Show that or can you prove that in a square of side 3 cm if 19 points are present then at least 3 points form a traingle of area lesse than or equal to 1

Let us imagine a square of side 3 cm so area 9 cm² ,it can be broken into 9 squares of area 1 cm² and each of 19 points should be located in these 9 squares of are 1cm² , now since 2*9+1=19 or 2x9+1=19 so not all squares could have 2 or less points in it each and at least 1 square would have 3 points in it and thatsit thats it !!!!!!!!!, In that square you have 3 points and they surely and niot probably make a triangle and are less than or equal to 1 cm²

#NumberTheory

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$