# Understanding Geometry – 4

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Aleksej Ivanovič Markuševič’s book, “Remarkable Curves” discusses the properties of ellipses, parabolas, hyperbolas, lemniscates, cycloids, brachistochrone, spirals and catenaries.  Among these “lemniscates” are the ones that I encountered only once before starting undergraduate education (all other curves appeared frequently in physics textbooks) and that too just to calculate the area enclosed by this curve. So I will discuss the properties of lemniscates in this post.

Let’s begin with the well-known curve, ellipse. An ellipse is the locus of points whose sum of distances from two fixed points (called foci) is constant. My favourite fact about ellipses is that we can’t find a general formula for the perimeter of an ellipse, and this little fact leads to the magical world of elliptic integrals. This, in turn, leads to the mysterious elliptic functions, which were discovered as inverse functions of elliptic integrals. Further, these functions are needed in the parameterization of certain curves, now called elliptic curves. For more details about this story, read the paper by Adrian Rice and Ezra Brown, “Why Ellipses are not Elliptic curves“.

Lemniscate is defined as the locus of points whose product of distances from two fixed points $F_1$ and $F_2$ (called foci) is constant. Lemniscate means, “with hanging ribbons” in Latin.  If the length of the segment $\overline{F_1F_2}$ is $c$ then for the midpoint of this line segment will lie on the curve if the product constant is $c^2/4$. In this case we get a figure-eight lying on its side.

Lemniscate of Bernoulli; By Kmhkmh (Own work) [CC BY 4.0], via Wikimedia Commons

The attempt to calculate the perimeter of the above curve leads to elliptic integral, hence can’t derive a general formula for its perimeter. Just like an ellipse!

If we equate the value of the constant product not to $c^2/4$ but to another value, the lemniscate will change its shape. When the constant is less than $c^2/4$, the lemniscate consists of two ovals, one of which contains inside it the point $F_1$, and the other the point $F_2$.

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700b.gif

When the product constant is greater than $c^2/4$ but less than $c^2/2$, the lemniscate has the form of a biscuit. If the constant is close to $c^2/4$, the “waist” of the biscuit is very narrow and the shape of the curve is very close to the figure-eight shape.

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700b.gif

If the constant differs little from $c^2/2$, the waist is hardly noticeable, and if the constant is equal or greater than $c^2/2$ the waist disappears completely, and the lemniscate takes the form of an oval.

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700a.gif

We can further generalize this whole argument to get lemniscate with an arbitrary number of foci, called polynomial lemniscate.

# Arithmetic & Geometry

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After a month of the unexpected break from mathematics, I will resume the regular weekly blog posts. It’s a kind of relaunch of this blog, and I  will begin with the discussion of an arithmetic problem with a geometric solution.  This is problem 103 from  The USSR Olympiad Problem Book:

Prove that
$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$
where $t_k$ is the number of divisors of the natural number $k$.

One can solve this problem by using the principle of mathematical induction or using the fact that the number of integers of the sequence $1,2,3,\ldots , n$ which are divisible by any chosen integer $k$ is equal to $\left\lfloor \frac{n}{k}\right\rfloor$. The second approach of counting suggests an elegant geometric solution.

Consider the equilateral hyperbola $\displaystyle{y = \frac{k}{x}}$ , (of which we shall take only the branches in the first quadrant):

We note all the points in the first quadrant which have integer coordinates as the intersection point of the dotted lines. Now, if an integer $x$ is a divisor of the integer $k$, then the point $(x,y)$ is a point on the graph of the hyperbola $xy=k$. Conversely, if the hyperbola $xy=k$ contains an integer point, then the x-coordinate is a divisor of $k$. Hence the number of integers $t_k$ is  equal to the number of the integer points lying on the hyperbola $xy=k$. The number $t_k$ is thus equal to the number of absciassas of integer points lying on the hyperbola $xy=k$. Now, we make use of the fact that the hyperbola $xy=n$ lies “farther out” in the quadrant than do $xy=1, xy=2, xy=3, \ldots, xy=n-1$. The following implication hold:

The sum $t_1+t_2\ldots + t_n$ is equal to the number of integer points lying under or on the hyperbola $xy=n$. Each such point will lie on a hyperbola $xy=k$, where $k\leq n$. The number of integer points with abscissa $k$ located under the hyperbola is equal to the integer part of the length of the segment $\overline{AB}$ [in figure above $k=3$]. That is $\left\lfloor\frac{n}{k}\right\rfloor$, since $\displaystyle{|\overline{AB}|=\frac{n}{k}}$, i.e. ordinate of point $A$ on hyperbola $xy=n$ for abscissa $k$. Thus, we obtain

$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$

Caution: Excess of anything is harmful, even mathematics.