Download Axiomatic Systems/Fe

Final Exam – Review Questions
Chapter 1
1) Kevin M.: Fe Fo System:
Ax 1- Exactly 3 distinct Fe’s exist.
Ax 2- Two Fe’s belong to exactly one Fo.
Ax 3- Not all Fe’s belong to the same Fo.
Ax 4- Any two Fo’s contain at least one Fe which belongs to both.
Prove the Theorem: For every set of two distinct Fe’s, every Fo in the system must
contain at least one of them.
2) Cheryl H.: Suppose that point P is not on line L. Prove that there is only one line
through P that is perpendicular to L
3) Jason et al:
In the triangle ABC below, segment AD is an extension of segment CA. Angle BAD is
considered an exterior angle of this triangle. In Euclidean Geometry, prove that the
measure of angle BAD is equal to the sum of the measure of angle BCA and angle CBA.
4) Jason et al:
Axiomatic Systems/Fe-Fo Review Problem
Consider the following Finite Geometry System:
Axiom 1. There exists exactly 6 Fo’s in the system.
Axiom 2. Every Fe contains exactly 3 Fo’s.
Axiom 3. No two distinct Fe’s share more than one Fo.
a) Model this system in a creative way and sketch. Label all parts of your system and
provide a key (i.e. Fe = Dog & Fo = Leash).
b) State 2 conjectures/theorems based on the axioms above that could most likely be
proven to be true.
c) Choose one of your conjectures and justify using the axioms above.
5) Danielle B.:Fe-Fo Type example
Undefined terms: point, line
Axiom 1: There exists exactly 5 points.
Axiom 2: Every set of two distinct points lies on exactly one line.
Axiom 3: Each line contains exactly two points.
Write a conjecture and justify it using only the axioms.
Chapter 2
6) Christyn C: Use the converse of Ceva’s Theorem to prove that the three medians of a
triangle are concurrent.
7) Sara B: Given M is the midpoint of BC, prove angle BCD > angle ABC.
B
E
M
A
C
D
8) Kate M.: Prove that every triangle is cyclic and that the circumcenter is the
concurrency of the perpendicular bisectors
9) Jason et al:
Given: Line l is parallel to line m. Q is a point on line l and P and R are points on line m.
Construct PQR and calculate the area of the interior of the triangle. Now move point Q
along line l. What do you notice about the area of the triangle? What is happening and why?
l
Q
m
P
R
10) Jason et al: Prove that the perpendicular bisectors of a triangle are concurrent. (Hint:
Let O be the intersection of two of the perpendicular bisectors. By finding congruent
triangles, prove that the line through O perpendicular to the third side is also a bisector
11) Prove that a triangle is isosceles if and only if the base angles are congruent.
Chapter 3
12) Chris P et al: Given: Chords AB and CD, intersecting at point E in the interior of
circle O. Prove: (AE)(EB) = (CE)(ED)
13) Chris P et al:
14) Chris P et al:
Complete the proof of Theorem 3.4 by considering the case where l intersects C.
We are given a circle C, with center O and radius r, Let l be a line that does not go through
O.
The inversion of l will be a circle through point O when l does intersect C.
P
P*
O
T'
T
r
15) Chris P et al:
Let T and U be the points where
and
Prove that PTRU is a parallelogram.
intersect the smaller arcs of the arbelos.
R
R
T
T
U
U
B
A
P
P
S
16) Alex B.:Given: Circle T and circle Y are orthogonal at point G. Prove that there are
perpendicular tangents at the other point of intersection of the circles.
17) Alex B: We talked a lot about the nine point circle in class. What are the nine points
located on the nine point circle?? There are 4 more points on the nine-point circle.
What are they??
18) Alex B: Given a circle C and a point P, state the definition of the Power (P, C).
19) Alex B: What will happen to the value of Power (P, C) when P is interior to C?
Exterior to the circle? On the circle?
20) Alex B: State the definition of the inversion of P with respect to C (denoted Inv (P, C))
given that C is a circle centered at O and P is a point in the plane (except O).
21) Alex B: Describe the location of P’ as P is exterior to the circle, interior to the circle,
and on the circle
22) Alex B: When talking about circles, what is the difference between a secant and a
chord?
23) Alex B: Write the definition and name of the center for the following circles:
A. Incircles
B. Excircles
C. Circumcircles
24) Alex B: What is the Inscribed Angle Theorem?
Chapter 4
25) Matt M.: Using coordinates, prove that the midpoint of the hypotenuse of a right
triangle is equidistant from all of the vertices
26) Laura F:
i) Construct an arbitrary kite on a Cartesian coordinate system. Prove that your
construction is a kite.
ii) (Using your figure and analytic geometry, prove that the longer diagonal of a
kite bisects the shorter one.
