Let there be two squares with side length 1 (in th...
Создано: 8 февраля 2025 г.
Отвечено с использованием O3 Mini High от Chat01
Создано: 8 февраля 2025 г.
Отвечено с использованием O3 Mini High от Chat01
Let there be two squares with side length 1 (in the same plane) ABCD and ABEZ. Both are rotated about their centers by 45 degrees. Find the area of their common area after their rotation. (The area of the intersection of their interiors).
We will show that if you start with two unit‐squares sharing a side and then rotate each one by 45° about its own center, the area where their interiors overlap is
which may also be written as (approximately 0.086).
In what follows we describe one way to arrive at this answer.
Initial placement before rotation.
Suppose we have two squares of side length 1 that share a side. For convenience choose coordinates so that:
The “top square” has vertices
so that its center is at .
The “bottom square” has vertices
so that its center is at .
Rotation by 45° about the centers.
Each square is rotated by 45° (say counter‐clockwise) about its center. (It does not matter which direction, as long as both are rotated by the same angle.) A standard fact is that a square of side 1 rotated by 45° becomes a “diamond” whose four vertices lie on the lines and (for the top square) and similarly for the bottom square. In other words, one may show that a square of side 1 rotated 45° about its center has the equation
where is its center.
Thus after rotation we have:
We wish to find the area of
A natural strategy is to “slice” the region horizontally (or vertically) and integrate.
Determine the vertical ( –) range for the intersection.
For a point to lie in the inequality
must hold. In particular, even if we choose (so that ), we need
i.e.
For the bottom square , setting we need
i.e.
Notice that
Hence the two vertical ranges intersect only when
(Since , these endpoints are approximately and , so the total vertical extent of the intersection is about .)
For a fixed in that range, find the allowed .
For a given in the common range, the two inequalities become
and
When is between and the expressions simplify (since then and ); one obtains
In fact one may show that the most restrictive condition is
(A short exercise in checking the two cases and confirms this.)
Set up the integral.
It is convenient to change variables by writing
Then for a given the allowed satisfies
Thus the horizontal “width” at height is
And the overall area of the intersection is
Since the integrand is even in (and the limits are symmetric about 0) we may write
Evaluating the integral gives
Thus,
Now, since
we have
But
so finally
The area common to the interiors of the two rotated squares is
This is the answer to the problem.