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In an acute angled triangle ABC, let H be the orthocenter, and let D, E, F be the feet of altitudes from A, B, C to the opposite sides, respectively. Let L, M, N be midpoints of segments AH, EF, BC, respectively. Let X, Y be feet of altitudes from L, N on to the line DF. Prove that XM is perpendicular to MY.

This is the fifth problem from RMO(Regional Mathematics Olympiad) 2019. I looked at the answer key and understood it here. However, being a geometry question the solution given seems to work for only one configuration.

Here is my configuration:

enter image description here

I tried to adapt the solution given to this configuration but I have not found anything yet.

Your inputs are most welcome. Thank you!

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  • $\begingroup$ Why did you make the $\triangle ABC$ acute? According to the source, it can be any scalene triangle. What is the configuration, for which the given solution works? $\endgroup$
    – YNK
    Commented Dec 5, 2023 at 7:50
  • $\begingroup$ @YNK In the question paper it is an acute triangle, though the answer key has omitted that condition. The configuration with its solution is linked in the question post. $\endgroup$
    – Shreya Mundhada
    Commented Dec 5, 2023 at 8:37
  • $\begingroup$ Thanks for replying quickly. Can you describe the difference between your configuration and theirs? I see that they have interchanged the positions of the points $X$ and $Y$ while drawing their configuration. That is a minor matter and does not affect the proof. $\endgroup$
    – YNK
    Commented Dec 5, 2023 at 9:48
  • $\begingroup$ You are right. Sorry for the diagram earlier. I have edited my question and added the configuration I meant. $\endgroup$
    – Shreya Mundhada
    Commented Dec 5, 2023 at 10:06
  • $\begingroup$ Solution posted by @Sahaj successfully addresses the difficulty you faced when you tried to adapt the solution given in the answer sheet to your configuration. As Sahaj has pointed out in his post, the solution given in the answer sheet can be used for your configuration only with modifications. This is because the cyclic quadrilaterals $YFLM$ and $FXNM$ appear as the cyclic quadrilateral $FXML$ and $FMYN$ respectively in your configuration. $\endgroup$
    – YNK
    Commented Dec 5, 2023 at 22:16

1 Answer 1

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The solution is almost correct and holds for this geometry too but with minor modification.

A brief idea of the proof:

First show that $N, \ M, \ L$ are collinear points.

Also, $FXML$ and $FMYN$ are cyclic quadrilaterals.

Then it is clear that $\angle FYM = \angle FNM$ and $\angle FLM = \pi - \angle FXM$.

Now it follows that $\angle MXY = \pi - \angle FXM = \angle FLN$ and $\angle XYM = \angle FNM$.

Hence $\Delta XMY \sim \Delta LFN$.

You need to show $\angle LFN = \frac{\pi}{2}$ which follows easily since $\angle AFL = \angle CFN$ and hence the result.

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  • $\begingroup$ Feel free to ask if you need hints/clarification/explanation for any part. $\endgroup$
    – Sahaj
    Commented Dec 5, 2023 at 15:56

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