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

  1. Solve the rational equation \frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=-23.
  2. (Golden quartic) Let r be the golden ratio. PROVE that r^4-3r^2+1=0.
  3. (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}.
  4. 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.)
  5. If r^2+7r+1=0, PROVE that q=-19.
  6. 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.
  7. Let a,ar,ar^2 be the slopes of sides AB,BC,CA in \triangle ABC. 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.
  8. Consider \triangle ABC with vertices at A(1,-2), B(2,0), C(-3,1). 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}.
  9. 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.
  10. 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}.