An isosixless triangle – High School Geometry

Doubtless, it’s quite careless — if not senseless — to name a triangle isosixless. Unless an attempt is made to buttress the use of such an address, one’ll never be impressed by the meaningless, lame name.

Given any triangle $ABC$ , take the quotient of the slope of each median and the slope of the opposite side, and form the sum

(1) $\begin{equation*} q=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}. \end{equation*}$

Among all triangles with one side parallel to the $x$ -axis (or $y$ -axis), it’s only the isosceles triangle that satisfies $q=-6$ . Thus, we thought it might be apt to adapt isosceles as isosixless, in view of $-6$ . Now, beat that.

For $\triangle ABC$ with vertices at $A(0,6)$ , $B(-4,0)$ , $C(4,0)$ , verify that $q=-6$ , where $q$ is given by equation (1).

A quick calculation shows that

$m_{AB}=\frac{3}{2},~m_{BC}=0,~m_{CA}=-\frac{3}{2},~m_{A}=\infty,~m_{B}=\frac{1}{2},~m_{C}=-\frac{1}{2}$

and so

$\begin{equation*} \begin{split} q&=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}\\ &=0+\left(-\frac{3}{2}\times\frac{2}{1}\right)+\left(\frac{3}{2}\times\frac{-2}{1}\right)\\ &=-6 \end{split} \end{equation*}$

No doubt about that.

Negative partsixtions

Normally, partitions are not defined for negative integers, but with your permission, we’re doing so for this discussion. Taking two/three numbers at a time, we can write

$-6=(-1)+(-2)+(-3)=-2-2-2=-3-3=-1-5=-1-1-4$

We’ll show that only the “partition” $(-3)+(-3)$ works, and yields a characterization of isosceles triangles.

Consider $\triangle ABC$ with vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_2)$ . PROVE that

$q=\frac{2(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}{(x_1-x_2)(x_1-x_3)},$

where $q$ represents the sum of quotients $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}$ .

We have:

$\begin{equation*} \begin{split} q&=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}\\ &=0+\left(\frac{y_1-y_2}{x_1-x_3}\right)\times\left(\frac{x_1+x_3-2x_2}{y_1-y_2}\right)+\left(\frac{y_1-y_2}{x_1-x_2}\right)\times\left(\frac{x_1+x_2-2x_3}{y_1-y_2}\right)\\ &=\frac{(x_1+x_3-2x_2)(x_1-x_2)+(x_1+x_2-2x_3)(x_1-x_3)}{(x_1-x_2)(x_1-x_3)}\\ &=\frac{2x_1^2+2x_2^2+2x_3^2-2x_1x_2-2x_1x_3-2x_2x_3}{(x_1-x_2)(x_1-x_3)} \end{split} \end{equation*}$

Consider $\triangle ABC$ with vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_2)$ . PROVE that

$q=\frac{2(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}{(x_1-x_2)(x_1-x_3)}=-6$

if and only if $2x_1=x_2+x_3$ .

First suppose that $\frac{2(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}{(x_1-x_2)(x_1-x_3)}=-6$ . After clearing fractions and combining like terms, we obtain

$4x_1^2+x_2^2+x_3^2-4x_1x_2-4x_1x_3+2x_2x_3=0,$

where the left side is precisely the expansion of $(x_2+x_3-2x_1)^2$ . Thus, $x_2+x_3=2x_1$ .

