Ugly-looking triangles – High School Geometry

If, for whatever reason, someone decides to produce something that’s weird, that thing may turn out “ugly” as well. But why will anyone embark on such an undertaking?

Well, in our case, the present reason is simply to counterbalance what we did in our previous post, where we so much “touted” (and then pressured you with) the two precious diagrams in that post. It was taken for granted that such adulation won’t trigger anything unexpected from our respected audience. Notwithstanding, we’ve decided to explore triangle diagrams that have looks that are not appealing, in a bid to restore “parity”. But look, don’t despise these triangles’ looks; they’re still cool.

Example 1

Find coordinates for the vertices of a triangle whose sides have slopes $1,~2,~3$ . Is the answer unique?

For this example, take $\triangle PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4)$ :

The slope of line segment $QR$ is $1$ ; the slope of $PR$ is $2$ ; the slope of $PQ$ is $3.$ Needless to say, the above triangle looks rather weird; but does anyone really care?

As we show below, there are other choices for the coordinates of $\triangle PQR$ , so the answer is not unique.

Example 2

Find a different set of coordinates (other than the ones in Example 1 above) for the vertices of a triangle whose sides have slopes of $1,~2,~3$ .

In $\triangle PQR$ shown above, $P$ has coordinates $(0,4)$ , $Q$ is located at $(-2,-2)$ , while $R$ is the point $(-4,-4)$ . So the slope of $QR$ is $1$ , the slope of $PR$ is $2$ , while the slope of $PQ$ is $3$ .

Another set of coordinates that works is: $P(0,0),~Q(-2,-4),~R(-1,-3)$ , which is essentially a rotation of the triangle in Example 1. The theory behind this will be explored later. For now, we’ll calculate the side lengths, centroids, circumcenters, orthocenters, and Euler slopes of the two examples above.

Side lengths and angles

Example 3

Find the lengths of the sides of $\triangle PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4).$

By the distance/length formula:

$\begin{equation*} \begin{split} PQ&=\sqrt{(0-1)^2+(0-3)^2}\\ &=\sqrt{10}\\ &\cdots\vdots\cdots\\ PR&=\sqrt{(0-2)^2+(0-4)^2}\\ &=\sqrt{20}\\ &\cdots\vdots\cdots\\ QR&=\sqrt{(2-1)^2+(4-3)^2}\\ &=\sqrt{2} \end{split} \end{equation}$

Observe that the longest side, $PR$ , is the side with the “median” slope (the statistical median of $1,2,3$ is $2$ , as you’ll recall; at the same time $2$ is the mean, since $\frac{1+2+3}{3}=2$ ). Furthermore

$\sqrt{20}=\sqrt{2}\times\sqrt{10}=\frac{\sqrt{2}\times\sqrt{10}}{\sqrt{\Big(\frac{2}{2}\Big)}}.$

Just saying.

The angles can be calculated using the cosine law:

$\cos P=\frac{7}{5\sqrt{2}},\quad \cos Q=-\frac{2}{\sqrt{5}},\quad \cos R=\frac{3}{\sqrt{10}}.$

Observe that one of the angles is obtuse. Also

$5\sqrt{2}=\sqrt{7^2+1},~\sqrt{5}=\sqrt{2^2+1},~\sqrt{10}=\sqrt{3^2+1}.$

Example 4

Find the lengths of the sides of $\triangle PQR$ with vertices at $P(0,4),~Q(-2,-2),~R(-4,-4).$

By the distance/length formula:

$\begin{equation*} \begin{split} PR&=\sqrt{(0+4)^2+(4+4)^2}\\ &=\sqrt{80}\\ &\cdots\vdots\cdots\\ PQ&=\sqrt{(0+2)^2+(4+2)^2}\\ &=\sqrt{40}\\ &\cdots\vdots\cdots\\ QR&=\sqrt{(-4+2)^2+(-4+2)^2}\\ &=\sqrt{8} \end{split} \end{equation}$

Again, the longest side, $PR$ , is the side with the median slope (namely $2$ ). Furthermore:

$\sqrt{80}=\frac{\sqrt{8}\times\sqrt{40}}{\sqrt{\Big(\frac{8}{2}\Big)}} .$

Just saying.

