Slopefessor Fr1day – Page 26 – High School Geometry

Sum of quotients of slopes

In the case of $\triangle ABC$ with side-slopes $a,ar,ar^2$ , the sum we formed in our last post becomes:

(1) $\begin{equation*} q=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}. \end{equation*}$

For certain, things always turn out interesting in the setting of geometric progressions. As ascertained in example 3 for instance, the order of each quotient in each summand can be reversed, while the final sum remains preserved:

(2) $\begin{equation*} q=\frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}}=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}. \end{equation*}$

Another interes-thing that pertains to this familiar terrain is how we obtain specific values when we restrict to equilateral and right isosceles triangles (spoiler alert: $-15$ and $-19$ ).

How come???

In $\triangle ABC$ , let sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . PROVE that the slopes of the medians from each vertex are: $m_A=-ar$ , $m_B=\left(\frac{r+2}{2r+1}\right)ar$ , and $m_C=\left(\frac{2r+1}{r+2}\right)ar$ .

Observe that the median slopes $m_B,m_A,m_C$ form a geometric progression. Except for the special case $r=-2,-\frac{1}{2}$ , the slopes of the sides of a triangle form a geometric progression if and only if the slopes of the medians form another geometric progression, and their common ratios are related.

Why does the case $r=-2,-\frac{1}{2}$ fail the equivalence? First, it is the required restriction in the median slopes stated above. Another reason is that any three-term geometric progression with common ratio $r=-2$ is, upon re-arrangement, an arithmetic progression (for example, $5,-10,20$ is geometric while $-10,5,20$ is arithmetic). And if the side-slopes of a triangle form an arithmetic sequence, then the triangle necessarily contains a vertical median.

In other to obtain $m_A=-ar$ , $m_B=\left(\frac{r+2}{2r+1}\right)ar$ , and $m_C=\left(\frac{2r+1}{r+2}\right)ar$ , we’ll place the vertices of $\triangle ABC$ at the points $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ . Since the slopes of sides $AB,BC,CA$ are $a,ar,ar^2$ respectively, the coordinates of the vertices are related according to:

(3) $\begin{equation*} x_1=x_3+\frac{y_2-y_3}{ar(r+1)},~y_1=\frac{ry_2+y_3}{r+1},~x_2=x_3+\frac{y_2-y_3}{ar} \end{equation*}$

Thus, the midpoint of $BC$ is $\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$ . In view of the above relations, we get $\left(x_3+\frac{y_2-y_3}{2ar},\frac{y_2+y_3}{2}\right)$ . The slope of the median from vertex $A$ can then be obtained:

$\begin{equation*} \begin{split} m_A&=\left[\left(\frac{y_2+y_3}{2}\right)-\left(\frac{ry_2+y_3}{r+1}\right)\right]\div\left[\left(x_3+\frac{y_2-y_3}{2ar}\right)-\left(x_3+\frac{y_2-y_3}{ar(r+1)}\right)\right]\\ &=\frac{(1-r)(y_2-y_3)}{2(r+1)}\div \frac{(r-1)(y_2-y_3)}{2ar(r+1)}\\ &=-ar. \end{split} \end{equation*}$

The midpoint of $CA$ is $\left(x_3+\frac{y_2-y_3}{2ar(r+1)},\frac{ry_2+(r+2)y_3}{2(r+1)}\right)$ , as per (3). The slope of the median from vertex $B$ is:

$\begin{equation*} \begin{split} m_B&=\left[\frac{ry_2+(r+2)y_3-2(r+1)y_2}{2(r+1)}\right]\div\left[\frac{(y_2-y_3)-(2r+2)(y_2-y_3)}{2ar(r+1)}\right]\\ &=\frac{-(r+2)(y_2-y_3)}{2(r+1)}\div\frac{-(2r+1)(y_2-y_3)}{2ar(r+1)}\\ &=\left(\frac{r+2}{2r+1}\right)ar. \end{split} \end{equation*}$

Similar calculation gives $m_C=\left(\frac{2r+1}{r+2}\right)ar$ . Easy stuff.

Derivations

Derive equation (1): $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ for triangle $ABC$ with sides $AB,BC,CA$ having slopes $a,ar,ar^2$ .

