Did I fake solve it? Sharygin P13 wrote: Eight points in a general position are given in the plane. The areas of all 56 triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols "+" and "−" between them in such a way that the obtained sum is equal to zero. We begin with a lemma. Lemma: Given $ABCD$ a quadrilateral, we consider areas of the triangle. Then it is possible to insert the symbols $+$ and $-$ between them in such a way that the obtained sum is equal to zero. Proof: Note that $[ABD]+[ABC]=[ACD]+[ABC].$ So assign $[ABD],[ABC]$ as positive and $[ACD],[ABC]$ negative. Claim: Given $8$ points $ABCDEFGH,$ we can choose $17$ tuples of four points, such no two tuples share three points. Proof: Just consider the following contruction. $$ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$$$$BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH $$ Let the eight points be $A,B,C,D,E,F,G,H.$ Then consider the following quadrilaterals $$ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$$$$BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH. $$Then note that by $\textbf{Lemma 1.3}$ we can assign sign to the areas of each of the triangles of any of quadrilateral such that sum is $0.$ Now note since no to quadrilateral share three vertices, we get that $$[ABCD]+[ABEF]+[ABGH]+[ACEG]+[ACEF]+[ACFH]+[ADEH]+[ADFG]+$$$$[BCEH]+[BCFG]+[BDEG]+[BDHF]$$$$+[CDEF]+[CDGH]+[EFGH]=\text{ Sum of all } 56 \text{ triangles.}$$Let the $56$ triangles arranged in a row be $T_1,T_2,\dots T_{56}.$ So the sign of $T_1$ is positive, then we can proceed by noticing where $T_1$ belongs to from the $8$ quadrilateral i.e $ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$ $BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH. $ And then proceed with assigning the signs ( using the two above lemmas). EDIT: Huge typos so really sorry for that

jelena_ivanchic wrote: Did I fake solve it? Sharygin P13 wrote: Eight points in a general position are given in the plane. The areas of all 56 triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols "+" and "−" between them in such a way that the obtained sum is equal to zero. I have also done the same but i am also in dilemma.

Definition: The sign of a triple $S(a, b, c)$ is $-$ if the value $|a-1| + |b| + |c-1|$ is odd and $+$ otherwise. So for each triangle $(x_i, y_i), (x_j, y_j), (x_k, y_k)$, give the area $[ P_iP_jP_k]$ the sign $S(i, j, k)$ (Note that here the area itself is also signed, i.e., if $P_iP_jP_k$ are clockwise, their area is negative after which the sign is multiplied). Now we write the areas in term of the points' $x$ and $y$ coordinates via the determinant form of the area. It suffices to show that all occurrences of a specific term (WLOG say $x_iy_j$ vanish). But this is easy as fixing $P_i$ and $P_j$ in our triangle $\triangle P_iP_jX$, and summing over all possible values of $X$ ($X = P_1, P_2, ... P_8$ as the degenerate triangles contribute no area), the sign alternates for adjacent points by definition. Because there are an even number of points ($8$), there are an equal number of $+x_iy_j$ as there are $-x_iy_j$. Doing this for all pairs $x_\alpha y_\beta$, we get the signed area sum is zero! ************************************* The only thing to take care of here is that all permutations of 3 points have the same signed area, which is easily shown. What I mean by this is the $S(a, b, c)[P_aP_bP_c]$ must be the same as $S(c, a, b)[P_cP_aP_b]$ and so on for all permutations of $a, b, c$

Same as @above, but the quadruplets have to be $14$, not $17$... Let's label the points as $A_{1},A_{2},\ldots ,A_{8}$. Consider four of those points, say, $A_{i_{1}}$,$A_{i_{2}}$,$A_{i_{3}}$,$A_{i_{4}}$ where the indices are different numbers from $1$ to $8$. We'll prove that we can put either a $+$ or a $-$ in front of the area of the 4 triangles, which are formed by three of those points, so that the sum of the thus formed numbers is $0$. We can examine two cases: $\textbf{Case 1:}$ The four points form a concave quadrilateral, therefore one of the $4$ points lies inside the triangle formed by the other three. WLOG let $A_{i_{4}}$ lie inside $\triangle A_{i_{1}}A_{i_{2}}A_{i_{3}}$, but then: $$S_{A_{i_{1}}A_{i_{2}}A_{i_{3}}}=S_{A_{i_{1}}A_{i_{2}}A_{i_{4}}}+S_{A_{i_{2}}A_{i_{3}}A_{i_{4}}}+S_{A_{i_{3}}A_{i_{1}}A_{i_{4}}}$$$$\Longrightarrow S_{A_{i_{1}}A_{i_{2}}A_{i_{3}}}-S_{A_{i_{1}}A_{i_{2}}A_{i_{4}}}-S_{A_{i_{2}}A_{i_{3}}A_{i_{4}}}-S_{A_{i_{3}}A_{i_{1}}A_{i_{4}}}=0$$$\textbf{Case 2:}$ If the $4$ points form a convex quadrilateral, then: $$S_{A_{i_{1}}A_{i_{2}}A_{i_{3}}}+S_{A_{i_{3}}A_{i_{4}}A_{i_{1}}}=S_{A_{i_{1}}A_{i_{2}}A_{i_{3}}A_{i_{4}}}=S_{A_{i_{2}}A_{i_{3}}A_{i_{4}}}+S_{A_{i_{4}}A_{i_{1}}A_{i_{2}}}$$$$\Longrightarrow S_{A_{i_{1}}A_{i_{2}}A_{i_{3}}}+S_{A_{i_{3}}A_{i_{4}}A_{i_{1}}}-S_{A_{i_{2}}A_{i_{3}}A_{i_{4}}}-S_{A_{i_{4}}A_{i_{1}}A_{i_{2}}}=0$$Therefore we've shown that in any quadruplet the $4$ areas of the triangles can be set with $+$ or $-$, so that the sum is zero. Now notice that the $56$ areas of the triangles can be partitioned into $14$ such quadruplets: $$\{1,2,3,4\} \text{ }\{5,6,7,8\} \text{ }\{1,2,5,6\} \text{ }\{2,3,6,7\} \text{ } \{3,4,7,8\} \text{ }\{1,2,7,8\} \text{ } \{3,4,5,6\}$$$$\{1,3,5,7\} \text{ }\{2,4,6,8\} \text{ }\{1,3,6,8\} \text{ }\{2,4,5,7\} \text{ }\{2,3,5,8\} \text{ }\{1,4,6,7\} \text{ }\{1,4,5,8\}$$Here the number $x$ represents the point $A_{x}$, so if $x,y,z$ are in a set, then we consider $\triangle A_{x}A_{y}A_{z}$. Every triangle appears exactly in one set, so we can divide the sum of all $56$ areas into $14$ quadruplets, for which we showed how to make the sum $0$ with the appropriate signs. Of course, we have to be careful, as in one group the signs will be predetermined because we can't put a minus before the first area ( in the problem statement we can see that we can put signs only between the numbers and not in front of the first number). This, however, is not a problem, because we can just change every sign for the quadruplet of the triangle, which area is in first if the sign we've given it in the process is a $-$ and since the sum of the areas of the $4$ triangles with the appropriate signs in this quadruplet is $0$ as we proved, then changing all the signs won't change this and therefore the whole sum of all $56$ areas will stay $0$ while the first sign would be a $+$, but we don't need to write a plus before the first area, because it's a plus by default, so we won't put a sign before the first area in the line. Thus the problem is solved.

This problem was also Iran MO 3rd Round 2018 C2 which has another thread here.