As expected, the two triangles in Examples 3 and 4 are similar, since

$\frac{\sqrt{80}}{\sqrt{20}}=\frac{\sqrt{40}}{\sqrt{10}}=\frac{\sqrt{8}}{\sqrt{2}}=2$

Slope of the Euler line

Since the Euler line connects the centroid, orthocenter, and circumcenter, we’ll first find these centers for the two triangles under consideration.

Example 5

Find the centroid of $\triangle PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4).$

Here, it is convenient to use the centroid formula, which involves taking the average of the $x$ -coordinates, and then taking the average of the $y$ -coordinates. Thus the centroid in this case is located at

$\Big(\frac{0+1+2}{3},\frac{0+3+4}{3} \Big)=\Big(1,\frac{7}{3}\Big).$

Observe that the $x$ -coordinate of the centroid is the $x$ -coordinate of one of the vertices.

Example 6

Find the circumcenter of $\triangle PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4).$

To do this, we’ll find the equations of any two right bisectors, then solve the equations simultaneously. Below are the equations of the three (instead of two) right bisectors:

$\begin{equation*} \begin{split} y&=-x+5\\ y&=-\frac{1}{2}x+\frac{5}{2}\\ y&=-\frac{1}{3}x+\frac{5}{3} \end{split} \end{equation}$

The solution is $x=5,~y=0$ . So the circumcenter is located at $(5,0)$ .

Observe that the $y$ -coordinate of the circumcenter is the $y$ -coordinate of one of the vertices.

Example 7

Find the orthocenter of triangle $PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4).$

To do this, we’ll find the equations of any two altitudes, then solve the equations simultaneously. Below are the equations of the three (instead of two) altitudes:

$\begin{equation*} \begin{split} y&=-x\\ y&=-\frac{1}{2}x+\frac{7}{2}\\ y&=-\frac{1}{3}x+\frac{14}{3} \end{split} \end{equation}$

The solution is $x=-7,~y=7$ . So the orthocenter is located at $(-7,7)$ .

Using Examples 5 to 7, we see that the Euler line for $\triangle PQR$ with vertices at $P(0,0),~Q(1,3),~R(2,4)$ contains the three important (collinear) points: $(1,\frac{7}{3})$ (the CENTROID), $(5,0)$ (the CIRCUMCENTER), and $(-7,7)$ (the ORTHOCENTER). So the slope of the Euler line is

$\frac{0-7}{5--7}=-\frac{7}{12}.$

Example 8

Find the centroid of $\triangle PQR$ with vertices at $P(0,4),~Q(-2,-2),~R(-4,-4).$

As before, the centroid is located at

$\Big(\frac{0+(-2)+(-4)}{3},\frac{4+(-2)+(-4)}{3} \Big)=\Big(-2,-\frac{2}{3}\Big) .$

Again, the $x$ -coordinate of the centroid is the $x$ -coordinate of one of the vertices. Don’t forget this.

Example 9

Find the circumcenter of $\triangle PQR$ with vertices at $P(0,4),~Q(-2,-2),~R(-4,-4).$

The equations of the right bisectors are:

$\begin{equation*} \begin{split} y&=-x-6\\ y&=-\frac{1}{2}x-1\\ y&=-\frac{1}{3}x+\frac{2}{3} \end{split} \end{equation}$

And the solution is $x=-10,~y=4$ . So, the circumcenter is located at $(-10,4)$ .

Observe that the $y$ -coordinate of the circumcenter is the $y$ -coordinate of one of the vertices.