Using our definition of $q$ and the preceding example:

$\begin{equation*} \begin{split} q&=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}\\ &=\frac{ar}{-ar}+\left[ar^2\div\left(\frac{r+2}{2r+1}\right)ar\right]+\left[a\div\left(\frac{2r+1}{r+2}\right)ar\right]\\ &=(-1)+\left(\frac{r(2r+1)}{r+2}\right)+\left(\frac{r+2}{r(2r+1)}\right)\\ &=\frac{-r(r+2)(2r+1)+r^2(2r+1)^2+(r+2)^2}{r(r+2)(2r+1)}\\ &=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}. \end{split} \end{equation*}$

Consequently, triangles having slopes in geometric progressions with the same common ratios have the same $q$ -values. And it is even possible to have different common ratios yielding the same $q$ -values.

PROVE that $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}}$ for $\triangle ABC$ with sides $AB,BC,CA$ having slopes $a,ar,ar^2$ respectively.

Thus, we can use either equation (1) or equation (2) to compute $q$ in the case of geometric progressions. Using example 1 again:

$\begin{equation*} \begin{split} \frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}}&=\frac{-ar}{ar}+\left[\left(\frac{r+2}{2r+1}\right)ar\div ar^2\right]+\left[\left(\frac{2r+1}{r+2}\right)ar\div a\right]\\ &=(-1)+\frac{r+2}{r(2r+1)}+\frac{r(2r+1)}{r+2}\\ &=\frac{-r(r+2)(2r+1)+(r+2)^2+r^2(2r+1)^2}{r(r+2)(2r+1)}\\ &=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}\\ &=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}. \end{split} \end{equation*}$

Notice that the coefficients in the numerator and denominator both add up to $9$ . Furthermore, the “symmetric” nature of these coefficients allows for easy calculations in some situations.

Demonstrations

Let $\triangle ABC$ be equilateral with side-slopes $a,ar,ar^2$ . PROVE that $q=-15$ .

There’s an important quadratic equation satisfied by the common ratio of the geometric progression formed by the slopes of an equilateral triangle:

$r^2+4r+1=0.$

It’s the key to obtaining $q=-15$ , together with equation (1) and simple algebraic manipulations.

$\begin{equation*} \begin{split} q&=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}\\ &=\frac{4(r^4+1)+2r(r^2+1)-3r^2}{2r(r^2+1)+5r^2}\\ &=\frac{4(14r^2)+2r(-4r)-3r^2}{2r(-4r)+5r^2}\\ &=\frac{56r^2-8r^2-3r^2}{-8r^2+5r^2}\\ &=\frac{45r^2}{-3r^2}\\ &=-15, \end{split} \end{equation*}$

where we’ve used the implications

$r^2+4r+1=0\implies r^2+1=-4r\implies (r^2+1)^2=(-4r)^2\implies r^4+1=14r^2.$

Let $\triangle ABC$ be right isosceles with side-slopes $a,ar,ar^2$ . PROVE that $q=-19$ .

Similar to the above, in a right isosceles triangle with side-slopes $a,ar,ar^2$ , the common ratio $r$ satisfies the quadratic equation

$r^2+3r+1=0.$

Using equation (1) and a couple of simplifications, we find:

$\begin{equation*} \begin{split} q&=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}\\ &=\frac{4(r^4+1)+2r(r^2+1)-3r^2}{2r(r^2+1)+5r^2}\\ &=\frac{4(7r^2)+2r(-3r)-3r^2}{2r(-3r)+5r^2}\\ &=\frac{28r^2-6r^2-3r^2}{-6r^2+5r^2}\\ &=\frac{19r^2}{-r^2}\\ &=-19, \end{split} \end{equation*}$

where the implications

$r^2+3r+1=0\implies r^2+1=-3r\implies r^4+1=7r^2$

were used in the simplifications.

Deductions

Suppose that a triangle with side-slopes $a,ar,ar^2$ satisfies $q=-15$ . Deduce that $r^2+4r+1=0$ .

Quite interesting. Set $q=-15$ in equation (1):

$\begin{equation*} \begin{split} \frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}&=-15\\ \implies 4r^4+2r^3-3r^2+2r+4&=-15(2r^3+5r^2+2r)\\ \implies 4r^4+32r^3+72r^2+32r+4&=0\\ \implies r^4+8r^3+18r^2+8r+1&=0 \end{split} \end{equation*}$

Let’s show a few steps leading to the factorization of the quartic $r^4+8r^3+18r^2+8r+1$ . If we apply the rational root theorem, the only values of $r$ worth testing are $\pm 1$ , none of which are roots of the quartic. However, the nature of the coefficients means that if two (quadratic) factors exist, they must both be monic and also have $1$ as the constant terms. Put

$r^4+8r^3+18r^2+8r+1=(r^2+\alpha r+1)(r^2+\beta r+1)$

and expand the right side:

