Approximate Pythagorean Identity – High School Geometry

Let $l_1,l_2,l_3$ be lengths of the sides of a triangle, and let the slopes of these sides be $a,ar,ar^2$ , respectively. Then, as we show in example 10:

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

This is analogous to the Pythagorean identity for right triangles. In theory, if we let $r\rightarrow\infty$ (or $r\rightarrow 0$ ), then $l_1^2+l_3^2\approx l_2^2$ . Thus, in a sense, a right triangle is a limiting case of a triangle with slopes in geometric progression.

Whatever a right triangle can do: $\ddots\ddots\ddots$

Whatever a right triangle can “do”, there is a triangle with slopes $a,ar,ar^2$ that can “do” the same thing. Did we just exaggerate?

Compare the hypotenuse of a right triangle with side $l_2$ of a triangle whose slopes are $a,ar,ar^2~~(a\neq 0, r\neq 0,\pm 1)$ .

By side $l_2$ , we mean the side whose slope is $ar$ , the geometric mean of the progression $a,ar,ar^2$ .

In terms of length: If $r> 0$ , then (in view of (1)) the side with length $l_2$ is the longest side of the triangle, just like the hypotenuse of a right triangle.
(Winner: Right triangle — because it always satisfies this property; our own triangle doesn’t always satisfy it if $r< 0$ .)
The median to the hypotenuse: In a right triangle, the length of the median to the hypotenuse is always half of the length of the hypotenuse. In a triangle with slopes $a,ar,ar^2$ , the length of the median to side $l_2$ has length $\frac{1}{2}\left(\frac{r-1}{r+1}\right)l_2$ (See the exercises). And so as $r\rightarrow 0,\infty$ , this becomes $\frac{1}{2}l_2$ .
(Winner: Right triangle again — because it always satisfies this property, while our own triangle satisfies it only when $r\rightarrow 0,\infty$ .)
Slope of the median to the hypotenuse: If the legs of a right triangle are parallel to the coordinate axes, then the slope of the median to the hypotenuse is the negative of the slope of the hypotenuse. On the other hand, in a triangle with slopes $a,ar,ar^2$ , the slope of the median to side $l_2$ is always the negative of the slope of side $l_2$ .
(Winner: The triangle with slopes $a,ar,ar^2$ wins here — because it always satisfies this property, whereas a right triangle satisfies it only when its legs are parallel to the coordinate axes.)

Final Score: Right triangle won $2$ to $1$ — courtesy of our generosity.

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

Observe that the slopes are $1,-3,9$ for sides $AC,CB,BA$ , respectively. They form a geometric progression in which $r=-3$ .

In (1), the side with length $l_2$ is the side whose slope is the geometric mean of the progression — in the present case, it is side $CB$ . This is important. $l_1$ and $l_3$ can then be chosen as the lengths of the other two sides.

Using the given vertices $A(1,9),~B(0,0),~C(-2,6)$ , let’s compute the side lengths:

$\begin{equation*} \begin{split} BA^2&=(0-1)^2+(0-9)^2\\ &=82\\ AC^2&=(-2-1)^2+(6-9)^2\\ &=18\\ CB^2&=(0+2)^2+(0-6)^2\\ &=40 \end{split} \end{equation*}$

Since $CB^2=40$ , we have to take $l_2^2=40$ . Let $l_1^2=AC^2=18$ and let $l_3^2=BA^2=82$ . Then:

$\begin{equation*} \begin{split} l_1^2+l_3^2&=18+82\\ &=100\\ \frac{1+r^2}{(1+r)^2}l_2^2&=\frac{1+(-3)^2}{(1+-3)^2}\times 40\\ &=\frac{10}{4}\times 40\\ &=100\\ &\vdots\ddots\ddots\vdots\\ \therefore l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^2}l_2^2 \end{split} \end{equation*}$

Consider $\triangle ABC$ with vertices $A(-1,2)$ , $B(-0.5,-2)$ , and $C(1.5,-3)$ . Verify that the side lengths satisfy equation (1).

The slopes of sides $BC,CA,AB$ are $-\frac{1}{2},-2,-8$ , respectively. They form a geometric progression with common ratio $r=4$ . Further, the slope of side $CA$ , namely $-2$ , is the geometric mean of the progression. Thus the length of side $CA$ will be taken as $l_2$ in (1).