Example 10

Find the orthocenter of $\triangle PQR$ with vertices located at $P(0,4),~Q(-2,-2),~R(-4,-4).$

The equations of the three altitudes are:

$\begin{equation*} \begin{split} y&=-x+4\\ y&=-\frac{1}{2}x-3\\ y&=-\frac{1}{3}x-\frac{16}{3} \end{split} \end{equation}$

And the solution is $x=14,~y=-10$ . So the orthocenter for this triangle is located at $(14,-10)$ .

Thus, for $\triangle PQR$ with vertices at $P(0,4),~Q(-2,-2),~R(-4,-4)$ , we have the three important (collinear) points: CENTROID $(-2,-\frac{2}{3})$ , CIRCUMCENTER $(-10,4)$ , and ORTHOCENTER $(14,-10)$ . Therefore, the slope of the Euler line is:

$\frac{-10-4}{14--10}=-\frac{7}{12}.$

The Euler lines for the two triangles have the same slopes.

Takeaway

We encourage you to play around with different triangle diagrams that have side slopes of $1,2,3$ . You’ll understand why we chose to call such triangles “ugly-looking”.

Tasks

Find a triangle whose sides have slopes of $1,2,4$ .
$\Big($ Note that the numbers $1,~2,~4$ form a geometric sequence. $\Big)$
PROVE that any triangle in which the sides have slopes of $1,2,3$ ALWAYS contains an OBTUSE angle.
PROVE that in any triangle in which the sides have slopes of $1,2,3$ , the side with slope of $2$ is ALWAYS the LONGEST side.
In a triangle with sides of slopes of $1,2,3$ , let $l_1$ be the length of the side with slope $1$ , let $l_2$ be the length of the side with slope $2$ , and let $l_3$ be the length of the side with slope $3.$ PROVE that $l_2=l_3\times \sqrt{2},~l_3=l_1\times\sqrt{5},~l_2=l_1\times\sqrt{10}$ .
$\Big($ Part of the beauty of an “ugly” triangle. Note that the third equation follows from the first two. Also, $\sqrt{2}=\sqrt{1^2+1},\quad \sqrt{5}=\sqrt{2^2+1},\quad \sqrt{10}=\sqrt{3^2+1}$ . $\Big)$
PROVE that in any triangle with sides having slopes of $1,2,3$ , the centroid shares the same $x$ -coordinate with one of the vertices.
$\Big($ In particular, it is the vertex opposite the side with slope of $2$ . $\Big)$
Let $x_1,~x_2,~x_3$ be the $x$ -coordinates of the vertices of a triangle. PROVE that the $x$ -coordinate of the centroid is $x_1$ if, and only if, the sequence $x_2,x_1,x_3$ is arithmetic.
$\Big($ Thus, if the centroid shares the same $x$ -coordinate with any of the triangle’s vertices, then those $x$ -coordinates form an arithmetic sequence. In particular, any triangle whose sides have slopes of $1,2,3$ satisfies this condition. $\Big)$
Let $x_1$ be the $x$ -coordinate of one of the vertices of a triangle. PROVE that the $x$ -coordinate of the centroid is $x_1$ if, and only if, the median through the vertex containing $x_1$ is vertical.
$\Big($ Similarly, if the centroid shares the same $y$ -coordinate with a vertex, then the median through that vertex is horizontal. In view of the preceding exercise, we have an additional characterization of centroids that share the same $x$ -coordinate (or $y$ -coordinate) with one of the triangle’s vertices. $\Big)$
PROVE that in any triangle with sides of slopes $1,2,3$ , the circumcenter shares the same $y$ -coordinate with one of the vertices.
$\Big($ In particular, it is the $y$ -coordinate of the vertex opposite the side with slope $1$ . $\Big)$
PROVE that any triangle with sides of slopes $1,2,3$ CANNOT be ISOSCELES.
$\Big($ Actually, there’s only “one” isosceles triangle in which the slopes of the sides (are all integers and) follow an arithmetic sequence. Can you guess which one it is? You’d be surprised. $\Big)$
PROVE that in any triangle with sides of slopes $1,2,3$ , the SHORTEST side is the side with slope $1$ .
$\Big($ Compare this with Exercise 3 above. $\Big)$