Collinearity with the Orthocenter
What Is This About?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet below provides an illustration to a problem from an outstanding collection by T. Andreescu and R. Gelca (also from an earlier Russian book by I. Sharygin):

Let ABC be an acute triangle. The Points M and N are taken on the sides AB and AC, respectively. The circles with diameters BN and CM intersect at points P and Q. Prove that P, Q, and the orthocenter H are collinear.


Let ABC be an acute triangle. The Points M and N are taken on the sides AB and AC, respectively. The circles with diameters BN and CM intersect at points P and Q. Prove that P, Q, and the orthocenter H are collinear

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

A clue to the solution lies in the observation that the feet of the altitudes from two vertices and these vertices themselves are concyclic; the circumcircle of the four points has the side joining the two vertices as a diameter.

So, let BHB and CHc be the altitudes to AC and AB, respectively. The orthocenter H is the intersection of the altitudes BHB and CHc. By the Intersecting Chords Theorem

BH × HHb = CH × HHc.

We have two circles, say, C(BN) and C(CM), with diameters BN and CM respectively. Assume line PH intersects C(BN) in Qb, and C(CM) in Qc. We want to show that Qb = Qc.

But in C(BN),

BH × HHb = PH × HQb

while in C(CM),

CH × HHc = PH × HQc.

By transitivity of equality, it follows that

PH × HQb = PH × HQc,

and subsequently, HQb = HQc, implying Qb = Qc, so that both coincide with Q.

Note that, as the solution shows, the requirement that ΔABC be acute is entirely spurious. Note also that the statement just proved is equivalent to the assertion that the radical axis of the circles on two cevians as diameters passes through the orthocenter of the triangle, making the orthocenter the radical center of three such circles.

References

  1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.3.8 (p. 12)
  2. I. F. Sharygin, Problems in Geometry. Plane geometry, Nauka, Moscow, 1982, p. 46, problem 188

Power of a Point wrt a Circle

  1. Power of a Point Theorem
  2. A Neglected Pythagorean-Like Formula
  3. Collinearity with the Orthocenter
  4. Circles On Cevians
  5. Collinearity via Concyclicity
  6. Altitudes and the Power of a Point
  7. Three Points Casey's Theorem
  8. Terquem's Theorem
  9. Intersecting Chords Theorem
  10. Intersecting Chords Theorem - a Visual Proof
  11. Intersecting Chords Theorem - Hubert Shutrick's PWW

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Copyright © 1996-2017 Alexander Bogomolny

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