$\begin{equation*} \begin{split} CA^2&=(-1-1.5)^2+(2+3)^2\\ &=31.25~\textrm{or}~\frac{125}{4}\\ \therefore l_2^2&=\frac{125}{4}\\ AB^2&=(-1+0.5)^2+(2+2)^2\\ &=16.25~\textrm{or}~\frac{65}{4}\\ BC^2&=(-0.5-1.5)^2+(-2+3)^2\\ &=5 \end{split} \end{equation*}$

Since $l_2^2$ has been chosen as $\frac{125}{4}$ , we can take $l_1^2=\frac{65}{4}$ and $l_3^2=5$ . Then, in (1):

$\begin{equation*} \begin{split} l_1^2+l_3^2&=\frac{65}{4}+5\\ &=\frac{85}{4}\\ \frac{1+r^2}{(1+r)^3}l_2^2&=\frac{1+4^2}{(1+4)^2}\times \frac{125}{4}\\ &=\frac{17}{25}\times \frac{125}{4}\\ &=\frac{85}{4}\\ &\ddag\ddots\ddag\\ \therefore l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^3}l_2^2 \end{split} \end{equation*}$

So the formula works for both negative and positive common ratios.

The slopes of the sides of $\triangle ABC$ are all non-zero and the side lengths satisfy $AB^2+CA^2=25BC^2$ . Find possible coordinates for the vertices $A,B,C$ .

The only challenge here is the non-zero slope specification; otherwise, we could just place two vertices along the $x$ -axis, making the third vertex easy to find.

Fortunately, we can always appeal to triangles whose slopes form geometric progressions under this situation. They’re cut out for these.

In (1), set $\frac{1+r^2}{(1+r)^2}=25$ and solve for $r$ :

$\begin{equation*} \begin{split} 25(1+2r+r^2)&=1+r^2\\ 24r^2+50r+24&=0\\ 12r^2+25r+12&=0\\ (4r+3)(3r+4)&=0\\ r&=-\frac{3}{4},-\frac{4}{3} \end{split} \end{equation*}$

The workload now reduces to finding coordinates for a triangle whose slopes form a geometric progression in which $r=-\frac{3}{4}$ .

Easy-peasy. We can use $B(0,0)$ , $A(1,r^2)$ , $C(1+r,r+r^2)$ and just put $r=-\frac{3}{4}$ .

We obtain $B(0,0)$ , $A\left(1,\frac{9}{16}\right)$ , and $C\left(\frac{1}{4},-\frac{3}{16}\right)$ . The side slopes are all non-zero: $m_{AB}=\frac{9}{16}$ , $m_{BC}=-\frac{3}{4}$ , and $m_{CA}=1$ .

For the side lengths, we have:

$\begin{equation*} \begin{split} AB^2&=\left(1-0\right)^2+\left(\frac{9}{16}-0\right)^2\\ &=\frac{337}{256}\\ CA^2&=\left(\frac{1}{4}-1\right)^2+\left(-\frac{3}{16}-\frac{9}{16}\right)^2\\ &=\frac{288}{256}\\ BC^2&=\left(\frac{1}{4}-0\right)^2+\left(-\frac{3}{16}-0\right)^2\\ &=\frac{25}{256}\\ &\ddots\ddots\ddots\\ AB^2+CA^2&=\frac{337}{256}+\frac{288}{256}\\ &=\frac{625}{256}\\ &=25\times\left(\frac{25}{256}\right)\\ &=25BC^2 \end{split} \end{equation*}$

DONE!!! $AB^2+CA^2=25BC^2$ , with $A\left(1,\frac{9}{16}\right)$ , $B(0,0)$ , and $C\left(\frac{1}{4},-\frac{3}{16}\right)$ .

Quantity digression

Let’s examine the quantity $\frac{1+r^2}{(1+r)^2}$ that appeared in formula (1).

For real $r\neq -1$ , PROVE that the minimum value of $\frac{1+r^2}{(1+r)^2}$ is $\frac{1}{2}$ .

Let’s do this for real. And, without using calculus. Set $\frac{1+r^2}{(1+r)^2}=k$ and clear fractions to obtain $k(1+r)^2=1+r^2$ .

