Two special quadratics

The chances that you’ve seen the quadratic equations

(1)   \begin{equation*} r^2+4r+1=0 \end{equation*}

(2)   \begin{equation*} r^2+3r+1=0 \end{equation*}

are not slim. We now show you that they spring from familiar scenes.

Equilateral triangles

Example

If the slopes of the sides of an equilateral triangle are a,ar,ar^2, PROVE that r^2+4r+1=0.

In a previous post we showed that the side lengths l_1,l_2,l_3 of any triangle with slopes a,ar,ar^2 are related via

(3)   \begin{equation*} l_1^2+l_3^2=\frac{r^2+1}{(r+1)^2}l_2^2. \end{equation*}

In the case of an equilateral triangle we put l_1=l_2=l_3 in (3) and obtain

    \[1+1= \frac{r^2+1}{(r+1)^2},\]

from which r^2+4r+1=0 follows after simplification.

That was too simple, aided by (3). Below we provide an alternate derivation.

Let r_1 be the common ratio of the slopes of the sides of an equilateral triangle, and let r_2 be the common ratio of the slopes of the medians. PROVE that r_1r_2=1.

To see this, first observe that if the slopes of the sides of a triangle are a,ar,ar^2, then the slopes of the medians to these sides are

(4)   \begin{equation*}\left(\frac{2r+1}{r+2}\right)ar,~-ar,~\left(\frac{r+2}{2r+1}\right)ar\end{equation*}

respectively. For an equilateral triangle, the three medians are perpendicular to the sides, so:

(5)   \begin{equation*} \left \begin{split} \left(\frac{2r+1}{r+2}\right)ar\times (a)&=-1\\ -ar\times (ar)&=-1\\ \left(\frac{r+2}{2r+1}\right)ar\times (ar^2)&=-1 \end{split}\right\} \end{equation*}

The first and third ones give \left(\frac{2r+1}{r+2}\right)a^2r=\left(\frac{r+2}{2r+1}\right)a^2r^3. Divide by the common factor a^2r:

(6)   \begin{equation*} \begin{split} \left(\frac{2r+1}{r+2}\right)&=\left(\frac{r+2}{2r+1}\right)r^2$ \end{split} \end{equation*}

From (4), the common ratio r_2 for the slopes of the medians is -\left(\frac{r+2}{2r+1}\right). Like so:

(7)   \begin{equation*} r_2=-\left(\frac{r_1+2}{2r_1+1}\right). \end{equation*}

where r_1 is the original common ratio. Equation (6) may then be re-written as

    \[-\frac{1}{r_2}=-r_2r_1^2\implies r_1^2r_2^2=1.\]

Square roots: r_1r_2=\pm 1. We claim that r_1r_2=-1 is not possible. Otherwise, suppose we have r_1r_2=-1. Use the fact that r_2=-\left(\frac{r_1+2}{2r_1+1}\right) to obtain r_2(2r_1+1)=-r_1-2.

    \begin{equation*} \begin{split} 2r_1r_2+r_2&=-r_1-2\\ 2(-1)+r_2&=-r_1-2\\ r_2&=-r_1 \end{split} \end{equation*}

Now use r_2=-r_1 and r_1r_2=-1 to get r_1^2=1, whence r_1=\pm 1. From slope consideration, this is impossible for the common ratio of a geometric progression, and so rules out the possibility of r_1r_2=-1.

This forces r_1r_2=1.

While having r_1r_2=1 is enough for what we want to achieve, it is important to point out that one can also have r_1=r_2, depending on the arrangement of the geometric progressions. Hopefully this doesn’t cause any confusion.

Let r_1 be the common ratio of the slopes of the sides of an equilateral triangle, and let r_2 be the common ratio of the slopes of the medians. PROVE that r_1+r_2=-4.

From the preceding example, r_1r_2=1. Use this in (7):

    \begin{equation*} \begin{split} r_2&=-\left(\frac{r_1+2}{2r_1+1}\right)\\ 2r_1r_2+r_2&=-r_1-2\\ 2(1)+r_2&=-r_1-2\\ \therefore r_1+r_2&=-4 \end{split} \end{equation*}

Simple.

Use examples 2 and 3 to deduce that in an equilateral triangle with side slopes a,ar_1,ar_1^2, the quadratic equation r^2+4r+1=0 is satisfied by the common ratio of the slopes of the sides and the common ratio of the slopes of the medians.

Since r_1r_2=1 (view this as product of roots) and r_1+r_2=-4 (sum of roots), we get r_1^2+4r_1+1=0. Similarly, r_2^2+4r_2+1=0.

Solve the quadratic equation r^2+4r+1=0 and interpret the solution(s) in the context of equilateral triangles.

