Download 3 Solution of Homework

Math 3181
Dr. Franz Rothe
April 23, 2012
Name:
All3181\3181_spr12h3.tex
Use the back pages for extra space
due date: April 19
3
Solution of Homework
10 Problem 3.1. Given is a triangle 4ABC. Construct with straightedge and
compass a square inside this triangle, as large as possible, one side of which is part of a
side of the triangle.
Answer. We erect a square onto one side a of the triangle. We connect the two new
Figure 1: Inscribing a square into a triangle.
vertices of the square 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 square to be inscribed.
Finally, we draw the inscribed square.
1
10 Problem 3.2. Is the area of the square from the problem above less, more
or equal to half of the area of the triangle. Give a simple folding argument to decide
without any calculation.
Answer. For the special case that the endpoints of one side of the square are midpoints
of two sides of the triangle, the area of the square is half of the area of the triangle.
In the figure on page 2 you see this borderline case. In all other cases, the area of the
Figure 2: For h = a, one can fold the triangle to cover the square twice.
square is less than half of the area of the triangle.To see this, we fold the triangle along
three sides of the square. We get three reflection images A0 , B 0 and C 0 of the vertices of
the triangle. Because of the right angles of the square, the folded sides touch along the
lines A0 B 0 and A0 C 0 . In the figure on pages 3, you see the case that triangle side a < h
is less than the altitude h, and in the figure on page 3, the case that triangle side a > h
is longer than the altitude. In the first case, one covers the square and an extra triangle
4A0 B 0 C 0 is left over in the opposite half plane. In the second case, one again covers the
square, and the part of the extra triangle 4A0 B 0 C 0 inside the square is covered twice.
2
Figure 3: For a < h, one can fold the triangle to cover the square twice, and triangle
4A0 B 0 C 0 left over.
Figure 4: For a > h, one can fold the triangle to cover the square twice, and part of triangle
4A0 B 0 C 0 is covered twice.
3
10 Problem 3.3. Given are two triangles 4ABC and 4A0 B 0 C 0 . We use the
standard notation from Euler for angles, sides and vertices. Assume that
α = α0 ,
a0
a
= 0
b
b
Are the two triangles always similar. Give examples, counterexamples, a general reason.
Answer. The two triangles need not always to be similar. If angle α = α0 is acute, and
a
< 1, there may be two solutions which are not similar. The example α = α0 = 30◦
b
0
and ab = ab0 = 35 gives two solutions, which are not similar. It is obtained from example
Figure 5: For the given data, there are two non-congruent solutions.
in the figure on page 4 with the permutation C 7→ A, A 7→ B, B 7→ C.
10 Problem 3.4. For the two triangles it is assumed
α = α0 ,
b0
b
= 0
c
c
Are the two triangles always similar. Give examples, counterexamples, a general reason.
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
pair are proportional, then the triangles are similar.
As assumed, the congruent angles are α = α0 . The sides adjacent to this angle are
b and c, with primes for the second triangle. These sides are proportional for the two
0
triangles by the assumption cb = cb0 . Hence Euclid VI.6 tells that the two triangles are
similar.
4
10 Problem 3.5. We want to define the area of a triangle as the product (from
segment arithmetic) half of base times height. Give a reason, based on similar triangles,
why it does not matter which side one chooses as base.
Figure 6: 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
|BE|
|AD|
=
= sin γ
|AC|
|BC|
By multiplication with the denominators, we obtain
|AD| · |BC| = |BE| · |AC| = sin γ|AC| · |BC|
We see that it does not matter which side one choses as base.
5