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Math 3181
Name:
Dr. Franz Rothe
March 27, 2012
All3181\3181_spr12h1.tex
Use the back pages for extra space
due date: January 20
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Solution of Homework
10 Problem 1.1 (A special triangle). Use the equilateral triangle and Thales’
Theorem to construct a triangle with angles of 30◦ , 60◦ and 90◦ . Describe your construction.
Figure 1: Construction of a triangle with angles of 30◦ , 60◦ and 90◦ .
Answer. On an arbitrary segment AB, an equilateral triangle is erected. This can be
done as in Euclid I.1 by finding the intersection point C of a circle around A through B
with a circle around B through A. We extend the segment AB on the side opposite to
B, and get the second intersection point D of the extension with the circle around A.
The triangle 4DBC has the angles 30◦ , 60◦ and 90◦ at its vertices D, B and C.
10 Problem 1.2 (Another special triangle). Describe the construction done
in the figure on page 2. Explain how you determine the three angles of the triangle
4ABC.
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Figure 2: What are the angles of 4ABC?
Answer. On an arbitrary segment OA, an equilateral triangle 4OAD is erected. This
can be done as described in Euclid I.1. We find the intersection point D 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◦ by the angle sum of triangle 4ABC.
10 Problem 1.3 (The midpoints of chords). Given is a circle C with center
O, and a point P inside C. Describe the location of the midpoints of all chords through
point P . Give a reason based on Thales’ Theorem or its converse.
Answer. The midpoints of all chords in the given circle C through the point P lie on a
circle with diameter OP . Here is the reason: The midpoints are the foot points F of
the perpendiculars dropped from the center O of the circle onto the respective chords.
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Figure 3: Construction of a triangle with angles of 45◦ , 60◦ and 75◦ .
By the converse of Thales’ Theorem, all the vertices F of the right angles with sides
through the two points O and P lie on Thales’ circle with diameter OP .
10 Problem 1.4 (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!
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 ?
. 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 ?
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 ?
.
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Figure 4: Where lie the midpoints of all chords through point P ?
Figure 5: The midpoints of chords lie on Thales’ circle with diameter OP .
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Figure 6: Tangents to a circle
Since T P is perpendicular to the radius OT , and hence to a diameter, it is a tangent of
circle C.
10 Problem 1.5. Construct a right triangle with projections p = 1 and q = 2 of
the legs onto the hypothenuse. Use a construction based on Thales’ theorem and describe
your construction.
Answer. We draw segments of the lengths as given, |AF | = q = 2 and |F B| = p = 1,
adjacent to each other on one line. Erect the perpendicular on line AB at point F . 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.
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Figure 7: Construction of a right triangle with projections p = 1, q = 2.
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