Geometry in Sports
Jonathan Choate
Groton School
Groton, MA 01450
jchoate@groton.org
www.zebragraph.com
I. The Oldest Max-Min Problem of the Modern
Age
D
1. Using right angle trig to play with the problem to play with
the problem
i. <APB = <APO - <BPD
<APD = tan-1(a/x) , <BPD = tan-1(b/x)
<APB = tan-1(a/x) - tan-1(b/x)
2. Use calculus to find the max of f(x) = tan-1(a/x) - tan-1(b/x)
but this is a secondary talk so can’t do that.
3.Think about the graph of <APB as a function of x, the
distance from the wall.
<APB
X
3. Use GSP to play with problem
Big Observation: Assume that the largest angle has measure
M. For any value Q less than M there must be two points on
line PD where the viewing angle is the same. What must be true
about these two points and the points A and B? What must be
true about the point P where the max occurs and points A and
B?
4. So what?
i) Implications for sports
ii) Nice way to generate a hyperbola
5. What about the general case?
Interesting geometric question. How do you construct the circle
through two given points tangent to a given line that contains
neither of the two points?
II. Playing The Bounce
P1
P3
P2
In hockey, you can use the boards to make bounce passes.
Consider the case above where P! has the puck and wants to pass
the puck to P2 but an opposing player P3 is in the way. P1 can pass
the puck off the boards to P2. Where should P1 aim?
Do basketball players face a similar problem?
How about tennis players?
How about squash players?
P4''
P4'
s ide wall
P3
P4
P2
front wall
P1
III: The Discovery of A New Molecule
Rice University’s Richard Smalley would have had a much easier
time determining the shape of the new Carbon molecule he had
just discovered if he knew the following. His discovery is now
known as a Buckminsterfullerene molecule or a Bucky Ball.
1.Some Polyhedral Preliminaries
a) Regular polyhedra are three dimensional shapes
consisting of vertices (V), edges (E) and faces (F). All
the faces are congruent regular polygons. There are five
of these and they are known as the Platonic Solids. The
table below gives information about each of the five. The
take-out angle (TOA) for a regular polyhedral is equal to
360 minus the sum of the face angles surrounding a
vertex. For example, in a tetrahedron, three equilateral
triangles meet at each vertex so the take-out angle for a
tetrahedron is equal to 360 – [60 + 60 + 60] = 180
degrees.
Name
V
E
F
Tetrahedron
4
6
Cube
8
12 6
90
Octahedron
6
12 8
120
4
TOA
180
Dodecahedron 20 30
12
36
Icosahedron
20
60
12 30
2. Tools for Polyhedral Detectives
a) Euler’s Theorem in 3- space
If a polyhedron has V vertices, E edges and F faces then
V–E+F=2
b) Some Basic Counting
If a polyhedra with V vertices has n edges meeting at each
vertex then nV = 2E.
c) Descartes’ Theorem
The total angular take-out for any polyhedron always equals
720 degrees.
d) Week’s Theorem
If a polyhedra has more than one kind of polygonal face
then how many of each face can be found by calculating
(n V)/ p where n is how many of that type of polygon are at
each vertex, V is the number of vertices in the polyhedra
and p is the number of edges in the polygonal face. To see
what this says consider the polyhedra shown below.
In this polyhedra, two triangles and two squares meet at each
vertex so Descartes Theorem tells us that the take out angle is
360 – 300 = 60 so there are 720/60 =12 Vertices. If S denotes
the number of square faces and T the number of triangular faces
then Week’s Theorem tells us that
S = (2)(12)/4 = 6
and
T = (2)(12)/3 = 8
So there are 6 square faces and 8 triangular faces
3. How Professor Smalley Could Have Made His Life Easier
He suspected that the molecule consisted of hexagonal and
pentagonal rings so one possibility was that each vertex or atom
belonged to two hexagons and one pentagon. In what follows llet
H equal the number of hexagonal faces and P the number of
pentagonal faces.
Step 1: Find V using Descartes’ Theorem
Since hexagons have an interior angle of 120 º and pentagons
108º, the takeout angle is equal to 360 -[ 120+120+108] = 12º.
