Let there be two squares with side length 1 (in th...

Создано: 8 февраля 2025 г.

Отвечено с использованием O3 Mini High от Chat01

Вопрос

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).

Мысль

Clarifying the rotation problem

We're asked about two squares, ABCD and ABEZ, both with side length 1, rotated 45 degrees about their centers. The phrasing is a bit unclear. They share vertices A and B, but the question doesn't specify if the rotation is relative to their original positions or relative to each other. This ambiguity makes it tough to understand the exact configuration of the squares and how to calculate the area of their intersection accurately. Let me figure out the most logical interpretation.

Clarifying square rotation and intersection

Let me think this through. The two squares, ABCD and ABEZ, each with side length 1, likely share side AB and are rotated by 45 degrees about their centers. The problem's asking for the area of their intersection after rotation, so I’m picturing both squares sharing the edge AB. This means that we need to consider how these squares' rotations affect their overlap, with their centers fixed and the squares rotated relative to that fixed position. I think I'll map this out in terms of coordinates.

Analyzing the configuration and intersection

I think the two squares are arranged such that ABCD is above AB and ABEZ is below AB. The question asks about the area of overlap after 45-degree rotations of both squares about their respective centers, (0.5, 0.5) and (0.5, -0.5). After rotation, they'll likely intrude into each other's space, around the shared side AB. The important thing is that the two squares seem to be reflections of each other, so the intersection should show symmetrical properties, which could make the calculation a bit more manageable.

Clarifying rotations and regions of overlap

I think the two squares share side AB, and after a 45-degree rotation, they become "tilted" or diamond-shaped squares. This seems like the squares rotate around their centers and end up angled relative to the axes, which places their vertices at 45 degrees from the horizontal. Since the squares are rotated in the same direction, they'll likely overlap near the edge AB, forming a quadrilateral intersection. The interpretation here seems to be that the squares share vertices A and B and are rotated at 45-degree angles from a standard position.

Determining the intersection of squares

The problem involves two squares, ABCD and ABEZ, that share side AB and are rotated 45 degrees about their centers. The initial configuration has one square above, the other below AB. After rotation, each square becomes tilted. I calculated the vertices of each square after rotation and identified their intersection region. The intersection, due to the rotation, isn’t a simple rectangle but a quadrilateral. I’m now working on finding the equations for the square edges to determine the exact area of the overlap.

Finding the intersection of rotated squares

To determine the intersection of the two rotated squares, I calculated the equations for the edges of both squares after each is rotated by 45 degrees about their center points. I then analyzed their overlap, which takes a polygonal form rather than a rectangle due to rotation. The intersection is symmetric, and I’m considering transforming the coordinates for simplicity. With those transformations, I intend to use integration to compute the area of the intersection, refining the bounds based on the vertical and horizontal ranges.

Calculating the intersection area

I originally calculated the intersection area of the two rotated squares by analyzing their properties and applying integration. After checking my formulas and integration, I find that the area is very small — around 0.086, which seems reasonable given the geometric overlap. The vertical and horizontal extents of the intersection are both 0.414, forming a diamond shape. This calculation matches the expected intersection and suggests the area is indeed small but plausible.

Recalculating intersection area

Let me check the geometric intersection of the two rotated squares. I calculated that their intersection exists within a diamond-shaped area with vertices determined by the boundaries of the squares. Through my integration process, I confirmed the intersection's area is indeed around 0.086. Though small, this is reasonable considering the squares initially only touched along a line before rotation, which resulted in a slight penetration. The final area is approximately 0.086, confirmed through both integration and geometry.

Reasoned for 1m 59s

Ответ

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

3222,\frac{3-2\sqrt{2}}{2}\,,

which may also be written as 322\frac{3}{2}-\sqrt2 (approximately 0.086).

In what follows we describe one way to arrive at this answer.


The Setup

  1. 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

      A=(0,0),B=(1,0),C=(1,1),D=(0,1)A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1)

      so that its center is at (0.5,0.5)(0.5,0.5).

    • The “bottom square” has vertices

      A=(0,0),B=(1,0),E=(1,1),Z=(0,1)A=(0,0),\quad B=(1,0),\quad E=(1,-1),\quad Z=(0,-1)

      so that its center is at (0.5,0.5)(0.5,-0.5).

  2. 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 x=0.5±22x=0.5\pm\frac{\sqrt{2}}{2} and y=0.5±22y=0.5\pm\frac{\sqrt{2}}{2} (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

xxc+yyc22,|x-x_c|+|y-y_c|\le \frac{\sqrt{2}}{2}\,,

where (xc,yc)(x_c,y_c) is its center.

Thus after rotation we have:

  • The top square becomes Stop={(x,y):x0.5+y0.52/2}.S_{\rm top}=\{(x,y): |x-0.5|+|y-0.5|\le \sqrt{2}/2\}\,.
  • The bottom square becomes Sbot={(x,y):x0.5+y+0.52/2}.S_{\rm bot}=\{(x,y): |x-0.5|+|y+0.5|\le \sqrt{2}/2\}\,.

