Download Section 1.1

§ 1.1 Homework
1. Use the following postulates to define a geometry. That is, list all of the points and all
of the line.
1. There exists at least one line.
2. Every line has exactly three distinct points.
3. Not all points are on one line.
4. There is at least one point on any two distinct lines.
5. Through any two distinct points there is
D
exactly one line.
Points: A, B, C, D, E, F, G.
E
G
th
Lines: ABC, need a 4 point giving ADE (but that requires
point E also) leading to BDF and CDG. Hence the 7 points
above. Now we need to complete all the lines using those
points using postulate 5. AFG, BEG, CEF.
(KNOWN AS FANO’S GEOMETRY.)
F
A
B
C
C
2. Every Triangle is Isosceles?
Here is a famous proof, first published by
G
F
W. W. Rouse Ball in 1892. It will show
E
that every triangle is isosceles.
A
D
B
Proof: Draw an arbitrary triangle ABC with AC > BC. We will prove that AC=BC.
Bisect the angle C, and draw the perpendicular bisector of side AB (bisecting AB at D).
These two lines (the bisector of angle C, and the perpendicular bisector of CD) cannot be
the same line or be parallel, as then the triangle ABC would be isosceles. So they must
meet at some point E. Draw the lines AE and BE. Draw the lines EF and EG,
perpendicular to sides AC and BC respectively, as shown in the diagram.
STATEMENT
REASON
Right triangles CEF and CEG are congruent.
HA
EF=EG and CF=CG.
CPCTE
Right triangles ADE and BDE are congruent.
SS
AE=BE.
CPCTE
Right triangles AEF and BEG are congruent.
SS
AF=BG.
CPCTE
AC=BC.
Equals added to equals.
Triangle ABC is isosceles.
THE PROBLEM IS WITH THE ORIGINAL CONSTRUCTION. POINT E IS
OUTSIDE THE TRIANGLE. THIS IS A WARNING TO BE VERY CAREFUL WHEN
YOU MAKE A DRAWING.
Why Do We Need Proof? (Mathematics Teacher, October 2013.)
3. Below are several circles with points on them.
a. In each circle draw all the chords possible.
b. Complete the following chart. For Number of Interior Regions we are looking for the
maximum. Equally spaced points will NOT give the maximum number.
Number of Points
Number of Chords
Number of Interior Regions
2
1
2
3
3
4
4
6
8
5
10
16
c. Generalize the chart by letting n be the number of points on the circle and
finding a formula for the number of chords and a formula for the maximum number of
interior regions.
If n is the number of points then the number of chords is (n(n + 1))/2. They are
triangular numbers. (See the first lesson for a proof of sum of the first n numbers by
induction.) The point of the exercise is that the number of interior regions at first
appears to be 2 n – 1 but indeed that is not the case for n = 6 (part d). One cannot do
proof from several cases.
d. Extend the chart by making a circle with 6 points on it and fill in the chart.
N = 6, number of chords = 15 and number of interior regions = 31. Count them
carefully. It is not 32!
4. The Centauri Challenge.
For the past sixty years, astronomers of the lunar colony have been communicating with
an intelligent life-form in the Centauri star-system. Because of the vast distances,
communication is very slow. It takes twenty years for a message to be sent and response
received. The most recent communication included The Centauri Challenge. The
Centauri Challenge was sent from logic students on the Centauri star system to students
at the lunar colony. The Centauri Challenge is shown below.
The Centauri Challenge
Centauri is a formal system of strings of the letters P, Q, R, S with four rules governing
their behavior. Using some combination of the four rules, it is possible to change one
string of letters to a different string. The four rules are:
Rule 1: any two adjacent letters in a string can change places with each other. (PQ>>QP)
Rule 2: If a string ends in the same two letters, then you may substitute a Q in place of
those two letters. (RSS>>RQ)
Rule 3: If a string begins in the same two letters, then you may add an S in front of those
two letters. (PPR>>SPPR)
Rule 4: If a string of letters starts and finishes with the same letter, then you may
substitute an R in place of all the letters between the first and last letters. (PQRSP>>PRP)
A theorem in this system is a string followed by >> followed by a string. For example,
PQQRSS>>QRQ says, if given string PQQRSS, then by using the rules you can arrive at
QRQ. A proof follows.
Example
Show:
PQQRSS>>QRQ
Proof:
PQQRSS
Given
PQQRQ
By Rule 2
QPQRQ
By Rule 1
QRQ
By Rule 4
Therefore, PQQRSS>>QRQ
Use the rules of Centauri to prove the following theorems.
1.
2.
3.
4.
5.
6.
PQPRQ>>RQ
PQRSSQR>>RQ
PSSRS>>RQ
PSRQQRSQPSSS>>RQ
PQQQQP>>RQ
QQQQQQ>>SRQ
Use the rules of Centauri to answer the questions below.
7. Can you find a string of five or more letters that cannot be reduced to RQ?
If yes, produce it. If not, prove you cannot.
8. One of the rules of Centauri can be removed without losing any of the first
five theorems proven above. Which one of the rules can be removed? Why
must this rule be used in the sixth theorem? Create another theorem that
must use this rule.
1. Show:
PQPRQ>>RQ
Proof:
PQPRQ
QPPRQ
Given
By Rule 1
QRQ
By Rule 4
RQQ
By Rule 1
RQ
By Rule 2
Therefore, PQPRQ>>RQ
2. Show:
Proof:
Proof:
PQRSSQR>>RQ
PQRSSQR
Given
QPRSSQR
By Rule 1
QPRSSRQ
By Rule 1
QRQ
By Rule 4
RQQ
By Rule 1
RQ
By Rule 2
Therefore, PQRSSQR>>RQ
PSRQQRSQPSSS
Given
SPRQQRSQPSSS
By 1
SRS
By Rule 4
RSS
By Rule 1
RQ
By Rule 2
Therefore,
PSRQQRSQPSSS >> RQ
5. Show:
PQQQQP>>RQ
Proof:
PQQQQP>>RQ
Given
PRP
By Rule 4
RPP
By Rule 1
RQ
By Rule 2
Therefore, PQQQQP>>RQ
6. Show:
QQQQQQ>>SRQ
Proof:
QQQQQQ
QRQ
QQR
SQQR
SQRQ
SRQQ
SRQ
3. Show: PSSRS>>RQ
Proof:
4. Show:
PSSRS
Given
SPSRS
By Rule 1
SRS
By Rule 4
RSS
By Rule 1
RQ
By Rule 2
Therefore, PSSRS>>RQ
PSRQQRSQPSSS>>RQ
Given
By Rule 4
By Rule 1
By Rule 3
By Rule 1
By Rule 1
By Rule 2
Therefore, QQQQQQ>>SRQ
7. Since there are only four letters, P, Q. R. S, one of them must repeat in any sequence of
five. Say it is P:
PxPxx
Given
PxxxP
By repeating Rule 1 – regardless of the location of the P’s.
PRP
By Rule 4
RPP
By Rule 1
RQ
By Rule 2
Therefore, xxxxx>>RQ
8. Rule 3 may be removed for the first six theorems.