Point within rectangle with integer distances to vertices.

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Point within rectangle with integer distances to vertices.

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#1 h Dec 7, 2010, 8:29 AM • 2 Y

Y by Adventure10, Mango247

The lengths of the sides of a rectangle are given to be odd integers. Prove that there does not exist a point within that rectangle that has integer distances to each of its four vertices.

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#2 h Dec 8, 2010, 1:56 AM • 1 Y

Y by Adventure10

Suppose the rectangle has side lengths $a,b$ . Let the rectangle be $ABCD$ with $AB=b$ . Let $P$ be the point inside. Let $X,Y$ be the feet of the perpendiculars from $P$ onto $CD,DA$ . Let $DX=x$ . Then $x^2-(b-x)^2=PD^2-PC^2$ is an integer, so $b(2x-b)$ is an integer, so $x$ is rational. Now let $x=\frac{c}{d}$ . Similarly let $DY=\frac{e}{f}$ , where $\gcd(c,d)=\gcd(e,f)=1$ . Then since $\frac{c^2}{d^2}+\frac{e^2}{f^2}$ is an integer, we must have $d=f$ . So now $\frac{\sqrt{c^2+e^2}}{d}$ is an integer. Since the sum of two odd squares cannot be a square, one of $c,e$ must be even; WLOG $c$ is even, so $d$ is odd since they are coprime. If $e$ is even, we consider the length $PB=\frac{\sqrt{(bd-c)^2+(ad-e)^2}}{d}$ , which is not an integer since both terms inside the square root are odd. If $e$ is odd, we consider the length $PC=\frac{\sqrt{(bd-c)^2+e^2}}{d}$ , again not an integer. Thus there is no such point $P$ .

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