Conversely, suppose that $x_2+x_3=2x_1$ . Then:

$\begin{equation*} \begin{split} q&=\frac{2(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}{(x_1-x_2)(x_1-x_3)}\\ &=\frac{2(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)}{x_1^2-x_1(x_2+x_3)+x_2x_3}\\ &=\frac{2\left(\frac{x_2+x_3}{2}\right)^2+2x_2^2+2x_3^2-2\left(\frac{x_2+x_3}{2}\right)(x_2+x_3)-2x_2x_3}{\left(\frac{x_2+x_3}{2}\right)^2-\left(\frac{x_2+x_3}{2}\right)(x_2+x_3)+x_2x_3}\\ &=\frac{2(x_2+x_3)^2+8x_2^2+8x_3^2-4(x_2+x_3)^2-8x_2x_3}{(x_2+x_3)^2-2(x_2+x_3)^2+4x_2x_3}\\ &=\frac{6x_2^2-12x_2x_3+6x_3^2}{-x_2+2x_2x_3-x_3^2}\\ &=\frac{6(x_2-x_3)^2}{-(x_2-x_3)^2}\\ &=-6. \end{split} \end{equation*}$

In any $\triangle ABC$ , PROVE that $m_{CA}=-3m_{B}$ and $m_{AB}=-3m_{C}$ if and only if $\triangle ABC$ is isosceles with side $BC$ parallel to the $x$ -axis.

Thus, $q=-6$ only when two of the summands in (1) are both $-3$ .

We show one direction of the proof — that both $m_{CA}=-3m_{B}$ and $m_{AB}=-3m_{C}$ yield an isosceles triangle with one side parallel to the $x$ -axis.

Place the vertices of $\triangle ABC$ at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . The first condition $m_{CA}=-3m_B$ amounts to

$\frac{y_1-y_3}{x_1-x_3}=-3\left(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\right)$

Clear fractions and combine similar terms:

(2) $\begin{equation*} 4x_1y_1-2x_3y_1-2x_2y_1+2x_1y_3-4x_3y_3+2x_2y_3-6x_1y_2+6x_3y_2=0 \end{equation*}$

The second condition $m_{AB}=-3m_C$ , in terms of the given coordinates, is:

$\frac{y_1-y_2}{x_1-x_2}=-3\left(\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}\right)$

Clear fractions and combine like terms:

(3) $\begin{equation*} 4x_1y_1-2x_2y_1-2x_3y_1+2x_1y_2-4x_2y_2+2x_3y_2-6x_1y_3+6x_2y_3=0 \end{equation*}$

Subtract equation (2) from equation (3):

$8x_1y_2-8x_1y_3-4x_2y_2+4x_3y_3+4x_2y_3-4x_3y_2=0$

This factors as $(y_2-y_3)(2x_1-x_2-x_3)=0$ . We claim that both $y_2-y_3=0$ and $2x_1-x_2-x_3=0$ must hold. If not, suppose that only $y_2-y_3=0$ holds and re-calculate the quotient $\frac{m_{CA}}{m_B}$ :

$\frac{y_1-y_3}{x_1-x_3}\div\left(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\right)=\frac{x_1+x_3-2x_2}{x_1-x_3}.$

This doesn’t equal $-3$ unless we impose $2x_1-x_2-x_3=0$ . Thus, we must have $y_2-y_3=0$ (side $BC$ parallel to the $x$ -axis) and $2x_1-x_2-x_3=0$ (which, in conjunction with $y_2-y_3=0$ yields an isosceles triangle).

For an equilateral triangle $ABC$ , PROVE that

$q=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=-(m_{AB}^2+m_{BC}^2+m_{CA}^2).$

This is a consequence of the fact that the medians in an equilateral triangle are perpendicular to the opposite sides. In particular:

$m_{A}=-\frac{1}{m_{BC}},~m_{B}=-\frac{1}{m_{CA}},~m_{C}=-\frac{1}{m_{AB}}$

ignoring any zero denominator. Then:

$\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=-(m_{AB}^2+m_{BC}^2+m_{CA}^2).$

The above example can be used to explain why the side-slopes of an equilateral triangle with side $BC$ parallel to the $x$ -axis are $-\sqrt{3}$ and $\sqrt{3}$ .