$r^4+8r^3+18r^2+8r+1=\Big(r^4+(\alpha+\beta)r^3+(\alpha\beta+2)r^2+(\alpha+\beta)r+1\Big).$

Compare coefficients of like terms:

$\alpha+\beta=8,~\alpha\beta+2=18.$

The above linear-quadratic system is satisfied only by $\alpha=\beta=4$ . So we have the factorization

$r^4+8r^3+18r^2+8r+1=(r^2+4r+1)(r^2+4r+1)=(r^2+4r+1)^2.$

In turn, $r^4+8r^3+18r^2+8r+1=0\implies r^2+4r+1=0$ , as desired.

Suppose that a triangle with side-slopes $a,ar,ar^2$ satisfies $q=-19$ . Deduce that $r^2+3r+1=0$ or $r^2+7r+1=0$ .

Put $q=-19$ in equation (1):

$\begin{equation*} \begin{split} \frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}&=-19\\ \implies 4r^4+2r^3-3r^2+2r+4&=-19(2r^3+5r^2+2r)\\ \implies 4r^4+40r^3+92r^2+40r+4&=0\\ \implies r^4+10r^3+23r^2+10r+1&=0 \end{split} \end{equation*}$

Set

$r^4+10r^3+23r^2+10r+1=(r^2+\alpha r+1)(r^2+\beta r+1)$

and expand the right side:

$r^4+10r^3+23r^2+10r+1=\Big(r^4+(\alpha+\beta)r^3+(\alpha\beta+2)r^2+(\alpha+\beta)r+1\Big).$

Compare coefficients of like terms:

$\alpha+\beta=10,~\alpha\beta+2=23.$

The above linear-quadratic system is satisfied only by $\alpha=3,~\beta=7$ (or $\alpha=7$ , $\beta=3$ ). So we have the factorization

$r^4+10r^3+23r^2+10r+1=(r^2+3r+1)(r^2+7r+1).$

In turn, $r^4+10r^3+23r^2+10r+1=0\implies r^2+3r+1=0$ or $r^2+7r+1=0$ .

Digression

A simple connection between the preceding discussion and something else.

Find six distinct non-zero rational numbers $x,y,z,t,u,v$ which satisfy the relation $\frac{x}{y}+\frac{z}{t}+\frac{u}{v}=\frac{y}{x}+\frac{t}{z}+\frac{v}{u}$ .

The non-zero requirement is somewhat redundant, being already forced by the fact that all six numbers appear in the denominators.

One way of solving this problem is to turn to slopes. In equations (1) and (2) we had

$\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}},$

and so we can set

$x=m_{BC},~y=m_{A},~z=m_{CA},~t=m_{B},~u=m_{AB},~v=m_{C}$

where $m_{AB},m_{BC},m_{CA}$ are the slopes of the sides of a triangle which form a geometric progression, and $m_A$ , $m_B$ , $m_C$ are the corresponding median slopes. We can find these slopes without specifying coordinates!

Any geometric progression (except the one with $r=-2$ ) will do. Choose

$m_{AB}=1,~m_{BC}=2,~m_{CA}=4.$

Then $m_A,m_B,m_C$ can be obtained (in view of example 1) as

$m_A=-2,~m_B=\frac{8}{5},~m_C=\frac{5}{2}.$

Thus

$x=2,y=-2,z=4,t=\frac{8}{5},u=1,v=\frac{5}{2}.$

The fractions now become

$\begin{equation*} \begin{split} \frac{x}{y}+\frac{z}{t}+\frac{u}{v}&=\frac{2}{-2}+\frac{4}{8/5}+\frac{1}{5/2}\\ &=\frac{19}{10}\\ \frac{y}{x}+\frac{t}{z}+\frac{u}{v}&=\frac{-2}{2}+\frac{8/5}{4}+\frac{5/2}{1}\\ &=\frac{19}{10} \end{split} \end{equation*}$

Equal.

Find three rational numbers $x,y,z$ , not all integers, which satisfy $\frac{9}{x}+\frac{27}{y}+\frac{3}{z}=\frac{x}{9}+\frac{y}{27}+\frac{z}{3}$ .

Were it not for the not all integers requirement, we could easily have chosen $x=9$ , $y=27$ , and $z=3$ . Instead, we turn to slopes again.

