Mathematics Magazine for Grades 112 



Grade 9 Issue 2/2005 



TheoryAnalytic
Geometry Analytic
geometry, also called coordinate geometry and earlier referred to as Cartesian
geometry, is the study of geometry using the principles of algebra. Usually the
Cartesian coordinate system is applied to manipulate equations for planes,
lines, curves, and circles, often in two and sometimes in three dimensions of
measurement. Some consider that the introduction of analytic geometry was the
beginning of modern mathematics. Analytic
geometry can be explained more simply: it is concerned with defining geometrical
shapes in a numerical way, and extracting numerical information from that
representation. The numerical output, however, might also be a vector or a
shape. René
Descartes introduced the foundation for the methods of analytic geometry in 1637
in the appendix titled GEOMETRY of the titled Discourse on the Method of
Rightly Conducting the Reason in the Search for Truth in the Sciences,
commonly referred to as Discourse on Method. Solutions from the Previous Issue:
1.
Angle bisectors in a quadrilateral form another quadrilateral that
is a quadrilateral that can be inscribed in a circle Solution:
Angle bisectors in a
quadrilateral form another quadrilateral that is cyclic, i.e. a quadrilateral
that can be inscribed in a circle. Indeed, the sum of opposite angles of the
latter is always 180^{o}. 2. In ABC, a line is drawn through centroid G. Assume the line intersects AB in M and AC in N. Then BM/MA + CN/NA = 1. Here, BM, MA, CN, NA are considered as signed segments. In a certain sense the identity even holds when the line in question is parallel to, say, AB. In this case, M is a point at infinity and BM/MA = 1 whereas CN/NA = 2. Conversely, if any line through a point P satisfies (1), then necessarily P = G.
Solution:
Let M_{a} be the midpoint of side BC. Drop perpendiculars BD, M_{a}E, and CF onto the given line. M_{a}E = (BD + CF)/2. (1) Let also AL be perpendicular to MN. Triangles ALG and M_{a}EG are similar and GA = 2·M_{a}G. Therefore, LA = 2M_{a}E, or LA = BD + CF. (2) Triangles BDM and ALM are similar, as are triangles CFN and ALN, from where we get BM/MA + CN/NA = BD/LA + CF/LA = (BD + CF)/LA = LA/LA = 1 from (2) Let's now tackle the converse. Assume point P is such that BM'/M'A + CN'/N'A = 1 (1’) holds for any line through P that intersects AB in M' and AC in N'. We have to show that P = G.. To this end assume that P is different from G and that the line is different from GP. Let MN passes through G parallel to M'N'. Then, with the reference to the diagram above, BM/MA + CN/NA < BM'/M'A + CN'/N'A. (3) This is because, in the diagram, BM/MA < BM'/M'A and CN/NA < CN'/N'A. For other locations of P or the straight line, the inequalities may have to be simultaneously reversed. (3) implies: BM/MA + CN/NA< 1 that contradicts the proven part.
3. Let M be the midpoint of a chord PQ of a circle, through which two other chords AB and CD are drawn; AD cuts PQ at X and BC cuts PQ at Y. Prove that M is also the midpoint of XY.  The Butterfly Theorem Solution:
Let's drop perpendiculars x_{1} and x_{2} from X and y_{1} and y_{2} from Y on AB and CD. Let's also introduce a = MP = MQ and x = XM and y = YM. Observe several pairs of similar triangles, which implies the following proportions:
From which x^{2}/y^{2} = x_{1}/y_{2}· x_{2}/y_{1} = x_{1}/y_{1}· x_{2}/y_{2} = AX·XD/ CY·YB = PX·XQ/ PY·YQ. So that x^{2}/y^{2} = (a  x)(a + x)/ (a  y)(a + y) = (a^{2}  x^{2})/(a^{2}  y^{2}) = a^{2}/a^{2} = 1 And finally x = y Proposed
Problems
What is the volume of the cone?


Read more on the written version of the publication. 
