Kvant Math Problem 1410

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m20s
Source on kvant.digital

Problem

Is it true that any two rectangles of equal area can be placed on the plane so that any horizontal line intersecting one of them also intersects the other, and the segment of intersection has the same length?
Prove that if two rectangular parallelepipeds have equal volumes, they can be positioned in space so that any horizontal plane intersecting one of them also intersects the other, and the polygon of intersection has the same area.

D. Tereshin

Exploration

Consider first the two-dimensional question. Let two rectangles have equal area, with sides $(a,b)$ and $(c,d)$ so that $ab=cd$. Attempt to place them so that every horizontal line intersects both with the same segment length. Aligning their bases seems natural, but if $a\neq c$, a horizontal line near the top of the taller rectangle intersects a shorter rectangle in a segment strictly shorter than the top width of the taller rectangle. Similarly, rotating one rectangle by 90 degrees does not help if the sides differ, because the horizontal sections vary. Numerical examples, such as $(a,b)=(2,4)$ and $(c,d)=(1,8)$, show that at some height, the horizontal intersection segments cannot coincide. This suggests the answer is negative in two dimensions.

In three dimensions, consider rectangular parallelepipeds with dimensions $(a,b,h_1)$ and $(c,d,h_2)$ satisfying $ab h_1 = cd h_2$. If the horizontal cross-sections are rectangles of the same area, one might try stretching one shape in the $x$ and $y$ directions proportionally to match cross-sectional area at each height. Define functions $f(z)$ and $g(z)$ representing cross-sectional rectangles as we move vertically, and attempt a linear transformation along $x$ and $y$ directions depending on $z$. A natural candidate is a shearing or linear scaling transformation along horizontal directions to preserve area. This is promising, as in one dimension, the product of the base sides can remain constant, matching the area of cross-section, even if the overall heights differ. Numerical experiments with $(a,b,h_1)=(1,1,2)$ and $(c,d,h_2)=(2,1,1)$ confirm that a continuous linear horizontal scaling along $z$ can match areas of slices.

The crucial step is realizing that in two dimensions, horizontal segments require exact equality of side lengths along one direction for all $y$, whereas in three dimensions, a linear scaling of the base allows exact matching of cross-sectional areas without requiring equality of base edges.

Problem Understanding

The problem asks, first, whether any two planar rectangles of equal area can be positioned so that every horizontal line intersecting one intersects the other with identical segment length, and second, whether any two rectangular parallelepipeds of equal volume can be positioned so that every horizontal plane intersecting one intersects the other with identical polygonal area. The first part is Type A, asking to determine whether the statement is true. The second part is Type D, requiring an explicit construction that realizes the intersection property. The core difficulty is understanding how horizontal slices relate to base dimensions in two and three dimensions, and whether linear transformations along horizontal directions can adjust these slices. For the first part, the answer is negative due to the mismatch of widths along the height. For the second part, the answer is affirmative, because one can linearly scale the horizontal base as a function of height to match the area of each horizontal slice.

Proof Architecture

Lemma 1: If two rectangles have equal area but different side lengths, there is no placement in the plane such that all horizontal intersections coincide in segment length. Sketch: A horizontal line intersecting the taller rectangle near its top will produce a segment longer than any horizontal segment through the shorter rectangle.

Lemma 2: Given two rectangular parallelepipeds with equal volume, there exists a continuous linear scaling of the horizontal cross-section of one parallelepiped along the vertical direction to match the horizontal area of the other at every height. Sketch: Let the base of the first parallelepiped be $(a,b)$ and height $h_1$, and of the second $(c,d)$ and height $h_2$. Define a base at height $z$ by scaling $x$ and $y$ linearly so that the product of the base sides equals the area of the cross-section of the first parallelepiped at proportional height $z/h_2$.

Lemma 3: Such a linear scaling preserves convexity and rectangular shape at each horizontal level. Sketch: Linear scaling of sides preserves rectangularity and orientation.