The discriminant of the quadratic is 4^2-4\times 1\times 1=12. Since it is positive but not a perfect square, we expect two real, irrational solutions. By the quadratic formula:

    \begin{equation*} \begin{split} r&=\frac{-4\pm\sqrt{12}}{2\times 1}\\ &=-2\pm\sqrt{3} \end{split} \end{equation*}

Should the slopes of the sides (and medians) of an equilateral triangle form a geometric progression, the common ratio of the progression is restricted to either -2+\sqrt{3} or -2-\sqrt{3}. Talk about restrictions.

Right isosceles triangles

If the slopes of the sides of a right isosceles triangle are a,ar,ar^2, PROVE that r^2+3r+1=0.

The side lengths l_1,l_2,l_3 will satisfy (3):

    \begin{equation*} l_1^2+l_3^2=\frac{r^2+1}{(r+1)^2}l_2^2, \end{equation*}

where l_1 is the length of the side with slope a, l_2 is the length of the side with slope ar, and l_3 is the length of the side with slope ar^2.

Since l_2 cannot be the hypotenuse, we’ll choose either l_1 or l_3 as the hypotenuse. To be specific, let l_1 be the hypotenuse.

The isosceles condition then forces l_2=l_3. And:

    \begin{equation*} \begin{split} l_1^2+l_2^2&=\frac{r^2+1}{(r+1)^2}l_2^2\\ l_1^2&=\frac{r^2+1}{(r+1)^2}l_2^2-l_2^2\\ \therefore l_1^2&=\frac{-2r}{(r+1)^2}l_2^2 \end{split} \end{equation*}

By the Pythagorean theorem, l_1^2=l_2^2+l_3^2. As l_2=l_3, this yields l_1^2=2l_2^2. Together with l_1^2=\frac{-2r}{(r+1)^2}l_2^2, we obtain

    \begin{equation*} \begin{split} 2l_2^2&=\frac{-2r}{(r+1)^2}l_2^2\\ 1&=\frac{-r}{(r+1)^2}\\ (r+1)^2+r&=0\\ \therefore r^2+3r+1&=0 \end{split} \end{equation*}

Solve the equation r^2+3r+1=0 and interpret the solution(s) in relation to right isosceles triangles.

The discriminant is 3^2-4\times 1\times 1=5. Since the discriminant is positive but not a perfect square, we expect two real, irrational solutions.

By the quadratic formula:

    \begin{equation*} \begin{split} r&=\frac{-3\pm\sqrt{5}}{2\times 1} \end{split} \end{equation*}

Should the slopes of the sides of a right isosceles triangle form a geometric progression, there are only two possible values for the common ratio of the progression, namely \frac{-3+\sqrt{5}}{2} or \frac{-3-\sqrt{5}}{2}.

Let a,ar,ar^2 be the slopes of the sides of a right isosceles triangle, with r=\frac{-3+\sqrt{5}}{2}. PROVE that a=\pm\frac{1}{2}(1+\sqrt{5}).

The side with slope ar cannot be the hypotenuse. Let the hypotenuse be the side with slope ar^2. Then the slopes of the legs are a and ar. As such a\times(ar)=-1:

    \begin{equation*} \begin{split} a^2&=-\frac{1}{r}\\ &=\frac{2}{3-\sqrt{5}}\\ &=\frac{2}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}\\ &=\frac{2(3+\sqrt{5})}{4}\\ \therefore a&=\pm\sqrt{\frac{3+\sqrt{5}}{2}}\\ &=\pm\frac{1}{2}(1+\sqrt{5}) \end{split} \end{equation*}

where \sqrt{\frac{3+\sqrt{5}}{2}}=\frac{1}{2}(1+\sqrt{5}) was obtained by first setting \sqrt{3+\sqrt{5}}=\sqrt{a}+\sqrt{b} then squaring both sides to get 3+\sqrt{5}=(a+b)+2\sqrt{ab}, and then finally solving the linear-quadratic system a+b=3, 4ab=5 to get a=\frac{5}{2}, b=\frac{1}{2} (or a=\frac{1}{2}, b=\frac{5}{2}).

Due to its connection to the golden ratio, we’ll have more to say about this example in a later post.

Let a=\frac{1+\sqrt{5}}{2} and r=\frac{-3+\sqrt{5}}{2}. PROVE that a-r=2 and a+r=-2ar.

By direct calculation:

    \begin{equation*} \begin{split} a-r&=\frac{1+\sqrt{5}}{2}-\left(\frac{-3+\sqrt{5}}{2}\right)\\ &=2\\ &\ddots\ddots\ddots\\ a+r&=\frac{1+\sqrt{5}}{2}+\left(\frac{-3+\sqrt{5}}{2}\right)\\ &=-1+\sqrt{5}\\ -2ar&=-2\times\frac{1+\sqrt{5}}{2}\times\frac{-3+\sqrt{5}}{2}\\ &=-2\times\frac{1}{4}\left(-3+\sqrt{5}-3\sqrt{5}+5\right)\\ &=-1+\sqrt{5}\\ \therefore a+r&=-2ar \end{split} \end{equation*}

Of course, both a-r=2 and a+r=-2ar yield r^2+3r+1=0.