27) Kristen et al: Situate a triangle on the coordinate plane and label appropriately. Ensure
that your labeling is general enough to apply to all triangles. Write a coordinate proof
that a midline of the triangle (formed by joining the midpoints of two sides) is parallel
to the other side of the triangle and has half the length of that side.
28) Kristen et al: Use coordinates to prove that the midpoint of the hypotenuse of a right
triangle is equidistant from all of the vertices.
29) Kristen et al: By using the xy-plane construct lengths of 1, 1/3, SQRT (2), SQRT (3).
30) Kristen et al: Use coordinates to prove that the diagonals of a rhombus are
perpendicular bisectors of each other (Note: There are two things to prove)
31) Britt G.:Use coordinates to prove that the midpoint of the hypotenuse of a right
triangle is equidistant from all vertices
32) Paul D.: What is the formula for distance between two points in R2 and from which
theorem is it derived?
Chapter 6
33) Brittany D: Consider a triangle ABC with the vertices (-1, 1), (2,0), (-2, -1). Perform
the following operations to these points. What happens to this triangle when it is put
through the functions? Calculate the vertices and tell what type of isometry it is, or
explain why they cannot be calculated
a) F(x,y) = f(y,x)
b) F(x,y) = f(x+5, y-2)
c) F(x,y) = f(2x, 2y)
34) Amanda H: Given the function f(x,y) = (3y, x-2), decide if f is a transformation and
whether or not f is an isometry.
35) Jacci: Let f(x,y) = (2x-3, ½y + 2). Determine whether this function is a transformation.
Is it an isometry?
36) Jacci: Prove the composition of translations is commutative.
37) Jacci: Develop a transformational Proof that the vertical angles formed by two
intersecting lines are congruent.
38) Iracé y Meaghan: Consider the functions
such that
and
Let
be such that
.
a) Is j a transformation?
b) Is j a similitude? If yes, which is the dilator factor?
c) Is j a dilation?
d) Is j an isometry? If yes, which one? Which are its elements?
39) Iracé y Meaghan: TRUE or FALSE? Give a proof or a counterexample when it
corresponds.
a) Let A’ be the image of A in a rotation center at O. The angle bisector of the
is included in the perpendicular bisector of the segment
.
b) The reflection conserves slope.
c) Consider a fixed lone a. Call T to the set of all the translations in the plane such as
the vector of translation is parallel to a. Then,
is a subgroup.
d) The composition of two reflections is a translation
40) Iracé y Meaghan: Given a line a and a point
Show: A Rotation centered at O
and with angle 30 composed with a reflection with mirror a, is a reflection that has as
mirror a line that intersects a at o and that determines an angle of 15 with a.
Chapter 9
41) Alex S.: Determine which of the following properties of lines in Neutral Geometry are
also true in Spherical Geometry:
True or False? If false, why?
a) Two points determine a line.
b) Lines contain an infinite number of points.
c) Lines have infinite length.
d) When two lines intersect, their intersection is exactly one point.
e) Parallel lines exist.
f) The angle sum of a triangle is < 180 degrees.
42) Alex S.: Neutral Geometry Question:
Prove: The diagonals of a Saccheri Quadrilateral are congruent.
43) Alex S.: Hyperbolic Geometry: Show that the defects of a regular polygon are
additive.
44) Alex S.: From a given line (semicircle) centered at (5,0) with radius 3 on the halfplane model, find the equation of this line and another line that intersects this line at
(7,
with radius 7.
45) Arlona/AshleyA.: Spherical Geometry Questions
1. What is the maximum number of lines you can have with no intersections?
a. With exactly one intersection?
b. With exactly two intersections?
c. With exactly three intersections?
2. What is the maximum sum of the degrees in a triangle?
3. How can you prove triangles are congruent in spherical geometry?
46) Arlona/Ashley A: Hyperbolic Geometry Questions
1. How can you prove triangles are congruent in hyperbolic geometry?
2. What is the maximum and minimum sum of the degrees in a triangle?
3. If line M and linen N are both parallel to the same line L, show M is parallel to N.
Final Exam – Review Question Solutions
Chapter 1
1) Proof: Pick two Fe’s call them A and B that belong to Fo X. By axiom two A and B
cannot both belong to another Fo. By axiom four, WLOG, we can say that Fo Y
contains A and the third Fe called C. We can then say the third Fo called Z must
contain B and C because Ax-2 states B and C must belong to a distinct Fo. Therefore
for every set of two distinct Fe’s, every Fo in the system must contain at least one of
them.
2)
Draw line L
Draw a point P not on line L
Draw a line through point P and line L, label line 1
Draw another line through P and line L, label line 2
Assume that line 1 and line 2 form perpendicular angles with line L. Call the intersection
of Line 1 and Line 2 with line L points A and B respectively.
Triangle ABP is formed by lines L, 1, and 2
Recall that a triangles three angles must sum to 180 degrees (in Euclidean Geometry)
Angle A =90, angle B=90, angle P=180-(A+B).