Finding the Intersection Area

We wish to find the area of

S=StopSbot={(x,y):x0.5+y0.52/2andx0.5+y+0.52/2}.S = S_{\rm top}\cap S_{\rm bot} = \{(x,y) : |x-0.5|+|y-0.5|\le \sqrt{2}/2 \quad\text{and}\quad |x-0.5|+|y+0.5|\le \sqrt{2}/2\}\,.

A natural strategy is to “slice” the region horizontally (or vertically) and integrate.

  1. Determine the vertical ( yy–) range for the intersection.

    For a point (x,y)(x,y) to lie in StopS_{\rm top} the inequality

    x0.5+y0.522|x-0.5|+|y-0.5|\le \frac{\sqrt{2}}{2}

    must hold. In particular, even if we choose x=0.5x=0.5 (so that x0.5=0|x-0.5|=0), we need

    y0.522,|y-0.5|\le \frac{\sqrt{2}}{2}\,,

    i.e.

    y[0.522,0.5+22].y\in\left[0.5-\frac{\sqrt{2}}{2},\,0.5+\frac{\sqrt{2}}{2}\right]\,.

    For the bottom square SbotS_{\rm bot}, setting x=0.5x=0.5 we need

    y+0.522,|y+0.5|\le \frac{\sqrt{2}}{2}\,,

    i.e.

    y[0.522,0.5+22].y\in\left[-0.5-\frac{\sqrt{2}}{2},\,-0.5+\frac{\sqrt{2}}{2}\right]\,.

    Notice that

    0.522=122and0.5+22=212.0.5-\frac{\sqrt{2}}{2}=\frac{1-\sqrt{2}}{2}\quad\text{and}\quad -0.5+\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-1}{2}\,.

    Hence the two vertical ranges intersect only when

    y[122,212].y\in\left[\frac{1-\sqrt{2}}{2},\,\frac{\sqrt{2}-1}{2}\right]\,.

    (Since 21.414\sqrt{2}\approx1.414, these endpoints are approximately 0.207-0.207 and 0.2070.207, so the total vertical extent of the intersection is about 0.4140.414.)

  2. For a fixed yy in that range, find the allowed xx.

    For a given yy in the common range, the two inequalities become

    x0.522y0.5|x-0.5|\le \frac{\sqrt{2}}{2}-|y-0.5|

    and

    x0.522y+0.5.|x-0.5|\le \frac{\sqrt{2}}{2}-|y+0.5|\,.

    When yy is between 122\frac{1-\sqrt{2}}{2} and 212\frac{\sqrt{2}-1}{2} the expressions simplify (since then y0.50y-0.5\le0 and y+0.50y+0.5\ge0); one obtains

    x0.522max{0.5y,  y+0.5}.|x-0.5|\le \frac{\sqrt{2}}{2}-\max\{0.5-y,\;y+0.5\}\,.

    In fact one may show that the most restrictive condition is

    x0.5212y.|x-0.5|\le \frac{\sqrt{2}-1}{2}-|y|\,.

    (A short exercise in checking the two cases y0y\ge0 and y0y\le0 confirms this.)

  3. Set up the integral.

    It is convenient to change variables by writing

    u=x0.5.u=x-0.5\,.

    Then for a given yy the allowed uu satisfies

    udywithd=212.|u|\le d-|y|\quad \text{with}\quad d=\frac{\sqrt{2}-1}{2}\,.

    Thus the horizontal “width” at height yy is

    2(dy).2\Bigl(d-|y|\Bigr)\,.

    And the overall area of the intersection is

    A=y=(12)/2(21)/22(dy)dy.A=\int_{y=(1-\sqrt{2})/2}^{(\sqrt{2}-1)/2}2\Bigl(d-|y|\Bigr)\,dy\,.

    Since the integrand is even in yy (and the limits are symmetric about 0) we may write

    A=220d(dy)dy=40d(dy)dy.A=2\cdot 2\int_{0}^{d}\Bigl(d-y\Bigr)\,dy=4\int_{0}^{d}(d-y)\,dy\,.

    Evaluating the integral gives

    0d(dy)dy=[dyy22]0d=d2d22=d22.\int_{0}^{d}(d-y)\,dy = \left[dy-\frac{y^2}{2}\right]_0^d = d^2-\frac{d^2}{2}=\frac{d^2}{2}\,.

    Thus,

    A=4d22=2d2.A=4\cdot\frac{d^2}{2}=2d^2\,.

    Now, since

    d=212so thatd2=(21)24,d=\frac{\sqrt{2}-1}{2}\quad\text{so that}\quad d^2=\frac{(\sqrt{2}-1)^2}{4}\,,

    we have

    A=2(21)24=(21)22.A=2\cdot\frac{(\sqrt{2}-1)^2}{4}=\frac{(\sqrt{2}-1)^2}{2}\,.

    But

    (21)2=222+1=322,(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\,,

    so finally

    A=3222.A=\frac{3-2\sqrt{2}}{2}\,.

Conclusion

The area common to the interiors of the two rotated squares is

3222or equivalently322.\boxed{\frac{3-2\sqrt{2}}{2}\quad\text{or equivalently}\quad\frac{3}{2}-\sqrt{2}\,.}

This is the answer to the problem.

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