There are 8 points in the plane.we write down the area of each triangle having all vertices amoung these points(totally 56 numbers).Let them be $a_1,a_2,\dots a_{56}$.Prove that there is a choice of plus or minus such that: $$\pm a_1 \pm a_2 \dots \pm a_{56}=0$$
Problem
Source: 2018 Iran national olympiad third round combinatorics exam
Tags: combinatorics
Taha1381
29.08.2018 14:56
I think the argument is true for any even number of points.I have an inductive proof which I am not sure that it is true.My main idea is to take two points on the convex hull and see what we are going to show in the inductive step and use a second induction to provethe new claim.Use the four points caseto deal with the induction and use the fact that you can let the sign of an arbitary triangle to be positive or negative in the case where points form a quadrilateral.And make sure you use the fact that the two points are on the convex hull.I will be thankful if someone checks my answer.
H.HAFEZI2000
30.08.2018 14:43
check here: https://artofproblemsolving.com/community/u366128h1699926p10915927
M.Sharifi
30.08.2018 17:58
If we have 3 points on coordinate plane, $A=(x_A,y_A)$ , $B=(x_B,y_B)$ and $C=(x_C,y_C)$ then: $$S_{\triangle{ABC}}=\frac{1}{2} |\begin{vmatrix}1 & x_{A} & y_{A} \\ 1 & x_{B} & y_{B} \\ 1 & x_{C} & y_{C} \\ \end{vmatrix}| =\frac{1}{2} |x_Ay_B-x_By_A-x_Ay_C+x_Cy_A+x_By_C-x_Cy_B|$$Knowing this identity makes this problem so easy!
bumjoooon
07.10.2019 15:37
To complete M. Sharifi's comment.. (with generalization to any even number) I will omit 'half' for the area of the triangle. Label points from 1 to 2n. Area of triangle with $i<j<k$ are $|(x_i y_j - x_j y_i)+(x_j y_k - x_k y_j) + (x_k y_i -x_i y_k)|$. For convenience, Let $b(i,j)=(x_i y_j - x_j y_i)$. Then, the Area is $|b(i,j)+b(j,k)+b(k,i)|$. Note that $b(i,j)+b(j,i)=0$. For triangle with even $i+j+k$, choose plus or minus so that $\pm a_{ijk}=b(i,j)+b(j,k)+b(k,i)$ and for triangle with odd $i+j+k$, choose plus or minus so that $\pm a_{ijk}=-b(i,j)-b(j,k)-b(k,i)$. Then for $b(i,j)$ with different parity, the numbers of odd and even $k$s between $i,j$ are equal and also the number of odd and even $k$s are equal outside of $(i,j)$. For $b(i,j)$ with same parity, the numbers of odd and even $k$s differ exactly one on both inside and outside of $(i,j)$. However, since $b(i,j)$ is counted as $b(i,j)$ in inside and $b(j,i)$ on outside, they cancel out. And thus every term cancels out. Therefore, we can always select $\pm$ for even number of points so that sum of areas are zero.
guptaamitu1
29.06.2022 07:13
This problem was also Sharygin Correspondence Round 2022 Problem 13.