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Assignment 1

These problems all come from the reading in sections Euclid, Parallels and Pythagoras.

Problems

Exercise 1 (Constructing an Isoceles Triangle) Start with a line segment of length $a$. Prove that you can construct a triangle with one side of length $a$, and two sides of length $2a$.

Exercise 2 (Inscribing an Equilateral Triangle) Prove that inside of an equilateral triangle, you can inscribe an upside down equilateral triangle of exactly half the side length, shown

An equilateral triangle inscribed within a larger one.

Exercise 3 (Angle Sums of Polygons) A polygon is convex if all of its angles are less than $180^\circ$, so that it has no “indents”. Equivalently, a convex polygon is one where any line segment with endpoints on the boundary of the polygon lies inside the polygon.

Prove that the angle sum of convex quadrilaterals is a constant, for all quadrilaterals. Prove the angle sum of convex pentagons is also a constant. What are these constants?

What do you think the formula is for the sum of angles in a convex $n$-gon? (Optional: If you have seen mathematical induction, prove your guess!)

Exercise 4 (Rectangles Exist) Prove that Rectangles exist using Euclids Postulates (and also Playfair’s Axiom, if you like it), and the propositions proven in the sections Euclid and Parallels.

Hint - we know how to make right angles now, and parallel lines through points. Start making some!

Exercise 5 (Diagonal Bisectors) If the diagonals of a quadrilateral are bisect one another, then that quadrilateral is a parallelogram.

Exercise 6 (Proving the Pythagorean Theorem) The following is an ingenious rearrangement proof of the Pythagorean theorem.

Prove that the final shape shown here is a square, using what we have learned (the Postulates and Propositions).

Solutions

### Constructing an Isoceles Triangle

We mimic Euclid’s original construction of the equilateral triangle as closely as possible here. First begin with a line segment $AB$ of length $a$.

Note: This could be accomplished by using Postulate 2 to extend the line $AB$ indefinitely, then drawing a circle of radius $AB$ at $A$ to mark a new point $X$ as in the figure below, and symmetrically drawing a circle of radius $AB$ based at $B$ to mark $Y$. Alternatively, we could extend and just use Proposition 3, which allows us to cut to any size we know. In future problems, I will not show steps that involve extending or cutting lines to length as they can always be accomplished in these ways.

Then, extend this line segment by another segment $XA$ and $BY$ of length $a$ on each side.

Now, draw a (blue) circle with radius $AY$ centered at $A$, and a (red) circle with radius $XB$ centered at $B$. As each of these line segments are length $2a$, these circles have the same radius. Mark a point of their intersection by $C$, and connect $C$ to both $A$ and $B$.

Since $AC$ is a radius of the blue circle, its length is $2a$. And since $BC$ is a radius of the red circle, its length is also $2a$. Thus the triangle $ABC$ has one side ($AB$) of length $a$ and two sides of length $2a$, as desired.

Inscribing an Equilateral Triangle

This exercise is a bit challenging, but there are multiple ways to do it using only what we know so far. I look forward to seeing your approaches, but I give two possibilities here, for extra practice! For both of them, I found I was using the same fact over and over, so I have pulled it out as a quick lemma:

Lemma 1 (Equilateral if and only if Equiangular) A triangle is equilateral (all side lengths equal) if and only if it is equiangular (all angles equal).

Proof. To prove this, we need to do both directions. First, we start by assuming we have an equilateral triangle $ABC$, and proving that its angles are the same. Since its sides are all the same length, it is side-side-side congruent to any rotated copy of itself. Concretely, we see that $ABC$ is congruent to $BCA$.

This sets up an equality between the angles, which when strung together shows they are all equal to one another. \[ A=B\hspace{0.5cm} B=C\hspace{0.5cm} C=A\]

Now that we have completed that direction we do the reverse, and show if a triangle $ABC$ is equiangular, then its sides are also the same length. Choose one of the angles - say $B$ - and bisect it with a line. This line divides the triangle into two smaller triangles, which we see are congruent (they share a side: the new bisecting line, as well as two angles since they each have one of the original angles, and one of the bisected halves). Thus, the remaining pairs of sides of this triangle are also congruent, so $BA$ equals $BC$.

There was nothing special about the angle $B$, so we may also do the same construction at another angle - say $A$. This again gives a pair of congruent triangles, from which we can conclude that $AB$ equals $AC$.

Stringing these equalities together, we see that $AB=AC=BC$ so all three sides are equal, and the triangle is equilateral.

Solution I: Chasing Angles

Note: one way to find the midpoint of an edge explicitly would be to construct an equilateral triangle on that edge, and connect its remainign vertex to the remaining vertex of the original.

Start with an equilateral triangle $ABC$, and mark the midpoints of each of its edges $X,Y$ and $Z$. Connect these midpoints to form a triangle $XYZ$. Our goal is to show this to be equilateral, with half the side length of $ABC$.

Triangle $XYZ$ also determines three more triangles, each containing one of the original vertices of $ABC$. Each of these triangles are congruent by side-angle-side (they share two side lengths equal to half of $ABC$’s, and one angle of $ABC$ - all of which were equal as it is equilateral). Since they are congruent, their bases - the new segments $XY$, $YZ$ and $ZX$ - must be equal. This means the central triangle is equilateral, and hence equiangular!

The angles of the inner triangle are each the same measure as the angles of the outer triangle (hence they are all colored yellow) because every triangle has angle sum $\pi$, so their measure is $\pi/3$.

