Math F651: Homework 1
Due: January 23, 2015
1. Read all the definitions of Euclid I. Then divide them into two categories.
1. Definitions that are descriptive: they are trying to give a the reader a sense of what the
object is in terms of their past experience.
2. Definitions that are logical: they define a new object in terms of previously defined objects.
In modern mathematics, we have logical definitions, but we dispense with descriptive definitions
in favor of accepting that some things just won’t be defined.
Solution:
Descriptive: 1, 2, 3, 4, 5, 6, 7, 8, 13, 14
Logical: all others.
Some students said 3 and 6 are logical. For example, definition 3 says the end of a line is a point.
I suppose if you know what a ’line’ is and what the ’end’ of a line is, this could be the definition
of a point. But that’s not what Euclid is doing here. He seems to be giving further insight about
how you should think about points. So I’d call this a descriptive definition. Similarly for 6.
Several people classified 10 and 23 as descriptive (the definitions of right angles and parallel lines).
But these are logical. For example, ”Two lines are parallel if they have no points in common” is
a fine modern logical definition of parallel, and this is essentially Euclid’s definition.
2. Use Euclid I-1 to construct a regular hexagon. Your solution must include a hexagon constructed
with a real straightedge and compass, as well as a description of the steps of the construction.
Solution:
Let AB be the given line segment and apply Euclid I-1 to construct an equilateral triangle ABO.
On AO repeat this procedure to construct an equlateral triangle AOC. Now form the circle with
center O and radius AO, which will pass through B and C. Now extend AO, BO, and CO until
Math F651: Homework 1
Due: January 23, 2015
these lines intersect the circle at D, E, and F. Then the vertices A through F when joined together
form a regular hexagon.
3. Read Euclid I-3 and discuss any gaps that you find.
Solution:
The main gap is that the circle with center A and radius AD may not intersect AB. Full marks
were given for spotting this.
4. Given a line segment, describe how to construct a square with that line segment as a side. You
must also hand in a square constructed this way. Note that to do this problem, you will likely
need to remind yourself how to draw a line perpendicular to another line (Euclid I-11).
Solution:
Let AB be the given line segment. Extend a perpendicular line from A to a point E. Consider
the circle with center A and radius AB; let C be the point of intersection of the circle with the
line AE, so AB equals AC. Finally, let D be the point of intersection of the circles of radius AB
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Math F651: Homework 1
Due: January 23, 2015
centered at B and C. So CD and BD both equal AB as well. Hence ABCD is a square.
5. A rectangle is a four-sided figure such that the four interior angles are all right angles.
a) Prove that the opposite sides of a rectangle have equal length. Hint: You can use ASA
(I-26).
b) Prove that the diagonals have equal length.
c) Prove that the diagonals bisect each other.
Solution:
Let ABCD be a rectangle. By I-27, the opposite sides of the rectangle are parallel. Form the
diagonal AC. By I-29, ∠BAC equals ∠DCA. Similarly, by I-29, ∠BCA equals ∠DAC. By ASA
with the common side AC we conclude that △BAC is congruent to △DCA. In particular, ∣DC∣ =
∣AB∣ and ∣AD∣ = ∣BC∣.
Note that triangle △BAD is congruent to △CDA by SAS: ∣AB∣ = ∣CD∣, ∣AD∣ = ∣AD∣, and ∠BAD
and ∠CDA are right. Hence the two diagonals AC and BD are equal. Also, ∠ABD equals
∠DCA.
Let E be the intersection of the two diagonals. Then by I-29, ∠BDC equals ∠ADB (which equals
∠DCA). Since △EDC has two equal angles, it is isosceles and ∣ED∣ = ∣EC∣. Similar arguments
show that ∣EB∣ = ∣EA∣, and ∣EB∣ = ∣EC∣. Hence the diagonals bisect each other.
6. Prove that the exterior angles of a regular pentagon add to four right angles.
Solution:
Pick a point in the interior of the pentagon and join it to each vertex. The sum of the interior
angles of the resulting triangles is 10 right angles (two for each triangle). Hence the sum of the
interior angles of the pentagon is 6 right angles (we get 4 right angles at the vertex in the interior
of the pentagon). The sum of the interior angles plus exterior angles is 10 right angles (2 right
angles for each vertex). Hence the sum of the exterior angles is 4 right angles.
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