Advanced Geometry/Solutions

Chapter 1

Problem 1

Let the three given lines be $a,b,c$. Construct $[A,a]$. On a arbitrary point $B$ on $[A,a]$, construct $[B,b]$. Construct $[A,c]$ and let $C=⟨[A,c],[B,b]⟩$ (any one will do). Connect $A,B,\text{and}C$. A triangle should form.

Problem 2

Construct two concentric circles. On one of the circles, mark an arbitrary point and connect it to the center.. Proceed to copy an angle on this radius. Connect the center with the other circle through the non-terminal point. Lastly, connect the two endpoints of the line segment. A triangle should form.

Problem 3

Construct a circle with the given radius on $A$. Construct an arbitrary point on the circle. Copy the two angles onto those points. Connect the vertexes of the triangle with the non-terminal points of the angle. Get the intersection of the two rays. A new triangle should form.

Problem 4

Construct a right triangle with the legs being the two squares. Construct a square from the new hypotenuse. This square, by the Pythagorean theorem, will be equal to the sum of the square of the two legs.

Problem 5

construct a circle with the given radius ($a$) on the given point ($A$). Draw a line ($b$) through the given point that is perpendicular to the given line. Let $B=⟨b,[A,a]⟩$. On this point, construct a circle with the same radius, thus satisfying the problem.

Problem 6

Let us call the given hypotenuse $a$. On an arbitrary point $A$, construct $[A,a]$. Construct $B$ on the circle and copy the angle on $\overline{AB}$. On the non-terminal point, draw a line that connects to the angle vertex. After that, construct a perpendicular that goes through the center and is perpendicular to the previous line you just made.

Problem 7

Construct point $A$. Create a line segment $\overline{AB}$ that has the length of the larger square's side (denoted by $a$). With the smaller square's side, construct a circle that has that side as the radius and a center of $A$. Bisect $\overline{AB}$ and construct a circle with the bisection point $C$ as the center, and A as the endpoint. Let $D=⟨[C,CA],[A,a]⟩$. This point will be the segment

Problem 8

Let us call the segments $a,b,\text{and}c$. Create an arbitrary point $A$ and create two concentric circles, one with a radius of $a$ and the other with a radius of $c$. Construct an arbitrary point on the circle with the smaller radius and call it $B$. Construct $[B,b]$. Intersect [B, b] with [A, c], and mark one of the points-- call it $C$. Through $C$ construct a line parallel to $\overleftrightarrow{AB}$ called $d$. Construct a line through $A$ which is parallel to $\overleftrightarrow{BC}$ called $e$. Call the intersection of $d\text{and}e$ D. The parallelogram ABCD should be constructed.

Problem 9

Suppose we have a right triangle with legs $c$ and $d$ and hypotenuse $e$. Suppose that the altitude perpendicular to the hypotenuse divides the triangle into the ratio $a$ and $b$. Then it follows that $c^2=ae$ and $d^2=be$. Therefore, the ratio of the squares, $c^2:d^2$, is equal to $ae:be$. By simplifying, this becomes $c^2:d^2=a:b$, thus concluding the proof.

Problem 10

Construct a right triangle with legs consisting of the two given segments. Unfortunately, you need the ratio of the squared segments, when you were given just the plain, normal segments. Once the right triangle is constructed, drop the altitude which is perpendicular to the hypotenuse. The hypotenuse will be divided into the square of the legs, which is the ratio you're looking for. Now that you have the ratio of the segments squared, you need to divide the first given segment into that ratio.

To do this, take the endpoints of the first segment, and connect them to the respective endpoints of the segment we just constructed. These lines should intersect. Remember the point where the altitude divides the hypotenuse? Connect that point with the intersection point we just constructed, and draw a line. The line will intersect the first given segment and solve the problem.

