Any two rectangles of equal area can be placed in the plane?

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Any two rectangles of equal area can be placed in the plane?

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Source: All Russian Olympiads 1993 - Grade 10 - Day 2 - Problem 2

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#1 h Jul 13, 2011, 11:26 AM • 1 Y

Y by Adventure10

Is it true that any two rectangles of equal area can be placed in the plane such that any horizontal line intersecting at least one of them will also intersect the other, and the segments of intersection will be equal?

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#2 h Jul 14, 2011, 6:10 AM • 1 Y

Y by Adventure10

I don't think this is true, if I understand the problem correctly. Assume one of the rectangles is a $1 \times 1$ square and the other is a $\frac{1}{10} \times 10$ rectangle. We can assume wlog that the square is positioned as a standard unit square in the plane, having vertices at $(0,0)$ , $(0,1)$ , $(1,1)$ and $(1,0)$ . Then every horizontal line intersects that square at segment of length $0$ or $1$ , the latter occurring precisely for horizontal lines $y=t$ with $0 \leq t \leq 1$ .

This forces the other rectangle to be placed such that the line $y=0$ , for example, intersectes it at a segment of length $1$ . This can certainly be done by rotating it appropriately, but then it necessarily extends below the line $y=0$ failing the desired behavior for horizontal lines just below $y=0$ . Quick proof: any line that meets the rectangle but avoids all its interior points must meet exclusively its boundary, which is the union of four segments, two of length $\frac{1}{10}$ and two of length $10$ . Thus any such line meets this rectangle in a segment of length $0$ , $10$ or $\frac{1}{10}$ , so a line that meets it in a segment of the desired length $1$ must meet an interior point. But every interior point is surrounded by a small circle contained in the interior, so there are interior points both above and below the line $y=0$ .

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#3 h Jul 14, 2011, 6:34 AM • 2 Y

Y by Adventure10, Mango247

You cannot assume any special positioning of one of the rectangles, like your square, because this forces the meaning of "horizontal". What if, for your very example, there exists a positioning where horizontal is not parallel with one of the sides?

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#4 h Jul 14, 2011, 7:01 AM • 2 Y

Y by Adventure10, Mango247

Of course - how silly of me. Will keep thinking.

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#5 h Jul 16, 2011, 11:14 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

If the two rectangles are congruent the the answer is obvious, so assume we have $ABCD$ and $XYZW$ of equal area and $|AB|>|XY|$ .

Lemma: there exists rotations of both rectangles such that four horizontal lines through the vertices $A,B,C,D$ also pass through the vertices $X,Y,Z,W$ respectiely.

Proof: Suppose $AB$ is horizontal with the x-axis and let the lines through the vertices be $\ell_A,...,\ell_D$ . Now if the distance between $\ell_A$ and $\ell_B$ is less than $XY$ then there exists a unique way to place $X$ on $\ell_A$ and $Y$ on $\ell_B$ (up to horizontal translation and reflection).

Since $|AB|>|XY|\Rightarrow |BC| < |YZ|$ the point $Z$ will lie below $\ell_C$ . Now rotate $ABCD$ arround $B$ while keeping $X$ on $\ell_A$ and $Y$ on $\ell_B$ . Eventually $XYZW$ will be rotated 90 degrees and since $|AB|>|XY|$ we now have $Z$ lying above $\ell_C$ . By continuity there will exist some angle such that $Z$ lies on $\ell_C$

Now we have $X,Y,Z$ lying on $\ell_A$ , $\ell_B$ and $\ell_C$ respectively. But by symmetry we must also have $W$ lying on $\ell_D$ so the lemma is proven $\square$

The problem now follows easily. Assume wlog that $\ell_A$ lies above $\ell_B$ lies above $\ell_D$ which lies above $\ell_C$ and let the distance between them be $d_1,d_2,d_3$ respectively. Then if the line segment through $ABCD$ and $XYZW$ lying on $\ell_B$ have length $b$ and $y$ respectively then the area of $ABCD$ is $\textstyle\frac{1}{2}d_1b + d_2b + \frac{1}{2}d_3b$ . And similarly for $XYZW$ , so we have $b=y$ and the rest follows by Cavalieri's Principle

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