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Math 3181
Dr. Franz Rothe
May 12, 2014
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18 problems are required for full credit
Solution of Final
10 Problem 1. For a right triangle with the angles 30◦ , 60◦ , 90◦ , the hypothenuse
has twice the length of the shorter leg. Give any convincing reason you want for this
fact.
Answer. Here are several possible answers;—a drawing instead of the exact explanation
would be acceptable, too. Note that one cannot get this simple result directly from
Pythagoras’ Theorem.
(i) Reflect the triangle across its longer leg. Together with the reflected image, one
obtains a triangle with three angles of 60◦ , which is known to be equilateral. Its
sides are congruent to the hypothenuse of the given triangle. The reflection axis
bisects one of the sides of the equilateral triangle. The shorter leg of the original
triangle is one half of this bisected side. Hence the shorter leg is half of the
hypothenuse.
(ii) We draw the semicircle with the hypothenuse AB as diameter. From Thales’
Theorem, we know that the vertex C with the right angle, and hence all three
vertices lie on the semicircle. Half of the hypothenuse and the shorter leg of the
given triangle are sides of an isosceles triangle 4OAC. This triangle has two
congruent base angles at vertices A and C, which we know to measure 60◦ . Hence
all three angles of triangle 4OAC measure 60◦ , and this triangle is equilateral.
For the original triangle, we see that the shorter leg AC is half of the hypothenuse
AB .
(iii) From the definition of the sin function, we see that sin 30◦ is the ratio of the shorter
leg across to the 30◦ angle to the hypothenuse. We know that sin 30◦ = 1/2. 1
Hence the shorter leg is half of the hypothenuse.
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We see that this answer depends on previous knowledge of trigonometry.
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10 Problem 2 (Another triangle). Construct a triangle with angles of 30◦ , 45◦
and 105◦ , using only compass and straightedge, but no protractor. Describe your construction.
Figure 1: Construction of a triangle with angles of 45◦ , 30◦ and 105◦ .
Answer. On an arbitrary segment OB, an equilateral triangle 4OBD is erected. As
described in Euclid I.1, D is an intersection point of a circle around B through O with
a circle around O through B. Let A be the second endpoint of diameter AOB.
At point O, we erect the perpendicular onto diameter AB. Let E be the intersection
of the perpendicular with the left circle, lying on the same side of AB as point D. The
−−→
−−→
rays AD and BE intersect in point C.
We thus get a triangle 4ABC with the angles 30◦ , 45◦ and 105◦ at its vertices A, B
and C. The angle ∠BAD = 30◦ since triangle 4OBD is equilateral and ∠ADB = 90◦ .
The angle ∠ABC = 45◦ is obtained from the right isosceles triangle 4OBE. Finally,
one calculates the third angle ∠BCA = 180◦ − 30◦ − 45◦ = 105◦ by the angle sum of
triangle 4ABC.
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10 Problem 3. Construct a right triangle with projections p = 5 and q = 4 of the
legs onto the hypothenuse. Use Thales’ theorem. Describe your construction. Measure
the lengths of the two legs of your triangle.
Figure 2: Construction of a right triangle with projections p = 5, q = 4.
Answer. Adjacent to each other on one line, we draw segments with the given lengths
|AF | = q = 4 and |F B| = p = 5. We erect the perpendicular on line AB at point F and
draw a semicircle with diameter AB. The semicircle and the perpendicular intersect
at point C. The triangle 4ABC is a right triangle with hypothenuse AB, and the
projections q = AF and F B = p have the lengths as required.
I measure the two legs as a = 6.7 and b = 6.
10 Problem 4. What are the lengths of the two legs of the triangle from the last
problem. Use the leg theorem to calculate exact expressions.
Answer. The leg theorem gives the squares a2 = (p + q)p = 45 and b2 = (p + q)q = 36.
Hence the lengths of the legs are
√
√
√
a = 45 = 3 5 and b = 36 = 6
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Angles in a circle
10 Problem 5 (A construction using an altitude). Using Euclid III.21,
construct a triangle 4ABC with the three given pieces: side c = |AB| = 6, opposite
angle γ = ∠BCA = 30◦ , and altitude hc = 4.
( hc is the altitude dropped from vertex C onto the opposite side AB).
Question (a). Do the construction as exact as possible.
Figure 3: A triangle construction
Answer.
Question (b). Estimate the length of the shortest side of the triangle.
between 3.0 and 4.0
between 5.0 and 6.0
X between 4.0 and 5.0
between 6.0 and 7.0
Question (c). Describe the steps for your construction.
