A special square (root) – High School Geometry

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.

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:

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

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.
Consider an equilateral triangle with vertices , , and . If the slope of side is , PROVE that:
- the slopes of sides $AB$ and $CA$ are reciprocals of each other;
- $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .
Consider an equilateral triangle with vertices , , and . If the slope of side is , PROVE that:
- the slopes of sides $AB$ and $CA$ are reciprocals of each other
- $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .
Let be a right triangle in which is the point and . Side makes angle with the positive -axis, and the lengths of and are and , respectively. PROVE that one can describe:
- $A$ as the point $(x_1+b\sin\theta,y_1-b\cos\theta)$
- $B$ as the point $(x_1+a\cos\theta,y_1+a\sin\theta)$ .
Given a vertex , two side lengths and , and a parameter :
- 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.
For $r\neq 0,\pm 1$ , PROVE that any triangle with side slopes $\frac{1}{r},1,r$ is isosceles.
Let , , , where , 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.
For a triangle with vertices , , , let , , and 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.
Suppose that an equilateral triangle has one side parallel to the -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.
Given , , and :
- 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 $(-2+\sqrt{2},-3+\sqrt{2})$ .