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).

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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:

  1. 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}.

  2. 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.

  3. 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

  1. (Minus one) Let k\neq 0. If, in \triangle ABC we have the relations m_{AB}=km_{C}, m_{BC}=km_{A}, m_{CA}=km_{B}:
    • 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.
  2. Let a and b 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.
  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.
  4. 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.
  5. PROVE that the relations m_{AB}=-m_{C}, m_{BC}=-5m_{A}, m_{CA}=0.m_{B} are not possible simultaneously.
  6. Solve the quartic equation r^4+8r^3+18r^2+8r+1=0.
  7. 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.)
  8. 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.)
  9. (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.)
  10. 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.)