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Let us consider rectangles.

Among the three numbers associated with rectangle is the perimeter, the long side, and the short side.

If a rectangle is a square, the ratio of the perimeter to the long side is
4, while the ratio of the long side to the short side is
1

For a rectangle with short side 0.5 and long side
100, then the perimeter is
201 The ratio of the perimeter to the long side is 2.01, and the long side to short side is 200.

Sometimes the ratio of the perimeter to the long side is greater than the ratio of the long side to the short side, like squares for example. Other times, the ratio of the long side to the short side is much greater than the ratio of the perimeter to the long side.

So it must be possible for a rectangle to have side lengths such that the ratio of the perimeter of the long side equal to he ratio of the long side to the short side.

Is this ratio rational?

Bill Dubuque
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Michael Ejercito
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2 Answers2

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Answer posted since the OP had questions about the answer of Michael Ejercito. My approach is almost exactly the same as his.

Assume that the long side is $~L,~$ and the short side is $~S.~$ Then the perimeter is $~2L + 2S,~$ and you are given that

$$\frac{2L+2S}{L} = \frac{L}{S} \implies L^2 - 2LS - 2S^2 = 0 \implies $$

$$L = \frac{1}{2} \left[ ~2S \pm \sqrt{12S^2} ~\right] = S \times \left[ ~1 \pm \sqrt{3} ~\right] \implies $$

$$[\text{since} ~L,S ~\text{must both be positive}] ~~L = S \times [ ~1 + \sqrt{3} ~]. \tag1 $$

Assume, without loss of generality, that $~S~$ is rational. Then, since the second factor on the RHS of (1) above is irrational, $~L~$ must also be irrational.

user2661923
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  • To construct an Ejercito rectangle, construct a square and a 30-60-90 right triangle using one of the square's sides as the short side. From there on only a straightedge is needed to finish the rectangle. – Michael Ejercito Jan 29 '24 at 19:38
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Let us choose $P$ as the perimeter and $L$ as the length of the long side. Because we are assuming this ratio is rational, we choose a common unit of measure such that
$P$ and $L$ are coprime positive integers. thus, the ratio, $\frac P L $, is a fraction in lowest terms.

The ratio of the long side to the short side can be expressed in terms of $P$ and $L$ as $\frac {2L}{P-2L}$ ( $\frac{P-2L}{2}$ is the length of the short side, and we know that a rectangle can have an integer perimeter and long side, while the short side is a non-integer rational.)

Note that $2L$ is a positive integer strictly less than
$P$. (The perimeter is always strictly greater than twice the length of the long side. ) As such, $P-2L$ is strictly less than L. And from the closure of integers under subtraction, we know that $P-2L$ is an integer.

Thus, $\frac{2L}{P−2L}$ is a fraction in lower terms than $\frac P L$. But we already assumed that $P$ and $L$ are coprime. We have a contradiction, and this ratio must be irrational.

So what is this ratio? We can deduce:

$$P(P-2L)=L(2L)=P^2-2LP=2L^2$$

Subtracting $2L^2$ from both sides:

$$P^2-2LP-2L^2=0$$

Now let $L=1$ and substitute:

$$P^2-2P-2=0$$

This is a univariate quadratic equation, so we can easily solve for
$P$ .

$$P=1+\sqrt{3}$$

. This number is approximately
$2.7321$

(This also establishes that $\sqrt{3}$ is irrational . If $\sqrt{3}$ were rational, then $1+\sqrt {3}$ would also be rational, because the rationals are closed under addition. But we have already proven that $1+\sqrt {3}$ is irrational, so $\sqrt {3}$ must be irrational as well.)

These rectangles have short side
$S=1$,L=$1+\sqrt {3}$, and
$P=4+2\sqrt {3}$ ( is about 7.4641 ) The diagonal has a length of √(5+2√3), about
2.9093.

The area is the square of the short side multiplied by 1+√3

I can find nothing else online about these type of rectangles (where the ratio of the perimeter to the long side equals the ratio of the long side to the short side). I propose these rectangles be called Ejercito rectangles.

Michael Ejercito
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  • Do you mean "As such, $\frac{P−2L}2$ is strictly less than $L$" in the third paragraph? – Arthur Jan 28 '24 at 17:11
  • While that is true, I did mean that $2L$ is strictly less than $P$. Recall that $P=2L+2S$, and since both $L$ and $S$ are strictly positive real numbers, $P>2L$. It was necessary to state this to prove the contradiction and the irrationality of this ratio. – Michael Ejercito Jan 28 '24 at 17:32
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    You first got $L=1, P=1+\sqrt 3$, then how did these change to $L=1+\sqrt3, P=4+2\sqrt 3$? – peterwhy Jan 28 '24 at 17:44
  • By setting $S=1$ after the ratio was determined. If $L$ is set as 1, $S=\frac{1}{1+\sqrt{3}}$ – Michael Ejercito Jan 28 '24 at 18:16
  • I understand and agree with $P>2L$. What I don't understand is why you say $P-2L<L$. – Arthur Jan 28 '24 at 18:24
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    $\frac{P}{L}=\frac{2L}{P-2L}$. Let $\frac{x}{y}=\frac{m}{n}$. If $x>m$, then $y>n$. – Michael Ejercito Jan 28 '24 at 18:27
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    @MichaelEjercito Right, I see. I had forgotten we had made assumptions about the rectangle. – Arthur Jan 29 '24 at 10:07
  • Making those assumptions was necessary to reach the contradiction that proves that this ratio is irrational. (It is very similar to one of the proofs that the golden ratio is irrational.) – Michael Ejercito Jan 29 '24 at 16:55