Download 2 Solution of Test

Math 3181
Dr. Franz Rothe
October 30, 2015
Name:
All3181\3181_fall15t2.tex
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Solution of Test
Figure 1: Central and circumference angle of a circular arc, both obtained as differences
Problem 2.1. Complete the following proof of Euclid III.20, referring to the figure on
page 1.
Reason for Euclid III.20. We deal with the central angle ω = ∠AOB and the circum_
ference angle γ = ∠ACB of the arc AB. We use the two isosceles triangles 4BOC
and 4COA. Each one of them has a pair of congruent base angles, called α and β,
−→
beyond the center O and let C 0 be any
respectively. One extends the !! segment CO
point on this extension. Now we see the exterior angle x = ∠AOC 0 of triangle 4AOC
equals !! 2α. Indeed, we use the congruence of the base angles of the isosceles triangle
4AOC together with Euclid I.32:
The exterior angle of a triangle is the sum
of !! the two nonadjacent interior angles.
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Similarly, we see that exterior angle y = ∠BOC 0 of triangle 4COB is !! 2β.
For the case drawn in the figure on page 1, points A and O lie on different sides of
line BC. In that case, angle subtraction is used at the vertices C and O. Indeed, the
The circumfercentral and circumference angle are both obtained as !! differences.
and the central angle is !! ω = x − y.
Together with
ence angle is !! γ = α − β
the equations
x = 2α and y = 2β
one gets
ω = x − y = 2α − 2β = 2γ
as to be shown.
Problem 2.2 (Construction with compass and straightedge). Given is a line l
and a point not lying on the line. Provide a drawing and explain how the perpendicular
is dropped with traditional Euclidean tools, using only two circles.
Figure 2: Construction of the perpendicular with just two circles.
Answer. We choose any two points A and B on the line l and draw the two circles with
these centers through the given point P . The two circles intersect at point P and a
second point Q. The line P Q is the perpendicular to the given line l through point P .
Remark. The two points may lie on the same or different sides of the perpendicular,
the construction works for any choice of the points A 6= B. The two circles may have
different radius. In any case, they indeed intersect at two points.
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Problem 2.3 (Construction with with compass and straightedge, in neutral
geometry). Give a construction of an equilateral triangle, and an angle of 60◦ with
compass and straightedge in neutral geometry.
(a) Do the construction and give a description.
(b) What can one say about the angles at the vertices of the triangle, in neutral geometry? Why are extra steps needed for the construction of the 60◦ angle?
(c) At which vertex can you get the angle of 60◦ even in neutral geometry, nevertheless?
(d) Convince yourself once more that all reasoning has been done in neutral geometry.
Additional to Hilbert’s axioms, which intersection property do you need.
Figure 3: The construction of a 60◦ angle at the center O is possible with straightedge
compass in neutral geometry.
Answer. (a) Description of the construction. Draw any segment AB. The circles
about A through B, and about B through A intersect in two points C and D. We
get an equilateral triangle 4ABC. Now we draw a third circle about C through
point A, which passes through point B, too. With the two circles drawn earlier,
the third circle has additional intersection points H and F . Finally, we draw the
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segments AF , BH and CD. These are the perpendicular bisectors of the sides
of triangle 4ABC. All three intersect in one point O inside triangle 4ABC. At
vertex O, one gets six congruent angles which add up to 360◦ , hence they are all
60◦ .
(b) The angles at the vertices may not be 60◦ . Extra steps are needed, because,
in neutral geometry, the angle sum of a triangle may be less than two right angles.
All we can conclude about the angles at the vertices A, B, C, is their congruence.
This follows because the angles opposite to congruent sides are congruent. Still
they may all three be less than 60◦ .
(c) The 60◦ angles appear at the center. At vertex O, one gets six congruent angles which add to 360◦ , hence they are all 60◦ .
(d) The circle-circle intersection property is needed. All justifications can be
given in neutral geometry. Besides Hilbert’s axioms, we only need the circle-circle
intersection property.
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Problem 2.4. Complete the proof of the exterior angle theorem given below.
Figure 4: A proof of the exterior angle theorem presupposing bisection of a segment.
Proof. In the given triangle 4ABC, we bisect the side AC and let M be the midpoint. 1
−−→
We extend the ray BM and construct on the extension a point F such that M B ∼
= MF .
As one easily checks via SAS congruence, one obtains congruent triangles
(2.1)
4M CB ∼
= 4M AF
Indeed, these triangles have congruent !! vertical angles
at vertex !! M , and
∼
and !! M B ∼
by construction two pairs of congruent sides !! M C = M A
= MF .
From the triangle congruence (2.1) we obtain congruent angles γ := ∠ACB ∼
= ∠CAF .
−→
We choose any point D on the ray opposite to AB. Since by construction A ∗ M ∗ C
side of line AB. Hence
and B ∗ M ∗ F , all three points C, M and F lie on the !! same
−→
the ray AF lies in the !! interior of the
angle δ := ∠CAD, which is an exterior angle
∼
of triangle 4ABC. One obtains γ = ∠CAF < δ ∼
= ∠CAD. As to be shown, the exterior
than the !! nonadjacent interior
angle γ.
angle δ is indeed !! greater
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This proof is presupposing a valid construction to bisect a segment.
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Proposition 1 (The Kite-Theorem). [Theorem 17 in Hilbert] Let Z1 and Z2 be two
points on different sides of line XY , and assume that XZ1 ∼
= XZ2 and Y Z1 ∼
= Y Z2 .
∼
Then the two triangles 4XY Z1 = 4XY Z2 are congruent.
Problem 2.5. Complete the proof of the kite-theorem and provide drawings for all three
cases.
Proof. The congruence of the !! base angles of the isosceles triangle 4XZ1 Z2 yields
∠XZ1 Z2 ∼
= ∠XZ2 Z1 . Similarly, one gets ∠Y Z1 Z2 ∼
= ∠Y Z2 Z1 . By !! angle addition
(or substraction, respectively, )
(*)
∠XZ1 Z2 ∼
= ∠XZ2 Z1 and ∠Y Z1 Z2 ∼
= ∠Y Z2 Z1 imply ∠XZ1 Y ∼
= ∠XZ2 Y
−−−→
Angle addition is needed in case ray Z1 Z2 lies inside the angle ∠XZ1 Y ,
−−−→
angle !! subtraction in case ray Z1 Z2 lies !! outside the angle ∠XZ1 Y .
In the special case that either point X or Y lies on the line Z1 Z2 , one gets the same
conclusion even easier.
One now applies SAS congruence to 4XZ1 Y and !! 4XZ2 Y . Indeed the angles at
Z1 and Z2 and the adjacent sides are pairwise congruent. Hence the assertion 4XY Z1 ∼
=
4XY Z2 follows.
Answer. Here are drawings for the three cases.
Figure 5: The symmetric kite
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Figure 6: Where on the boundary can guys A and B meet the quickest?
Problem 2.6. Two guys who are equally good runners stand at points A and B on the
border of a place with the shape shown in the figure on page 7. They are allowed to run
across the place, but they want to meet at any point on the boundary. Which point do
they need to choose to meet as quick as possible. Give a construction and explain your
reasoning.
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Figure 7: They meet at the intersection of the perpendicular bisector of AB with the
boundary marked nearest.
Answer. The runners pass congruent segments AM and BM until they meet at some
point M . Hence they meet where the perpendicular bisector of AB intersects the boundary. There are two such points, of which they choose the nearer one.
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