P=0 degrees when angle A and B are both 90 degrees, which means that triangle ABP can
not be formed with two perpendicular lines
Therefore, there exists only one line perpendicular to line L through point P.
3) Let x represent x be the measure of angle BAC. Angle BAC and angle BAD are
supplementary angles, which means that their total sums to 180 degrees. Therefore
the measure of angle BAD can be represented as 180 degrees – x (the total sum of the
supplementary angles minus the measure of the other angle.) Angles BAC, CBA, and
CAB make up the tree angles of a triangle. In Euclidean geometry, the sum of the
three angles of a triangle is 180 degrees. Let y be the measure of angle CBA and z be
the measure of angle BCA. The sum of the three angles can be expressed as
x+y+z=180 degrees and simplified to y+z (the sum of angle CBA and CAB)= 180-x.
Therefore angle BAD and the sum of angle CBA and CAB both equal 180 degrees- x.
4)
Dog 1
Dog 6
Dog 5
Dog 2
Dog 4
Dog 3
a) Each line above represents a leash (or Fo) “Dog 1-6” represents the “Fe’s”
b) 1. There are exactly 9 leashes (Fo’s) in the system. 2. No dog (Fe) has every leash
in the system connected to it.
c) Justification for Conjecture 1:
By Axiom 1, there are exactly 6 dogs in the system. Let’s label them {Fido (Dog 1),
Rover (Dog 2), Fluffy (Dog 3), Prince (Dog 4), Buddy (Dog 5), and Louis (Dog 6)}. By
Axiom 3, we know that no two dogs share more than one leash. This explains that every
combination of two dogs have either 1 or 0 leashes connecting them.
Let’s display the possible combinations of dogs:
Fido ---- Rover
Rover ---- Fluffy
Fluffy ---- Buddy
Fido ---- Fluffy
Rover ---- Prince
Fluffy ---- Louis
Fido ---- Prince
Rover ---- Buddy
Prince ---- Buddy
Fido ---- Buddy
Rover ---- Louis
Prince ---- Louis
Fido ----Louis
Fluffy ---- Prince
Buddy ---- Louis
By Axiom 2, every dog has exactly 3 leashes connected to it. So, in the above list some of
these combinations must be eliminated. Therefore, there are exactly 9 leashes in the
system. (6 x 3 / 2)
5) Conjecture: There are exactly 10 lines.
Proof:
By axiom 1 we know that there are 5 distinct points (A,B,C,D and E). By axiom 2 we
know that every set of two distinct points lies on exactly one line.
So we have lines AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.
This is exactly ten lines because by axiom 3 we know each line contains exactly two
points
Chapter 2
6) Construct an arbitrary triangle ABC.
B
X
C
Z
Y
A
Construct the midpoint of each edge BC, AC, and BA. Label these points X, Y, and Z
respectively. Construct the segments AX, BY, and CZ. A median of a triangle is a
segment whose endpoints are a vertex of the triangle and the midpoint of the opposite
side. Thus, segment AX, BY, and CZ are the medians of triangle ABC.
According to Ceva’s Theorem and its Converse, AX, BY, and CZ are concurrent if and
only if
*
*
= 1.
Because a midpoint is a point that divides a segment into two congruent segments, AZ
ZB, BX
XC, and CY
YA.
By using substitution in the converse of Ceva’s Theorem,
*
*
= 1.
By cancelling we have 1*1*1 = 1. This results in 1 =1 which is a true statement. This
means that AX, BY, and CZ are concurrent. These segments were defined as the medians
of the triangle ABC. Thus, the three medians of a triangle are concurrent.
7) We are told that M is the midpoint of BC (which means segments BM and MC are
congruent) and segments AM and AE are congruent. Since angle BMA and angle
CME are vertical angles, they are congruent. We can then say that triangle BMA is
congruent to triangle CME by SAS. From this we can conclude that angle ABM is
congruent to angle ECM. Angle BCD is the sum of angle ECM and angle ECD.
Since angle ABM and angle ECM are congruent, it must be smaller than this sum.
Therefore, angle BCD > angle ABM.
To prove the other part of the Exterior Angle Theorem, one might use this set up: since
angle ACB and angle DCB form a straight line, angle ACB + angle DCB=180 degrees.
Because they are in a triangle, angle BAC+ angle ABC+ angle BCA=180 degrees. Using
substitution, angle DCB=angle BAC+ angle ABC. Since neither angle BAC nor angle
ABC are 0 degrees, angle DCB is greater than angle BAC (and angle ABC).
8) Start with three noncollinear points A, B, and C. In order to form a circle through all
of these points, there must be a point, O, such that the distance from O to each of the
three points (A, B, and C) are equivalent. Therefore, by the definition of a
perpendicular bisector, O must lie on the perpendicular bisector of segments AB and
BC. In order for this to be possible, the perpendicular bisectors of segments AB and
BC must not be parallel, and therefore segments AB and BC must not be parallel.
