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).

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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:

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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 overkill 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:

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

  1. 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.
  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.
  3. 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.
  4. Derive equation (2).
  5. 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.
  6. Consider the rational function y=\frac{(x+2)^2+(2x+1)^2}{(x-1)^2}:
    • 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.
  7. Verify equation (2) for a triangle with vertices at (0,0), (2,36), and (8,48).
  8. 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.
  9. 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?
  10. 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.