Circles, Arcs and Angles
... • A, B, C are points on a circle centered at O. • Angle BAC an inscribed angle • Angular measure of arc BC is the measure of the central angle BOC, where the angle is measured on the same side of O as the arc. B • BAC subtends the arc BC. O A 14-Sept-2011 ...
... • A, B, C are points on a circle centered at O. • Angle BAC an inscribed angle • Angular measure of arc BC is the measure of the central angle BOC, where the angle is measured on the same side of O as the arc. B • BAC subtends the arc BC. O A 14-Sept-2011 ...
Geometry Honors - Santa Rosa Home
... Use a variety of problem-solving strategies, such as drawing a diagram, making a chart, guess-and-check, solving a simpler problem, writing an equation, and working ...
... Use a variety of problem-solving strategies, such as drawing a diagram, making a chart, guess-and-check, solving a simpler problem, writing an equation, and working ...
Solutions #6
... through P that are parallel to the horizontal line; this means that they should be arcs of Euclidean-circles that are tangent to M N at the points M and N , as in Figure 1. If we draw the Euclidean-tangent-lines to these Euclidean-circles at P (say one of them intersects the horizontal line at Q, th ...
... through P that are parallel to the horizontal line; this means that they should be arcs of Euclidean-circles that are tangent to M N at the points M and N , as in Figure 1. If we draw the Euclidean-tangent-lines to these Euclidean-circles at P (say one of them intersects the horizontal line at Q, th ...
Three Famous Problems of Antiquity Historical Context: Suggested
... Consider this Lemma: Given a cubic equation p(x) with integer coefficients p(x) = a3x3 + a2x2 + a1x + a0 = 0, either one of the equation’s roots is rational or none of its roots are constructible. It will be assumed true here, but can be proven using extension fields in abstract algebra. 3. Use this ...
... Consider this Lemma: Given a cubic equation p(x) with integer coefficients p(x) = a3x3 + a2x2 + a1x + a0 = 0, either one of the equation’s roots is rational or none of its roots are constructible. It will be assumed true here, but can be proven using extension fields in abstract algebra. 3. Use this ...
POP Geometric Terms
... A line that touches a curve at a point without crossing over. Formally, it is a line which intersects a differentiable curve at a point where the slope of the curve equals the slope of the line. Note: A line tangent to a circle is perpendicular to the radius to the point of tangency. ...
... A line that touches a curve at a point without crossing over. Formally, it is a line which intersects a differentiable curve at a point where the slope of the curve equals the slope of the line. Note: A line tangent to a circle is perpendicular to the radius to the point of tangency. ...
Geometry 12.2 ‐ chords and arcs A. Chord A chord is a segment
... (a) Chords equidistant from the center are congruent (b) Congruent chords are equidistant from the center C. More Theorems (a) Theorem 12 ‐ 6: In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs (b) Theorem 12 ‐ 7: In a circle, a diameter that bisects a chord i ...
... (a) Chords equidistant from the center are congruent (b) Congruent chords are equidistant from the center C. More Theorems (a) Theorem 12 ‐ 6: In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs (b) Theorem 12 ‐ 7: In a circle, a diameter that bisects a chord i ...
Topic VII Properties of Circles Topic VII Properties of Circles Topic VII
... When the center of dilation is the origin, a = 0 and b = 0, so the rule simplifies to: Dk (x, y) = (kx, ky) As the radius of the circle increases from 1 to 3, the scale factor is 3, then the function for the dilation is: D3 (x, y) = (3x, 3y). The second transformation is a translation. The function ...
... When the center of dilation is the origin, a = 0 and b = 0, so the rule simplifies to: Dk (x, y) = (kx, ky) As the radius of the circle increases from 1 to 3, the scale factor is 3, then the function for the dilation is: D3 (x, y) = (3x, 3y). The second transformation is a translation. The function ...
Problem of Apollonius
In Euclidean plane geometry, Apollonius's problem is to construct circles that are tangent to three given circles in a plane (Figure 1). Apollonius of Perga (ca. 262 BC – ca. 190 BC) posed and solved this famous problem in his work Ἐπαφαί (Epaphaí, ""Tangencies""); this work has been lost, but a 4th-century report of his results by Pappus of Alexandria has survived. Three given circles generically have eight different circles that are tangent to them (Figure 2) and each solution circle encloses or excludes the three given circles in a different way: in each solution, a different subset of the three circles is enclosed (its complement is excluded) and there are 8 subsets of a set whose cardinality is 3, since 8 = 23.In the 16th century, Adriaan van Roomen solved the problem using intersecting hyperbolas, but this solution does not use only straightedge and compass constructions. François Viète found such a solution by exploiting limiting cases: any of the three given circles can be shrunk to zero radius (a point) or expanded to infinite radius (a line). Viète's approach, which uses simpler limiting cases to solve more complicated ones, is considered a plausible reconstruction of Apollonius' method. The method of van Roomen was simplified by Isaac Newton, who showed that Apollonius' problem is equivalent to finding a position from the differences of its distances to three known points. This has applications in navigation and positioning systems such as LORAN.Later mathematicians introduced algebraic methods, which transform a geometric problem into algebraic equations. These methods were simplified by exploiting symmetries inherent in the problem of Apollonius: for instance solution circles generically occur in pairs, with one solution enclosing the given circles that the other excludes (Figure 2). Joseph Diaz Gergonne used this symmetry to provide an elegant straightedge and compass solution, while other mathematicians used geometrical transformations such as reflection in a circle to simplify the configuration of the given circles. These developments provide a geometrical setting for algebraic methods (using Lie sphere geometry) and a classification of solutions according to 33 essentially different configurations of the given circles.Apollonius' problem has stimulated much further work. Generalizations to three dimensions—constructing a sphere tangent to four given spheres—and beyond have been studied. The configuration of three mutually tangent circles has received particular attention. René Descartes gave a formula relating the radii of the solution circles and the given circles, now known as Descartes' theorem. Solving Apollonius' problem iteratively in this case leads to the Apollonian gasket, which is one of the earliest fractals to be described in print, and is important in number theory via Ford circles and the Hardy–Littlewood circle method.