Let $ABCD$ be a square with side length $1$. How many points $P$ inside the square (not on its sides) have the property that the square can be cut into $10$ triangles of equal area such that all of them have $P$ as a vertex? Proposed by Josef Tkadlec - Czech Republic
Problem
Source: 2023 IGO Elementary P3
Tags: geometry, combinatorics, combinatorial geometry, square
EquationTracker
29.01.2024 11:59
$16$ Points.
By $[S]$, we denote the area of the polygon $S$.
We declare a new unit for the areas. Let say $\text{(tu)}$. The area of this square is $10$ tu .
For any point $P$ inside $ABCD$, we will must get $4$ triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$ and $\triangle PDA$. We have to choose $P$ such that each of these $4$ triangles have integer areas (This is obvious).
Let $a_1,a_2,a_3,a_4$ be the areas of these four triangles. If $P$ lies on $AB$, then $[\triangle PCD]=5\text{ tu}$. But $P$ lies inside $ABCD$, therefore we can say that $1\le a_1,a_2,a_3,a_4\le 4$.
Now, considering all such quadruples $(a_1,a_2,a_3,a_4)$ gives us $16$ points. In fact, these $16$ points makes a grid inside the square. Thus, we are done.
sami1618
15.02.2024 03:33
We claim $16$ points satisfy this. First we claim that lines $AP$, $BP$, $CP$, and $DP$ all must be drawn. If $P$ doesn't lie on a diagonal this is obvious otherwise the area of one of the triangles bounded by the diagonal and the square would be too large. Notice that $$[ABP]+[CDP]=[BCP]+[ADP]=\frac{1}{2}$$. And each of those triangles area must be an integer multiple of $\frac{1}{10}$. More over this uniquely determines $P$ so some calculation gives an answer of $16$.
RedFireTruck
27.07.2024 07:04
is 16 basically if u have a tringels above P, b tringels right P, c tringels below P, d tringels left P then a+c=b+d=5 and u get the picture above depending on (a,c) and (b,d)