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An a × b rectangle can be packed with 1 × n strips if and only if n divides a or n divides b. [ 15 ] [ 16 ] de Bruijn's theorem : A box can be packed with a harmonic brick a × a b × a b c if the box has dimensions a p × a b q × a b c r for some natural numbers p , q , r (i.e., the box is a multiple of the brick.) [ 15 ]
The five-room puzzle is a classical, [1] popular puzzle involving a large rectangle divided into five "rooms". The objective of the puzzle is to cross each "wall" of the diagram with a continuous line only once.
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent.
The first nine blocks in the solution to the single-wide block-stacking problem with the overhangs indicated. In statics, the block-stacking problem (sometimes known as The Leaning Tower of Lire (Johnson 1955), also the book-stacking problem, or a number of other similar terms) is a puzzle concerning the stacking of blocks at the edge of a table.
An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch uniformly at a constant rate of 1 km per second, so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc.
Although there are 10 mm in 1 cm, there are 100 mm 2 in 1 cm 2. Calculation of the area of a square whose length and width are 1 metre would be: 1 metre × 1 metre = 1 m 2. and so, a rectangle with different sides (say length of 3 metres and width of 2 metres) would have an area in square units that can be calculated as: 3 metres × 2 metres ...
In 1968, John Hammersley stated a lower bound of / + /. [4] This can be achieved using a shape resembling an old-fashioned telephone handset , consisting of two quarter-disks of radius 1 on either side of a 1 by 4 / π {\displaystyle 4/\pi } rectangle from which a half-disk of radius 2 / π {\displaystyle 2/\pi } has been removed.
Let x and θ be as in the illustration in this section. Placing a needle's center at x, the needle will cross the vertical axis if it falls within a range of 2θ radians, out of π radians of possible orientations. This represents the gray area to the left of x in the figure. For a fixed x, we can express θ as a function of x: θ(x) = arccos(x).