However, since segments AB and BC intersect each other at point B, AB and BC are
not parallel and their corresponding perpendicular bisectors must intersect. These two
perpendicular bisectors intersect at O, making O equidistant from A, B, and C.
Construct the circumcircle of triangle ABC with center at O. It is clear that O is the
concurrency point of segments AB and BC. O is equidistant from points A and C,
thus by the definition of a perpendicular bisector, O must lie on the perpendicular
bisector of segment AC. Therefore, every triangle is cyclic and the circumcenter will
be the point of concurrency of the three perpendicular bisectors.
9) Sketchpad problem (Chapter 2 problem number 5):
The area of the triangle remains constant no matter where Q is on line l. This occurs
because the area of the triangle is ½(base)(height). Because P and R are staying constant
on line m, the length of the base (segment PR) remains constant. Also the height, by
definition, of the triangle is the perpendicular distance from a vertex to the opposite side
of the triangle (or the line containing the triangle). Since the points of the triangle lie on
two parallel lines, the altitude (height of the triangle) will always be the perpendicular
distance from one line to the other. Thus the height remains constant no matter where Q is
on line l. Therefore, since the length of the base and height remain constant, the area
remains constant as well.
10)
A
R
B
O
T
S
C
Construct arbitrary triangle ABC like shown. Construct perpendicular bisectors to sides
AB and BC, at points R and S respectively. The two perpendicular bisectors will intersect
at a point in the interior of the triangle, say O. Thus, segments AR and BR are congruent,
and likewise, segments BS and CS are congruent since the sides were bisected. In
addition, angles ORA, ORB, OSB, and OSC are all right angles because of the
perpendicular bisectors, and are likewise congruent to one another. By reflexivity, OR =
OR, and OS = OS. Thus, by SAS congruency, we can say that triangles ORA and ORB
are congruent, as well as triangles OSB and OSC. Therefore, by CPCTC, AO = BO, and
BO = CO. By transitivity, AO = CO. Now, construct a line segment perpendicular to AC
that passes through point O intersecting AC at point T. We want to show that this line is
also a perpendicular bisector of AC. From earlier, we stated that AO = CO. By looking at
triangle AOC, since two adjacent sides are congruent, the triangle is isosceles. Therefore,
since AOC is isosceles, we can say that angles OAC and OCA are congruent. In addition,
since OT is perpendicular to AC, we know angles OTA and OTC are right angles, and are
thus congruent. Furthermore, OT = OT by reflexivity. We can also say that angles AOT
and COT are congruent since the other two pairs of angles of each corresponding triangles
are congruent. Thus, triangles OTA and OTC are congruent. By CPCTC, AT = CT.
Therefore, OT is also a perpendicular bisector. Finally, this shows that the three
perpendicular bisectors of a triangle are concurrent.
11) First assumer triangle ABC is isosceles; we wish to prove that the base angles are
congruent. By definition AB is congruent to AC and Ac ids congruent to AB. Angle
BAC is congruent to Angle CAB, since they are the same angle, so triangles ABC and
ACB are congruent by SAS. Angle ABC is congruent to Angle ACB since they are
corresponding angles and therefore the base angles in an isosceles triangle are
congruent.
Next assume that the base angles are congruent, we wish to prove that the triangle is
isosceles. Since the base angles are congruent, Angle ABC and Angle ACB are
congruent. Side BC is congruent to itself, so triangle ABC is congruent to triangle
ACB by ASA. Therefore by CPCTC, AB is congruent to AC. Thus triangle ABC is
an isosceles triangle
Chapter 3
12) Construct ACE and BED. We can now conclude that m<CEA = m<BED because
they are vertical angles. We can also say that m<DBA = m<DCA because they
intercept the same arc, AD. The measure of angle BDC = measure of angle CAB
because they both intercept the same arc, CB. Now ACE ~ BED by Angle-AngleAngle theorem. Since these triangles are similar, we can say that corresponding sides
AE CE
=
are proportional. Thus,
. We can rewrite this proportion as (AE)(EB) =
ED EB
C
(CE)(ED).
A
E
O
D
B
13)
14) We want to show that for any point P on l, point P’=Inversion(P,C) lies on some circle
that goes through point O.
•
•
•
We start by constructing a line l that intersects Circle C. We then construct a point P
on line l. Next we will create a perpendicular line from line l that goes through point
O. We will label this intersection point T. Next, we construct T’ = Inversion(T, C).
Then we will construct the circle with diameter OT’.
We label the intersection of the circle and the ray OP as P*
Next we will show that the triangle OPT is similar to OT’P* by Angle-Angle.
A1
: OPT and OT’P* both share Angle O, and by reflexive property, Angle
O is congruent to itself.