Answer. Draw side AB = 6 and its perpendicular bisector p. Let M be the midpoint
of AB. The center O of the circum circle lies on the perpendicular bisector. The
center angle is double the circumference angle γ. Hence ∠AOB = 2γ = 60◦ , and
∠AOM = 30◦ . For the example given, the point O is especially easy to find because
the 4AOB is equilateral. Next, we draw the circle around O through A and B. This
is the circum circle of 4ABC, on which vertex C lies.
Secondly, vertex C lies on a parallel q to AB of distance |M D| = hc = 4, because of
the given altitude hc = 4. Hence vertex C is an intersection point of this parallel with
the circum circle. One may choose any one of these two intersection points.
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A proof of commutativity of segment arithmetic
Figure 4: Prove commutativity ab = ba.
The following proof of commutativity of segment arithmetic does not refer to Pappus’
theorem. Instead, we use angles in a circle directly. Indeed, only one circle is needed. We use
measurements along the horizontal and vertical axes crossing at point O perpendicularly.
The given positive segment |OA| = a > 0 is transferred to the vertical axis. The
segments |OE| = 1 and |OB| = b > 0 are transferred to the horizontal axis.
10 Problem 6. Answer the following questions:
Question (a). Explain how the product ab has been constructed.
Answer. We draw the line through B parallel to EA, and get its intersection point with
the vertical axis at point ab with distance |Oab| = ab from the origin. As next step, the
triangle 4OAE is reflected across the −45◦ line.
Question (b). Explain why, because of these constructions, there appear three congruent
angles.
Answer. The angles ∠OAE ∼
= ∠OabB are congruent because the vertical axis traverses
a pair of parallel lines. The angles ∠OAE ∼
= ∠OA0 E 0 are congruent by SAS congruence
of the corresponding triangles. We have marked the three congruent angles by α.
Question (c). We now draw a circle through points A0 , E 0 and B. Explain why this
circle goes through the point ab.
Answer. The angles ∠E 0 A0 B ∼
= ∠E 0 abB are congruent, and lie both on the same side
of E 0 B. They are hence circumference angles in the circle drawn. Hence ab lies on this
circle. The triangle 4OBE 0 is reflected across the +45◦ line, as needed in the next step.
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Question (d). Explain how the product ba has to be constructed, in the left upper
quadrant and complete the drawing.
Answer. We draw the line through A0 parallel to E 00 B 0 , and get its intersection point
with the vertical axis at point ba with distance |Oba| = ba from the origin.
Question (e). Explain why, because of these constructions, there appear three congruent
angles.
Answer. The angles ∠OB 0 E 00 ∼
= ∠ObaA0 are congruent because the vertical axis traverses a pair of parallel lines. The angles ∠OBE 0 ∼
= ∠OB 0 E 00 are congruent because of
SAS congruence of the corresponding triangles. We have marked the three congruent
angles by β.
Question (f). Explain why the circle through A0 , E 0 and B—as already drawn—goes
through the point ba.
Answer. The angles ∠E 0 BA0 ∼
= ∠E 0 baA0 are congruent, and lie both on the same side
0 0
of E A . They are hence circumference angles in the circle drawn. Hence ba lies on this
−→
circle. Since the circle has only one intersection point with the vertical ray OA, and we
have obtained that both ab and ba is this intersection point, we conclude that ab = ba.
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Figure 5: Commutativity ab = ba follows from both lying on the circle.
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In some former drawing—which is not from Leibniz’ ”best of all worlds”—the construction does
not come out right. Would Leibniz have believed the proof or the construction?
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Theorems of Pappus
10 Problem 7 (Theorem of Pappus). Formulate the Theorem of Pappus and
explain:
• Explain in general words what the Pappus configuration is.
• Provide a drawing, using a straightedge.
• State the Theorem of Pappus in parallel setting. What does it say about three pairs
of opposite sides.
• Name the points of the relevant hexagon in your drawing.
• Restate the Theorem of Pappus using the names of the points as you have put them
into your drawing.
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Answer.
• We call the system of two intersecting lines, with three points on both of
them, and the M-W hexagon zig-zaging between those triplets the configuration
of Pappus.
Figure 6: For Pappus’ configuration: If BC 0 k B 0 C and AC 0 k A0 C, then AB 0 k A0 B.
• If the hexagon of Pappus’ configuration contains two pairs of opposite parallel
sides, the third pair of opposite sides is parallel, too.
• In the figure on page 8, we have two lines intersecting at point O, we have put
three points A, B, C on the horizontal line, and three points A0 , B 0 , C 0 on the other
line
• there are three pairs of lines
AB 0 and A0 B
BC 0 and B 0 C
C 0 A and CA0
the parallelism of which has to be investigated. With the names from the figure
on page 8, we state:
Theorem 1 (Pappus Theorem). Let A, B, C and A0 , B 0 , C 0 be both three points
on two intersecting lines, all different from the intersection point. If the lines BC 0
and B 0 C are parallel, and the lines AC 0 and A0 C are parallel, then the lines AB 0
and A0 B and parallel, too.
