Slopefessor Fr1day – Page 30 – High School Geometry

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 $r$ ules 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 $r$ est $r$ ictions.

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

(T w o nice) Let 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, $(a\times ar)+(ar\times ar^2)+(ar^2\times a)=2r$ .
In , suppose that the slopes of sides are , 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 $\left(\frac{2r+1}{r+2}\right)ar$ .
In , let be the slopes of sides , 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).)
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.)
Suppose that the slopes of the sides of a triangle form a geometric progression with common ratio , and that the slopes of the medians form a geometric progression with common ratio . 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.
In , suppose that sides have slopes , 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?
Consider with vertices at , , and . 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.
Find coordinates for the vertices of 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.
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).)
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?

A relationship between medians

For $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , respectively, the equation connecting their side lengths is

(1) $\begin{equation*} AB^2+AC^2=\frac{r^2+1}{(r+1)^2}BC^2, \end{equation*}$

as we saw a post ago. The equation connecting their median lengths is

(2) $\begin{equation*} m_{B}^2+m_{C}^2=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}m_{A}^2, \end{equation*}$

and is today’s goal.

Compare the relationship between the medians of a right triangle with the medians of a triangle with slopes $a,ar,ar^2~(a\neq 0,~r\neq\pm 1)$ .

This is the final, knockout round of our compare-contrast-contest with the right triangle. The contest started on June 28, with the right triangle currently having an edge.

In a right triangle, the medians are related via $m_a^2+m_b^2=5m_c^2$ , where $m_c$ is the length of the median to the hypotenuse (see the exercises). The triangle with slopes $a,ar,ar^2$ doesn’t satisfy this relation, unless we put $r=0$ in (2).
Winner: Right triangle; Score is $1:0$ .
The previous score was $2:1$ , in favour of the right triangle.

Final verdict: Right triangle wins $3:1$ on aggregate. Case closed.

We still celebrate the triangle with slopes $a,ar,ar^2$ , despite been rightly trounced by the right triangle.

Consider $\triangle ABC$ with vertices $A(1,4)$ , $B(3,6)$ , and $C(0,0)$ . Verify that three medians satisfy equation (2).

Note that the slopes of sides $AB,BC,CA$ are $1,2,4$ , respectively. They form a geometric progression with $r=2$ . The midpoints of these sides have been marked. Thus, by the distance formula:

$\begin{equation*} \begin{split} m_A^2&=(1-1.5)^2+(4-3)^2\\ &=1.25\\ m_B^2&=(3-0.5)^2+(6-2)^2\\ &=22.25\\ m_C^2&=(0-2)^2+(0-5)^2\\ &=29\\ &\vdots\ddots\vdots\cdots\\ m_B^2+m_C^2&=22.25+29\\ &=51.25\\ \left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\right)m_A^2&=\left(\frac{(2+2)^2+(4+1)^2}{(2-1)^2}\right)\times 1.25\\ &=\frac{16+25}{1}\times 1.25\\ &=51.25\\ &\vdots\ddots\vdots\cdots\\ \therefore m_B^2+m_C^2&=\left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\right)m_A^2. \end{split} \end{equation*}$

YEA.

Consider $\triangle ABC$ with vertices at $A(1,9)$ , $B(-2,6)$ , and $C(0,0)$ . Verify that the three medians satisfy equation (2).

Easy-peasy. Observe that the slopes of sides $AB,BC,CA$ are $1,-3,9$ . They form a geometric progression with common ratio $r=-3$ . Let’s draw a diagram:

The median lengths then follow by the distance formula. We have:

$\begin{equation*} \begin{split} m_A^2&=(1--1)^2+(9-3)^2\\ &=40\\ m_B^2&=(-2-0.5)^2+(6-4.5)^2\\ &=8.5\\ m_C^2&=(0--0.5)^2+(0-7.5)^2\\ &=56.5\\ &\vdots\ddots\vdots\cdots\\ m_B^2+m_C^2&=8.5+56.5\\ &=65\\ \left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\right)m_A^2&=\left(\frac{(-3+2)^2+(-6+1)^2}{(-3-1)^2}\right)\times 40\\ &=\frac{1+25}{16}\times 40\\ &=65\\ &\vdots\ddots\vdots\cdots\\ \therefore m_B^2+m_C^2&=\left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\right)m_A^2. \end{split} \end{equation*}$

There we go. The formula works for both positive and negative common ratios.