A2 : OTP is a right angle because OT is perpendicular to TP and OP*T’ is a
right angle because P* lies on the circle and is opposite the diameter making P*
a right angle.
•
Because OPT is similar to OT’P*, we can say that
•
By Cross-Multiplication
(Equation for Inversion)
Therefore, by substitution
•
and we know that
•
15) We can show that ARB is a right angle because R lies on the circle with diameter AB
and has line segments going from R to the endpoints of the diameter.
Similarly, we can show that ATP is a right angle because T lies on the circle with
diameter AP and has line segments going from T to the endpoints of the diameter.
Because Angle ATP is a linear pair with PTR on the line
, the angles must be
supplementary. Because ATP equals 90 degrees, then PTR, must equal 90 degrees.
Similarly, we can show that PUB is a right angle because U lies on the circle with
diameter PB and has line segments going from U to the endpoints of the diameter.
Because Angle BUP is a linear pair with PUR on the line BR, the angles must be
supplementary. Because BUP equals 90 degrees, then PUR, must equal 90 degrees.
By the parallel postulate, we can say that
and
are parallel because the sum of the
interior angles at the transversal RT sum to 180 degrees.
Similarly, by the parallel postulate, we can say that
and
are parallel because the
sum of the interior angles at the transversal
sum to 180 degrees.
A Quadrilateral is a Parallelogram if and only if there exists two pairs of parallel line
segments.
We have shown that
and
is one pair of parallel lines and that
pair, therefore we have two pairs of parallel line segments.
and
is another
16) Let H be the other point of intersection of circles T and Y. Because segment YG and
segment YH are both radii of circle Y, YG is congruent to YH. Similarly, TY is
congruent to TG because they are both radii of circle T. By the reflexive property, we
know that TY is congruent to TY. By combining these three pieces of information, we
know that triangle TGY is congruent to triangle THY by SSS. By CPCTC, we know
that angle TGY is congruent to angle THY. Thus they are both right angles proving
that there are perpendicular tangents at both intersections of orthogonal circles.
17) The 3 feet of the altitudes
The midpoints of the 3 sides of the triangle
The 3 midpoints of the segments joining the vertices of the triangle to the orthocenter
The 4 points where the NPC is tangent to the incircles and the excircles
18) The Power (P, C) is a function that depends on the radius r of C and d, which is the
distance of P from the center of C. The function is defined by Power (P, C) = d² - r².
19) The Power (P, C) will become negative since d will become a smaller distance than
the length of r when P is interior to C. When P is exterior to C, this value will be
positive since d will become larger than the length of r. When P is on the circle, then
the Power (P, C) will be zero since the length of d will be the same as the length of r.
20) For any point P, the inversion of P with respect to C is the point P’ on ray OP such
that |OP| * |OP’| = r², where r is the radius of C.
21) When P is exterior to the circle, P’ will be interior. Since P is exterior, than r²/|OP| < r.
Then P’ will be interior of the circle. When P is interior to the circle, P’ will be
exterior. When P is on the circle, then P’=P.
22) A secant is a line that passes through two points on a circle. A chord is a line segment
that connects two points on a circle
23) Solutions
A. Incircles are circles that are inscribed inside of a polygon (tangent to the sides).
Incircles are centered at the incenter, which is found by the intersection of the
angle bisectors of a polygon.
B. Excircles are external to polygons. They are formed by extending the sides of a
polygon and then creating a circle tangent to two extended sides.
C. Circumcircles are circles that have all points of a given polygon on the
circumference. The center of a circumcircle is the circumcenter which is located
at the intersection of the perpendicular bisectors of a polygon.
24) The Inscribed Angle Theorem states that the measure of an inscribed angle (an angle
formed by two secant lines that intersect on the circumference of a circle) of a circle
equals ½ the measure of its arc
Chapter 4
25) Solution:
B(0,b)
B(0,b)
m
m
A(0,0)
C(a,0)
D(a,b)
A(0,0)
C(a,0)
Diagram 1
Diagram 2
By constructing a right triangle with vertices at (0,0), (a,0), and (0,b), we need to prove
that the distance from the midpoint, M, of the hypotenuse of triangle ABC is equidistant
from all of the vertices.
Because M is the midpoint of the hypotenuse (segment BC) of triangle ABC, we know by
the definition of a midpoint that the measure of segment BM is equal to the measure of
segment CM, so therefore m is equidistant from B and C.
To prove that the midpoint of the hypotenuse of triangle ABC is also equidistant from
point A we need to find the measure of segment AM and compare this measurement with
the measure of segment BM or CM, which we already know are the same distance.
