Grade Level: Middle School/High School Class Title: Geometry
... 1. Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Translate between the geometric description and the equat ...
... 1. Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Translate between the geometric description and the equat ...
Geometry
... 2. Derive the equation of a parabola given a focus and directrix. 3. (+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. 3.1 Given a quadratic equation of the form ax2 + by2 + cx + dy + e = 0, use the m ...
... 2. Derive the equation of a parabola given a focus and directrix. 3. (+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. 3.1 Given a quadratic equation of the form ax2 + by2 + cx + dy + e = 0, use the m ...
Task - Illustrative Mathematics
... 180 degrees about M and then shows that the quadrilateral, obtained by taking the union of these two triangles, is a rectangle. The basic steps in this alternative argument are: This quadrilateral can be shown to be a parallelogram because of the 180 degree rotations used in its construction. Its di ...
... 180 degrees about M and then shows that the quadrilateral, obtained by taking the union of these two triangles, is a rectangle. The basic steps in this alternative argument are: This quadrilateral can be shown to be a parallelogram because of the 180 degree rotations used in its construction. Its di ...
Additional Exercises for Chapter 2 All angles in these exercises are
... A car moves along a straight stretch of highway. A passenger P looks out at an approaching billboard. Concentrate on the angle ∠ AP B between the passenger’s lines of sight to the left and right edges of the billboard. A look at the diagram below confirms that ∠ AP B first increases, but then decrease ...
... A car moves along a straight stretch of highway. A passenger P looks out at an approaching billboard. Concentrate on the angle ∠ AP B between the passenger’s lines of sight to the left and right edges of the billboard. A look at the diagram below confirms that ∠ AP B first increases, but then decrease ...
Geometry - Mountain Brook Schools
... reflections that carry it onto itself. (G-CO3) 4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. (G-CO4) 5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed ...
... reflections that carry it onto itself. (G-CO3) 4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. (G-CO4) 5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed ...
handout - English for Maths
... line of reflection symmetry and it has rotational symmetry around the centre for every angle. Its symmetry group is the orthogonal group O(2,R). The group of rotations alone is the circle group. ...
... line of reflection symmetry and it has rotational symmetry around the centre for every angle. Its symmetry group is the orthogonal group O(2,R). The group of rotations alone is the circle group. ...
Chapter 13 Geometry
... Failing this, the Man of Steel would like to fly between his current position and the target. He can see through objects, but not fly through them. His desired path flies straight to the goal, until it bumps into an object. At this point, he flies along the boundary of the circle until he returns to ...
... Failing this, the Man of Steel would like to fly between his current position and the target. He can see through objects, but not fly through them. His desired path flies straight to the goal, until it bumps into an object. At this point, he flies along the boundary of the circle until he returns to ...
Geometry 10-1 Circles and Circumference
... A. Parts of Circles 1. A circle is the locus of all points equidistant from a given point called the center of the circle. 2. A circle is usually named by its ________ point. 3. The circle below is called circle K, written ○K ...
... A. Parts of Circles 1. A circle is the locus of all points equidistant from a given point called the center of the circle. 2. A circle is usually named by its ________ point. 3. The circle below is called circle K, written ○K ...
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.