(2) $\begin{equation*} \begin{split} k(r^2+2r+1)&=1+r^2\\ kr^2+2kr+k&=1+r^2\\ (k-1)r^2+(2k)r+(k-1)&=0 \end{split} \end{equation*}$

Now comes the main point. Since $k$ is a real number, the discriminant of the above quadratic must be greater than or equal to zero. So:

$\begin{equation*} \begin{split} (2k)^2-4\times(k-1)\times(k-1)&\geq 0\\ 4k^2-4(k^2-2k+1)&\geq 0\\ 8k-4&\geq 0\\ \implies k&\geq \frac{1}{2} \end{split} \end{equation*}$

Thus, $\frac{1+r^2}{(1+r)^2}=k\geq \frac{1}{2}$ .

Let $r$ be an integer. PROVE that $L(r)=\frac{1+r^2}{(1+r)^2}$ is also an integer if and only if $r=-2,0$ .

If you’ve been following our posts, you may have noticed our public, profuse admiration for $r=-2$ . It is a peach.

Put $r=-2$ to obtain $L(-2)=\frac{1+(-2)^2}{(1+-2)^2}=5$ ; put $r=0$ to get $L(0)=\frac{1+0^2}{(1+0)^2}=1$ .

Conversely, suppose that $L(r)=\frac{1+r^2}{(1+r)^2}$ is an integer, with $r$ an integer too. We’ll show that $r$ must be $-2$ or $0$ . Set $\frac{1+r^2}{(1+r)^2}=k$ , then we obtain what we had in (2):

$\begin{equation*} \begin{split} (k-1)r^2+(2k)r+(k-1)&=0\\ r&=\frac{-2k\pm\sqrt{(2k)^2-4\times (k-1)\times(k-1)}}{2(k-1)}\\ &=\frac{-2k\pm\sqrt{8k-4}}{2(k-1)}\\ r&=\frac{-k\pm\sqrt{2k-1}}{k-1} \end{split} \end{equation*}$

To obtain an integer solution, the quantity under the square root, namely $2k-1$ , must first of all be a perfect square. Since $2k-1$ is an odd number, it must be congruent to $1$ modulo $8$ if it’s a perfect square (a bit of number theory here $\cdots$ no worries $\ddots\ddots\ddots\vdots\vdots\vdots\ddots\ddots\vdots$ ). Basically, the difference $(2k-1)-1$ must be divisible by $8$ .

Continuing, let’s set $(2k-1)-1=8t$ , for some integer $t$ . As this leads to $2k-2=8t$ , we obtain $k=4t+1$ . Using this in the expression we obtained for $r$ :

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

Now $8t+1$ is a perfect square if and only if $t=\frac{n(n+1)}{2}$ for some positive integer $n$ (see the exercises at the end). Thus we have

$\begin{equation*} \begin{split} r&=\frac{-4\times\left(\frac{n(n+1)}{2}\right)-1\pm\sqrt{8\times \left(\frac{n(n+1)}{2}\right)+1}}{4\times\left(\frac{n(n+1)}{2}\right)}\\ &=\frac{-2n(n+1)-1\pm(2n+1)}{2n(n+1)}\\ r&=-\frac{n}{n+1},~-\frac{n+1}{n} \end{split} \end{equation*}$

Consider $r=-\frac{n}{n+1}$ . It is an integer only when $n=-2,0$ — giving $r=-2,0$ , respectively.

Consider $r=-\frac{n+1}{n}$ . It is an integer only when $n=\pm 1$ . Put $n=1$ to obtain $r=-2$ . (Using $n=-1$ gives $r=0$ again. However, because of the discriminant of the quadratic equation we solved, we don’t want $n=-1$ . Similarly, we don’t want $n=-2$ above.)

Another way of proceeding is to use polynomial division to write $\frac{r^2+1}{(r+1)^2}$

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

. Since $r$

is an integer, the consecutive integers $r$

and $r+1$

have no common factor. Thus the denominator of $\frac{2r}{(r+1)^2}$

must be $1$

. In turn, $(r+1)=\pm 1$

, giving $r=-2,0$

, as before.

Our decision to follow a longer procedure is that it also yields the form of rational numbers — not just integers — that make the expression $L(r)=\frac{r^2+1}{(r+1)^2}$ an integer.

If a right triangle satisfies (1), PROVE that the side whose length is $l_2$ cannot be the hypotenuse.

We can use slopes to do this, but let’s use lengths instead.