By constructing point D given by the coordinates (a,b) we now have a rectangle,
quadrilateral ABDC by our construction. (diagram 2) Because ABDC is a rectangle we
know that the diagonals, represented by segment DC and the hypotenuse of triangle ABC,
segment BC, intersect at the midpoints of one another and they are all congruent in length,
meaning that the measure of segments AM, BM, DM, and CM are all equal to one
another. This means that the vertices of right triangle ABC are all equidistant from the
midpoint, M, of the hypotenuse.
If we calculate these distances using the distance formula and the given coordinates we
prove this once again.
We know that the measure of segment BM is equal to the measure of segment CM and
that, because M is the midpoint of BC, this measurement equals:
Because M is the midpoint of segment AD, the measure of segment AM is equal to half of
the measure of segment AD, which equals:
This measurement of segment AM is equal to that of segments BM and CM, so we have
proved that the midpoint of a right triangle is equidistant from all of the vertices.
26) (i) ANSWER:
To prove this figure is a kite, we must show that each pair of adjacent sides is…
(ii)Create the diagonals of kite ABCD. We must draw segments from A to C and B to D.
27) Solution
The midpoint of two points can
be expressed as ((x1 + x2)/2,
(y1 + y2)/2)
Thus, the midpoint of BA, D, is ((x1 + 0)/2, (y1 + 0)/2) which is ((x1)/2, (y1)/2)
Similarly, the midpoint of CA, E, is ((x2)/2, (y2)/2)
The slope of BC is (y2 - y1)/(x2 - x1)
The slope of DE is ((y2 - y1)/2)/((x2 - x1)/2)
Simplified, the slope of DE is (y2 - y1)/(x2 - x1) which is the slope of BC
Thus, the midline DE is parallel to the other side, BC, of the triangle
By the distance formula, d(B,C) = sq. rt[(x1 - x2)^2 + (y2 - y1)^2]
Since we are working in the 90 degree coordinate plane, the term -2|x1 - x2||y1 y2|cos(90) = 0. Therefore it is not factored into the equation.
d(D,E) = sq. rt[((x1-x2)/2)^2 + ((y1 - y2)/2)^2]
= sq. rt[((x1 - x2)^2)/4 + ((y1 - y2)^2)/4]
= sq. rt[(1/4){(x1 - x2)^2 + (y1 - y2)^2}]
= sq. rt[1/4] sq. rt[(x1 - x2)^2 + (y1 - y2)^2]
= (1/2)sq. rt[(x1 - x2)^2 + (y1 - y2)^2]
= (1/2) d(B,C)
Therefore the midline is also half the length of the parallel side.
28)
(0,a )
A
M (1/2 b , 1/2 a)
(0,0 )
B (b,0 )
C
Prove:
The distances from each vertex
(A,B,C) to the midpoint of the hypotenuse are equidistant.
Let ∆ABC be a right triangle with the vertices located at the following with vertex A
located at (0,a), vertex B located at (b,0), and vertex C located at (0,0). Let AB be the
hypotenuse of ∆ABC. Let M be the midpoint of hypotenuse AB. Therefore, the
coordinates of point M are
( ½ b, ½ a). So, using the distance formula, the distances
from each vertex to midpoint M can be calculated as follows:
Distance from vertex A to midpoint M is:
SQRT( ( ½ b - 0)^2 + ( ½ a – a)^2) = SQRT( ¼ b^2 + ¼ a^2)
Distance from vertex B to midpoint M is:
SQRT( ( ½ b – b)^2 + ( ½ a – 0)^2) = SQRT( ¼ b^2 + ¼ a^2)
Distance from vertex C to midpoint M is:
SQRT( ( ½ b – 0)^2 + ( ½ a – 0)^2) = SQRT( ¼ b^2 + ¼ a^2)
So, since the distances from A,B,C to midpoint M are all equal to SQRT( ¼ b^2 + ¼ a^2),
then the midpoint of the hypotenuse of a right triangle is equidistant from all of the
vertices.
29)
30)
6
4
B (a , b -a)
C (b+a , b -a )
2
-10
-5
A (0 ,0)
D ( b, 0 )
5
-2
-4
First, we will prove that the diagonals bisect each other.
Using the midpoint formula,
Midpoint AC= (0+(b+a), 0+(b-a)) = (a+b, b-a)
2
2
2 2
Midpoint BD= (a+b), (b-a)+0) = (a+b, b-a)
2
2
2 2
Since the midpoints of AC and BD are equal, the two lines must intersect at that point and
thus bisect each other.
Next we will prove that the diagonals are perpendicular.
Using the slope formula,
Slope AC= (b-a)-0 = b-a = b-a
(b+a)-0 b+a
b+a
Slope BD= 0-(b-a) = -(b+a)
b-a
b-a
Since the slopes are opposite reciprocals, AC and BD are perpendicular
31) Solution
Let ∆ABC be a right triangle located on the coordinate plane with the following vertices.
A= (0,0)
B
B= (0,b)
C= (c,0)
A
C
10
Construct the midpoint of the hypotenuse BC and let that point be point D.
