Apologies if this isn't the right place to answer this. What are the HM points and their isogonal conjugates, also known as the Humpty and Dumpty points? I found some information via a google search, but there wasn't much on line that was a comprehensive overview. I ask, now, for a short list of properties of these points.

I am asking this because I came across an old USAMO problem (2008/2) and a friend told me it was very easy by noticing that a point (F) was in fact that A-dumpty point. This made me curious, so I decided to ask for some resources here.

thanks for your time.


1 Answer 1


Taken from Kapil Pause's note "On Two Special Points in a Triangle" (PDF link via wordpress.com)

Humpty point.

Definition 1. In $\triangle ABC$ the $A$-Humpty point $P_A$ is defined to be a point inside triangle such that $\angle P_ABC = \angle P_AAB$ and $\angle P_ACB = \angle P_AAC$.

Facts about $P_A$:

  1. lies on $A$ median of $\triangle ABC$
  2. lies on $A$ appolonius circle of $\triangle ABC$; that is, $\frac{AB}{AC} =\frac{P_AB}{P_AC}$.
  3. $B$, $P_A$, $H$, $C$ are concyclic, where $H$ is the orthocentre of $\triangle ABC$.
  4. $HP_A \perp AP_A$

Dumpty point.

Definition 2. In $\triangle ABC$, the $A$-Dumpty point $Q_A$ is defined to be a point inside [the] triangle such that $\angle Q_ABA = \angle Q_AAC$ and $\angle Q_AAB = \angle Q_ACA$.

Facts about $Q_A$:

  1. lies on $A$ symmedian
  2. It is the centre of spiral symilarity [sic] sending $\triangle A_QAC$ to $\triangle CQ_AB$; ie, sending $AC$ to $BA$.
  3. $B$, $Q_A$, $O$, $C$ are concyclic where $O$ is circumcentre of $\triangle ABC$.
  4. $OQ_A \perp AQ_A$

Also notice that the Humpty Dumpty points are isogonal conjugates of each other.

We recommend you to try to prove these facts by yourself. This will help you in befriending the humpty-dumpty points well

Back to your question, which is also addressed in Pause's note:

Question: (USAMO 2008). $ABC$ is a triangle. $M$, $N$ are respectively the midpoints of $AB$, $AC$. The perpendicular bisector of $AB$, $AC$ meet the $A$ median at $D$, $E$ respectively. $BD$ and $CE$ meet at $F$. Prove that $A$, $M$, $F$, $N$ are concyclic.

Proof: Ah, yes $F$ is none other than the $A$−Dumpty point of $ABC$. To reach the desired conclusion, see that $OF\perp AF$, $ON \perp AC$, $OM \perp AB$ directly implies $A$, $M$, $O$, $N$, $F$ are concyclic with $AO$ as diameter. Another way to finish would have been to see that since a spiral similiarity centered at $F$ maps $AC$ to $BA$, it also maps the midpoint of $AB$ to midpoint of $AC$ that is, $M$ to $N$. The angle of spiral similarity is $\angle AFC$ which is $180^\circ − A$, and this must also be the $\angle MFN$ and we finish.

  • 2
    $\begingroup$ I have added a link to the source of the text of your solution, to provide proper credit. (I also cleaned-up some formatting.) Please refrain from simply copy-pasting other people's work and presenting it as if it is your own (as you also appear to have done with this answer). Plagiarism is bad! $\endgroup$
    – Blue
    Nov 5, 2020 at 10:51

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