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 with respect to which you’re applying the polar map. Now, drop a perpendicular to the given line from the centre of to the given line. Name that point . Now, invert with respect to the circle. We get a point which lies on the same line as Then, is the Pole of the given line.
Polar of a point
Take the circle with respect to which you’re applying the polar map. Let the centre of the circle be Now, invert the given point with respect to the circle and name the image as
Now, Draw the line perpendicular to which passes through the point Then, this line is the Polar of the point
Now let us look into some theorems and lemmas that we will be using while solving problems.
(i) Again, direct from the definition.
(ii) is a point on the polar of First, extend to the polar of and name that point If then we are through. Otherwise, let the inverse of the point wrt the circle be Now, Now, by similar triangles, we have and so we are through.
Three points are collinear if and only if their polars are concurrent.
I shall prove only the forward direction. The reverse direction is entirely analogous.
Let the three points be and their polars be Now, let the line through be and let the pole of be . Since, lie on the polar of therefore, must lie on the polar of i.e. lines Thus, the lines are concurrent at
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 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 w.r.t. is The line perpendicular to and passing through will intersect the circle in two points which will be symmetric w.r.t So let one of these two intersections be and since we note that so this leads to ie:
The polar of a point lying outside the circle is actually the chord of contact of the point with respect to
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 , find their poles and join the two corresponding points to obtain the polar of that point
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.
(The fundamental theorem of poles and polars, or The theorem of two pascals)
Take a quadrilateral which is cyclic. If then the polar of with respect to is the line joining and
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 we obtain are collinear. Again using the same in we obtain that are collinear. Therefore it is noted that the points are collinear. Denote this line as
Since passes through and which is the polar of we are done.
This uses harmonic divisions.
Let us take If the tangent to at touches it at respectively such that , denote and
It can be shown easily that and are harmonic. Then using we note that concur; and are collinear. Similarly we obtain that and are collinear. Since therefore are collinear.
So lies on the polar of w.r.t By symmetry lies on the polar of w.r.t So the intersection of these two polars will be , and the polar of this pole will be We are done!
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 is the pole of a line w.r.t then any line through is cut harmonically by and
Let be the centre of and let
Let be the circle passing through Then its centre is the midpoint of If then is the pole of So Then and are orthogonal circles, which means that is tangent to Therefore we obtain and since is the midpoint of from Theorem 1 – Harmonic Divisions, we are done.
Theorem 5 (b).
If two lines are self-conjugate lines and if then are harmonic with the two tangents drawn to the circle from
Let the poles of be if the two tangents from meet the circle in
Since the polar of passes through from La Hire’s theorem we see that the polar of passes through and similarly, Then are collinear. Now, from Theorem we are done.
Finally, if four points form a harmonic range then their polars will create a harmonic pencil.
Let be the pole of and let
If we draw four perpendicular lines perpendicular to respectively, then we have an interesting result. The polar of passes through so from La Hire’s, the polar of passes through , and therefore is the polar of with respect to our aforementioned circle. It is obvious that the pencils and are equiangular pencils, and since are harmonic, therefore are also harmonic.
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!
Let be the diameter of a circle with two points on its circumference such that is closer to than If and intersect at outside the circle, and if the tangents at to the circle meet at then show that
Let We already know that(from Theorem 4), lies on the polar of The polar of is we see that lies on the polar of So is the polar of , and therefore we are done.
Let be a tangential quadrilateral with incentre Let the opposite sides of meet each other at ie let Let its incircle touch the sides at respectively. If then show that
Note that are tangents to the incircle of so that is the polar of Similarly the polar of is Therefore the pole of the line is and we are done.
(China 2006 Western MO) is the diameter of a circle with centre is a point on extended. A line through cuts the circle with centre at is a diameter of which has centre Let Prove that lie on a circle.
Let So the polar of with respect to the circle passes through and So and it is obvious that are collinear.
So by considering the power of we have and so are concyclic points.
(IMO 1985) A circle with center passes through the vertices and of triangle and intersects segments and again at distinct points and respectively. The circumcircles of triangles and intersects at exactly two distinct points and Prove that
Note that the spiral similarity centered at which sends to and to sends the midpoint of (say, ) to that of (say, ). So is the center of unique spiral similarity that sends to
and to , and thus it follows that are concyclic. Again since so are concyclic, and is the diameter of the common circle. So, we are done.
Circles and meet at points and Circle centered at meets circles and in four distinct points and such that is a convex quadrilateral. Lines and meet at Lines and meet at Prove that
Actually this is equivalent to our last problem. Note that if then from simple angle chasing we can show that Also it is obvious that passes through All of this implies that as desired.
Let be a triangle with incentre . Reflections of in are Prove that are concurrent.
