Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon

Q: Wait, why do we use combinations (${}^nC_r$) instead of permutations (${}^nP_r$)?

That is a brilliant question! In geometry, a triangle defined by vertices A, B, and C is exactly the same as one defined by B, C, and A. Since the order of picking the vertices does not change the resulting shape, we use combinations to count unique sets of vertices, not ordered arrangements.

Q: Is there a faster way to solve $(n+1)n(n-1)(n-2) = 3024$ without expanding it into a quartic equation?

Absolutely! Solving a quartic equation is a nightmare. Instead, look at the product of four consecutive integers. Since $3024$ is roughly $9^4$ (which is $6561$) or $7^4$ (which is $2401$), you can estimate the value of $n$ by testing integers around $8$. It is much faster and less prone to algebraic errors.

Q: Why is the eccentricity formula $e = \sqrt{1 - \frac{b^2}{a^2}}$ and not $e = \sqrt{1 - \frac{a^2}{b^2}}$?

The eccentricity $e$ must always be between $0$ and $1$. Therefore, the fraction subtracted from $1$ must be less than $1$. This means we must always divide the square of the semi-minor axis by the square of the semi-major axis. Always check which denominator is larger!

Q: Does the value of $n$ have to be an integer?

Yes, it must! Since $n$ represents the number of sides of a polygon, it must be a positive integer greater than or equal to $3$. If you ever calculate a non-integer $n$, it is a clear sign to re-check your combination arithmetic.

Q: What if the ellipse equation was $\frac{x^2}{8} + \frac{y^2}{16} = 1$? Would the eccentricity change?

The eccentricity value itself would remain the same, but the orientation of the ellipse would flip! The major axis would now be along the y-axis. The formula for $e$ remains $\sqrt{1 - \frac{\text{minor}^2}{\text{major}^2}}$, so the ratio of the squares of the axes is what truly dictates the 'flatness' of the ellipse.

Question

Let $p$ be the number of all triangles that can be formed by joining the vertices of a regular polygon $P$ of $n$ sides and $q$ be the number of all quadrilaterals that can be formed by joining the vertices of $P$. If $p + q = 126$, then the eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{n} = 1$ is :

Options:

(A) $\frac{3}{4}$

(B) $\frac{1}{2}$

(C) $\frac{\sqrt{7}}{4}$

(D) $\frac{1}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a regular polygon with $n$ vertices. It is a perfect, symmetric structure. You are tasked with a simple yet profound challenge: how many ways can we draw triangles and quadrilaterals using only these vertices?

To form a triangle, we must choose any $3$ vertices from the $n$ available. Since the order of selection does not matter, we use the combination formula: $p = {}^nC_3$.

Similarly, to form a quadrilateral, we select $4$ vertices, giving us $q = {}^nC_4$. We are told that the sum of these possibilities is $126$, or mathematically:

$${}^nC_3 + {}^nC_4 = 126$$

### The Elegance of Pascal’s Identity

When you see the sum of two combinations with consecutive indices, your mind should immediately jump to the legendary Pascal's Identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$. This is one of the most powerful tools in a mathematician's arsenal.

By applying this to our equation, we collapse the two terms into a single, elegant expression: ${}^{n+1}C_4 = 126$. Suddenly, the complexity vanishes.

Expanding this, we get:

$$\frac{(n+1)n(n-1)(n-2)}{4!} = 126$$

Since $4! = 24$, we arrive at the product of four consecutive integers:

$$(n+1)n(n-1)(n-2) = 126 	imes 24 = 3024$$

### The Hunt for $n$

Do not rush to expand this into a quartic polynomial, as that is a trap that leads to messy calculations. Instead, let us test values for four consecutive integers that multiply to $3024$.

We quickly find that $9 	imes 8 	imes 7 	imes 6 = 3024$. By comparing this to our expression $(n+1)n(n-1)(n-2)$, we immediately see that $n+1 = 9$, which means **$n = 8$**.

### The Final Transformation: Into the Ellipse

With $n = 8$ in hand, we turn our attention to the ellipse:

$$\frac{x^2}{16} + \frac{y^2}{8} = 1$$

We identify the semi-axes by comparing this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here, $a^2 = 16$ and $b^2 = 8$.

Since $a^2 > b^2$, the major axis lies along the x-axis. The eccentricity $e$ measures how 'stretched' our ellipse is, defined by the formula:

$$e = \sqrt{1 - \frac{b^2}{a^2}}$$

Substituting our values, we get:

$$e = \sqrt{1 - \frac{8}{16}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

The final eccentricity of the ellipse is **$\frac{1}{\sqrt{2}}$**.

Select Answer:

Visualized Solution

Problem Overview

Defining $p$ (Triangles)

Defining $q$ (Quadrilaterals)

Setting up the Equation

Pascal's Identity

Expanding the Combination

Simplifying the Equation

Factorization Trick

Finding $n$

The Ellipse Equation

Identifying Semi-Axes

Eccentricity Formula

Calculating $e$

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Elegance of Pascal’s Identity

The Hunt for $n$

The Final Transformation

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