Let $ABC$ be a triangle with orthocenter $H$. Let $AB = 84$, $AC = 32\sqrt{3}$ and $\angle{BAC} = 60^\circ$. Let $D, E, F$ be points on $AH, BH, CH$, respectively, such that: $7AD = AH$, $7BE = BH$, $7CF = CH$. Find $AB\cdot CF + BC\cdot AD + AC\cdot BE$.

This problem is taken from a math competition and you should be able to solve it without the aid of a calculator; I managed to find both $CF$ and $BE$, using similar triangles. However I'm stuck with $AD$: the expression that leads to the evaluation of $AD$ is simply too difficult (a lot of square roots), so I was wondering if someone could give me a hint or sketch the solution. Any help would be greatly appreciated.


On each point $A, B, C$ of the triangle, pass through it a line parallel to its opposite side. Then, these 3 new lines form a new triangle, similar to the original one. the orthocenter of $ABC$ coincide with the circumcenter of the new triangle. Notice that $$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = AB \cdot CH + BC \cdot AH + AC \cdot BH$$ $$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = \frac{2AB \cdot CH}{2} + \frac{2BC \cdot AH}{2} + \frac{2AC \cdot BH}{2}$$ Now, the sides of the new triangle are given by $2AB$, $2BC$ and $2CA$, so the expression above corresponds to its area. Thus, we have $$7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = 4 \text{Area}(ABC) = 4 \frac{84 \cdot 32\sqrt{3} \sin 60^\circ}{2} = 2 \cdot 84 \cdot 32 \cdot \frac{3}{2}$$ implying that $$ AB\cdot CF + BC\cdot AD + AC\cdot BE = 12 \cdot 32 \cdot 3 = 9 \cdot 128 = 1152.$$

  • $\begingroup$ Thank you. Why did you think about constructions? $\endgroup$ – Dude Dec 5 '16 at 20:47
  • $\begingroup$ The construction I described is rather classical... I have seen it before, but I am not sure where $\endgroup$ – Daniel Dec 5 '16 at 20:55
  • $\begingroup$ Seeing the expression I just guessed that there would be an area relation, since there were perpendiculars to the sides and the area of ABC was easy to compute $\endgroup$ – Daniel Dec 5 '16 at 20:56

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