.
Let (I) be the incircle of \triangle ABC and \triangle DEF is its intouch triangle.D', E', F' are the touch points of (I) with \omega_a, \omega_b, \omega_c respectively.Now let tangents at D', E', F' meets BC, CA, AB at X,Y,Z respectively.Then its easy to show that XYZ is the radical axis of (I) and \odot ABC.So X is the pole of DD' wrt (I) and similar for others.So DD', EE', FF' concur at the pole of XYZ wrt (I).Now consider the circles (I), \omega_b, \omega_c.Then by radical axis theorem the lines AA', E'E', F'F' are concurrent ,say at X_1.Then X_1 is the pole of E'F' wrt (I).Since A, X_1, A' are collinear, their polars i.e EF,E'F',\text{Polar of A'} are concurrent.So \triangle DEF and the triangle formed by the polars of A', B', C' are perspective wrt the perspective axis of \triangle DEF, \triangle D'E'F'.But according to our lemma this perspective axis is the polar of the perspective point of \triangle DEF, \triangle D'E'F', i.e the radical axis of (I), \odot ABC.So we conclude that AA', BB', CC' concur at the pole of the radical axis of (I), \odot ABC wrt (I).Since IO\perp \text{Radical axis of (I),circumcircle of ABC} we conclude that the pole of the radical axis of (I), \odot ABC lies on IO.Hence AA', BB', CC' concur on IO.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...
\Box

As for exercises, I won’t give any for this post. I will just write down some references from where you can get plenty of problems. ^_^

References.
1. Cyclic Quadrilaterals – The Big Picture by Yufei Zhao.
2. Power of a Point by Yufei Zhao.
3. Circles by Yufei Zhao.
4. Poles and Polars by Kin Y. Li.
5. Mathscope Topic on Poles and Polars by Hoàng Quốc Khánh.
6. Introduction to the geometry of triangle by Paul Yiu.

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7 thoughts on “Poles and Polars – Another Useful Tool!

  1. Thank you very much, Potla, and I have enjoyed it. Anyway, I doubt that Th2(ii) is typo maybe…
    Bigwood

    • Haha, thanks for pointing out the typo. 😀
      ありがとう ^_^
      I had written the first part of this post 6 months ago. 🙂

  2. Little Brother is awesome and cool banner! The math made me feel stupid for not understanding, but it’s okay!

  3. Pingback: Two similar geometry problems based on perpendiculars to cevians | Chaitanya's Random Pages

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