##### Theory

Analytic 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 180o.

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 Ma be the midpoint of side BC. Drop perpendiculars BD, MaE, and CF onto the given line.

MaE = (BD + CF)/2.         (1)

Let also AL be perpendicular to MN. Triangles ALG and MaEG are similar and GA = 2·MaG. Therefore, LA = 2MaE, 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 x1 and x2 from X and y1 and y2 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:

 Triangles Proportion Mx1 and My1 x/y = x1/y1 Mx2 and My2 x/y = x2/y2 Ax1 and Cy2 x1/y2 = AX/CY Dx2 and By1 x2/y1 = XD/YB

From which

x2/y2 = x1/y2· x2/y1 = x1/y1· x2/y2 = AX·XD/ CY·YB = PX·XQ/ PY·YQ.

So that

x2/y2 = (a - x)(a + x)/ (a - y)(a + y) = (a2 - x2)/(a2 - y2) = a2/a2 = 1

And finally x = y

##### Proposed Problems
1. The cone shown below has a radius of 8 cm and a height of 12 cm.

What is the volume of the cone?

1. What is the value of the expression 3|2 - 4| - 7?
2. Solve the inequality |x – 7 | < 8 for x.
3. Solve the following equation for x.  3x - (2x - 3) = 2x + 9

Read more on the written version of the publication.