Therefore, the total takeout is equal to 12V so
12V = 720
V = 60
Therefore, Smalley’s polyhedron has 60 vertices.
Step 2: Find E using nV = 2E where n is the number of edges
per vertex
Since 3 edges meet at each vertex,
3V = 2E
3(60) = 2E
E = 90
Therefore, Smalley’s polyhedra has 90 edges.
Step 3: Find F using Euler’s Theorem
Since we know V and E now Euler’s Theorem tells us that
V –E + F =2
60 – 90 + F = 2
F = 32
and
H + P = 32
Step 4: Use Nu’s Theorem
Since the ratio of hexagonal faces to pentagonal faces equals
2:1, by Nu’s Theorem we have
6H:5P = 2:1
or
6H = 10P
We now have two equations in two unknowns. Solving these you
get H = 20 and P = 12.
Therefore a bucky ball has
60 Vertices
90 Edges
32 Faces
20 Hexagonal faces
12 Pentagonal faces
The Bucky Ball’s technical name is a truncated icosahedron. It is
an example of a semi-regular polyhedra. A semi-regular
polyhedron is one where all the faces are regular polygons and
each vertex has the same configuration of faces. The semiregulars are known as the Archimedian Solids and there are
thirteen of them. Here they are
Eric W. Weisstein. "Archimedean Solid." From MathWorld-A Wolfram Web Resource.
http://mathworld.wolfram.com/ArchimedeanSolid.html
References:
•The article on Regiomontanus can be found at
http://pup.princeton.edu/books/maor/sidebar_c.pdf.
•The article about buckminsterfullerenes can be found at
http://www.uh.edu/admin/engines/epi563.htm
•Where to sit in a movie theatre can be found at
http://www.gvsu.edu/math/calculus/M201/pdf/movie.pdf
The Problem
A Function Approach
In the above, <APB = <APO - <BPO
<APO = tan-1( a /x )
<BPO = tan-1( b /x )
Therefore,
<APB = tan-1( a /x )- tan-1( b /x )
Now you can find a solution in a variety of
ways: bisection method, calculus, etc.
Regiomontanus’s Problem
An Interesting Graph
Viewing Angle
Distance from Wall
•
An Application to Sports
Playing The Bounces
P1
P3
P2
P4''
P4'
s ide wall
P3
P4
P2
front wall
P1
A Quick and Easy Conic Generator
How about a proof?
In the diagram above PQ = y, QB=x-a,
AQ = x+a. Since PQ2 = AQ•QB,
y2 = (x+a)(x-a)
y2 = x2 - a2
and
x2 - y2 = a2
which is the equation of a rectangular
hyperbola with center at the origin.
A Very Interesting Table
Name
V
E
F
Tetrahedron
4
6
Cube
8
12 6
90
Octahedron
6
12 8
120
4
TOA
180
Dodecahedron 20 30
12
36
Icosahedron
20
60
12 30
Tools for Polyhedral Detectives
a) Euler’s Theorem in 3- space
If a polyhedron has V vertices, E
edges and F faces then
V–E+F=2
b) Some Basic Counting
If a polyhedra with V vertices has n
edges meeting at each vertex then nV
= 2E.
c) Descartes’ Theorem
The total angular take-out for any
polyhedron always equals 720 degrees.
e) Week’s Theorem
If a polyhedra has more than one
kind of polygonal face then how
many of each face can be found by
calculating
(n V)/ p where n is how many of
that type of polygon are at each
vertex, V is the number of vertices
in the polyhedra and p is the
number of edges in the polygonal
face. To see what this says
consider the polyhedra shown
below.
How Professor Smalley Could Have
Made His Life Easier
Step 1: Find V using Descartes’
Theorem
Step 2: Find E using nV = 2E where n
is the number of edges per vertex
Step 3: Find F using Euler’s Theorem
Step 4: Use Week’s Theorem
Therefore a Bucky Ball has
______Vertices
______________Edges
______Faces
______Hexagonal faces
______ Pentagonal faces