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$H$ is a point in non-isoceles triangle $\triangle ABC$. The intersections of $AH$ and $BC$, $BH$ and $CA$, $CH$ and $AB$ are respectively $D$, $E$, $F$. $AD$, $BE$ and $CF$ cuts $(A, B, C)$ respectively at $M$, $N$ and $P$. Is there a point $H$ such that the following equality is satisfied? $$\large \frac{AH \cdot DM}{HD^2} = \frac{BH \cdot EN}{HE^2} = \frac{CH \cdot FP}{HF^2}$$

  • If there is not, prove why.

  • If there is, illustrate how to put down point $H$.

Of course, point $H$ should be one of the triangle centres identified in the Encyclopedia of Triangle Centers. But I don't which one it is.

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  • $\begingroup$ Can you tell the origin of the issue ? As it is presented, it looks a little artificial... $\endgroup$
    – Jean Marie
    Commented May 20, 2019 at 13:20
  • $\begingroup$ It's a rough translation of a problem in my class. You can edit the problem however you want, just accordingly to the figure. $\endgroup$
    – Lê Thành Đạt
    Commented May 20, 2019 at 13:28
  • $\begingroup$ Although laborious, it could be possible to place the structure onto the standard Cartesian plane centred at the origin and work out the coordinates of each point and thus their required distances. $\endgroup$
    – TheSimpliFire ♦
    Commented May 26, 2019 at 18:33

1 Answer 1

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Unfortunately, this point is probably not in ETC. If you consider the triangle with $B=(0,0)$, $C=(6,0)$, $(BC,CA,AC)=(6,9,13)$ and $A$ above the $x$ axis you will find that the point you are looking for has coordinates equal to (approximately) $(5.13198,1.12946)$ and $6$, $9$, $13$ search doesn't find the value $1.1294$.(https://faculty.evansville.edu/ck6/encyclopedia/Search_6_9_13.html)

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  • $\begingroup$ Is it unique? Because if not, it might not even be a triangle center. $\endgroup$
    – darij grinberg
    Commented May 30, 2019 at 20:39
  • $\begingroup$ I think it is. For a fixed $a$ the locus of points for which $\frac{AH \cdot DM}{HD^2}=a$ is the parabolic-looking curve through $B$ and $C$. It seems like there is only one value of $a$ for which all three curves intersect at one point and for such a value of $a$ the point is unique. But I don't have any rigorous proof nor any idea on how it can look like. $\endgroup$
    – Bartek
    Commented May 30, 2019 at 21:13
  • $\begingroup$ Now I see that it is actually a conic. $\endgroup$
    – Bartek
    Commented May 30, 2019 at 21:37
  • $\begingroup$ You should submit the new "triangle centre" to ETC (how to construct the point in the answer). $\endgroup$
    – Lê Thành Đạt
    Commented May 31, 2019 at 9:29

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