Problem 164 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(T_{n}\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(T_{n+1}-T_{n}=10\), then the value of \(n\) is (A) 5 (B) 10 (C) 8 (D) 7

Step-by-Step Solution

Answer

The value of \(n\) is 5.

Step 1: Understand the Problem

The problem requires us to find the value of \(n\) for which the difference in the number of triangles formed by the vertices of two consecutive regular polygons (one with \(n+1\) sides and one with \(n\) sides) is 10.

Step 2: Recall the Formula for Triangles in a Polygon

The number of triangles that can be formed by the vertices of an \(n\)-sided polygon is given by the combination formula \( \binom{n}{3} \). This represents the number of ways to choose 3 vertices out of \(n\) to form a triangle.

Step 3: Express the Given Condition Mathematically

We have the expression \( T_{n+1} - T_{n} = 10 \). Using the formula from Step 2, this becomes \( \binom{n+1}{3} - \binom{n}{3} = 10 \).

Step 4: Calculate \( \binom{n+1}{3} \) and \( \binom{n}{3} \)

The expression for \( \binom{n+1}{3} \) is \( \frac{(n+1)n(n-1)}{6} \), and for \(\binom{n}{3}\), it is \( \frac{n(n-1)(n-2)}{6} \).

Step 5: Simplify the Difference Expression

Subtract \(\binom{n}{3}\) from \(\binom{n+1}{3}\), which gives:\[ \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 10 \].Simplify it to obtain:\[ \frac{n(n-1)((n+1)-(n-2))}{6} = 10 \].

Step 6: Further Simplify to Find \(n\)

The simplified form is \[ \frac{n(n-1)(3)}{6} = 10 \], which reduces to:\[ \frac{n(n-1)}{2} = 10 \].Multiply both sides by 2 to get:\[ n(n-1) = 20 \].

Step 7: Solve the Quadratic Equation

The equation \( n^2 - n - 20 = 0 \) is a quadratic equation. Solve using the quadratic formula, factoring, or inspection to find root candidates. Factoring gives \( (n-5)(n+4)=0 \), so \( n=5 \) or \( n=-4 \). Since \( n \) must be positive, \( n=5 \).

Key Concepts

Combinatorics

In combinatorics, we explore how to count, arrange, and select objects. When it comes to forming triangles from the vertices of a polygon, combinatorics plays a crucial role. The number of triangles that can be formed from a polygon's vertices is determined by selecting any three out of the total number of vertices. This is where the binomial coefficient or combination formula is handy. It's denoted as \( \binom{n}{k} \), which reads as "n choose k." Here, \( n \) is the total number of objects, and \( k \) is the number of objects to choose.
For triangles in a polygon, \( k = 3 \), since we're choosing 3 vertices. So, for an \( n \)-sided polygon, the formula is \( \binom{n}{3} \). This formula calculates how many different ways you can pick 3 vertices from \( n \) vertices. The formula is:

• \( \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \)

This equation tells you the total number of triangles possible from an \( n \)-sided polygon's vertices.

Regular Polygons

A regular polygon is a geometric figure with equal sides and angles. These polygons are significant in both geometry and combinatorics because they offer symmetry and uniformity, simplifying calculations. If you consider an \( n \)-sided regular polygon, it implies that all its sides and its internal angles are equal.
When analyzing problems involving regular polygons, like finding the number of triangles that can be formed by their vertices, the polygons’ regular nature ensures that each vertex and edge are identical in characteristic. This simplifies solving problems and ensures consistent results. Regular polygons exhibit a natural order that makes calculating properties such as the number of triangles or diagonals straightforward.

• The formula to calculate triangles using regular polygon vertices is derived from combina.
• Properties like rotational and reflective symmetries simplify complex geometric problems.

Using these properties, complex problems can be broken down into simpler components, making mathematical analysis more manageable.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, commonly written in the form \( ax^2 + bx + c = 0 \). Solving these equations is a fundamental skill in algebra, and they frequently appear in various mathematical contexts, including geometry and combinatorics.
In solving combinatorial problems involving regular polygons, quadratic equations arise naturally. For example, determining \( n \) such that the number of triangles formed by a polygon’s vertices meets a certain condition leads to forming and solving a quadratic equation. In our case, we derived the equation \( n^2 - n - 20 = 0 \) from combinatorial calculations.
There are multiple strategies to solve a quadratic equation:

• Factoring: If the equation can be factored easily, like \( (n-5)(n+4) = 0 \), it's a straightforward process to find \( n = 5 \) or other roots.
• Quadratic Formula: If factoring is not feasible, the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides a reliable method.
• Completing the Square: This method involves rearranging and modifying the equation to create a perfect square trinomial.

Understanding these solutions can help you tackle not only geometric and combinatorial problems but many areas in mathematics. Quadratic equations are a versatile tool in the mathematician's toolkit.