Note: This is a direct consequence of the isogonic theorem. For details, see here.
I will use a lemma.
If is the intouch triangle of and if is the incentre of and if is the triangle directly homothetic to such that ; then is perspective to
Now let us come back to our original problem.
Let be, as usual, the intouch triangle of Then let be the incircle of Since we see that the polar of with respect to is the perpendicular bisector of Similarly, the polar of w.r.t is the perpendicular bisector of and the polar of w.r.t is the perpendicular bisector of
Let these perpendicular bisectors intersect each other to form a triangle
Denote by the circles Then is the intersection of and Since is the radical axis of and and since is the radical axis of and therefore is the radical centre of
Since where is the inradius of we note that has the same power with respect to and So all lie on the radical axis of these two circles, and therefore, collinear.
From here it is also obvious that and so and are directly similar and perspective around So they are homothetic.
Applying the aforementioned lemma, we note that the intersections are collinear.
Since and are tangents to we again note that is the pole of w.r.t Also since the polar of w.r.t is it follows from the La Hire’s theorem that the pole of w.r.t is
Due to symmetry we note that are respectively the poles of We already have seen that are collinear. Therefore it is obvious that meet at the pole of w.r.t We are done.
(Polish MO Second Round 2012) Let be a triangle with and , -incentre, -circumcentre. Prove that perpendicular bisector of , line and line have a common point.
Assume that the incircle touches at and that Since therefore and are equilateral. Also therefore and the incircle of bisects
Now, we claim that is the perpendicular bisector of
Proof of claim.
Let be the reflection of in Then note that we have therefore lies on Now we already have which give leading to the fact that bisects This readily implies that bisects
Coming back to our main proof, note that we have as the perpendicular bisector of so passes through the pole of wrt Since pole of is and because is the perpendicular bisector of we are done.
(2012 Indonesia Round 2 TST 4: P2) Let be a circle with center , and let be a line not intersecting . is a point on such that is perpendicular with . Let be an arbitrary point on different from . Let and be distinct points on the circle such that and are tangents to . Let and be the foot of perpendiculars from to and respectively. Let be the intersection of and . As moves, determine the locus of .
Let be the inverse of wrt I claim is the midpoint of . Clearly, since is in the polar of then is in the polar of , and therefore are collinear. It is easy to see that lie on a circle with diameter . From this, it’s very natural to think of droping a perpendicular from to . Let be the foot of this perpendicular. By Simson’s line theorem, we have are collinear. Also, using that is cyclic we easily get The last because . Hence and we get that is midpoint of since is a right triangle. We are done.
Let incircle of be tangent to in . A line from that is parallel to meets in and a line from that is parallel to meets in . If be midpoint of , show that is on
From simple angle chasing, we see that This leads to the fact that lie on a circle with as diameter. Thus, leading to being collinear. Let be the orthocentre of Then note that polar of passes through the pole of wrt which is Also since it follows that Similarly, Therefore, the pole of wrt is Thus, and leading to
Let then note that we have is isosceles, and because therefore is the midpoint of Thence Similarly Using all these, we see that lie on a line parallel to
This problem teaches us that the polar of the orthocentre of wrt bisects and This is a very useful fact. Don’t ask me why, because even I don’t know where it may come in handy.
(Romania MOM 2012)Let be a triangle and let and denote its incentre and circumcentre respectively. Let be the circle through and which is tangent to the incircle of the triangle ; the circles and are defined similarly. The circles and meet at a point distinct from ; the points and are defined similarly. Prove that the lines and are concurrent at a point on the line .
Firstly, we’ll cite a lemma:
Let be a line and be circle.Suppose is the pole of wrt .Now let be inscribed in .Let be the circumcevian triangle of wrt .Then is the perspective axis of and .
Proof of lemma.
Now applying Pascal’s theorem on the hexagon we get are collinear.
Again and are perspective.So are collinear.Hence we conclude that are collinear.But this line is nothing but the polar of i.e .Hence we conclude that
Coming back to the main proof,
Clearly is the radical axis of , is the radical axis of and is the radical axis of .So by radical axis theorem are concurrent.
Let be the incircle of and is its intouch triangle. are the touch points of with respectively.Now let tangents at meets at respectively.Then its easy to show that is the radical axis of and .So is the pole of wrt and similar for others.So concur at the pole of wrt .Now consider the circles .Then by radical axis theorem the lines are concurrent ,say at .Then is the pole of wrt .Since are collinear, their polars i.e are concurrent.So and the triangle formed by the polars of are perspective wrt the perspective axis of .But according to our lemma this perspective axis is the polar of the perspective point of , i.e the radical axis of .So we conclude that concur at the pole of the radical axis of wrt .Since we conclude that the pole of the radical axis of lies on .Hence concur on .
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. ^_^
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.