Download Math 3181 Name: Dr. Franz Rothe May 9, 2012 All3181

Math 3181
Dr. Franz Rothe
May 9, 2012
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10 Problem 1 (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.
At point O, we erect the perpendicular onto this diameter. Let E be the intersection
of the perpendicular with the left circle, lying on the same side of AB as point D. Let
−−→
−−→
A be the second endpoint of diameter AOB.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.
1
10 Problem 2 (Tangents to a circle). Given is a circle C with center O, and
a point P outside of C. Construct the tangents from point P to the circle C. Actually
do and describe the construction!
Figure 2: Tangents to a circle
Answer.
Construction 1 (Tangents to a circle). One begins by constructing a second circle
T with diameter OP . (I call this circle the Thales circle over the segment OP ). The
Thales circle intersects the given circle in two points T and S . The lines P T and P S
are the two tangents from P to circle C.
Validity of the Construction. By Thales theorem, the angle ∠P T O is a right angle,
because it is an angle in the semicircle over diameter OP . Since point T lies on the
circle C, too, the segment OT is a radius of that circle. By Euclid III. 16,
The line perpendicular to a diameter is tangent to a circle.
Since T P is perpendicular to the radius OT , and hence to a diameter, it is a tangent of
circle C.
2
10 Problem 3. Construct a right triangle with projections p = 3 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 3: Construction of a right triangle with projections p = 3, q = 4.
Answer. Adjacent to each other on one line, we draw segments with the given lengths
|AF | = q = 4 and |F B| = p = 3. 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 = 4.6 and b = 5.2.
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 = 21 and b2 = (p + q)q = 28.
Hence the lengths of the legs are
√
√
a = 21 and b = 28
3
10 Problem 5 (Another special triangle). Describe the construction done in
the figure on page 4. Explain how you determine the three angles of the triangle 4ABC.
Figure 4: Construction of a triangle with angles of 45◦ , 60◦ and 75◦ .
Answer. On an arbitrary segment OA, an equilateral triangle 4OAD is erected. As
described in Euclid I.1, D is an intersection point of a circle around A through O with
a circle around O through A.
Let B be the second endpoint of diameter AOB. At point O, we erect the perpendicular onto this diameter. 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 60◦ , 45◦ and 75◦ at its vertices A, B
and C. The angle ∠CAB = 60◦ is obtained from the equilateral triangle 4OAD. The
angle ∠ABC = 45◦ is obtained from the right isosceles triangle 4OBE. Finally, one
calculates the third angle ∠BCA = 180◦ − 60◦ − 45◦ = 75◦ by the angle sum of triangle
4ABC.
4
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 6. For any two triangles it is assumed
α = α0 ,
b
b0
= 0
c
c
Are the two triangles always similar or not? Give examples, counterexamples, or a
general reason depending on what is appropriate.
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.
10 Problem 7. Given is a triangle 4ABC with side length |BC| = 6. Construct,
with straightedge and compass, a rectangle inside this triangle with width three times of
its height. One side of the rectangle is part of a side of the triangle, and the two
remaining vertices lie on the two other sides of the triangle.
Answer. We erect a rectangle with side lengths a = 6 and
triangle.
a
3
= 2 onto the side a of the
Figure 5: Inscribing a rectangle x × x/3 into a triangle.
5
We connect the two new vertices of the rectangle to the third vertex A of the triangle.
The connecting lines intersect the side a in two points, which are the endpoints of one
side of the rectangle to be inscribed. The two other vertices are obtained by erecting
the perpendicular onto side a at these vertices, and intersecting them with the two
remaining sides of the triangle.
Trigonometry
10 Problem 8. For a triangle are given
4
c
= , α < 90◦ and b = 20
a
5
Calculate the angles α, β, and sides a and c.
γ = 45◦ ,
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:
20
5
b
= √ ·
= 18.497
sin β
4 2 sin 72.89◦
1
20
b
=√ ·
= 14.798
c = sin γ
sin β
2 sin 72.89◦
a = sin α
10 Problem 9. Calculate the largest angle of a triangle with sides a = 2, b = 4
and c = 5. Is the triangle acute or obtuse? Use the extended sin theorem to calculate
the diameter 2R of the circum circle.