Native progressions

The side-slopes of an equilateral triangle are precisely:

$m_1=m,~m_2=\frac{m-\sqrt{3}}{1+\sqrt{3}m},~m_3=\frac{m+\sqrt{3}}{1-\sqrt{3}m},$

where $m$ is the slope of one of the sides. The fact that the slopes can be expressed in terms of the slope of one side leads to some interesting consequences.

PROVE that the slopes of the sides of an equilateral triangle form an arithmetic progression if, and only if, one of the slopes is zero.

First suppose that one of the slopes is zero. There are three possibilities:

$m_1=0$ . That is, $m=0$ (since $m_1=m$ as per above). In this case we have:
$m_2=\frac{m-\sqrt{3}}{1+\sqrt{3}m}=-\sqrt{3},~m_3=\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\sqrt{3},$

and so we obtain the arithmetic progression $m_2,m_1,m_3$ corresponding to $-\sqrt{3},0,\sqrt{3}$ .
$m_2=0$ . This implies $\frac{m-\sqrt{3}}{1+\sqrt{3}m}=0$ , and so $m=\sqrt{3}$ . In other words, $m_1=\sqrt{3}$ . This also gives
$m_3=\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\frac{\sqrt{3}+\sqrt{3}}{1-\sqrt{3}\times\sqrt{3}}=-\sqrt{3}.$

We obtain an arithmetic progression $m_3,m_2,m_1$ this time around.
$m_3=0$ . This implies $\frac{m+\sqrt{3}}{1-\sqrt{3}m}=0$ , and so $m=-\sqrt{3}$ . That is, $m_1=-\sqrt{3}$ . With this
$m_2=\frac{m-\sqrt{3}}{1+\sqrt{3}m}=\frac{-\sqrt{3}-\sqrt{3}}{1+\sqrt{3}\times(-\sqrt{3})}=\sqrt{3},$

giving the arithmetic progression $m_1,m_3,m_2.$

For the converse, suppose that the side-slopes form an arithmetic progression. There are three cases to consider: the progression could be $m_1,m_2,m_3$ , or $m_1,m_3,m_2$ or $m_2,m_1,m_3$ .

PROVE that the slopes of the sides of an equilateral triangle form an arithmetic progression if, and only if, $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=-6$ .

If the slopes form an arithmetic progression, then one of them must be zero, in view of example 6. Since an equilateral triangle is also isosceles, we have (by example 3) that $q=-6$ .

On the other hand, suppose that $q=-6$ . Using example 5 and the expressions for the slopes of the sides of an equilateral triangle, we have:

$m^2+\left(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\right)^2+\left(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right)^2=6.$

Afer simplifications, this leads to

$m^2(m^2-3)^2=0\implies m=0,\pm\sqrt{3}.$

Taking each of these three values of $m$ in turn, we see that the side-slopes form an arithmetic progression (similar to example 6).

Solve the sixth-degree polynomial equation $9m^6-135m^4+135m^2-9=0$ .

By the way, such an equation is also sometimes called “bicubic” (just like “biquadratic”), since the odd degree terms are missing.

The form of the coefficients in this polynomial equation means that we can re-arrange the terms and factor systematically:

$\begin{equation*} \begin{split} 9m^6-135m^4+135m^2-9&=0\\ (9m^6-9)+(-135m^4+135m^2)&=0\\ 9(m^6-1)-135m^2(m^2-1)&=0\\ 9[(m^2)^3-1]-135m^2(m^2-1)&=0\\ 9(m^2-1)(m^4+m^2+1)-135m^2(m^2-1)&=0\\ 9(m^2-1)(m^4+m^2+1-15m^2)&=0\\ (m^2-1)(m^4-14m^2+1)&=0 \end{split} \end{equation*}$

The first quadratic factor, $m^2-1$ , has $m=\pm 1$ as solutions. The second “biquadratic” factor, $m^4-14m^2+1$ , has solutions that can be obtained from the quadratic formula:

$\begin{equation*} \begin{split} m^4-14m^2+1&=0\\ (m^2)^2-14m^2+1&=0\\ m^2&=\frac{14\pm\sqrt{196-4}}{2}\\ &=7\pm 4\sqrt{3}\\ \therefore m&=\pm\sqrt{7\pm 4\sqrt{3}}\\ &=\pm\sqrt{(2\pm\sqrt{3})^2}\\ &=\pm(2\pm\sqrt{3}) \end{split} \end{equation*}$

Thus, we obtain six solutions

$m=\pm 1,~ m=2\pm\sqrt{3},~m=-2\pm\sqrt{3}.$

PROVE that the slopes of the sides of an equilateral triangle form a geometric progression if, and only if, one of the slopes is $\pm 1$ .

This is the geometric progression version of example 6. First suppose that the slopes form a geometric progression. There are three cases to consider:

$m$ is the geometric mean of the progression. In this case we have:

$\begin{equation*} \begin{split} m^2&=\left(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\right)\left(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right)\\ m^2&=\frac{m^2-3}{1-3m^2}\\ m^4&=1\\ m&=\pm 1 \end{split} \end{equation*}$

$\frac{m-\sqrt{3}}{1+\sqrt{3}m}$ is the geometric mean of the progression. In this case we have:

$\begin{equation*} \begin{split} \left(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\right)^2&=m\left(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right)\\ (m-\sqrt{3})^2(1-\sqrt{3}m)&=(m^2+\sqrt{3}m)(1+\sqrt{3}m)^2\\ 3m^4+6\sqrt{3}m^3+6\sqrt{3}m-3&=0\\ (m^2+1)(3m^2+6\sqrt{3}m-3)&=0\\ m&=-\sqrt{3}\pm 2 \end{split} \end{equation*}$

If $m=-\sqrt{3}+2$ , then $\frac{m-\sqrt{3}}{1+\sqrt{3}m}$ becomes $-1$ , and if $m=-\sqrt{3}-2$ , then $\frac{m-\sqrt{3}}{1+\sqrt{3}m}$ is $1$ . So we get that one of the slopes is $\pm 1$ .

$\frac{m+\sqrt{3}}{1-\sqrt{3}m}$ is the geometric mean of the progression. In this case we have:

$\begin{equation*} \begin{split} \left(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right)^2&=m\left(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\right)\\ 3m^4-6\sqrt{3}m^3-6\sqrt{3}m-3&=0\\ (m^2+1)(3m^2-6\sqrt{3}m-3)&=0\\ m&=\sqrt{3}\pm 2 \end{split} \end{equation*}$

If $m=\sqrt{3}+2$ , then $\frac{m+\sqrt{3}}{1-\sqrt{3}m}$ is $-1$ , and if $m=\sqrt{3}-2$ , then $\frac{m+\sqrt{3}}{1-\sqrt{3}m}$ evaluates to $1$ .

Conversely, if one of the slopes is $\pm 1$ , it is also easy to see that the other slopes combine to form a geometric progression.

PROVE that the slopes of the sides of an equilateral triangle form a geometric progression if, and only if, $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=-15$ .

First suppose that $q=-15$ . Using the expression for $q$ in example 5 and the expressions for the side-slopes of an equilateral triangle, we have:

$m^2+\left(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\right)^2+\left(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right)^2=15,$

which, after simplifications, leads to the sixth-degree equation

$9m^6-135m^4+135m^2-9=0,$

same as the one in example 8. Since the solutions are $m=\pm 1$ , $m=2\pm\sqrt{3}$ , $m=-2\pm\sqrt{3}$ , the side-slopes form a geometric progression.

Conversely, if the side-slopes form a geometric progression, all possiblities can be enumerated. For example, we can have $m_1=-1$ , $m_2=2+\sqrt{3}$ , $m_3=2-\sqrt{3}$ . Using example 5 again we have:

$q=-\left((-1)^2+(2+\sqrt{3})^2+(2-\sqrt{3})^2\right)=-\left(1+7+4\sqrt{3}+7-4\sqrt{3}\right).$

That is, $q=-15$ . Other combinations of $m_1,m_2,m_3$ also yield $q=-15$ .

