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In Euclidean geometry, Apollonius' problem is to construct all the circles that are tangent to three given circles. Special cases of Apollonius' problem are those in which at least one of the given circles is a point or line, i.e., is a circle of zero or infinite radius.
Case 2: a and a' are not perpendicular to each other. Using a hyperbolic ruler, construct a line BI such that BI is perpendicular to a and parallel to a'. Also, construct a line CI' such that CI' is perpendicular to a and parallel to a' but in the opposite direction of BI. Now draw a line II" so that II" is the common parallel to BI and I'C.
Angle trisection is the construction, using only a straightedge and a compass, of an angle that is one-third of a given arbitrary angle. This is impossible in the general case. For example, the angle 2 π /5 radians (72° = 360°/5) can be trisected, but the angle of π /3 radians (60°) cannot be trisected. [8]
The segment AB is perpendicular to the segment CD because the two angles it creates (indicated in orange and blue) are each 90 degrees. The segment AB can be called the perpendicular from A to the segment CD, using "perpendicular" as a noun. The point B is called the foot of the perpendicular from A to segment CD, or simply, the foot of A on CD ...
adventitious quadrangles problem. A quadrilateral such as BCEF is called an adventitious quadrangle when the angles between its diagonals and sides are all rational angles, angles that give rational numbers when measured in degrees or other units for which the whole circle is a rational number. Numerous adventitious quadrangles beyond the one ...
Carnot's theorem: if three perpendiculars on triangle sides intersect in a common point F, then blue area = red area. Carnot's theorem (named after Lazare Carnot) describes a necessary and sufficient condition for three lines that are perpendicular to the (extended) sides of a triangle having a common point of intersection.
Three problems proved elusive, specifically, trisecting the angle, doubling the cube, and squaring the circle. The problem of angle trisection reads: Construct an angle equal to one-third of a given arbitrary angle (or divide it into three equal angles), using only two tools: an unmarked straightedge, and; a compass.
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