Since the numbers $3,9,27$ form a geometric progression, we can associate them with the slopes of the sides of a triangle. Like so:

$m_{AB}=3,~m_{BC}=9,~m_{CA}=27.$

Then the slopes of the medians would be:

$m_{A}=-9,~m_B=\frac{45}{7},~m_C=\frac{63}{5}\implies x=-9,~y=\frac{45}{7},~z=\frac{63}{5}.$

The fractions become

$\begin{equation*} \begin{split} \frac{9}{x}+\frac{27}{y}+\frac{3}{z}&=\frac{9}{-9}+\frac{27}{45/7}+\frac{3}{63/5}\\ &=(-1)+\frac{21}{5}+\frac{5}{21}\\ &=\frac{361}{105}\\ \frac{x}{9}+\frac{y}{27}+\frac{z}{3}&=\frac{-9}{9}+\frac{45/7}{27}+\frac{63/5}{3}\\ &=(-1)+\frac{5}{21}+\frac{21}{5}\\ &=\frac{361}{105}. \end{split} \end{equation*}$

Is the triple $x=-9,~y=\frac{45}{7},~z=\frac{63}{5}$ the only not all integers solution?

Find the slopes of the sides of a triangle if the median slopes are $1,2,4$ .

Since the median slopes form a geometric progression (with common ratio $\neq -2$ ), so must the side-slopes.

From example 1, if the side slopes are $a,ar,ar^2$ , then the geometric mean of the median slopes is $-ar$ ; that is,

$-ar=2\implies ar=-2.$

Also, the common ratio $R$ of the median slopes is related to the common ratio $r$ of the side-slopes via

$R=-\frac{r+2}{2r+1}\implies 2=-\frac{r+2}{2r+1}\implies r=-\frac{4}{5}.$

The side-slopes are then:

$\frac{5}{2},-2,\frac{8}{5}.$

Takeaway

The following statements are equivalent for a triangle with slopes in geometric progression:

$q=-15$ ;
the common ratio $r$ satisfies $r^2+4r+1=0$ ;
the common ratio is that of an equilateral triangle.

Similarly, the following statements are equivalent for a triangle with slopes in geometric progression:

$q=-19$ ;
the common ratio $r$ satisfies $r^2+3r+1=0$ or $r^2+7r+1=0$ .

Finally:

$\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}}.$

Remember the name: exquisite equation.

Tasks

Solve the rational equation $\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=-23$ .
(Golden quartic) Let $r$ be the golden ratio. PROVE that $r^4-3r^2+1=0$ .
(Related ratios) Suppose that the slopes of the sides of $\triangle ABC$ form a geometric progression with common ratio $r_1$ . PROVE that the slopes of the medians form another geometric progression with common ratio $r_2$ , where $r_2=-\frac{r_1+2}{2r_1+1}$ .
Let $r$ be an integer. PROVE that $q$ is also an integer if, and only if, $r=\pm 1$ .
(If $r$ is not an integer, $q$ can still be an integer, as in the next exercise.)
If $r^2+7r+1=0$ , PROVE that $q=-19$ .
Let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that $(2r^2+2r+2)\frac{m_{BC}}{m_A}+(r+2)\frac{m_{CA}}{m_B}+(2r^2+r)\frac{m_{AB}}{m_C}=0$ .
Let be the slopes of sides in . PROVE that:
- $m_{CA}\times m_{AB}=m_C\times m_B$
- $\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{m_B}{m_{CA}}+\frac{m_C}{m_{AB}}$
- $\frac{m_{CA}\times m_B}{m_C\times m_{AB}}=\frac{m_{CA}^2-m_B^2}{m_C^2-m_{AB}^2}$
- $\frac{m_{CA}\times m_B}{m_C\times m_{AB}}$ is a ratio of perfect squares if $r$ is a rational number.
Consider with vertices at , , . Verify that:
- $\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=-\frac{222}{25}$
- $\frac{m_{A}}{m_{BC}}+\frac{m_{B}}{m_{CA}}+\frac{m_{C}}{m_{AB}}=\frac{71}{9}$
- the side-slopes do not form a geometric progression
- $\frac{m_{AB}}{m_C}=\frac{m_{CA}}{m_B}$ .
Find the slopes of the sides of a triangle, if:
- the slopes of the medians are $1,3,9$ ;
- the slopes of the medians are $2,-6,18$ ;
- the slopes of the medians are $-\frac{4}{3},2,-3$ .
Find distinct rational numbers $x,y,z,t,u,v$ which satisfy the relation $\frac{x}{y}+\frac{z}{t}+\frac{u}{v}=\frac{y}{x}+\frac{t}{z}+\frac{v}{u}.$

An isosixless triangle

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.)