In $\triangle ABC$ , the median lengths are related via $m_B^2+m_C^2=13m_A^2$ . Find (possible) coordinates for the vertices $A,B,C$ .

Try solving this problem in a way other than what we present below.

In (2), set $\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}=13$ and solve for $r$ :

$\begin{equation*} \begin{split} (r+2)^2+(2r+1)^2&=13(r-1)^2\\ 8r^2-34r+8&=0\\ 4r^2-17r+4&=0\\ (r-4)(4r-1)&=0\\ r&=4,~\frac{1}{4} \end{split} \end{equation*}$

The workload reduces to finding coordinates for a triangle whose slopes form a geometric progression with common ratio $r=4$ .

Easy-peasy.

Among the many options we have for the vertices of a triangle with side slopes $1,r,r^2$ , we can take $(0,0)$ , $(1,r^2)$ , and $(1+r,r+r^2).$ With $r=4$ , these become $(0,0)$ , $(1,16)$ , and $(5,20)$ . Use these as $C,A,B$ . That is, take $C(0,0)$ , $A(1,16)$ , and $B(5,20)$ as the triangle’s vertices. A quick calculation shows that $m_A^2=38.25$ , $m_B^2=164.25$ , and $m_{C}^2=333$ (as in, triple three).

$\begin{equation*} \begin{split} m_B^2+m_C^2&=164.25+333\\ &=496.25\\ &=13\times 38.25\\ &=13 m_{A}^2. \end{split} \end{equation*}$

Watch what happens when $r=-2$ in both (1) and (2).

Put $r=-2$ in (1). Obtain $AB^2+AC^2=5BC^2$ .

Put $r=-2$ in (2). Obtain $m_B^2+m_C^2=m_A^2$ .

These are well-known, equivalent conditions for the medians from $B$ and $C$ to be perpendicular.

Is it surprising that it’s our favourite $r$ -value that saved the day?

For $r\neq 1$ , let $M(r)=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}$ . Evaluate $M(-2),M(0),M(2),M(4)$ , and show that they are all sums of two consecutive squares.

Observe that the only integer values of $r$ that produce integer values of $M(r)$ are $\{-2,0,2,4\}$ .

$\begin{equation*} \begin{split} M(-2)&=\frac{(-2+2)^2+(-4+1)^2}{(-2-1)^2}\\ &=1\\ &=0^2+1^2\\ M(0)&=\frac{(0+2)^2+(0+1)^2}{(0-1)^2}\\ &=5\\ &=1^2+2^2\\ M(2)&=\frac{(2+2)^2+(4+1)^2}{(2-1)^2}\\ &=41\\ &=4^2+5^2\\ M(4)&=\frac{(4+2)^2+(8+1)^2}{(4-1)^2}\\ &=13\\ &=2^2+3^2 \end{split} \end{equation*}$

In addition, $M(r)$ defines a bijection $\{-2,0,2,4\}\rightarrow\{1,5,13,41\}$ .

In $\triangle ABC$ , we have $AB^2+AC^2=kBC^2$ and $m_{B}^2+m_{C}^2=km_{A}^2$ . Find the value of $k$ .

You can suspect what $k$ will be. From (1) and (2), set

$\frac{r^2+1}{(r+1)^2}=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}.$

Clear fractions, and simplify to obtain the quartic polynomial equation

$\begin{equation*} \begin{split} r^4+5r^3+6r^2+5r+1&=0\\ \textrm{Factor:}~(r^2+4r+1)(r^2+r+1)&=0 \end{split} \end{equation*}$

The second quadratic factor, $r^2+r+1$ , is really irreducible (its discriminant is $-3< 0$ ). So the only real roots come from the first one:

$r^2+4r+1=0\implies r=-2\pm\sqrt{3}.$

Thus, for $r=-2+\sqrt{3}$ , we have:

$\begin{equation*} \begin{split} k&=\frac{r^2+1}{(r+1)^2}\\ &=\frac{(-2+\sqrt{3})^2+1}{(-2+\sqrt{3}+1)^2}\\ &=\frac{7-4\sqrt{3}+1}{4-2\sqrt{3}}\\ &=\frac{8-4\sqrt{3}}{4-2\sqrt{3}}\\ &=2 \end{split} \end{equation*}$

Similarly, for $r=-2-\sqrt{3}$ , we have:

$\begin{equation*} \begin{split} k&=\frac{r^2+1}{(r+1)^2}\\ &=\frac{(-2-\sqrt{3})^2+1}{(-2-\sqrt{3}+1)^2}\\ &=\frac{7+4\sqrt{3}+1}{4+2\sqrt{3}}\\ &=\frac{8+4\sqrt{3}}{4+2\sqrt{3}}\\ &=2 \end{split} \end{equation*}$

The value of $k$ is $2$ , as you may have correctly guessed.

OK, the above approach was definitely an over $k$ ill for something that can be obtained in a simpler way, but it just confirms that our theory conforms to what is more established. See the next example for a better strategy.

In any $\triangle ABC$ , PROVE that the following two statements are equivalent:

the relations $AB^2+AC^2=kBC^2$ and $m_{B}^2+m_{C}^2=km_{A}^2$ hold simultaneously;
$k=2$ .

Consider $(1)\implies (2)$ and let’s default to standard notation. Use the fact that the median lengths are related to the side lengths via:

(3) $\begin{equation*} \begin{split} m_a^2&=\frac{2b^2+2c^2-a^2}{4}\\ m_b^2&=\frac{2a^2+2c^2-b^2}{4}\\ m_c^2&=\frac{2a^2+2b^2-c^2}{4} \end{split} \end{equation*}$

from which $m_b^2+m_c^2=\frac{4a^2+b^2+c^2}{4}$ . Now, since $b^2+c^2=ka^2$ , this becomes $m_b^2+m_c^2=\frac{4a^2+ka^2}{4}=\frac{(k+4)a^2}{4}$ . Also, $m_a^2=\frac{2b^2+2c^2-a^2}{4}$ becomes $m_a^2=\frac{2(b^2+c^2)-a^2}{4}=\frac{(2k-1)a^2}{4}$ . Since $m_b^2+m_c^2=km_a^2$ , we have:

$\begin{equation*} \begin{split} \frac{(k+4)a^2}{4}&=k\frac{(2k-1)a^2}{4}\\ (k+4)&=k(2k-1)\\ 2k^2-k-k-4&=0\\ k^2-k-2&=0\\ (k-2)(k+1)&=0\\ \implies k&=2,-1 \end{split} \end{equation*}$

In view of $b^2+c^2=ka^2$ , we see that $k$ can’t be negative (in fact, $k> 0.5$ : example 10), so discard $k=-1$ , and take $k=2$ as the only valid solution. Thus, $(1)\implies (2)$ .

To see that $(2)\implies (1)$ , suppose that $k=2$ . Assume that $AB^2+AC^2=2BC^2$ . We’ll show that $m_B^2+m_C^2=2m_A^2$ as well. Indeed, from (3):

$\begin{equation*} \begin{split} m_B^2+m_C^2&=\left(\frac{2a^2+2c^2-b^2}{4}\right)+\left(\frac{2a^2+2b^2-c^2}{4}\right)\\ &=\frac{4a^2+b^2+c^2}{4}\\ &=\frac{4a^2+2a^2}{4}\\ &=\frac{3}{2}a^2\\ \textrm{Also,}~2m_A^2&=2\times\left(\frac{2b^2+2c^2-a^2}{4}\right)\\ &=2\times\left(\frac{2(b^2+c^2)-a^2}{4}\right)\\ &=2\times\frac{4a^2-a^2}{4}\\ &=\frac{3}{2}a^2\\ \therefore m_B^2+m_C^2&=2m_A^2. \end{split} \end{equation*}$

Similarly, if we started with $m_B^2+m_C^2=2m_A^2$ , we’ll again arrive at $AB^2+AC^2=2BC^2$ . This proves $(2)\implies (1)$ .