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Figure 7: Explain the Theorem of Pappus, in the general setting.
Theorem 2 (Pappus’ Theorem in general setting). Let A, B, C and A0 , B 0 , C 0
be three points on two intersecting lines, respectively. The three intersection points
X = BC 0 ∩ B 0 C, Y = AC 0 ∩ A0 C and Z = AB 0 ∩ A0 B of three pairs of opposite sides of
hexagon AB 0 CA0 BC 0 lie on a line.
10 Problem 8. Mark the three points X, Y, Z and put the line through them into
the figure on page 9.
Answer.
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Figure 8: The Theorem of Pappus in the general setting.
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Similarity
We use Euler’s notation: angle α lies opposite to side a, angle β lies opposite to side b,
and angle γ lies opposite to side c.
10 Problem 9. For any two triangles it is assumed
α = α0 ,
b
b0
= 0
c
c
Are the two triangles always similar or not? Depending on what is appropriate, give
examples, counterexamples, or a reason based on my script, the lecture, or Euclid’s book
VI.
Answer. As stated in the section about similar triangles, Euclid VI.6 tells us:
If two triangles have one pair of congruent angles, and the sides containing these
pairs are proportional, then the triangles are similar.
For the present example, the angles α = α0 are congruent. The sides adjacent to
angle α are b and c, and to angle α0 are b0 and c0 . These sides are proportional for the
0
two triangles due to the assumption cb = cb0 . Hence Euclid VI.6 tells the two triangles
are similar.
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10 Problem 10. In triangle 4ABC, the altitudes are dropped from vertices A
and B and have the lengths ha and hb , respectively. Use similar triangles to show
ha
hb
=
b
a
Figure 9: For the calculation of its area, any side of a triangle may be used as its base.
Answer. For the triangle 4ABC, we can take side BC as base. The corresponding
altitude is AD, were D is the foot-point of the perpendicular dropped from vertex A
onto side BC.
As a second possibility, we can take side AC as base. The corresponding altitude
is BE, were E is the foot-point of the perpendicular dropped from vertex B onto side
AC.
The two triangles 4CAD and 4CBE are equiangular, and hence similar. We get
the proportion
|AD|
|BE|
hb
ha
=
=
=
= sin γ
b
|AC|
|BC|
a
Remark 1. By multiplication with the denominators, we obtain
aha = |AD| · |BC| = |BE| · |AC| = bhb = ab sin γ
which is both the double area. We see that the area of a triangle is well defined. For the
product of half base times height, it does not matter which side one chooses as base.
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10 Problem 11. A candle is standing at distance a = 3 from a lens with focus
length f = 2. At which distance b behind the lens has one to put a screen to get a sharp
image of the candle flame. Give a construction and calculate the distance b.
Answer. The construction is shown in the figure on page 13. One measures that about
b = 6.
Figure 10: Construct the appropriate position of the screen.
From the lens equation one gets 1/b = 1/f − 1/a = 1/2 − 1/3 = 1/6 and hence b = 6 is
the appropriate distance from lens to screen.
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Trigonometry
10 Problem 12. For a triangle are given
γ = 45◦ ,
c
4
= , α < 90◦ and b = 20
a
5
Calculate the angles α, β, and sides a and c.
Answer. The sin Theorem yields
sin α =
a sin γ
5
= √ = 0.8839
c
4 2
We get α = 62.11◦ since this angle is acute. The angle β is obtained from the angle
sum. One gets
β = 180◦ − α − γ = 72.89◦
Now we determine a and c by the sin theorem:
b
5
20
= √ ·
= 18.497
sin β
4 2 sin 72.89◦
b
1
20
c = sin γ
=√ ·
= 14.798
sin β
2 sin 72.89◦
a = sin α
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Figure 11: The common notation for a right triangle
Figure 12: How does Euclid III.36 imply the leg theorem?
Pythagoras group
Theorem 3 (Theorem of chord and tangent (Euclid III.36)). From a point
outside a circle, a tangent and a second are drawn. The square of the tangent segment
equals the product of the segments on the chord, measured from the point outside to the
two intersection points with the circle.
10 Problem 13 (From Euclid III.36 to the leg theorem). Complete the
drawing on page 15, and explain how you get the leg theorem a2 = pc.
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Figure 13: The leg theorem.
Answer. We choose point A to be the other endpoint of diameter CA. We see that
4ACB is a right triangle, because tangent and radius are !! perpendicular to each
other. The segment AB intersects the circle in a second point F . The triangle 4ACB
has altitude CF , because !!
