Let T_n denote the number of triangles which can be formed using the vertices of a regular polygon o

Q: Wait, why do we use combinations instead of permutations? Doesn't the order of vertices matter?

That is a fantastic question! In geometry, a triangle defined by vertices $A, B, C$ is the exact same triangle as one defined by $B, A, C$ or $C, B, A$. Since the order of selection does not change the resulting shape, we use combinations, where order is irrelevant. If we used permutations, we would be overcounting by a factor of $3! = 6$.

Q: Why is $n=7$ the only answer? What about $n=-6$?

In the world of pure algebra, $n^2 - n - 42 = 0$ yields both $7$ and $-6$. However, we are dealing with a physical polygon. The number of sides $n$ must be a positive integer, and specifically, you need at least $3$ sides to form a polygon. Thus, we reject the negative root as it has no physical meaning in this context.

Q: Is there a faster way to solve $\binom{n}{2} = 21$ without the quadratic formula?

Absolutely! You can use logical estimation. We know $\binom{n}{2} = \frac{n(n-1)}{2} = 21$, so $n(n-1) = 42$. We are looking for two consecutive integers that multiply to $42$. Since $6 \times 7 = 42$, it is immediately clear that $n=7$. Developing this 'number sense' is a great way to save time in the exam hall!

Question

Let $T_n$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n+1} - T_n = 21$, then $n$ equals

Options:

(A) 5

(B) 7

(C) 6

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open space, and before you lies a circle. On the boundary of this circle, you place $n$ distinct points. These points are the vertices of a regular polygon.

Your task is to determine how many unique triangles you can form by connecting these vertices. To form a triangle, you need exactly three vertices.

Since this is a regular polygon, no three vertices are collinear. Every single combination of three vertices will yield a valid, non-degenerate triangle. Mathematically, we express this as:

$$T_n = \binom{n}{3}$$

### The Ripple Effect of a New Vertex

Now, let us introduce a change. Suppose we add one more vertex to our polygon, transforming it from an $n$-sided polygon to an $(n+1)$-sided one. The number of triangles we can now form is:

$$T_{n+1} = \binom{n+1}{3}$$

The difference $T_{n+1} - T_n$ represents the number of **new** triangles created by the addition of that single vertex. Every one of these new triangles must include the new vertex.

To complete such a triangle, we only need to select two more vertices from the original $n$ vertices. Thus, the number of new triangles is simply $\binom{n}{2}$.

### The Algebraic Bridge

We are given that the difference $T_{n+1} - T_n = 21$. By our geometric reasoning, this means:

$$\binom{n}{2} = 21$$

We can also arrive at this using the powerful tool of **Pascal's Identity**, which states that $\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}$. If we rearrange this, we get:

$$\binom{n+1}{r} - \binom{n}{r} = \binom{n}{r-1}$$

Setting $r=3$, we see that $\binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$. The algebra confirms our geometric intuition perfectly.

### The Quadratic Resolution

Now, we descend into the final phase of our journey: solving for $n$. We know that:

$$\binom{n}{2} = \frac{n(n-1)}{2}$$

Setting this equal to $21$, we get:

$$\frac{n(n-1)}{2} = 21 \Rightarrow n(n-1) = 42$$

Expanding this, we arrive at the quadratic equation:

$$n^2 - n - 42 = 0$$

We need two numbers that multiply to $-42$ and add to $-1$. Those numbers are $-7$ and $6$. Thus, the equation factors into:

$$(n-7)(n+6) = 0$$

This gives us two potential roots: $n=7$ and $n=-6$. Since a polygon cannot have a negative number of sides, we discard $n=-6$.

We are left with the triumphant result: **$n=7$**. Our polygon is a heptagon.

Animated Solution for Mathematics - Permutations and Combinations: Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals

Select Answer:

Visualized Solution

Visualizing the Polygon and Triangles

The Combination Formula for $T_{n}$

Adding a Vertex to get $T_{n + 1}$

Geometric Meaning of $T_{n + 1} - T_{n}$

Setting up the Equation

Applying Pascal's Identity

Simplifying the Equation

Expanding the Combination

Forming the Quadratic Equation

Factorizing the Quadratic Equation

Finding the Roots

Selecting the Valid Solution

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Ripple Effect of a New Vertex

The Algebraic Bridge

The Quadratic Resolution

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