Answer. The greatest angle lies across the longest side c. The cos theorem yields
c2 = a2 + b2 − 2ab cos γ
a2 + b 2 − c 2
4 + 16 − 25
cos γ =
=
<0
2ab
16
−5
γ = cos−1
= 108.21◦
16
Hence the triangle is obtuse. The extended sin theorem gives the diameter of the circum
circle
c
5
80
2R =
=p
=√
= 5.2636
sin γ
231
1 − (5/16)2
6
Area
10 Problem 10. Let r be the radius of the in-circle of triangle 4ABC. Prove
with a dissection that the area is
area(4ABC) =
(a + b + c) r
2
Answer. As explained in an earlier exercise, we construct the in-circle of a triangle.
The perpendiculars are dropped from its center I onto the three sides. The foot points
Figure 6: An arbitrary triangle is partitioned into six right triangles.
F, G, H are the touching points of the in-circle to the three sides. The six segments
from the center of the in-circle to the vertices and the touching points partition the given
triangle into six right triangles. All of them have the radius of the in-circle as one of their
legs. The other legs have lengths x = |AH|, y = |BF |, and z = |CG|, each one occurring
twice for two congruent triangles. Since the circumference is a + b + c = 2x + 2y + 2z,
the total area of the given triangle 4ABC is
area(4ABC) = 2 ·
yr
zr
(a + b + c) r
xr
+2·
+2·
= (x + y + z) r =
2
2
2
2
10 Problem 11. Given is a right triangle 4ABC with angle α = 60◦ and smaller
leg b = 1. Squares are erected on its three sides. As done in the figure on page 8, one
can construct dissections of them which prove the Theorem of Pythagoras. What kind
of quadrilateral is ALKB. Determine the segment lengths x = |LE| and y = |BL|.
Answer. The quadrilateral ABLK is a parallelogram. Its opposite sides are congruent.
Hence y = |AK| = |AC| + |CK| = |AC| + |LE| = b + x. Secondly, a = |BE| =
7
Figure 7: Dissection proof of the Pythagorean theorem in a special case
|BL| + |LE| = y + x. From the equations
√
x + y = a = 3 and y − x = 1
one concludes
√
√
3−1
a+b
1+ 3
a−b
=
and y =
=
x=
2
2
2
2
10 Problem 12. Draw two parallelograms on the same base with congruent altitudes, and explain why they are equidecomposable . Clearly draw the dissections you
use.
Figure 8: A square and a parallelogram that are easily seen to be equidecomposable .
8
U
Answer. In the figure on page 8, the square
is
decomposed
as
1
2. The parallelogram
U
∼
on the same base is decomposed as 2 3. Since the triangles 1 = 3 are congruent, the
square and the parallelogram as equidecomposable .
10 Problem 13. Explain why the two parallelograms, on the same base with
congruent altitudes, drawn in the figure on page 9, are equicomplementable .
Figure 9: A square and a parallelogram that are only shown to be equicomplementable .
U
Answer. In the figure on page 9, the square
is
decomposed
as
1
2. The parallelogram
U
on the same base is decomposed as 2 3. But this time 1 and 3 are not congruent.
To see that the square and the parallelogram are equicomplementable , we use the
triangle 4 as additional figure. Since the unions
]
]
1 4∼
=3 4
are indeed congruent, we conclude
] ] ] ] ] ] ] square
4∼ 1 2
4∼2
1 4 ∼2
3 4
] ]
]
∼ 2 3
4 ∼ parallelogram
4
Hence the square and the parallelogram—with the same base and congruent altitudes—
are equicomplementable . Similarly, we prove that any two parallelograms with the same
base and congruent altitudes are equicomplementable .
Angles in a circle
10 Problem 14 (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.
9
Figure 10: 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.
10
A direct proof of commutativity of segment arithmetic
Figure 11: 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 15. 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.
11
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.
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
of E 0 A0 . 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.
1
Figure 12: Commutativity ab = ba follows from both lying on the circle.
1
(In the drawing—which is not from Leibniz’ ”best of all worlds”—the construction does not come
out right. Would Leibniz believe the proof or the construction?)
12