In a future post we may form the sum:

(4) $\begin{equation*} p=m_{A}\times m_{BC} + m_{B}\times m_{CA} + m_{C}\times m_{AB}. \end{equation*}$

If an equilateral triangle does not have any of its sides parallel to the coordinate axes, then $p=-3$ . Accordingly, any triangle that satisfies (4) will be called equilessthree, a name as lame as isosixless.

Takeaway

For an equilateral $\triangle ABC$ , the following statements are equivalent:

one of the side-slopes is zero;
the sum of the side-slopes is zero;
the side-slopes form an arithmetic progression;
$q=-6$ , where $q$ is the sum of quotients $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}$ .

Similarly, the following conditions are equivalent for an equilateral $\triangle ABC$ :

one of the side-slopes is plus/minus one;
the sum of the side-slopes is plus/minus three;
the side-slopes form a geometric progression;
$q=-15$ , where $q$ is the sum of quotients $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}$ .

So much about slopes in our posts, eh? Probably a reason we tend to be sloppy with words and stuff?

Tasks

(Minus one) Let . If, in we have the relations , , :
- PROVE that $(3k+3)\Big(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Big)=0$ .
- Discuss what happens when $k=-1$ . Interesting case.
Let and be real numbers.
- PROVE that the quadratic equation $x^2-(a+b)x+(a^2+b^2-ab)=0$ has no real solutions.
- Deduce that $x_1^2+x_2^2+x_3^2=x_1x_2+x_2x_3+x_3x_1$ if, and only if, $x_1=x_2=x_3$ .
PROVE that $m_{A}\times m_{BC} + m_{B}\times m_{CA} + m_{C}\times m_{AB}=m_{AB}\times m_{BC}+m_{BC}\times m_{CA}+m_{CA}\times m_{AB}$ in an equilateral triangle $ABC$ with non-zero side-slopes.
In $\triangle ABC$ , let the slopes of sides $AB,BC,CA$ be $a,-3a,9a$ , respectively. PROVE that $15\left(\frac{m_{AB}}{m_C}\right)+14\left(\frac{m_{BC}}{m_A}\right)-\left(\frac{m_{CA}}{m_B}\right)=0$ .
PROVE that the relations $m_{AB}=-m_{C}$ , $m_{BC}=-5m_{A}$ , $m_{CA}=0.m_{B}$ are not possible simultaneously.
Solve the quartic equation $r^4+8r^3+18r^2+8r+1=0$ .
In $\triangle ABC$ , let the slopes of sides $AB,BC,CA$ be in geometric progression $a,ar,ar^2$ . PROVE that $q=\frac{4r^4+2r^3-3r^2+2r+1}{2r^3+5r^2+2r}$ .
(Things always become interesting in the setting of geometric progressions. Always.)
In $\triangle ABC$ , let the slopes of sides $AB,BC,CA$ be in geometric progression $a,ar,ar^2$ . PROVE that $p=m_{A}\times m_{BC} + m_{B}\times m_{CA} + m_{C}\times m_{AB}=\frac{a^2r(r^2+r+1)^2}{2r^2+5r+2}$ .
(Things always become interesting in the setting of geometric progressions. Always.)
(Simplified area) In $\triangle ABC$ , let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ , where $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ are the coordinates of the vertices. PROVE that the area of $\triangle ABC$ can be given by $\Delta = \frac{(x_2-x_3)(y_2+y_3-2y_1)}{2}$ .
(Absolute value may be needed in the final result.)
PROVE that there is no triangle $ABC$ for which the relations $m_{AB}=-2m_{C}$ , $m_{BC}=-2m_A$ , and $m_{CA}=-2m_B$ hold simultaneously.
(You can use exercise 1 above.)