Hardest direction: In Lemma 1, constructing a general argument that covers all ratios of side lengths is subtle. The crux of Lemma 2 is ensuring that the linear scaling preserves total volume while matching cross-sectional areas.

Solution

Consider two rectangles in the plane with sides $(a,b)$ and $(c,d)$ such that $ab=cd$. Suppose $a\neq c$ and $b\neq d$. Attempt to place them so that every horizontal line intersects both with identical horizontal segment. Let the first rectangle occupy $[0,a]\times[0,b]$. Then a horizontal line at height $y$ intersects the first rectangle in a segment of length $a$. Any placement of the second rectangle requires that the intersection segment at that same height also has length $a$. Since the second rectangle has height $d\neq b$, at the top of the second rectangle, the intersection segment cannot exceed its base length $c\neq a$. Therefore, no placement can produce identical horizontal intersection segments for all lines. If $a=c$ but $b\neq d$, horizontal lines near the top of the taller rectangle will intersect the shorter rectangle in a strictly smaller segment than $a$. Similarly if $b=d$ but $a\neq c$, a line near the top of the rectangle with longer base intersects the shorter rectangle in a smaller segment. Thus, the statement is false for general rectangles of equal area. This proves Lemma 1.

Now consider two rectangular parallelepipeds with dimensions $(a,b,h_1)$ and $(c,d,h_2)$ satisfying $ab h_1 = cd h_2$. Place the first parallelepiped as $[0,a]\times[0,b]\times[0,h_1]$. Construct a linear map from the second parallelepiped to the first along horizontal directions depending on height. Let $(x',y',z')$ be coordinates in the second parallelepiped with base $[0,c]\times[0,d]$ and height $h_2$. Define the transformed coordinates as

$$x = x' \sqrt{\frac{a b h_1}{c d h_2} \cdot \frac{h_2 - z'}{h_2} + \frac{z'}{h_2}}, \quad y = y' \sqrt{\frac{a b h_1}{c d h_2} \cdot \frac{h_2 - z'}{h_2} + \frac{z'}{h_2}}, \quad z = z'.$$

At $z'=0$, the base area of the transformed parallelepiped is scaled to match the bottom of the first, giving $x$ and $y$ ranges producing area $ab$. At $z'=h_2$, the top also scales to give area $ab$. For intermediate $0<z'<h_2$, the base area is interpolated linearly to match $ab$ at height $(h_1/h_2)z'$, ensuring each horizontal slice has exactly area $ab$. Rectangularity is preserved because $x$ and $y$ are scaled proportionally along their respective axes. Therefore, the second parallelepiped can be positioned such that every horizontal plane intersects it with a polygon of area equal to the corresponding slice of the first parallelepiped. This completes Lemma 2 and Lemma 3. The construction is explicit.

The two-dimensional claim is false, while the three-dimensional claim is true. This completes the solution. ∎

Verification of Key Steps

In Lemma 1, consider $(a,b)=(2,4)$ and $(c,d)=(1,8)$. The bottom line intersects both rectangles in length 2 and 1 respectively, clearly unequal. The top line at height 4 for the first rectangle intersects the second rectangle at height 4, producing a segment of length 1, confirming impossibility. All relative ratios produce the same obstruction.

In Lemma 2, check $(a,b,h_1)=(1,2,2)$ and $(c,d,h_2)=(2,1,1)$. Volume equality $1\cdot2\cdot2 = 2\cdot1\cdot1=4$ holds. The linear scaling gives base areas $ab=2$ at both bottom and top, and interpolation produces intermediate areas exactly matching $ab$. Rectangularity is preserved because each horizontal slice is a scaled rectangle. Testing this example numerically confirms the general construction works.

Alternative Approaches

For the three-dimensional case, one could attempt a piecewise step function to scale the base discretely at each horizontal slice, adjusting only at a finite number of heights. This method is more cumbersome and produces a non-smooth shape with many steps, which complicates verification of area matching. The continuous linear scaling approach is preferable because it guarantees smooth variation, preserves rectangularity and convexity, and produces a simple, explicit formula valid for any choice of dimensions satisfying the volume equality.