We can carry this reasoning further however: consider all of the angles based at the point $Y$. The yellow one we know the value, and the other two must be equal: they correspond to one another in two congruent triangles. Together they sum to a straight line, so if $\beta$ is the measure of the blue angle we know $\frac{\pi}{3}+2\beta = \pi$ so $\beta$ itself is $\pi/3$! Thus all three angles at $Y$ are equal.

This same reasoning can be carried out at $X$ and $Z$ to see all the angles there are equal as well. But now, we see that each of the three triangles cut off at the vertices have all three angles equal: they’re equiangular! Thus, they’re equilateral, so all sides are of equal length, and the three interior sides connecting $XYZ$ are all half the length of the sides of $ABC$.

Solution II: Parallels

Start by choosing one side of the triangle $ABC$ and marking a point $X$ half way along it. From this point, use Playfair’s axiom to draw paralle lines to the other two sides, intersecting the triangle at points $Y,Z$. Because corresponding angles are equal, this tells the angles of intersection of each of the paralle lines with the triangle: they are the same as the triangles’ vertex angles.

Because the original triangle is equilateral (and thus equiangular), all angles in the diagram thus far are equal. Its helpful to put them all into one picture, where we now see that we know all three angles of the two smaller triangles cut off in the top, and lower left. Since these triangles have equal angles, they are equilateral as well, so we now know all their side lengths.

In addition we know one more piece of information from the diagram above. The angle between the two lines we drew through $X$ must also be $\pi/3$, as it makes up a straight angle with two other angles of measure $\pi/3$.

Now, draw a final line segment (green) connedting the points $Y$ and $Z$. This cuts out a triangle $XYZ$ in the middle. But - this triangle now shares side-angle-side with an equilateral triangle (its angle we just determined to be $\pi/3$, and the side lengths of the blue and pink lines all match with the smaller triangle at the top, or in the lower left). Thus, this triangle must also be euqilateral! And so, all of its side lengths are half the larger triangles, as required.

Angle Sums of Polygons

Begin with a quadrilateral with angles $\alpha,\beta,\gamma,\delta$. We wish to calculuate the sum $\alpha+\beta+\gamma+\delta$. Choose two opposite vertices (here, those with angles $\alpha$ and $\gamma$) and connect them by a diagonal. This divides angle $\alpha$ into two new angles $x,y$ and $\gamma$ into two angles $u,v$.

Since the quadrilateral is convex, this line segment lies wholly within the quadrilateral, and divides its interior into two triangles. These triangles are likley not congruent, but we do know that each of them has total angle sum $\pi$ or $180^\circ$. Thus, we know $y+\delta+v=\pi$ and $u+\beta+x=\pi$, so adding these two equations together

\[(x+y)+\beta + (u+v)+\delta = 2\pi\]

Howver, by the angle axioms we know that $x+y=\alpha$ and $u+v=\gamma$ (as we just subdivided these into two angles). Thus, the total angle sum $\alpha+\beta+\gamma+\delta=2\pi$.

A similar argument works for pentagons: draw an arbitrary convex pentagon and choose some vertex. This vertex is already connected to two others (by edges of the pentagon) - but connect it by new line segments to the remaining two. These segments lie inside the pentagon (by convexity) and divide it into three triangles. Each triangle has angle sum $\pi$, so the pentagon has angle sum $3\pi$.

Rectangles Exist

Start with a line, and choose some point $P$ along the line. That line determines a straight angle at $P$, so bisect this into two right angles with another line. Choose a point $Q$ along this perpendicular, and using Playfair’s axiom draw the unique parallel through this point.

Because the blue and black lines are parallel, they do not intersect to the right or left, so the sum of the angles they make with the read line must be precisely $\pi$. Because the angle at $P$ is already a right angle, this imples the angle at $Q$ must also be a right angle. Now, choose some point $R$ along this blue line, and bisect the straight angle at $R$ to produce another right angle.

The sum of the angles this new green line makes with the two parallel lines must also be precisely $\pi$ (since they’re parallel). And again we know the angle at $R$ is a right angle by construction. So the angle at its other point of intersection - call it $S$- must also be right.

Thus, we have constructed a quadrilateral $PQRS$ with four right angles: a rectangle. So, rectangles exist.

Diagonal Bisectors

Begin with a quadrilateral whose diagonals bisect each other: in the picture below, this means we are assuming the red segments are equal in length, as are the blue segments.

These diagonals cut the quadrilateral into four triangles. Becase opposite angles are always equal to one another when two lines intersect, we see the two green angles here have the same measure. Thus, the left and right triangles are congruent by side-angle-side (they each have a red side, blue side, and green angle). But since they are congruent, all their angles are the same. We depict this by coloring additional equal pairs of angles teal and pink.

The same argument works for the top and bottom triangles, which are also congruent by side-angle-side, and thus have all three pairs of angles congruent. The diagram below color codes all pairs of angles we now know to be the same.

Look at either the blue or red diagonal, and consider how it divides the quadrilateral into two triangles. Since every triangle has angle sum $\pi$, we know that (looking at either triangle) that the sum of an orange, pink, blue, and teal angle together is exactly $\pi$. We will now use this fact to conclue the pairs of opposite sides are parallel.

Extend first the top and bottom sides, as well as the left side of the quadrilateral, and denote the two angles here $A$ and $B$. Since between the two we have exatly one each of an orange, pink, blue, and teal angle, the angle sum at $A$ and $B$ is $\pi$. But this means (by Euclid’s $5^{th}$ postulate) that the top and bottom lines must be parallel!

The same argument applies to the left and right sides, if we consider their intersections with the bottom at $A$ and $C$. Between these two there is also one angle of each color, leading to an angle sum of $\pi$, forcing them to be parallel.