Problem 11

Now you're given a line segment which is divided into a ratio. At the point where the ratio is, construct a perpendicular line. Now, take the midpoint of the given segment and construct a circle with a radius being the distance from one of the endpoints to the midpoint. Intersect this circle with the perpendicular line, and construct a right triangle. Connecting the intersection of the perpendicular line and the circle with the endpoints of the given segment will create a right triangle.

Problem 12

Construction: Let us call the center of the larger circle $A$ and the radius of the larger circle $r_a$ and let's call the center of the smaller circle $B$ and the radius of the smaller circle $r_b$. Create an arbitrary point called $C$ which lies on the larger circle. Construct $[C,r_b]$ and line $\overline{AC}$. Let $D=⟨[C,r_b],\overline{AC}⟩$ north. The direction is important; using the wrong direction solves Problem 13 instead. Construct $[A,AD]$. Now, construct two tangent lines to the circle $[A,AD]$ which intersect at B; call the points of tangency $E$ and $F$. Construct $\overrightarrow{AE}$ and $\overrightarrow{AF}$. Call the points of intersection with the original, smaller circle $G$ and $H$. Construct lines that go through $G$ and $H$ and are perpendicular to the two rays.

Proof: Pretty much the same as in 13, except with r_a + r_b instead of r_a - r_b.

Problem 13

Construction: Let us call the center of the larger circle $A$ and the radius of the larger circle $r_a$ and let's call the center of the smaller circle $B$ and the radius of the smaller circle $r_b$. Create an arbitrary point called $C$ which lies on the larger circle. Construct $[C,r_b]$ and line $\overline{AC}$. Let $D=⟨[C,r_b],\overline{AC}⟩$ south. Construct $[A,AD]$. Now, construct two tangent lines to the circle $[A,AD]$ which intersect at B; call the points of tangency $E$ and $F$. Construct $\overrightarrow{AE}$ and $\overrightarrow{AF}$. Call the points of intersection with the original, larger circle $G$ and $H$. Construct lines that go through $G$ and $H$ and are perpendicular to the two rays.

Proof: The circle [A, AD] has the property that its radius is $r_a-r_b$. That means that if there is a tangent point on [A, AD], by drawing a ray I will be able to find a tangent point on [A, r_a] (aka the larger given circle). However, since the tangent lines intersect at point $B$, if we increase the radius by this amount, we also get a tangent line for $[B,r_b]$! By doing this, we construct lines which are tangent to two circles.

Problem 14

Let us call the height of the triangle $a$ and let us call the base of the triangle $b$. Let us construct an angle with vertex $A$. On the first leg, create an arbitrary distance from A, and on this distance, construct the point $B$. Remember the segment $\overline{AB}$ for later. On the other leg, construct a point $C$ such that $\overline{AC}$ is the height of the triangle. Back at the first leg, construct a point $D$ so that $\overline{AD}$ is equal to the base of the triangle. Construct a parallel line from D so that it intersect the second leg at E. This point will be equal to the product of the base and the height.

Construct a circle with a center and a radius of $\overline{AE}$. construct a diameter from this circle, and let us call the antipodes of this circle $F\text{and}G$. From this, construct a 60 degree angle at one of the antipodes, and bisect it. (Hint: you can construct a 60 degree angle by partially constructing an equilateral triangle.) At the other antipode, construct the tangent to the circle. Find the intersection between the bisection line and the tangent. Let us call this point $H$. Take the distance from H and the antipode (abbreviated as $c$), and find the geometric mean between $c\text{and}\overline{AB}$. This length will be the side of the equilateral triangle.

An equilateral triangle with the previous side length will have the same area as the given triangle.

(Outline of the) Proof: The first operation with the parallel lines was multiplying the base by the height. After that, the operation with the circle was multiplying the previous product by two. After that, the thing with the triangle was dividing it by sqrt(3). Lastly, the geometric mean operation was taking the square root.

By virtue of the fact that a triangle's area is $\frac{bh}{2}$ and the area of an equilateral triangle is $\frac{s^2\sqrt{3}}{4}$. one can quickly see that we were using simple algebra to solve for $s$.