A special square (root)

The process of squaring (and taking square root) can sometimes preserve the operation of addition, as in this equation:

(1)   \begin{equation*}\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}. \end{equation*}

We show that if we take (\alpha_1,\beta_1) as the centroid, (\alpha_2,\beta_2) as the circumcenter, and (\alpha_3,\beta_3) as the orthocenter of any triangle, then the equation always holds — evaluating to \frac{1}{4} in every case.

Interestingly, if we append absolute values, then the equation holds twice for a right triangle — with (\alpha_1,\beta_1) as the vertex containing angle 90^{\circ}, while (\alpha_2,\beta_2) and (\alpha_3,\beta_3) are the other two vertices:

(2)   \begin{equation*}\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}. \end{equation*}

The latter equation is preferred to the former one, and is behind a little geometric application that we present.

Consider \triangle ABC with vertices at A(1,-3)=(\alpha_1,\beta_1), B(-3,-1)=(\alpha_2,\beta_2), and C(5,5)=(\alpha_3,\beta_3). VERIFY that equation (2) holds in this case.

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Using (\alpha_1,\beta_1)=(1,-3), (\alpha_2,\beta_2)=(-3,-1), and (\alpha_3,\beta_3)=(5,5), we have, from equation (2) that:

    \begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-3-1)^2+(-1--3)^2}{(5-1)^2+(5--3)^2}\\ &=\frac{(-4)^2+2^2}{4^2+8^2}\\ &=\frac{20}{80}\\ &=\frac{1}{4}\\ \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-3-1|+|-1--3|}{|5-1|+|5--3|}\right)^2\\ &=\left(\frac{|-4|+2}{4+8}\right)^2\\ &=\left(\frac{6}{12}\right)^2\\ &=\frac{1}{4} \end{split} \end{equation*}

CL!!!!!!!!!!!

Consider the same \triangle ABC in the previous example: A(1,-3), B(-3,-1), C(5,5). Using (\alpha_1,\beta_1) as the centroid, (\alpha_2,\beta_2) as the circumcenter, and (\alpha_3,\beta_3) as the orthocenter, VERIFY that equation (1) holds.

Since \triangle ABC is a right triangle, its circumcenter is the midpoint of the hypotenuse (in this case the midpoint of side BC), and its orthocenter is at the vertex containing angle 90^{\circ} (vertex A). The centroid is obtained as in any other triangle, using \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) or \left(\frac{\alpha_1+\alpha_2+\alpha_3}{3},\frac{\beta_1+\beta_2+\beta_3}{3}\right) in the present case.

So we have (\alpha_1,\beta_1)=\left(1,\frac{1}{3}), (\alpha_2,\beta_2)=(1,2), and (\alpha_3,\beta_3)=(1,-3). In equation (1):

    \begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{0^2+(5/3)^2}{0^2+(-10/3)^2}\\ &=\frac{25}{100}\\ &=\frac{1}{4}\\ \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{0+(5/3)}{0+(-10/3)}\right)^2\\ &=\frac{1}{4} \end{split} \end{equation*}

CL!!!

Right company:

In our next example we show that equation (2) doesn’t hold for a triangle that’s not right.

Consider a triangle with vertices (0,0), (1,4), (3,6). PROVE that equation (2) fails in this case.

We will select (\alpha_1,\beta_1) as any of the three vertices. This leads to three possibilities:

  • Case I: let (\alpha_1,\beta_1)=(0,0), (\alpha_2,\beta_2)=(1,4), and (\alpha_3,\beta_3)=(3,6).

        \begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{1+4}{3+6}\right)^2\\ &=\frac{25}{81}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{1^2+4^2}{3^2+6^2}\\ &=\frac{17}{45} \end{split} \end{equation*}

  • Case II: let (\alpha_1,\beta_1)=(1,4), (\alpha_2,\beta_2)=(0,0), and (\alpha_3,\beta_3)=(3,6).

        \begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-1|+|-4|}{2+2}\right)^2\\ &=\frac{25}{16}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-1)^2+(-4)^2}{2^2+2^2}\\ &=\frac{17}{8} \end{split} \end{equation*}

  • Case III: let (\alpha_1,\beta_1)=(3,6), (\alpha_2,\beta_2)=(0,0), and (\alpha_3,\beta_3)=(1,4).

        \begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-3|+|-6|}{|-2|+|-2|}\right)^2\\ &=\frac{81}{16}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-3)^2+(-6)^2}{(-2)^2+(-2)^2}\\ &=\frac{45}{8} \end{split} \end{equation*}

Although equation (2) fails for this triangle, equation (1) still holds, so long as we take (\alpha_1,\beta_1) as the centroid, (\alpha_2,\beta_2) as the circumcenter, and (\alpha_3,\beta_3) as the orthocenter.

