« Cube du prince Rupert » : différence entre les versions

Un cube unité comportant un trou assez large pour permettre le passage du cube du prince Rupert.

En géométrie, le Cube du prince Rupert (nommé d'après le Prince Rupert du Rhin) est le plus grand cube pouvant passer à travers un trou pratiqué dans un cube unitaire, i.e. un cube d'arête 1, sans séparer le cube en deux parties. La longueur de son arête est approximativement 6% plus longue que celle du cube au travers duquel il passe. Le problème consistant à trouver le plus grand carré tenant entièrement dans un cube unitaire est directement lié et possède la même solution.

Solution

Si deux points sont placés sur deux arêtes adjacentes D'un cube unité, chacun à une distance de 3/4 du point d'intersection de ces arêtes, alors la distance entre ces points est

${\frac {3{\sqrt {2}}}{4}}\approx 1.0606601.$

Ces deux points, avec un second couple de points placés symétriquement sur la face opposée du cube, forment les quatre sommets d'un carré entièrement contenu dans le cube unité. Ce carré, prolongé perpendiculairement dans les deux directions, forme le trou au travers duquel un cube plus grand que le cube original (avec une longueur allant jusqu'à ${\tfrac {3{\sqrt {2}}}{4}}$) peut passer.

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The parts of the unit cube that remain, after emptying this hole, form two triangular prisms and two irregular tetrahedra, connected by thin bridges at the four vertices of the square. Each prism has as its six vertices two adjacent vertices of the cube, and four points along the edges of the cube at distance 1/4 from these cube vertices. Each tetrahedron has as its four vertices one vertex of the cube, two points at distance 3/4 from it on two of the adjacent edges, and one point at distance 3/16 from the cube vertex along the third adjacent edge.

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Histoire

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Prince Rupert's cube is named after Prince Rupert of the Rhine. According to a story recounted in 1693 by English mathematician John Wallis, Prince Rupert wagered that a hole could be cut through a cube, large enough to let another cube of the same size pass through it. Wallis showed that in fact such a hole was possible (with some errors that were not corrected until much later), and Prince Rupert won his wager.

Wallis assumed that the hole would be parallel to a space diagonal of the cube. The projection of the cube onto a plane perpendicular to this diagonal is a regular hexagon, and the best hole parallel to the diagonal can be found by drawing the largest possible square that can be inscribed into this hexagon. Calculating the size of this square shows that a cube with side length

${\sqrt {6}}-{\sqrt {2}}\approx 1.03527$,

slightly larger than one, is capable of passing through the hole.

Approximately 100 years later, Dutch mathematician Pieter Nieuwland found that a better solution (in fact, the optimal solution) may be achieved by using a hole with a different angle than the space diagonal. Nieuwland died in 1794 (a year after taking a position as a professor at the University of Leiden) but his solution was published posthumously in 1816 by Nieuwland's mentor, Jean Henri van Swinden.

Since then, the problem has been repeated in many books on recreational mathematics, in some cases with Wallis' suboptimal solution instead of the optimal solution.

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Modèles

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The construction of a physical model of Prince Rupert's cube is made difficult by the accuracy with which such a model would need to be measured, and the thinness of the connections between the remaining parts of the unit cube after the hole is cut through it; for this reason, the problem has been called "mathematically possible but practically impossible". Nevertheless, in a 1950 survey of the problem, D. J. E. Schrek published photographs of a model of a cube passing through a hole in another cube. Martin Raynsford has designed a template for constructing paper models of a cube with another cube passing through it; in order to account for the tolerances of paper construction and not tear the paper at the narrow joints between parts of the punctured cube, the hole in Raynsford's model is slightly smaller than the cube it lets pass through. On 10 September 2015 David Howarth, during a talk on recreational mathematics to the Rotary Club of Turton, Bolton England, demonstrated taking a cube from a box, removing a precut square section from the cube and passing the box through it before reassembling the smaller cube and replacing it back in the box. This may be the first practical demonstration of a larger cube passing through a smaller cube.

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Généralisations

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The cube is not the only body that can pass through a hole cut into a copy of itself; the same is true for the regular tetrahedron and octahedron.

Another way to express the same problem is to ask for the largest square that lies within a unit cube. More generally, Jerrard & Wetzel (2004) show how to find the largest rectangle of a given aspect ratio that lies within a unit cube. As they show, the optimal rectangle must always pass through the center of the cube, with its vertices on edges of the cube. Based on this, they show, depending on the desired aspect ratio, that the optimal rectangle must either lie on a plane that cuts diagonally through four corners of the cube, or it must be formed by an isosceles right triangle on one corner of the cube and by the two opposite points, as in the case of Prince Rupert's problem. If the aspect ratio is not constrained, the rectangle with the largest area that fits within a cube is the one that has two opposite edges of the cube as two of its sides, and two face diagonals as the other two sides.

Alternatively, one may ask for the largest $m$-dimensional hypercube that may be drawn within an $n$-dimensional unit hypercube. The answer is always an algebraic number. For instance, the problem for $(m,n)=(3,4)$ asks for the largest cube within a four-dimensional hypercube. After Martin Gardner posed this question in Scientific American, Kay R. Pechenick DeVicci and several other readers showed that the answer for the (3,4) case is the square root of the smaller of two real roots of the polynomial $4x^{4}-28x^{3}-7x^{2}+16x+16$, which works out to approximately 1.007435. For $m=2$, the optimal side length of the largest square in an $n$-dimensional hypercube is either ${\sqrt {n/2}}$ or ${\sqrt {n/2-3/8}}$, depending on whether $n$ is even or odd respectively.

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Références

(en) Cet article est partiellement ou en totalité issu de l’article de Wikipédia en anglais intitulé « Prince Rupert's cube » (voir la liste des auteurs).

Lien externe

• (en) Eric W. Weisstein, « Cube du prince Rupert », sur MathWorld