Bottom line: the two statements are equivalent as claimed.

Kindly bear with us for switching back and forth between the standard notation $m_a$ and $m_A$ . Hopefully this doesn’t cause any confusion.

In any $\triangle ABC$ , if $AB^2+AC^2=kBC^2$ and $m_{B}^2+m_{C}^2=km_{A}^2$ , PROVE that $m_{A}=\frac{\sqrt{3}}{2}BC$ .

The relationship

$m_{A}^2+m_{B}^2+m_{C}^2=\frac{3}{4}\left(AB^2+BC^2+CA^2\right)$

holds in any triangle. Using the fact that $AB^2+AC^2=kBC^2$ and $m_{B}^2+m_{C}^2=km_{A}^2$ , we have:

$\begin{equation*} \begin{split} m_{A}^2+(km_{A}^2)&=\frac{3}{4}\left(BC^2+kBC^2\right)\\ \left(k+1\right)m_{A}^2&=\frac{3}{4}(k+1)BC^2\\ \therefore m_{A}&=\frac{\sqrt{3}}{2}BC \end{split} \end{equation*}$

Although we freely cancelled out the common factor $(k+1)$ , it shouldn’t obscure the fact that the relations $AB^2+AC^2=kBC^2$ and $m_{B}^2+m_{C}^2=km_{A}^2$ hold simultaneously only when $k=2$ .

If $AB^2+AC^2=kBC^2$ , PROVE that $k>\frac{1}{2}$ .

Compare with exercise 5 below.

For the proof, begin with one of the expressions in (3):

$\begin{equation*} \begin{split} m_a^2&=\frac{2b^2+2c^2-a^2}{4}\\ &=\frac{2(b^2+c^2)-a^2}{4}\\ &=\frac{2(ka^2)-a^2}{4}\\ &=\frac{(2k-1)a^2}{4} \end{split} \end{equation*}$

Since the left side $m_a^2$ is positive, so must the right side be, and we only have to impose this restriction on the nonkonstant $2k-1$ .

$2k-1> 0\implies k>\frac{1}{2}$ , as desired.

Takeaway

In the equation

$m_{B}^2+m_{C}^2=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}m_{A}^2,$

the median from vertex $B$ and the median from vertex $C$ are perpendicular only when $r=-2$ . If the focus shifts to the medians from $A$ and $B$ , or the medians from $A$ and $C$ , then other values of $r$ can still ensure perpendicularity.

Tasks

Find possible coordinates for the vertices of $\triangle ABC$ if its three medians are related via $m_{B}^2+m_{C}^2=25m_{A}^2$ .
Find possible coordinates for the vertices of $\triangle ABC$ if its three medians are related via $m_{B}^2+m_{C}^2=41m_{A}^2$ .
PROVE that the medians in a right triangle are related via $m_{A}^2+m_{B}^2=5m_{C}^2$ , where $m_{C}$ is the length of the median to the hypotenuse.
Derive equation (2).
Let $L(r)=\frac{r^2+1}{(r+1)^2}$ and $M(r)=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}$ . PROVE that both $L(r)$ and $M(r)$ have the same minimum value, but that their common minimum occurs at different values of $r$ .
Consider the rational function :
- Find the equation of its vertical asymptote. In light of equation (2), what is the significance of this vertical asymptote?
- Find the equation of its horizontal asymptote and state its significance in relation to equation (2).
- Find the axes intercepts (i.e $x$ and $y$ intercepts).
- Sketch the graph.
Verify equation (2) for a triangle with vertices at $(0,0)$ , $(2,36)$ , and $(8,48)$ .
PROVE that if a right triangle has slopes in geometric progression with common ratio $r=-2$ , then both the squares of the median lengths and the squares of the side lengths form arithmetic progressions.
PROVE that the “mapping” $M:\mathbb{Z}-\{1\}\rightarrow\mathbb{Z},~r\mapsto\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}$ , is not well-defined. What modifications should be made to the domain or range, so that the mapping can be well-defined?
If $1\neq r\in\mathbb{Q}$ and $M(r)=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\in\mathbb{Z}$ , PROVE that $M(r)$ is then a sum of two consecutive squares.