Download I. Fill-in the blanks. II. True or False III. Problem Solving

Institute of Mathematics
University of the Philippines-Diliman
Math 140: Introduction to Modern Geometries
Second Exam A
MHX
14 February 2008
Name:
Student I.D.:
I. Fill-in the blanks.
Fill-up the blank in each question below with,Euclidean geometry, Hyperbolic geometry, or Elliptic geometry so that the resulting statements will be true.
1. In
, the sum of the measures of the three angles of a triangle is π.
2. In
, all Saccheri quadrilaterals have an acute angle.
3. In
, there exists two triangles which are similar but not congruent
4. In
, a Lambert quadrilateral has an obtuse angle
5. In
, it is possible for a third line to intersect two parallel lines and
yet not form a pair of congruent alternate interior angles.
II. True or False
Write “TRUE” if the statement is true in the specified geometry. Otherwise, write “FALSE”.
If S ABCD is a Saccheri quadrilateral in a hyperbolic plane then |BC| < |AD|.
In neutral geometry, if A and D are points on the opposite sides of ℓ(B, C) and ∠ABC ∼
= ∠BCD
then the lines ℓ(A, B) and ℓ(C, D) are parallel.
In neutral geometry, if A and D are points on the same side of ℓ(B, C) and m∠ABC +m∠BCD < π
then the lines ℓ(A, B) and ℓ(C, D) intersect at some point.
In neutral geometry, if two lines are parallel then they have a common perpendicular.
In neutral geometry, if A and D are points on the same side of ℓ(B, C) and m∠ABC +m∠BCD = π
then the lines ℓ(A, B) and ℓ(C, D) are parallel.
III. Problem Solving
Answer the following question in details. (5 points each).
The following questions are to be answered in the setting of Euclidean Geometry.
1. Consider a triangle △ABC which is not isosceles. Let D be a point on ℓ(B, C) such that C is between
B and D and let the line t bisect angle ∠ACD and intersect ℓ(A, B) at F . The line t is called the
|
|BC|
external angle bisector of △ABC at C. Prove that |BF
|AF | = |AC| .
Hint: Draw a line through A parallel to ℓ(C, F ).
1
B
A
∠ACF ∼
= ∠DCF
C
D
F
2. Use the preceeding problem(question 1) and Ceva’s theorem to prove that the external angle bisector
at C and A and the internal angle bisector at B intersect at a point.
Note the conclusion in the preceeding problem(question 1) holds tivially true if △ABC is isosceles.
3. Prove that for any triangle, the radius of the nine point circles of is half of that of the circumcircle.
4. Suppose that l and m are parallel lines in hyperbolic geometry. Prove that there exists a unique line
perpendicular to l and m.
Total points: 30 ; passing: 18
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Answer Key for Exam A
I. Fill-in the blanks.
Fill-up the blank in each question below with,Euclidean geometry, Hyperbolic geometry, or Elliptic geometry so that the resulting statements will be true.
1. In
Euclidean geometry
2. In
Hyperbolic Geometry
3. In
Euclidean geometry
4. In
Elliptic geometry
, the sum of the measures of the three angles of a triangle is π.
, all Saccheri quadrilaterals have an acute angle.
, there exists two triangles which are similar but not congruent
, a Lambert quadrilateral has an obtuse angle
5. In
Hyperbolic Geometry
, it is possible for a third line to intersect two parallel lines and
yet not form a pair of congruent alternate interior angles.
II. True or False
Write “TRUE” if the statement is true in the specified geometry. Otherwise, write “FALSE”.
FALSE If S ABCD is a Saccheri quadrilateral in a hyperbolic plane then |BC| < |AD|.
TRUE
In neutral geometry, if A and D are points on the opposite sides of ℓ(B, C) and ∠ABC ∼
= ∠BCD
then the lines ℓ(A, B) and ℓ(C, D) are parallel.
FALSE In neutral geometry, if A and D are points on the same side of ℓ(B, C) and m∠ABC+m∠BCD < π
then the lines ℓ(A, B) and ℓ(C, D) intersect at some point.