B
D
D= [(c +0)/2, (0+b)/2]
D= (c/2, b/2)
A
C
Now show that the segments DB, DC and DA are all congruent using the distance
formula.
DB=
DB =
DB= 1/2
Similarly show that DC and DA also equal 1/2
. Therefore points A, B, and C are
all equidistant from the midpoint of BC. Thus proving the midpoint of the hypotenuse is
equidistant from each of the vertices of a right triangle.
32) dist[(x1,y1), (x2,y2)] = sqrt[ (x1 - x2)^2 + (y1 - y2)^2 ] is derived from Pythagorean’s
Theorem.
Chapter 6
33) Solution
a) F(-1, 1) = (1, -1)
f(2,0)= (0,2)
f(-2, -1) = (-1, -2)
The triangle will be reflected to a congruent triangle
b) b) F(-1, 1) = (4, -1)
f(2,0)= (7,-2)
f(-2, -1) = (3, -3)
The triangle will be translated right 5 and down 2
c) This functions cannot be performed because the is it clear to see that distance will
not be preserved there this cannot be an isometry.
34) Transformation?: (onto and one-to-one)
Onto: Can I obtain an arbitrary point (a,b)? Plug in (b+2, 1/3a) =>
(3(1/3a), (b+2)-2)= (a,b), therefore, f is onto!
One-to-One: Assume f is not one-to-one, then f(a,b)=(m,n)=(3b, a-2)
f(c,d)=(m,n)=(3b, a-2)
with (a,b) /= (c,d)
therefore f(c,d)=(3d, c-2) which must equal (3b, a-2)
however, 3d=3b
c-2=a-2
d=b
c=a
contradiction b/c (a,b) /= (c,d)!!!
therefore f is one-to-one!!!
Because f is onto and one-to-one f is a transformation
Isometry?: (distance preserving)
Distance from (a,b) to (c,d)
D1=√(a-c)^2+(b-d)^2
Distance from (3b, a-2) to (3d, c-2)
D2=√(3b-3d)^2+([a-2]-[c-2])^2
D2= √(3(b-d))^2+(a-c)^2
Distance is not preserved, therefore f is NOT an isometry!
35) Solution
Yes. It is a transformation. First check if it is one-to-one: Given 2x-3 and 2z-3,
assume x ≠ z. Then f will not be one-to-one. However, given 2x-3 = 2z-3  x = z.
This contradicts that f is not one-to-one, so f must be one-to-one.
Next, check if f is onto:
Let a=2x-3........ let b= 1/2y+2. x = (a+3)/2 y= 2(b-2). f(x,y) = (2(a+3)/2)-3)
, 1/2(2(b-2) +2))
= (a,b). Since (a,b) is mapped then f is onto. Hence f is a transformation.
Is f an isometry? No f is not an isometry. Let x = (1,0) and let y = (1,2) .........
distance of xy = 2
x' = (-1, 2) and y' = (-1, 3)............... distance of x'y' = 1. Since distance isn’t
preserved f isn’t an isometry.
36) Let f represent a translation by vector
vector
. Then f ◦ g =
. Let g represent a translation by
Similarly, g ◦ f =
The image under f ◦ g = the image under , g ◦ f, hence a composition of
translations is commutative.
37) Two intersecting lines have a line of reflection through their intersection point. Based
on the fact that a reflection is an isometry and that by definition isometries preserve
angle measure, the two vertical angles are congruent.
38) Iracé y Meaghan:
a) Is j a transformation?
We know from homework that s is a reflection and t is a translation. Then, because the set
of all the isometries with the composition form a group, the composition is closed in that
set. Then, j is an isometry. Therefore, j is a transformation.
b) Is j a similitude? If yes, which is the dilator factor?
Since a), j is an isometry. Then, j is a similitude with dilator factor 1
c) Is j a dilation?
j will be a dilation if and only if j preserves slopes, if and only if, for any two distinct
points A and B,
Consider
. Since
, the slope is not preserved. Then, j is NOT a dilation.
d) Is j an isometry? If yes, which one? Which are its elements?
We know from a) that j is an isometry. Since the translation is direct and the reflection is
reverse, j will be reverse. Then, j is a reflection or a glide reflection.
If
is a fixed point in j, then
. Then we can say that:
Therefore,
Then, all the points in the line
are fixed points.
In conclusion, j is a reflection with mirror line
39) Iracé y Meaghan:
a) TRUE!
By definition of rotation
, therefore, the triangle
isosceles at O. Then, the angle bisector of the
is included in
the perpendicular bisector of the segment
.
b) FALSE! See ex. 1 for a counterexample.
c) TRUE!
Call
class that:
to the translations with vectors
1) Consider any
, respectively. We know from
that belong to T.
Since
belong to T, the vectors
are parallel to a. Then, because sum
of vectors, the vector sum (
) is parallel to a. Therefore
belongs to
T.