Let a=\frac{-1-\sqrt{5}}{2} and r=\frac{-3+\sqrt{5}}{2}. PROVE that a+r=-2 and a-r=-2ar

As in the previous example, proceed by direct calculation:

    \begin{equation*} \begin{split} a+r&=\frac{-1-\sqrt{5}}{2}+\left(\frac{-3+\sqrt{5}}{2}\right)\\ &=-2\\ &\ddots\ddots\ddots\\ a-r&=\frac{-1-\sqrt{5}}{2}-\left(\frac{-3+\sqrt{5}}{2}\right)\\ &=1-\sqrt{5}\\ -2ar&=-2\times\frac{-1-\sqrt{5}}{2}\times\frac{-3+\sqrt{5}}{2}\\ &=-2\times\frac{1}{4}\left(3-\sqrt{5}+3\sqrt{5}-5\right)\\ &=1-\sqrt{5}\\ \therefore a-r&=-2ar \end{split} \end{equation*}

Several other nice relations exist between a and r.

Takeaway

At the level of slopes, equilateral triangles and right isosceles triangles are very close, as these equations show:

    \[r^2+4r+1=0,\qquad r^2+3r+1=0.\]

The slopes must form a geometric progression though.

Tasks

  1. (T w o nice) Let a,ar,ar^2 be the slopes of the sides of a triangle.
    • If the triangle is equilateral, PROVE that the sum of product of the slopes, taken two slopes at a time, is always negative three. That is, (a\times ar)+(ar\times ar^2)+(ar^2\times a)=-3;
    • If the triangle is right isosceles, PROVE that the sum of product of the slopes, taken two slopes at a time, is always twice the common ratio. That is, (a\times ar)+(ar\times ar^2)+(ar^2\times a)=2r.
  2. In \triangle ABC, suppose that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively. PROVE that:
    • the slope of the median from A is -ar;
    • the slope of the median from B is \left(\frac{r+2}{2r+1}\right)ar;
    • the slope of the median from C is \left(\frac{2r+1}{r+2}\right)ar.
  3. In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA, in that order.
    • If AB=BC, PROVE that \left(\frac{r+2}{2r+1}\right)a^2r^3=-1 (Use exercise 2 above together with the fact that the median from vertex B to side CA is actually an altitude due to the fact that AB=BC);
    • If AB=BC and \angle B=90^{\circ}, deduce that r^2+3r+1=0. (This provides another means of deriving equation (2).)
  4. PROVE that if the slopes of the sides of a triangle are a,-2a,4a (a\neq 0), then the slopes of the medians do not form a geometric progression. (Use exercise 2 above.)
  5. Suppose that the slopes of the sides of a triangle form a geometric progression with common ratio r_1, and that the slopes of the medians form a geometric progression with common ratio r_2. PROVE that:
    • r_1+r_2\neq 0;
    • r_1-r_2\neq 0;
    • r_1 and r_2 cannot have the same signs;
    • r_1 and r_2 cannot both be integers.
  6. In \triangle ABC, suppose that sides AB,BC,CA have slopes a,ar,ar^2, respectively. PROVE that:
    • the median from vertex A and the median from vertex B are perpendicular if and only if a^2=\frac{r+2}{r^2(2r+1)}, r\neq -2;
    • the median from vertex B and the median from vertex C are perpendicular if and only if (ar)^2=-1. Since this equation has no real solutions, does it imply that the median from vertex B and the median from vertex C can never be perpendicular?
  7. Consider \triangle ABC with vertices at A(0,0), B\left(1,\frac{4\sqrt{5}}{5}\right), and C\left(3,\frac{6\sqrt{5}}{5}\right). PROVE that:
    • the slopes of the sides form a geometric progression in which r=2;
    • the median from vertex A and the median from vertex B are perpendicular.
  8. Find coordinates for the vertices of \triangle ABC in which:
    • the slopes of sides AB,BC,CA form a geometric progression with r=-3;
    • the median from vertex A and the median from vertex B are perpendicular.
  9. Give an example of an isosceles triangle with slopes a,ar,ar^2 in which equation (1), namely r^2+4r+1=0, is satisfied. (There are isosceles triangles (that are not equilateral) that satisfy equation (1).)
  10. Let n\in\mathbb{Z}. PROVE that n^2+48 is a perfect square for n=1,4,11. Are these the only n\in\mathbb{Z} for which n^2+48 is a perfect square?