1. Let $ABC$ be an acute-angled triangle. The feet of the altitudes from $A,B$ and $C$ are $D, E$ and $F$ respectively. Prove that $DE +DF \le BC$ and determine the triangles for which equality holds. The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined.

Thanks in advance for any contributions.

  • $\begingroup$ Did you mean $DE +DF \ge BC$? $\endgroup$
    – Stefan4024
    Commented Mar 9, 2015 at 19:27

1 Answer 1


$\Delta DEF$ is the orthic triangle of $\Delta ABC$, and it is well-known that $DF=b\cos B$ and $DE=c\cos C$. Also using the well-known projection formulae, $a=b\cos C+c\cos B$. So the inequality is equivalent to: $$b\cos B + c\cos C \le b\cos C+c\cos B$$ $$(b-c)(\cos B - \cos C) \le 0$$

This inequality is easily proved because $b\le c \implies \sin B \le \sin C \implies \cos B \ge \cos C$. Similar procedure applies for the $b \ge c$ case.

The equality clearly holds in isosceles triangles. $A,B,C$ and $a,b,c$ have usual meaning.

For info on orthic triangle, click here

  • $\begingroup$ Many thanks. Do you think perhaps though there is another solution? I had made some progress in fashioning a cyclic quadrilateral with its base as the diameter of its circumcircle and the BMO1 does not tend to demand any theorems that are not taught at secondary school. $\endgroup$ Commented Mar 9, 2015 at 22:25
  • $\begingroup$ Can you tell me what all you know? I'll try to use only that $\endgroup$
    – najayaz
    Commented Mar 10, 2015 at 3:52
  • $\begingroup$ ukmt.org.uk/docs/BMO%20Preparation%20Sheet.pdf Here is what they state is satisfactory to know for the tests. As you can see the requirements for BMO1 are very basic, e.g. alternate angle theorem, opposite angle theorem. It does mention that you should know some properties of orthic circles for the BMO2 and it may well be the case that since this is the hardest question on the paper they require a bit extra knowledge. Thank you anyway though. $\endgroup$ Commented Mar 10, 2015 at 7:29
  • $\begingroup$ @MadChickenMan , So should I or should I not think of something else? $\endgroup$
    – najayaz
    Commented Mar 10, 2015 at 10:22
  • $\begingroup$ It's up to you, it seems unlikely that what you gave wouldn't be the solution, but I am curious as to whether there is another solution. If you think that your solution is probably the only one then don't bother. $\endgroup$ Commented Mar 10, 2015 at 17:12

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