Points contained in a bounded area
- Phlembac Adib Hasan
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$n$ points lie on a plane so that the triangle formed by any three of them has an area of at most $1\;\text{unit}^2$. Prove that all the points are contained in a triangle with area of at most $4\;\text{unit}^2$.
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- nahin munkar
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Re: Points contained in a bounded area
Let, within $n$ points, $\triangle XYZ$ is a triangle formed by such $3$ points where the area is maximal. Here,surely, $[XYZ] \le 1$. Now, we construct a triangle $\triangle PQR$ whose medial triangle is $\triangle XYZ$. So, now, $[PQR]=4[XYZ] \le 4$. We claim, all the points are contained in $\triangle PQR$ .
We suppose, A point named $A$ lies outside $\triangle PQR$. If we add $'A'$ with two vertices of $\triangle XYZ$, then that will form a triangle larger than $\triangle XYZ$, contradicts the maximality of $[XYZ]$.
We suppose, A point named $A$ lies outside $\triangle PQR$. If we add $'A'$ with two vertices of $\triangle XYZ$, then that will form a triangle larger than $\triangle XYZ$, contradicts the maximality of $[XYZ]$.
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