Consider ellipses Ek : kx² + k²y² = 1, k = 1, 2, ..., 20. Let Ck be the circle which touches the fou

Q: Wait, why is the center of the circle at the origin? Can't it be somewhere else?

That is a fantastic question! Because the ellipse E_k is centered at the origin and its axes are aligned with the coordinate axes, the four chords are symmetric with respect to both the x and y axes. A circle that touches all four of these symmetric chords must share that same center of symmetry. Therefore, the center must be (0,0).

Q: Why can't we just use the area of the rhombus to find the radius?

You absolutely can! The area of a rhombus is (1/2) * d_1 * d_2, where d_1 and d_2 are the lengths of the diagonals. Alternatively, the area is also equal to (Perimeter * r) / 2. Both methods will lead you to the same result. The perpendicular distance method is just a more direct path for this specific coordinate geometry setup.

Q: Is there a trick to summing k + k^2 from 1 to 20?

There is no 'trick' per se, but there is a very elegant way to handle it! You simply split the summation into two parts: the sum of the first n natural numbers, which is n(n+1)/2, and the sum of the squares of the first n natural numbers, which is n(n+1)(2n+1)/6. Just be careful with your arithmetic when plugging in n=20!

Question

Consider ellipses $E_k : kx^2 + k^2y^2 = 1, k = 1, 2, \dots, 20$. Let $C_k$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $E_k$. If $r_k$ is the radius of the circle $C_k$, then the value of $\sum_{k=1}^{20} \frac{1}{r_k^2}$ is

Options:

(A) 3080

(B) 2870

(C) 3210

(D) 3320

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a collection of twenty distinct ellipses, all born from the same elegant algebraic seed: $kx^2 + k^2y^2 = 1$. As $k$ ranges from $1$ to $20$, these ellipses shift and morph, yet they all share a common, beautiful structure.

Our journey today is to uncover the hidden circle $C_k$ that nestles perfectly within each of these ellipses, touching the chords that connect their major and minor axes. This is not just a problem of calculation; it is a problem of symmetry and discovery.

### Standardizing the Ellipse

To understand the soul of our ellipse $E_k$, we must first bring it into the standard form. The equation $kx^2 + k^2y^2 = 1$ can be rewritten by dividing both sides by the coefficients to isolate the variables:

$$
\frac{x^2}{1/k} + \frac{y^2}{1/k^2} = 1
$$

Comparing this to the classic form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we immediately see that our semi-major axis $a$ is $1/\sqrt{k}$ and our semi-minor axis $b$ is $1/k$. These are the boundaries of our world, with vertices lying at $(\pm 1/\sqrt{k}, 0)$ and $(0, \pm 1/k)$.

### The Rhombus and the Chord

When we connect these four vertices, we create a rhombus. This shape is the stage for our inscribed circle $C_k$. Let us focus on the chord in the first quadrant.

Using the intercept form of a line, $\frac{x}{a} + \frac{y}{b} = 1$, we substitute our values:

$$
\frac{x}{1/\sqrt{k}} + \frac{y}{1/k} = 1
$$

With a little algebraic grace, this simplifies to $\sqrt{k}x + ky = 1$, or in the standard form for distance calculations:

$$
\sqrt{k}x + ky - 1 = 0
$$

### The Radius of the Inscribed Circle

Now, we seek the radius $r_k$ of the circle $C_k$ that touches this chord. Because of the perfect symmetry of the rhombus, the center of this circle is the origin $(0,0)$.

The radius is simply the perpendicular distance from the origin to our line $\sqrt{k}x + ky - 1 = 0$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, we find:

$$
r_k = \frac{|\sqrt{k}(0) + k(0) - 1|}{\sqrt{(\sqrt{k})^2 + k^2}} = \frac{1}{\sqrt{k + k^2}}
$$

This is the geometric heart of the problem. Squaring this radius gives us $r_k^2 = \frac{1}{k + k^2}$, which implies that:

$$
\frac{1}{r_k^2} = k + k^2
$$

### The Grand Summation

We have arrived at the final act. We must calculate the sum $S = \sum_{k=1}^{20} \frac{1}{r_k^2} = \sum_{k=1}^{20} (k + k^2)$. By the linearity of summation, we can split this into two manageable parts:

$$
\sum_{k=1}^{20} k + \sum_{k=1}^{20} k^2
$$

The first part is the sum of the first $20$ natural numbers:

$$
\frac{20(21)}{2} = 210
$$

The second part is the sum of the squares of the first $20$ natural numbers:

$$
\frac{20(21)(41)}{6} = 2870
$$

Adding these two results together, $210 + 2870$, we reach our destination. The final result is **3080**.

Select Answer:

Visualized Solution

The Ellipse Equation $E_{k}$

Standard Form of Ellipse

Identifying Semi-axes $a$ and $b$

Drawing the Four Chords

Equation of the Chord

The Inscribed Circle $C_{k}$

Radius $r_{k}$ as Perpendicular Distance

Calculating Radius $r_{k}$

Simplifying $\frac{1}{r _{k}^{2}}$

Summation Setup

Sum of First $n$ Natural Numbers

Sum of Squares of First $n$ Numbers

The Final Result

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Standardizing the Ellipse

The Rhombus and the Chord

The Radius of the Inscribed Circle

The Grand Summation

Similar Questions

The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre $C$ of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius $r$ . Let $f_{1}, f_{2}$ be the focal distances of the point $C$ on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to

The ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0, 4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is