Thales’ theorem shows that the angle !!
∠AF C
is right. Now Euclid III.36 tells that the square of the leg CB equals the product of the
hypothenuse BA time the projection BF of that leg onto the hypothenuse. One gets
the statement in its usual form a2 = pc.
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Figure 14: How does Euclid III.36 imply the Pythagorean Theorem?
10 Problem 14 (From Euclid III.36 to the Pythagorean Theorem). Complete the drawing on page 17, and explain how you get the Pythagorean Theorem.
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Figure 15: The Pythagorean Theorem
−−→
Answer. We draw the ray BO from the point B outside the given circle through the
center O. We put A = O. We get a right angle at the touching point C of the tangent
since tangent and radius are !! perpendicular to each other. Let points E and D
be the endpoints of the diameter on the ray already drawn. Now Euclid III.36, as given
in theorem 3, tells that
|BC|2 = |BD| · |BE|
In terms of the sides of the triangle 4ABC, this shows that
a2 = (c + b)(c − b) = c2 − b2
We got the Pythagorean Theorem.
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and a2 + b2 = c2
Figure 16: Two equally good runners start at O and R. Do they better meet at point S or
point T .
10 Problem 15. Two equally fast runners start at the opposite points O and R
of the place shown in the figure on page 19. They are allowed to run across the place,
but cannot enter any space outside the place. They want to meet on the boundary. Can
they meet quicker at point S or at point T .
(i) Calculate the distance |OS| for which |OS| = |SR|.
(ii) Calculate the point T such that |OT | = |T C| + |CR|.
(iii) Calculate the distance |OT | and decide whether which distance is shorter |OT | or
|OS|.
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Answer. (i) Let S = (x, 0) be the coordinates of the meeting point S. Since the distances |OS| = |SR| are equal, one calculates
p
x = (3 − x)2 + 22
x2 = 9 − 6x + x2 + 4
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x=
6
(ii) Let T = (x, 1) be the coordinates of the meeting point T . Since the distances
|OT | = |T C| + |CR| are equal, one calculates
√
√
x2 + 1 2 = 2 − x + 2
√
√
x2 + 1 = (2 + 2)2 − 2(2 + 2)x + x2
√
√
2(2 + 2)x = (2 + 2)2 − 1
√
2+3 2
x=
= 1.56066
4
(iii) We calculate the distance
3
|OT | = |T C| + |CR| = (2 − x) +
√
2 = 1.85355
Since |OS| = 2.1667, we see that point T is where to meet quicker.
10 Problem 16. Continuing the last problem, I have shown a purely geometrical
solution. Describe and justify the construction of the two possible meeting points S and
T done in the figure on page 21.
Answer. The point S is the intersection of the perpendicular bisector of OR with the
lower horizontal boundary of the place.
To construct the point T , we extend the middle horizontal boundary of the place,
starting at the corner C by a segment CE ∼
= CR . The point T is the intersection of
the middle horizontal boundary of the place with the perpendicular bisector of OE.
3
Note that it is not sufficient to just compare the values of x from parts (i) and (ii).
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Figure 17: Two equally good runners start at O and R. Construction of the meeting points
S and T .
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10 Problem 17. A parallelogram has sides of length 3 and 4, and one diagonal
has length 6.
• Construct the parallelogram and measure its second diagonal.
• Calculate the length of the second diagonal exactly.
Answer. The construction is shown in the figure on page 22. One measures that the
second diagonal has length about 3.7. By the parallelogram equation the length x of
Figure 18: A parallelogram with sides 3 and 4 and one diagonal of length 6.
the second diagonal satisfies 2 · 32 + 2 · 42 = 62 + x2 and hence x2 = 14 and x =
22
√
14.
10 Problem 18. In the coordinate plane are given: the circle with radius 3 and
center (0, 0) and the line through points P = (−1, 0) and Q = (0, 1).
1. What are the equations for the circle and the line.
2. What are the coordinates (x, y) for the intersection points of the circle and the
line.
3. Give the exact root expressions for both coordinates of one intersection point. For
the construction of which regular polygon could this be a useful first step?
Answer. The equations for the circle and the line are
x2 + y 2 = 9 and y = x + 1
To get the coordinates (x, y) of the intersection points of the circle and the line, we plug
y = x + 1 into the equation of the circle and get
x2 + (x + 1)2 = 9
2x2 + 2x − 8 = 0
x2 + x − 4 = 0
x1,2
√
−1 ± 17
=
2
The two intersection points have coordinates
√ !
√
−1 − 17 1 − 17
,
and
2
2
√
√ !
−1 + 17 1 + 17
,
2
2
These values could be used as a first step in a construction of a regular 17-gon.
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