Approximate Pythagorean Identity

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).

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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).

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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

  1. (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.)
  2. For r\geq 0, PROVE that \frac{1}{2}\leq \frac{1+r^2}{(1+r)^2}\leq 1.
  3. 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.
  4. 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.
  5. 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}.)
  6. 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.
  7. (Quadratic proximity) Let a,ar,ar^2 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.
  8. 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.)
  9. Let M(r)=\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}.
    • 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.
  10. 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.)