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Here's the sketch:

enter image description here

From the inner point P of a triangle ABC the three connecting lines to the corner points are drawn. In addition, the lines PE, PD and PF are each drawn parallel to a median of ABC. Show that the grey areas cover half of the triangular area.

I thought about using ceva's theorem, but I don't really know how to start... any ideas?

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    $\begingroup$ I don't see any parallel line in this triangle!!! $\endgroup$
    – BarzanHayati
    Commented Aug 21, 2019 at 20:49
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    $\begingroup$ It's a sketch... sorry! $\endgroup$
    – Analysis
    Commented Aug 21, 2019 at 21:07
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    $\begingroup$ An ugly solution can be obtained by (1) applying an affine transformation that makes the triangle $ABC$ equilateral (thus $PD$, $PE$ and $PF$ become orthogonal to $BC$, $CA$, $AB$, respectively), and then (2) arguing that the alternating sum of the six areas is a quadratic function in the Cartesian coordinates of $P$ and therefore must vanish for all points $P$ if it vanishes for the vertices of the triangle $ABC$ and the midpoints of its sides (which is easy to check). I like neither of these two steps, though. $\endgroup$
    – darij grinberg
    Commented Aug 23, 2019 at 11:37
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    $\begingroup$ @ParabolicAlcoholic: Consider a quadratic function $f$ on the plane (i.e., a map $f$ from the plane to the real numbers that is given by a quadratic polynomial when $P$ is written in Cartesian coordinates). Assume you want to prove that $f = 0$ on the whole plane. Since $f$ is quadratic, you know that if $f$ vanishes on three points on a given line, then it must vanish on the whole line. Thus, it suffices to prove that $f$ vanishes on points on the lines $BC, CA, AB$ (because for any point $P$ not on these lines, you can ... $\endgroup$
    – darij grinberg
    Commented Aug 24, 2019 at 7:33
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    $\begingroup$ ... find a line through $P$ that intersects the three lines $BC, CA, AB$ in distinct points, and then $f$ must vanish on this line by the argument above and therefore, a fortiori, vanish on $P$). But this, in turn, only requires showing that $f$ vanishes on three points on each of these lines $BC, CA, AB$. One simple choice for these points is $A, B, C$ and the midpoints of $BC, CA, AB$. $\endgroup$
    – darij grinberg
    Commented Aug 24, 2019 at 7:33

3 Answers 3

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Ronel Leker's solution.

Let $\{F_1,F_2\}\subset AC$, $\{E_1,E_2\}\subset BC$ and $\{D_1,D_2\}\subset AB$ such that

$F_1E_2||AB$, $D_2E_1||AC$, $D_1F_2||BC$ and $F_1E_2\cap D_2E_1\cap D_1F_2=\{P\}.$

Thus, since $$\Delta PE_2E_1\sim\Delta F_1PF_2\sim\Delta D_2D_1P\sim\Delta ABC,$$ we obtain that $PE$, $PF$ and $PD$ are medians of $\Delta PE_1E_2,$ $\Delta PF_1F_2$ and $\Delta PD_1D_2$ respectively.

Also, since $AF_1PD_2$, $BD_1PE_2$ and $CE_1PF_2$ are parallelograms, we obtain: $$S_{\Delta PAD}+S_{\Delta PBE}+S_{\Delta PCF}=$$ $$=\left(S_{\Delta PAF_1}+S_{\Delta PDD_1}\right)+\left(S_{\Delta PBD_1}+S_{\Delta PEE_1}\right)+\left(S_{\Delta PCE_1}+S_{\Delta PFF_1}\right)=$$ $$=\left(S_{\Delta PAF_1}+S_{\Delta PFF_1}\right)+\left(S_{\Delta PBD_1}+S_{\Delta PDD_1}\right)+\left(S_{\Delta PCE_1}+S_{\Delta PEE_1}\right)=$$ $$=S_{\Delta PAF}+S_{\Delta PBD}+S_{\Delta PCE}$$ and we are done!

enter image description here

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    $\begingroup$ +1. I thought it could use a picture. I hope you don't mind. :) $\endgroup$
    – Blue
    Commented Aug 26, 2019 at 5:07
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    $\begingroup$ Thank you dear @Blue ! $\endgroup$
    – Michael Rozenberg
    Commented Aug 26, 2019 at 5:10
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COMMENT:

Let the perpendiculars from p on AB , BC and AC be $i, j , k$ respectively. Using relation:

$\frac{i}{h_C}+\frac{j}{h_B}+\frac{k}{h_A}=1$

where $h_c$, $h_B$ and $h_A$ are hights of triangle from vertices C, B and A respectively; we can find following relations:

$S_{ABC}=S_{PAD}.\frac{AB}{AD}+S_{PFC}.\frac{AC}{FC}+S_{PEB}.\frac{BC}{EB}$

$S_{ABC}=S_{PDB}.\frac{AB}{DB}+S_{PFA}.\frac{AC}{AF}+S_{PEc}.\frac{BC}{EC}$

If P is the intersection of medians then the question is answered. If P is not intersection of medians then we conclude that RHS of these relations is symmetric and therefore they must be equal, hence the question is answered.

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    $\begingroup$ What do the perpendiculars have to do with anything? and induction?? $\endgroup$
    – darij grinberg
    Commented Aug 23, 2019 at 11:30
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The way I would approach this problem is to do an affine transformation into an equilateral triangle, then use Carnot's theorem$$AD^2 - DB^2 + BE^2 - EC^2 + CF^2 - FA^2 = 0$$to get$$AD - DB + BE - EC + CF - FA = 0,$$which is the necessary and sufficient condition for the existence of $P$. Then we just have to express the ratio $PD:PE:PF$ in terms of the above lengths (which frankly I did not bother, but know it is possible). It is probably equivalent to darij grinberg's solution in the comments.

Because this and darij grinberg's solution makes use of the non-affine property of equilateral triangles, while the question is affine-invariant, the solution seems "morally wrong". That is why Ronel Leker's solution, by way of Michael Rozenberg here, seems more elegant. I also think that in darij grinberg's solution, observing that the expression is quadratic without actually computing the messy algebra is very elegant.

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