Class Notes
EMAT 6600
February 28, 2007
Maximum Area of a Rectangle Inscribed a Circle
 Follow-up was to determine the maximum area of an ellipse inscribed in a circle
o (a = length of semi-major axis, b = length of semi-minor axis)
o Use the equation from Conic Sections (Standard form of an ellipse with
center at the origin) and pick some point from the first quadrant (m,n).
o Must it be case that the rectangles have sides parallel to the major and
minor axes? (Diagonals, or at least one, should lie along the major axis.)
o Maybe a GSP sketch would be useful to proving our conjectures. (Can
construct an ellipse from a circle construction or from script tools.)
o From GSP exploration, we have decided that the maximum area for a
rectangle inscribed in an ellipse is not a square and does not contain the
focal points
o We could also change the question to ask about rectangles that are
circumscribed about an ellipse.
o Conjecture: Square provides maximum area, rectangle with sides parallel
to the minor axis would have minimum area
o Follow-up: What fraction of the circumscribed rectangle’s area is covered
up by the ellipse? (Answer: pi/4) Similar question appears on the web-site.
o Could we intuitively derive the formula for an area of an ellipse?
Discussions on Previous Problems
 Euler’s Formula for three-dimensional objects
o Debate: What happens if we consider a cone a solid? What would be
considered an edge?
o Classroom Activity: Counting exercise with concrete examples.
(Example: a triangular pyramid—tetrahedron—has 4 vertices, 4 faces, and
6 edges, a cube has 6 faces, 8 vertices, and 12 edges, a square pyramid
would have 5 faces, 5 vertices, and 8 edges)
o Extension: How could we determine the number of edges, vertices, and
faces of an n-sided pyramid? (Answer: faces = n + 1, vertices = n + 1,
edges = 2n).
o Look at the “unusual solid” of triangles and squares. It has 14 faces, 12
vertices, and 24 edges. (Solid is the yellow shape in room).
o A possible hypothesis would be: Vertices + Faces = Edges + 2.
o How could demonstrate/justify/prove Euler’s Formula?
 Note: For any n-sided polygon, the sum of the angles is invariant
and equal to 180*(n – 2), if measuring in degrees, or pi*(n – 2), if
measuring in radians. (Could base the claim on the fact at that
there could be n – 2 number of triangles that could be made.)
 Each edge is counted twice since an edge of a polygon is contained
in the edges of the two polygons it is adjacent to.
 Let s1, s2, …, sm be the number of sides on the F faces. The sum of
the angles of this polygon is pi*(s1 – 2) + pi*(s2 – 2) + … + pi*(sm
– 2) = pi*(s1 + s2 + … + sm) - 2*pi*F = pi*(2E) – 2*pi*(F) =
2*pi*(E – F).
 Let one of the faces be a base of the solid. Collapse the rest of the
solid onto the base by stretching the base long enough. The sum of
the angles of the new shape is made up of the base, the measure of
the vertices that form the “rim” of the collapsed points, and the
angles in the middle of the base. Now the sum of the angles is now
pi*(r – 2). The rim would be pi*(r – 2). The middle angles have a
sum of 2pi*(V – r). This should also give us the total of the
measurement of the angles. 2*pi*r-4*pi+2*pi*V-2*pi*r =
2*pi*(V – 2).
 Now set the last two equations equal to each other: 2*pi*(E – F) =
2*pi*(V – 2), or E – F = V – 2. (Can be made to look any version
of Euler’s Formula.)
 Goal: Re-produce the proof of the formula to gain familiarity.
o Applications of Euler’s Formula
 Look at the Euclidean solids (Regular polyhedra). Look at the
different ways the regular shapes that could be formed. To do that
look at the possibilities that could occur at each vertex. There are
only five such possibilities: 3 triangles, 4 triangles, 5 triangles, 4
squares, and 3 pentagons.
 For 3 triangles, we could replace both the number of
vertices and faces with 2E/3 for each. Making the
appropriate substitutions, we would know that this regular
polyhedron will have exactly 6 edges, 4 vertices, and 4
faces (the tetrahedron).
 For 4 triangles, we could replace both the number of
vertices and faces with 2E/4 and 2E/3 respectively.
Making the appropriate substitutions, we would know that
this regular polyhedron would have exactly 12 edges, 6
vertices, and 8 faces.
 For 5 triangles, we could replace both the number of
vertices and faces with 2E/5 and 2E/3 respectively.
Making the appropriate substitutions, we would know that
this regular polyhedron would have exactly 30 edges, 12
vertices, and 20 faces.
 For 3 squares, we could replace both the number of vertices
and faces with 2E/3 and 2E/4 respectively. Making the
appropriate substitutions, we would know that this regular
polyhedron would have exactly 12 edges, 8 vertices, and 6
faces.
 For 3 pentagons, we could replace both the number of
vertices and faces with 2E/3 and 2E/5 respectively.
Making the appropriate substitutions, we would know that
this regular polyhedron will have exactly have 30 edges, 20
vertices, and 12 faces.
Polya Problems

Starting with a tautology, like the Binomial Theorem.
n 1  n
2
2
n 1  n
2
2
 2n 1 or
 2n 1, which tells us that the difference between consecutive
integers is always odd. Replacing n – 1 with n yields n2  n 1  2n 11,
 2
2
continually replacing yields the following n 1  n  2  2n  21all the
way down to 22 – 12 = 2*1 + 1. Adding up all of the different equations yields
2
n  1 12  21 2  ...  n n . Taking thebottom and solving for S yields the
1
2

2
n  1 1  2S1 S0
2

n 2  2n 11  2S1  n
n 2  2n  2S1  n

following calculations: n 2  n  2 S
 1
n2  n
 S1
2
o What is valuable in this calculation is the generalizations to other
exponents, like what is the sum of the first n squares:

n  1  n  3n  3n  1
n  n 1  3n 1  3n 1 1
n 1  n  2  3n  2  3n  2 1
3
3
2
3
3
2
3
3
2
23 13  3
1  3
1 1
n  1 1
3
3
 3S2  3S1  S0
n 2  n 
n 3  3n 2  3n  3S2  3
 n
 2 
2n 3  6n 2  6n  2n  6S2  3n 2  3n
2n 3  3n 2  n  6S2 
n 2n  1n  1

 S2
6
o What about the sum of the cubes and the fourth powers and so on. Find
the expression for S3 and S4.
Let’s look at the ways that points, lines, planes, and space divide each other. 0
 the line to create 1 line, 1 point will cut a line into 2 parts, 2 points
points will cut
will cut a line into 3 parts, 3 points will cut a line into 4 parts, and so on. Moving
onto lines, 0 lines will create 1 plane, 1 line will cut the plane into 2 parts, 2 lines
will cut the plane into 4 parts, 3 lines will cut the plane into 7 parts, 4 lines will
cut the plane into 11 parts, 5 lines will cut the plane into 16 parts, and so on.
Moving onto planes, 0 planes will create 1 space, 1 plane will cut space into 2
regions, 2 planes will cut space into 4 regions, 3 planes will cut space into 8
regions, and so on. Is there a way to make a connection to the point and the
number of parts of a line it creates?/the line and the number of parts of a plane it
cuts?/the plane and the number of regions of space it cuts? (Where would look in
Pascal’s triangle to find the numbers for this problem?)