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Quadrilaterals that are both orthodiagonal and equidiagonal, and in which the diagonals are at least as long as all of the quadrilateral's sides, have the maximum area for their diameter among all quadrilaterals, solving the n = 4 case of the biggest little polygon problem. The square is one such quadrilateral, but there are infinitely many others.
Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. The two diagonals of a rhombus are perpendicular; that is, a rhombus is an orthodiagonal quadrilateral. Its diagonals bisect opposite ...
A square can also be defined as a parallelogram with equal diagonals that bisect the angles. If a figure is both a rectangle (right angles) and a rhombus (equal edge lengths), then it is a square. A square has a larger area than any other quadrilateral with the same perimeter. [7]
A perfect parallelepiped is a parallelepiped with integer-length edges, face diagonals, and space diagonals. In 2009, dozens of perfect parallelepipeds were shown to exist, [3] answering an open question of Richard Guy. One example has edges 271, 106, and 103, minor face diagonals 101, 266, and 255, major face diagonals 183, 312, and 323, and ...
In geometry, a golden rectangle is a rectangle with side lengths in golden ratio +:, or :, with approximately equal to 1.618 or 89/55. Golden rectangles exhibit a special form of self-similarity : if a square is added to the long side, or removed from the short side, the result is a golden rectangle as well.
If you've been having trouble with any of the connections or words in Sunday's puzzle, you're not alone and these hints should definitely help you out. Plus, I'll reveal the answers further down ...
Vectors involved in the parallelogram law. In a normed space, the statement of the parallelogram law is an equation relating norms: ‖ ‖ + ‖ ‖ = ‖ + ‖ + ‖ ‖,.. The parallelogram law is equivalent to the seemingly weaker statement: ‖ ‖ + ‖ ‖ ‖ + ‖ + ‖ ‖, because the reverse inequality can be obtained from it by substituting (+) for , and () for , and then simplifying.
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