Co-linear points

PROVE that equation (1) holds if the points (\alpha_1,\beta_1), (\alpha_2,\beta_2), and (\alpha_3,\beta_3) lie on the same straight line (co-linear).

Ignoring degeneracies (like \alpha_1=\alpha_3 etc), we’re to prove that

    \[\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\]

based on the co-linearity of the given points. [In the case where the slope of the line is -1, absolute values will be necessary on the left. As a result of this, it is generally preferable to use equation (2).]

Co-linearity ensures that the slope between any two points will be the same, so we have:

    \[\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}=\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1},\implies \beta_2-\beta_1=\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1).\]

Then

    \begin{equation*} \begin{split} \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{(\alpha_2-\alpha_1)+\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\alpha_3-\alpha_1}\right)^2\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\alpha_3-\alpha_1)^2} \end{split} \end{equation*}

In any triangle, let (\alpha_1,\beta_1), (\alpha_2,\beta_2), and (\alpha_3,\beta_3) be the centroid, circumcenter, and orthocenter, respectively. PROVE that equation (1) is satisfied, and that both sides evaluate to \frac{1}{4}.

This follows from the preceding example together with the well-known fact that these three principal triangle centers are co-linear.

To show that both sides evaluate to 1/4, consider the right member of equation (1):

    \[\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\]

where the numerator represents the square of the circumcenter-centroid distance, and the denominator represents the square of the centroid-orthocenter distance. Since the centroid divides the circumcenter-orthocenter distance in the ratio 1:2, the right side always reduces to (1/2)^2. And so is the left side.

Square root:

Equation (1) can be re-written as (if one wants, one can place absolute values on each term on the right):

    \[\sqrt{\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}}=\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\]

and so even square roots preserve addition in this situation.

Opposite slopes

In any triangle in which two sides have opposite slopes, PROVE that equation (2) is satisfied.

Let (\alpha_1,\beta_1) be the vertex from which the sides with opposite slopes originate, and let (\alpha_2,\beta_2) and (\alpha_3,\beta_3) be the two other vertices. We show that:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.\]

By assumption

    \[\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}=-\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1},\implies \beta_2-\beta_1=-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1).\]

The remainder of the proof is akin to that of the previous example; the only difference is the inclusion of absolute values:

    \begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|\alpha_2-\alpha_1|+\left |-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right |}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\alpha_3-\alpha_1}\right)^2\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\alpha_3-\alpha_1)^2} \end{split} \end{equation*}

Reciprocal slopes

In any triangle in which two sides have reciprocal slopes, PROVE that equation (1) is satisfied.

Let (\alpha_1,\beta_1) be the common vertex from which the sides with reciprocal slopes originate. Let the two other vertices be (\alpha_2,\beta_2) and (\alpha_3,\beta_3). We have:

    \[\left(\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}\right)\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)=1,\implies \beta_2-\beta_1=\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}.\]

Then

    \begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\beta_3-\beta_1)^2}\\ \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{(\alpha_2-\alpha_1)+\left[\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\beta_3-\beta_1}\right)^2 \end{split} \end{equation}

Favorite triangle:

Hardly would we talk about triangles without mentioning the ones we both cherish and relish — those whose slopes form geometric progressions. In today’s context, the ones with slopes \frac{1}{r},1,r satisfy equation (1).

Right triangles

PROVE that any right triangle satisfies equation (2).

Let (\alpha_1,\beta_1), (\alpha_2,\beta_2), and (\alpha_3,\beta_3) be the vertices of a right triangle, where (\alpha_1,\beta_1) is the vertex containing angle 90^{\circ}. We show that:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.\]

The proof is much like that of , with a slight modification.

Since the slopes of the sides originating from vertex (\alpha_1,\beta_1) are negative reciprocals of each other, we have:

    \[\left(\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}\right)\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)=-1,\implies \beta_2-\beta_1=-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}.\]

Then

    \begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\beta_3-\beta_1)^2}\\ \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|\alpha_2-\alpha_1|+\left |\left[-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]\right |}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\beta_3-\beta_1}\right)^2 \end{split} \end{equation}

In the end, only one of the four absolute values became absolutely valuable.

Absolute need: Here’s an example to show the difference an absolute value can make.

Let (\alpha_1,\beta_1)=(7,5), (\alpha_2,\beta_2)=(5,7), and (\alpha_3,\beta_3)=(8,4). Using these values in the equation

    \[\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\]

the left side evaluates to the indeterminate form (\frac{0}{0})^2, while the right side is 4. However, if we use absolute values on the left, then both sides evaluate to 4.

This is an exceptional case that happens whenever the slope of the line joining the points is -1. For this reason we generally recommend using equation (2).

