Miquel point of a quadrilateral formed by the H-tangent to (BHC)

Posted on by Lasitha Jayasinghe

Today we shall have a look at another orthocentric configuration, that of the Miquel point of the quadrilateral formed by the intersections of the tangent to circle (BHC) at H with the sides of \triangle ABC. Many thanks to Yangqin Yang for contributing several proofs and for saving me from crashing out.

As usual, points follow their standard labels (except for the renaming of the A- Humpty point to H_M) and a GeoGebra link is embedded to the diagram below for closer inspection.

A summary of the key results pertaining to this configuration,

  1. First, an easy angle chase shows that quadrilateral BCPQ is cyclic. Hence, by Radical center theorem on circles (ABC), (BCPQ) and (APQ) it follows that lines AK, PQ and BC concur at Z.
    • Since it is well known that the A-diameter of (ABC) and H-diameter of (BHC) are parallel, it follows that AA' \perp PQ.
  2. Reflection across the line \overline{BC} implies that ZH_a is tangent to (ABC) at H_a and since it is well known that (Q_aH_a;BC)=-1 we have that Z is the pole of Q_aH_a with respect to (ABC).
    • In particular, this implies that quadrilateral AQ_aKH_a is also harmonic and hence projecting onto line \overline{BC} through A we have that the A-tangent to (ABC) intersects BC at precisely the reflection of D across Z.
  3. It is not hard to see that the Miquel point of \triangle DPQ with respect to \triangle ABC must lie on the A-altitude, so quadrilaterals BDSQ and CDSP are also cyclic.
    • Since BCPQ is cyclic, it is well known that R lies on the rectangular circumhyperbola of \triangle ABC passing through A'. Pascal’s theorem on coconic hexagon A'RBHAC implies that lines \overline{AH \cap RA' , P}, BH and CA' concur, which since the latter two are both perpendicular to side AC implies that S = AH \cap RA' and hence points R, S and A' must be collinear.
    • It is not hard to see that point K must also lie on this line due to the right angles. Further, if J = (PQD) \cap BC \ne D then AKDJ is also cyclic by PoP at Z so it follows that J also lies on this line due to the right angles.
  4. Quadrilaterals BH_aHP and CH_aHQ are also cyclic which by an easy angle chase shows that the midpoint N of segment AS lies on (PQH_a).
  5. Now, consider the inversion centered at A with radius \sqrt{AS \cdot AD}. This inversion swaps the pairs (Q, B), (P, C). (K,Z) and (H, H_a) due to the cyclic quadrilaterals noted above.
    • Thus, this inversion swaps line \overline{PQ} with circle (ABC) so AU=AV = \sqrt{AS \cdot AD} which solves the following easy problem on AoPS.
    • Also, this inversion swaps circles (BHC) and (PQH_a) so the intersections of these circles remain fixed under inversion, implying that AE=AF = \sqrt{AS \cdot AD}. Combining this with the previous result it follows that quadrilateral UVFE is cyclic with center A. Also, Radical center now implies that line EF passes through Z.
    • Since HH_MMD is well known to be cyclic, it follows that HH_aWH_M is too by Reim’s theorem. Thus, this inversion swaps points H_M and W. In particular, this implies that W lies on (PQH_a) since it is well known that H_M lies on (BHC).
    • As we noted before that AA' \perp PQ, it is clear that this inversion also swaps A' and the point AA' \cap PQ. Thus, \overline{A'WH_a} which is the line through A' parallel to BC maps to the circle with diameter AH implying that Y = A'H_a \cap PQ must map to the A-Queue point.
  6. Note that WP and WQ are tangent to (APQ). Thus, W lies on the polar of H with respect to (APQ). Furthermore, since we showed that Y lies on AQ_a we have (YH;QP)=-1 implying that Y also lies on the polar of H and thus, \overline{H_aA'} is the polar of H with respect to (APQ).
    • But then L = AK \cap H_aA' lies on the polars of both T and H with respect to (APQ) which by La Hire’s implies that \overline{TH} is the polar of L with respect to this circle.
    • It is not hard to check that \frac{H_aN}{AH_a} = \frac{DZ}{DX_a} so it follows that lines NL and AY are parallel. In particular, this means that NL \perp HQ_a which must imply that \overline{QH_a} is the polar of L with resepct to (APQ). This implies that TH bisects segment BC, which solves 2023 Kazakhstan Junior MO Grade 9 Problem 6.
  7. Let \mathcal{H} denote the rectangular circumhyperbola of \triangle ABC passing through A'. We noted before that R lies on \mathcal{H}. The parallelogram point A_1 also lies on this hyperbola.
    • It is easy to show via lengths that the line through AQ_a \cap CH and AB \cap Q_aH is parallel to BC. Hence, points YA' \cap BC, AY \cap CH and AB \cap HA' are collinear which by the converse of Pascal’s theorem implies that Y must also lie on \mathcal{H}.
    • It is also not hard to see that points A, BT \cap QQ_a and CT \cap PQ_a are collinear implying that the points T, B, C, P, Q and Q_a are coconic.
  8. Since (AA';BC)= (T_aH_a;CB) by reflection, it is clear that lines KA' and H_aT_a intersect on BC. Thus, points H_a, J and T_a are collinear. By reflection across the line BC it also follows that points H, J and A_1 are collinear.
    • The inversion we mentioned before shows that KZH_aH is cyclic, on which J must also lie since ZH_aJH is a kite due to symmetry.
    • By PoP at J it then follows that quadrilateral KHA'A_1 is cyclic. Further, the above observation implies that Y must also lie on this circle by Reim’s theorem.
    • Also by Reim’s it further follows that Q_aJMT_a is cyclic, which implies that lines Q_aT_a, BC and YA_1 concur.
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Published by Lasitha Jayasinghe

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