TRUE
TRUE
In neutral geometry, if two lines are parallel then they have a common perpendicular.
In neutral geometry, if A and D are points on the same side of ℓ(B, C) and m∠ABC+m∠BCD = π
then the lines ℓ(A, B) and ℓ(C, D) are parallel.
III. Problem Solving
Answer the following question in details. (5 points each).
The following questions are to be answered in the setting of Euclidean Geometry.
1. Consider a triangle △ABC which is not isosceles. Let D be a point on ℓ(B, C) such that C is between
B and D and let the line t bisect angle ∠ACD and intersect ℓ(A, B) at F . The line t is called the
|
|BC|
external angle bisector of △ABC at C. Prove that |BF
|AF | = |AC| .
Hint: Draw a line through A parallel to ℓ(C, F ).
1
B
A
∠ACF ∼
= ∠DCF
C
D
F
Answer: Let l be the line through A parallel to t and let G be the point of intersection of l and
|BC|
|
ℓ(B, C, D). Then |BF
|AF | = |GC| . Since l and t are parallel and ℓ(A, C) is a transveral then
∠GAC ∼
= ∠F CA ∼
= AGC. Hence, △CGA is isosceles with |AC| = |GC|. The conclusion follows.
B
G
l
A
∠GAC ∼
= ∠F CA ∼
= AGC
C
D
F
2. Use the preceeding problem(question 1) and Ceva’s theorem to prove that the external angle bisector
at C and A and the internal angle bisector at B intersect at a point.
Note the conclusion in the preceeding problem(question 1) holds tivially true if △ABC is isosceles.
Answer: Let E be the intersection of the internal angle bisector at ∠B and ℓ(A, C). Let D be the point
of intersection of the external angle bisector at ∠A and ℓ(B, C) and F be the intersection of the
|AE|
|AB|
AE
external angle bisector at ∠C. Applying a theorem discussed in class we have EC
= |EC|
= |BC|
.
From the previous problem we have
AF BD CE
F B DC EA
=
|BC|
(− |DC|
)(− |AB|
|AC|
=
AF
FB
|AF |
|AC|
= − |F
B| = − |BC| and
BD
DC
|BC|
= − |DC|
=
|AB|
|AC| .
Hence,
|CE|
|EA| .
3. Prove that for any triangle, the radius of the nine point circles of is half of that of the circumcircle.
Answer: Let K be the center of the circumcircle of △ABC and let D, E and F be the respective
midpoints of the sides BC, AC and AB. Then ℓ(K, D), ℓ(K, E) and ℓ(K, F ) are the respective
perpendicular bisectors of the sides BC, AC and AB.
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Let O be the orthocenter of △ABC. Then ℓ(O, A) is an altitude, so ℓ(O, A) ⊥ ℓ(B, C). Since,
ℓ(K, D) is also perpendicular to ℓ(B, C), then ℓ(O, A) is parallel to ℓ(K, D).
Let P be the midpoint of OA P . and let H be the center of the nine-point circle. Thus, P is on
the nine point circle and HP is a radius of the nine-point circle.
It was discussed in class that |OP | = |KD| and since ℓ(A, O) k ℓ(K, D) then OP KD is a
parallelogram and the diagonal P D is a diameter of the nine-point circle.
Since P is the midpoint of AO the |AP | = |P O| = |KD|. Hence, AP DK is also a parallelogram
and therefore |KA| = |P D| = 2|P H|. Since, KA is a radius of the circumcircle, we obtain the
desired result.
4. Suppose that l and m are parallel lines in hyperbolic geometry. Prove that there exists a unique line
perpendicular to l and m.
Answer: The existence of the common perpendicular was already shown in class. Suppose there are
two common perpendiculars. Let A and B be the feet of these common perpendicular on line l
and let C and D be the feet of these common perpendiculars at line m.
Since the line perpendicular to a given point on a line is unique,A = B or c = D implies l = m.
If A 6= B and C 6= D then L ABDC is a Lambert quadrilateral with four right angles which
cannot occur in hyperbolic geometry. Therefore, we must have A = B and C = D and l = m.
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