Geometric application

The following shapes

  • squares
  • isosceles right triangles
  • equilateral triangles with one side having slope of 1
  • equilateral triangles with one side parallel to the x-axis
  • isosceles triangles with one side parallel to the x-axis

have something in common.

Let (\alpha_2,\beta_2) and (\alpha_3,\beta_3) be the endpoints of one diagonal of a square, and let (\alpha_1,\beta_1) be any of the two remaining vertices. PROVE that |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.

Since a diagonal splits a square into two right triangles, we can apply equation (2):

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\]

noting that the right side evaluates to 1 for our square. In turn:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=1\implies \frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=\pm 1.\]

In view of the absolute values, we only take \frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=1, which gives |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.

In an equilateral triangle with one side parallel to the x-axis, let (\alpha_1,\beta_1) be the “apex”, while (\alpha_2,\beta_2) and (\alpha_3,\beta_3) are the two other vertices. PROVE that |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.

In such an equilateral triangle, the slopes of the sides are of the form -a,0,a. Since two of these are opposite values, we can use a previous example to deduce that equation (2) holds:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\]

The right side evaluates to 1 for our equilateral triangle, so:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=1\implies \frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=\pm 1.\]

In view of the absolute values, we only take \frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=1, which gives |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.

Takeaway

Out of the two equations we considered, we recommend the second one:

    \[\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.\]

In practice, the equation works because of the ratio.

Tasks

  1. In any isosceles right triangle, PROVE that the equation |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1| holds, if (\alpha_1,\beta_1) is the vertex containing angle 90^{\circ}, while (\alpha_2,\beta_2) and (\alpha_3,\beta_3) are the other two vertices.
  2. Consider an equilateral triangle with vertices A(\alpha_1,\beta_1), B(\alpha_2,\beta_2), and C(\alpha_3,\beta_3). If the slope of side BC is 1, PROVE that:
    • the slopes of sides AB and CA are reciprocals of each other;
    • |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.
  3. Consider an equilateral triangle with vertices A(\alpha_1,\beta_1), B(\alpha_2,\beta_2), and C(\alpha_3,\beta_3). If the slope of side BC is -1, PROVE that:
    • the slopes of sides AB and CA are reciprocals of each other
    • |\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|.
  4. Let \triangle ABC be a right triangle in which C is the point (x_1,y_1) and \angle C=90^{\circ}. Side BC makes angle \theta with the positive x-axis, and the lengths of BC and AC are a and b, respectively. PROVE that one can describe:
    • A as the point (x_1+b\sin\theta,y_1-b\cos\theta)
    • B as the point (x_1+a\cos\theta,y_1+a\sin\theta).
  5. Given a vertex (x_1,y_1), two side lengths a and b, and a parameter \theta:
    • PROVE that any triangle with vertices (x_1,y_1), (x_1+a\cos\theta, y_1+a\sin\theta), and (x_1+b\sin\theta,y_1-b\cos\theta) is a right triangle
    • Using (x_1,y_1)=(1,-3), a=2\sqrt{5}, b=4\sqrt{5}, and \tan\theta=-\frac{1}{2}, verify that one obtains (-3,-1) and (5,5) as the remaining two vertices.
  6. For r\neq 0,\pm 1, PROVE that any triangle with side slopes \frac{1}{r},1,r is isosceles.
  7. Let (a,b), (b,a), (c,c), where a\neq b\neq c, be the vertices of a triangle. PROVE that:
    • the triangle is isosceles
    • the slopes of the sides form a geometric sequence with -2 as common ratio
    • equation (2) is satisfied in this case.
  8. For a triangle with vertices (0,0), (1,4), (3,6), let (\alpha_1,\beta_1), (\apha_2,\beta_2), and (\alpha_3,\beta_3) be the centroid, circumcenter, and orthocenter, respectively.
    • calculate (\alpha_1,\beta_1), (\apha_2,\beta_2), and (\alpha_3,\beta_3)
    • verify that equation (2) is satisfied.
  9. Suppose that an equilateral triangle has one side parallel to the y-axis (so that its slope is undefined). PROVE that:
    • the slopes of the other two sides are opposites of each other
    • equation (2) is satisfied.
  10. Given (x_1,y_1), a, and \theta:
    • PROVE that any quadrilateral with vertices (x_1,y_1), (x_1+a\cos\theta,y_1+a\sin\theta), (x_1+a(\cos\theta+\sin\theta),y_1+a(\sin\theta-\cos\theta)), and (x_1+a\sin\theta,y_1-a\cos\theta) is a square
    • determine a and \theta for the square with vertices (-2,-3), (-2+\sqrt{2},-3-\sqrt{2}), (-2+2\sqrt{2},-3), and (-2+\sqrt{2},-3+\sqrt{2}).