In conclusion the composition is closed in T.
2) Define the translation
such that
Since
belongs to the set T.
Pick up any
such that belongs to the set T.
.
In conclusion,
is the identity of
.
.
because
3) Consider any
that belongs to T. And define
.
Since
belongs to T, the vector
parallel to a. Then
is parallel to a, so
belongs to T. Also,
.
In conclusion, for any element of T, there exist an inverse of it.
4) Since the set of all translations in the plane with the composition form a group, the
associative property is hold for all translations in the plane, in particular to the
translations whose vectors are parallel to the line a.
From 1, 2, 3 and 4, we can say that
al the translations in the plane,
is a group. And, since T is a subset of the set of
d) FALSE! Consider two reflection whose mirrors intersect in a point O. Then O is
fixed in the composition because is fixed in each one of the reflections. Therefore,
the composition cannot be a translation because it has a fixed point. (NOTE: It is
not the identity translation neither because there exist points that are not fixed.)
40) Since the rotation is direct and the reflection is reverse, the composition of both will
be reverse. Therefore, this composition can be a reflection or a glide-reflection.
Since O is the center of rotation, O is fixed in the rotation centered at O and with angle 30.
Then, since O belongs to the mirror of reflection, O is fixed I the reflection. Therefore, O
is fixed in the composition. Therefore, there exist a point that is fixed in the composition.
Then, the composition cannot be a glide reflection.
In conclusion, the composition is a reflection.
Consider the line m such that O belongs to m and the angle
has a measure of 15
and its direction is the same as the angle of rotation (it means that if the angle of rotation
is clockwise or counter clockwise, then the angle
will be counterclockwise or
clockwise, respectively.)
Pick up any point A such that A belongs to the line m. and consider it image in the
rotation A’. Call B to the image of A’ in the reflection. T hen, B is the image of a in the
composition.
Since A’ is the image of a in the rotation, the angle bisector of the angle AOA’ is included
in the perpendicular bisector of the segment AA’. Since the angle of rotation is 30, and
mOa determine an angle of 15, then the perpendicular bisector of AA’ is the line a.
Therefore, the image of A’ (B) in the reflection has to be A.
In conclusion B=A.
Therefore, All the points that belong to the line m are fixed in the composition. Therefore,
the line m is the mirror of the reflection determined by the composition.
Chapter 9
41) Solution
a) True
b) True
c) False, lines in Spherical Geometry have finite length since the length of a line is
the equator of a sphere.
d) False, when two lines intersect on a sphere they will intersect on exactly two
points.
e) False, no lines on a sphere are parallel since all line will intersect the two global
points.
f) False, the angle sum of a triangle in Spherical Geometry is between 180 degrees
and 900 degrees.
42)
A
B
C
D
First, construct the diagonals AD and BC. We know that AB is congruent to CD and
angles ABC and CDB are congruent right angles because ABCD is a Saccheri
Quadrilateral. Next, we know that BD is congruent to itself. This is true because of the
reflexive property. Thus, triangles ABD and CDB are congruent because of SAS.
Therefore, by CPCTC, AD is congruent to BC and the diagonals of a Saccheri
Quadrilateral are congruent.
43) Solution:
B
[
T
A
F
[
T
G
[
T
C
H
[
T
N
[
T
[
T
m
I
[
T
l
[
T
J
E
[
T
k
D
Solution:
The generalized defect equation is (n-gon)=180(n-2) – (Sum of interior
angles)
The regular polygon defect is: 180(5-2)-(<a-<b-<c-<d-<e)
Now to cut the regular polygon in 3 triangles and add their defects together.
= 180- (<n-<f-<m) + 180-(<g-<h-<L) + 180-(<I-<J-<K)
= 540 -<n-<f-<M-<g-<H-<L-<I-<J-<K
= 540-<a-(<f+<G)- (<H+<I)- <d – (<M+<L+<K)
= 540 -<a-<b-<c-<d-<e
Thus the two defects are equal and therefore the defects are additive for a
regular polygon.
44)
_____(2,0)_____________
(5,0)_____________(8,0)_________
+
The equation of this line is
.
The equation of another line that intersects this line at (7,
with radius 7 is
To find a we use the quadratic equation
We take the positive one
So the equation of another line that intersects this line at (7,
with radius 7 is
45) Answers to Spherical Geometry Questions
1. You can have one line; any other great circle will intersect it.
2. The maximum sum of the degrees in a triangle is 900.
3. To prove triangles are congruent, you can use ASA, SSS, or SAS
46) Answers to Hyperbolic Geometry questions
1. AAA, ASA, SSS.
2. The maximum degrees for a triangle is 180, the minimum is zero.
3. This is not necessarily true; the proof is not possible without Euclid’s 5th postulate,
which is assumed False in Hyperbolic geometry