Suppose the length of the hypotenuse is $l_2$ . Then we have $l_1^2+l_3^2=l_2^2$ by the Pythagorean theorem. Together with (1) we have the following quadratic-quadratic system:

$\begin{equation*} \begin{split} l_1^2+l_3^2&=l_2^2\\ l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^2}l_2^2\\ &\ddots\vdots\ddots\\ \therefore 1&=\frac{1+r^2}{(1+r)^2}\\ (1+r)^2&=1+r^2\\ 1+2r+r^2&=1+r^2\\ \implies r&=0 \end{split} \end{equation*}$

Since $r=0$ is not allowed for a geometric progression, we conclude that $l_2$ cannot be the hypotenuse.

The peach of the bunch

Any triangle whose slopes form a geometric sequence with common ratio $r=-2$ is a peach. If it’s also a right triangle, then it becomes the peach of the pick of the bunch.

If a right triangle has $r=-2$ , PROVE that the $x$ -coordinates, the $y$ -coordinates, and the squares of the lengths form arithmetic progressions.

Our first post of this year showed that if $r=-2$ , then the $x$ -coordinates and the $y$ -coordinates form (different) arithmetic progressions — even when the triangle is not right.

Put $r=-2$ in (1) to obtain

$l_1^2+l_3^2=5l_2^2$

By the way, this is a well-known condition for perpendicular medians.

Since we’re dealing with a right triangle this time (and $l_2$ cannot be the hypotenuse, in view of the previous example), we can take $l_1$ as the hypotenuse, so that

$l_1^2=l_2^2+l_3^2$

Eliminate $l_1$ from the two equations: $(l_2^2+l_3^2)+l_3^2=5l_2^2\implies l_3^2=2l_2^2$ .

Then $l_1^2=l_2^2+l_3^2=l_2^2+2l_2^2=3l_2^2$ .

The squares of the lengths, in terms of $l_2$ , are: $l_2^2,2l_2^2,3l_2^2$ . They form an arithmetic sequence.

If a right triangle has $r=-2$ , PROVE that $a=\pm\frac{\sqrt{2}}{2}$ , or $a=\pm\frac{\sqrt{2}}{4}$ .

Assume the progression of the slopes is $a,ar,ar^2$ . If $r=-2$ , then this becomes $a,-2a,4a$ .

We must have $a\times (-2a)=-1\implies a=\pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2}$ .

Or we must have $4a\times (-2a)=-1\implies a=\pm\frac{\sqrt{2}}{4}$ .

Derivation of the formula

We now derive equation (1).

In $\triangle ABC$ , let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ , respectively. If the lengths of these sides are also denoted by $l_1,l_2,l_3$ , PROVE that $l_1^2+l_3^2=\frac{r^2+1}{(r+1)^2}l_2^2$ .

Let the vertices be $A(x_1,y_1),~B(x_2,y_2),C(x_3,y_3)$ . Then the following relations exist among the coordinates:

(3) $\begin{equation*} \begin{split} x_1&=x_3+\frac{y_2-y_3}{ar(r+1)}\\ y_1&=\frac{ry_2+y_3}{r+1}\\ x_2&=x_3+\frac{y_2-y_3}{ar} \end{split} \end{equation*}$

Using the distance formula:

$\begin{equation*} \begin{split} l_1^2&=AB^2\\ &=\left[x_3+\frac{y_2-y_3}{ar(r+1)}-\left(x_3+\frac{y_2-y_3}{ar}\right)\right]^2+\left[\frac{ry_2+y_3}{r+1}-y_2\right]^2\\ &=\left(-\frac{y_2-y_3}{a(r+1)}\right)^2+\left(-\frac{y_2-y_3}{r+1}\right)^2\\ &=\left(\frac{a^2+1}{a^2(r+1)^2}\right)(y_2-y_3)^2\\ l_2^2&=BC^2\\ &=\left(\frac{a^2r^2+1}{a^2r^2}\right)(y_2-y_3)^2\\ l_3^2&=CA^2\\ &=\left(\frac{a^2r^4+1}{a^2r^2(r+1)^2}\right)(y_2-y_3)^2\\ &\ddots\ddots\ddots\\ &\ddots\ddots\ddots\\ l_1^2+l_3^2&=\left(\frac{a^2+1}{a^2(r+1)^2}\right)(y_2-y_3)^2+\left(\frac{a^2r^4+1}{a^2r^2(r+1)^2}\right)(y_2-y_3)^2\\ &=\frac{(r^2+1)(a^2r^2+1)}{a^2r^2(r+1)^2}(y_2-y_3)^2\\ \frac{r^2+1}{(r+1)^2}l_2^2&=\frac{r^2+1}{(r+1)^2}\times\left(\frac{a^2r^2+1}{a^2r^2}\right)(y_2-y_3)^2\\ &=\frac{(r^2+1)(a^2r^2+1)}{a^2r^2(r+1)^2}(y_2-y_3)^2\\ &=l_1^2+l_3^2\\ &\ddots\ddots\ddots\ddots\ddots\ddots\ddots\\ &\ddots\ddots\ddots\ddots\ddots\ddots\ddots\\ \therefore l_1^2+l_3^2&=\frac{r^2+1}{(r+1)^2}l_2^2. \end{split} \end{equation*}$

