A rotating square

Geometry Level 2

A unit square is put in every possible configuration such that one vertex remains on the $x$ axis and one remains on the $y$ axis.

Do any vertices ever get further than $\sqrt{2}$ from the origin?

No Yes

2 solutions

Henry U
Jan 23, 2019

When the square lies such that its sides form 45°-angles with the axes, the distance from the origin to both of the outer vertices is $\sqrt{x^2 + y^2}$ , where $x$ and $y$ are the x- and y-coordinate of these vertices respectively.

Since a diagonal of the square is parallel to one axis, the respective distance is $\sqrt 2$ . The other distance is positive, so the overall distance is greater than $\sqrt 2$ .

The distance to the "outer" vertex nearest (but not on) the x-axis can be written as $D(x) = \sqrt{1 + x^{2} + 2x\sqrt{1 - x^{2}}}$ , where $0 \le x \le 1$ is the position of the vertex that lies on the x-axis. This function has a maximum of $\phi = \dfrac{1 + \sqrt{5}}{2}$ at $x = \sqrt{\dfrac{5 + \sqrt{5}}{10}}$ .

@Geoff Pilling The golden ratio strikes again! :)

Brian Charlesworth - 2 years, 4 months ago

Hahaha... Yup!

Makes you wonder if $\phi$ should take it's rightful place alongside $\pi$ and $e$ as the most influential figures in mathematics! A triumvirate of sorts! 😎

Geoff Pilling - 2 years, 4 months ago

Edwin Gray
Feb 8, 2019

Let two vertices of the square have coordinates (0,sqrt(2)/2) and (sqrt(2)/2,0). Then the other vertices have coordinates (sqrt(2)/2,sqrt(2)) and (sqrt(2),sqrt(2)/2). The distance from the origin to either of tha latter 2 vertices is d^2 = 2 + 1/2 = 2.5. So d > sqrt(2). Ed Gray

A rotating square

2 solutions

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