Tangents to the Incircle

Posted on by Lasitha Jayasinghe

Today we shall look at a rather interesting configuration which has appeared in only a handful of problems to date with rather surprising and occasionally obscure results. Our work shall explore the tangents from the point T which is defined as the intersection of the tangents from B and C to the circumcircle of \triangle ABC.

As usual, notation is borrowed from the diagram attached above, with most familiar points carrying their standard labels. In case of an unclear definition, the above diagram is linked to a GeoGebra file which can be referred.

A summary of key results pertaining to this configuration,

  1. First, the points P = \omega \cap DM and Q = \omega \cap DN are the tangency points from T to the incircle \omega.
    • Pascal’s Theorem on concyclic hexagons AACMBA and ACCMBB show that points S, T , AC \cap MB and CM \cap AB all lie on the same line.
    • Pascal’s Theorem on concyclic hexagons DFFEPP and DDFEEP show that points M, C , EP \cap DF and FF \cap PP are collinear. Thus, the line through P and CM \cap AB is tangent to \omega at P.
    • Finally, Pascal’s Theorem on concyclic hexagons DEEFPP and DDPFFE implies that points M , B , PP \cap AC and DE \cap FP are collinear.
    • Combining all these applications we conclude that the tangent to \omega at P contains points S and T and passes through BM \cap AC.
  2. Let T_1 and T_2 be the tangents from M to \omega. Since M lies on the polar of A with respect to \omega, by La Hire’s A lies on line \overline{T_1T_2}.
    • Projecting from A onto the circumcircle we note that MBK_nC is harmonic, so points M , K_n and T are collinear.
    • Further, since EF \cap BC is the harmonic conjugate of D with respect to segment BC a projection at N implies that K_n lies on \overline{ND}.
  3. Pascal’s Theorem on concyclic hexagon DDEFFP yields that R_b = PF \cap DE lies on \overline{MB}.
    • By Brokard’s Theorem on cyclic quadrilateral DEFP it follows that R_b lies on the polar of M with respect to \omega. Thus, lines \overline{DE} , \overline{PF}, \overline{MB} and \overline{T_1T_2} all concur at R_b.
    • This implies that (R_bR_c;T_1T_2)=-1 and projecting through M onto the circumcircle, we have that BL_1CL_2 is harmonic.
    • In particular, we have that \overline{AK_n} and \overline{AK_m} are the polars of M and N with respect to \omega respectively, which is the key result in Shortlist 2019 G6.
  4. By Poncelet’s Porism it follows that L_1L_2 is tangent to \omega. Combining this with our previous observation it follows that L_1 and L_2 lie on the tangent from T to \omega, solving this intriguing unsourced problem.
  5. Since (B,C;A,AT \cap (ABC))=-1, projecting through A we have that Y = AT \cap EF lies on the polar of Z – the intersection of the tangent to (ABC) at A and EF, with respect to the incircle.
    • Since T lies on the polar of Z, by La Hire’s it follows that Z lies on the polar of T with respect to \omega which implies that the A-tangent to (ABC), line \overline{EF} and \overline{PQ} concur at Z.
    • Thus, Y also lies on the polar of S with respect to the incircle, implying that P', the second tangency point from S to \omega lies on \overline{PY}.
    • Hence, projecting through P we note that P'EQF is harmonic which also implies that points A, P' and Q are collinear. This immediately also implies that lines SP' , TQ and EF concur at R which solves 2025 December MEOW Problem 6.
  6. Applying Pascal’s Theorem yet again on concyclic hexagon AACBBM and ACCBMM implies that points AS \cap BT , SM \cap TC and MB \cap AC are collinear and thus points A,C,T,B,M and S are conconic.
    • A further application of Pascal’s Theorem shows that R and MD \cap AQ also lie on this conic.
    • Further, sets of points A,B,C,Q,K_n,T and A,B,C,P,K_m,T are also conconic which is easy to confirm via Pascal’s Theorem.
  7. Let H denote the foot of the altitude from I to the A-tangent. Since \overline{AS} is the polar of Y with respect to the incircle, H is the inverse of Y under an inversion about the incircle.
    • Hence, YH \cdot YI = YE \cdot YF which implies that quadrilateral AHEF is cyclic. Further, ZH \cdot ZA = ZF \cdot ZE = ZP \cdot ZQ which implies that quadrilateral AHPQ is also cyclic.
    • Since the pedal circles of isogonal conjugates are well known to coincide, this implies that J = TI \cap AO is the isogonal conjugate of I with respect to \triangle TSS' since \overline{TI} is clearly the internal \angle STS'-bisector. This wraps up 2024 Vietnam TST Problem 5.
    • Incidentally, this also implies that S is the center of spiral similarity mapping segment AJ to P'I.
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Published by Lasitha Jayasinghe

3 thoughts on “Tangents to the Incircle”

  1. Nice work! I was somehow thinking to one day have an own problem using tangents of an incircle since I haven’t saw it on the problems I solved. I still couldn’t understand some of them (like Poncelet or things like that), but its beautiful config I think.

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  3. Though the config is somewhat purely projective, this is surely an interesting exercise to practice using and combining various tools in the field of projective geometry, also for the eyes of the students to look at bigger picture and learn to add points.

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