Takeaway

In the equation

$l_1^2+l_3^2=\frac{r^2+1}{(r+1)^2}l_2^2,$

the side whose length is $l_2$ plays a crucial role in the theory of triangles with slopes $a,ar,ar^2$ . In particular, it acts like the hypotenuse of a right triangle.

(For something totally irrelevant, we would like to point out that the title of this post — approximate pythagorean identity — reminds us of APIs, a truly endearing term in programming that abbreviates Application Programming Interface. We lve APIs!!!!!!!!!!)

Tasks

(Number theory) A triangle has non-zero side slopes, and its side lengths $l_1,l_2,l_3$ are related via $l_1^2+l_3^2=841 l_2^2$ . Find possible coordinates for the vertices of the triangle.
(Observe that $841=29^2=20^2+21^2$ . And, in our example, we had $25=5^2=3^2+4^2$ . The quantity $\frac{1+r^2}{(1+r)^2}$ that appeared in (1) is always a sum of two consecutive squares, if it is an integer. But it is not always a perfect square. Thus, this closely relates to a problem in number theory where integer solutions to equations of the form $n^2+(n+1)^2=u^2$ are sought.)
For $r\geq 0$ , PROVE that $\frac{1}{2}\leq \frac{1+r^2}{(1+r)^2}\leq 1$ .
Let $-1\neq r\in\mathbb{Q}$ . PROVE that $\frac{1+r^2}{(1+r)^2}$ is an integer if and only if $r$ is of the form $-\frac{n}{n+1}$ or $-\frac{n+1}{n}$ , for some (positive) integer $n$ .
Let $-1\neq r\in\mathbb{Q}$ . PROVE that the integer values of $\frac{1+r^2}{(1+r)^2}$ are always of the form $n^2+(n+1)^2$ .
PROVE that the roots of the quadratic equation $(k-1)r^2+(2k)r+k-1=0$ are reciprocals of each other.
(Alternatively, this means that if $r\neq 0$ , then both $r$ and $\frac{1}{r}$ yield the same values for the quantity $\frac{1+r^2}{(1+r)^2}$ .)
Let $t\geq 0$ be an integer. PROVE that $8t+1$ is a perfect square if and only if $t=\frac{n(n+1)}{2}$ for some (positive) integer $n$ .
(Quadratic proximity) Let be the slopes of the sides of a triangle.
- If the triangle is right isosceles, PROVE that $r^2+3r+1=0$ ;
- If the triangle is equilateral, PROVE that $r^2+4r+1=0$ .
In $\triangle ABC$ , let sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , respectively. PROVE that $m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC$ , where $m_A$ is the length of the median from vertex $A$ and $BC$ is the length of side $BC$ .
(In particular, when $r=0$ , we obtain $m_A=\frac{1}{2}BC$ , ignoring the negative sign. Thus, the length of the median to side $BC$ is half the length of $BC$ , similar to the median to the hypotenuse in a right triangle. But $r=0$ is not allowed here.)
Let .
- If $r\neq 0,1$ , PROVE that $M(r)=M\left(\frac{1}{r}\right)$ .
- If $r\in\mathbb{Z}$ , PROVE that $M(r)\in\mathbb{Z}$ if and only if $r\in\{-2,0,2,4\}$ .
- If $r\in\mathbb{Q}$ , PROVE that $M(r)\in\mathbb{Z}$ if and only if $r$ has the form $\frac{n+2}{n-1}$ or $\frac{n-1}{n+2}$ for some integer $n$ .
Find exact solutions to the equation $\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}=\frac{r^2+1}{(r+1)^2}$ .
(Later on we’ll show how $\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}$ appears in median lengths of triangles with slopes in geometric progression.)