A relationship between medians – High School Geometry

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.