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Let \(\mathrm{T}_{n}\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(\mathrm{T}_{n+1}-\mathrm{T}_{n}=10\), then the value of \(n\) is : (a) 7 (b) 5 (c) 10 (d) 8

Short Answer

Expert verified
The value of \(n\) is 5.

Step by step solution

01

Understand the Problem

We are given an equation involving the difference in the number of triangles that can be formed from a regular polygon with \(n+1\) sides and \(n\) sides. We need to find \(n\) for which this difference is 10.
02

Determine the Expression for \(\mathrm{T}_n\)

The number of triangles that can be formed using the vertices of an \(n\)-sided polygon is given by \(\mathrm{T}_n = \binom{n}{3}\), where \(\binom{n}{3}\) is the number of combinations of \(n\) vertices taken 3 at a time.
03

Set Up the Equation

We are given that \(\mathrm{T}_{n+1} - \mathrm{T}_n = 10\). Therefore, we have the equation: \[ \binom{n+1}{3} - \binom{n}{3} = 10. \]
04

Simplify the Binomials

Substitute the binomial expressions: \(\binom{n+1}{3} = \frac{(n+1)n(n-1)}{6}\) and \(\binom{n}{3} = \frac{n(n-1)(n-2)}{6}\).
05

Calculate the Difference

Simplifying the equation: \[ \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 10. \] This gives: \( \frac{n(n-1)((n+1)-(n-2))}{6} = 10 \).
06

Solve the Simplified Equation

Further simplification gives: \[ \frac{n(n-1) \cdot 3}{6} = 10. \] Which reduces to \[ \frac{n(n-1)}{2} = 10. \] Then multiply both sides by 2: \[ n(n-1) = 20. \]
07

Find the Integer Solution for \(n\)

Solve the quadratic equation \(n^2 - n - 20 = 0\) by factoring it as \((n-5)(n+4)=0\). This results in solutions \(n=5\) and \(n=-4\). Only the positive integer \(n=5\) is valid as \(n\geq3\) for a polygon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
The binomial theorem is a cornerstone of combinatorics and provides a way to expand expressions that are raised to a power.
In its simplest form, it is expressed as \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This expression shows how to expand powers of binomials with coefficients provided by the binomial coefficients, \(\binom{n}{k}\), also known as "n choose k." The coefficient represents the number of ways to choose \(k\) elements from a set of \(n\) elements.
In the problem of triangle formation in polygons, binomial coefficients are used to determine the number of triangles that can be formed from an \(n\)-sided polygon by selecting 3 vertices at a time.
Thus, the binomial expression \(\mathrm{T}_n = \binom{n}{3}\) gives us a quick and efficient method to compute the number of such triangles without listing each combination.
  • This reduces the problem-solving to arithmetic using coefficients rather than counting every possible triangle individually.
Understanding this theorem aids immensely in combinatorial counting problems and provides a bridge between algebra and geometry.
Polygons
A polygon, simply put, is a flat shape with straight sides. It is an essential concept in geometry, especially when dealing with figures like triangles, squares, hexagons, and more.
A regular polygon has all sides and angles equal, which gives it symmetry and simplicity. The sides determine the name of the polygon: a 5-sided polygon is a pentagon, a 6-sided one is a hexagon, etc.
When considering the challenge of forming triangles within a polygon, the regular nature ensures that any three chosen vertices will always appear to form a legitimate triangle as long as \(n \geq 3\).
  • This concept is powerful as it translates geometric shapes into combinatorial problems, facilitating calculations like those using binomial coefficients.
  • The unique property of choosing any three vertices to form distinct triangles is only possible because of the regular polygon's equal distribution of vertices.
This fork of mathematics serves as the basis for further geometric exploration and learning about angles, symmetry, and mathematical aesthetics.
Triangle Formation
The formation of triangles within a polygon involves selecting three vertices that do not lie on a single straight line.
For an \(n\)-sided polygon, combinations of vertices can be calculated using binomial coefficients, specifically \(\binom{n}{3}\).
  • Since a triangle is a three-sided polygon, every combination of three vertices forms one triangle.
  • The calculation ensures that all possible triangles are unique and covers every possible choice of three points.
Using a regular polygon, every distinct triad of vertices reliably forms a triangle. This method excludes degenerate cases where three points might be collinear, as happens in other geometric arrangements.
This process highlights the intersection of geometry and combinatorics, providing insight into both visible patterns and abstract calculations.
By using the principles of triangle formation, concepts of geometry become quantifiable, allowing for algebraic manipulation and solutions to geometric problems.

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Most popular questions from this chapter

Consider three boxes, each containing 10 balls labelled \(1,2, \ldots, 10 .\) Suppose one ball is randomly drawn from each of the boxes. Denote byn \(_{i}\), the label of the ball drawn from the \(l^{\text {th }}\) box, \((i=1,2,3)\). Then, the number of ways in which the balls can be chosen such that \(\mathrm{n}_{1}<\mathrm{n}_{2}<\mathrm{n}_{3}\) is : (a) 120 (b) 82 (c) 240 (d) 164

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