Euclidian vs. Manhattan

Geometry Level 3

Alice and Beth are playing in a rectangular field. Alice bets Beth that she can beat Beth in a race from one corner of the field to the other. Alice is so confident that she'll win that she says she can still win if Beth races straight to the other point and Alice races along the perimeter.

The race results in a tie: Alice and Beth reach the other corner at the exact same time. If Beth ran 50 metres, Alice ran $A$ metres. What is possible interval of $A$ ?

Inspired by some kids I saw doing this.

[\ 50, 25\sqrt{2}\ ]

(\ 50, 50\sqrt{2}\ ]

(\ 50, 25\sqrt{2}\ ]

50^2

Not enough information

1 solution

Julian Yu
Apr 5, 2019

Clearly, by the triangle inequality, $A>50.$

Let the length and width of the field be $x$ and $y$ . Beth ran $50$ meters, so $\sqrt{x^2+y^2}=50\implies x^2+y^2=2500.$

Note that by the AM-GM inequality, $x^2+y^2\geq 2xy$ , so $2xy\leq 2500$ .

$\implies (x^2+y^2)+2xy=(x+y)^2\leq 5000 \implies x+y\leq 50\sqrt{2}.$

Since $A=x+y$ , we can conclude that $\boxed {50<A\leq 50\sqrt{2}}.$ We can make $A$ very close to 50 by making the width of the field very close to 0 and making the length very close to 50. We can make $A$ equal to $50\sqrt{2}$ by making the length and width both $25\sqrt{2}$ .

Euclidian vs. Manhattan

Inspired by some kids I saw doing this.

1 solution

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