/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Show that \(n\) lines separate t... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that \(n\) lines separate the plane into \(\left(n^{2}+n+2\right) / 2\) regions if no two of these are parallel and no three pass through a common point.

Short Answer

Expert verified
n lines separate the plane into \(\frac{n^2 + n + 2}{2}\) regions.

Step by step solution

01

Understanding the Problem

We need to show that if we have n lines on a plane where no two lines are parallel and no three lines intersect at the same point, they will divide the plane into \(\frac{n^2 + n + 2}{2}\) regions.
02

Base Case

For the base case, consider n = 0 line. If there are no lines, the whole plane is one region. Thus, when n = 0, the number of regions is 1.
03

Second Case

Consider n = 1 line. A single line divides the plane into 2 regions.
04

Third Case

For n = 2 lines, they intersect at one point (since no two lines are parallel). These two lines divide the plane into 4 regions.
05

Using Induction

We will use mathematical induction to prove the formula for general n. Assume the formula holds for n lines; that is, the plane is divided into \(\frac{n^2 + n + 2}{2}\) regions. We need to prove it for n + 1 lines.
06

Adding the (n+1)th Line

Adding the (n+1)th line will intersect all n existing lines at n points. Each intersection creates a new region, thus adding n + 1 new regions.
07

Regions Calculation

If the number of regions with n lines is given by \(\frac{n^2 + n + 2}{2}\), then when adding another line, we must add n+1 regions: \(\frac{n^2 + n + 2}{2} + (n + 1)\).
08

Simplifying the Expression

Simplify \(\frac{n^2 + n + 2}{2} + n + 1\) to get: \[ \frac{n^2 + n + 2 + 2n + 2}{2} = \frac{n^2 + 3n + 4}{2} \]. Notice that this matches the original formula with n replaced by n+1.
09

Conclusion

By induction, the formula holds for all n. Therefore, n lines separate the plane into \(\frac{n^2 + n + 2}{2}\) regions if no two of these are parallel and no three pass through a common point.

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Key Concepts

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

induction in mathematics
Mathematical induction is a powerful method of proof. It involves two main steps: the base case and the inductive step.
The base case verifies that the statement holds true for the initial value, usually n = 0 or n = 1.
The inductive step assumes the statement is true for some n = k and then proves it for n = k + 1.
This chain of reasoning ensures that if the statement is true for one case, it must be true for the next, and so on.
In our problem, we start by proving the base cases for n = 0, 1, and 2 lines.
We then show that if our formula holds for some number k lines, it must also hold for k + 1.
This completes the proof by induction.
plane geometry
Plane geometry deals with shapes and figures on a flat surface, called a plane.
In our problem, we consider the geometric effect of lines on a plane.
When we add lines to a plane, we are interested in how these lines divide the plane into smaller areas.
This division depends heavily on the arrangement of the lines.
For simplicity, the problem assumes no parallel lines and no three lines intersecting at a single point.
These constraints ensure the maximum number of regions are formed.
combinatorial geometry
Combinatorial geometry studies the combination of different geometric objects.
Our problem falls into this category because we are combining lines in certain ways to determine the number of regions.
When lines are added, the number of regions changes based on the intersection points.
By understanding how these combinations work, we can derive formulas to describe their behavior.
The formula \(\frac{n^2 + n + 2}{2}\) precisely captures the combinatorial complexity of this problem.
non-parallel lines
In plane geometry, lines can be parallel or non-parallel. Parallel lines never meet, while non-parallel lines intersect at one point.
Our problem focuses on non-parallel lines to maximize intersections and, thus, the number of regions.
If two lines are non-parallel, they intersect at exactly one point.
Adding more non-parallel lines increases the number of intersection points and thereby increases the number of regions.
These intersections split existing regions into smaller ones, explaining the increase in regions with each additional line.
intersection points
Intersection points are where two or more lines meet.
A key part of solving our problem is understanding how new lines create new intersection points.
Each new non-parallel line added to the plane intersects all existing lines, creating new regions.
Specifically, adding the \(n + 1\)-th line intersects n existing lines, creating \ n + 1 \ new regions.
The number of these intersection points and regions can be calculated accurately using combinatorial geometry principles.

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Most popular questions from this chapter

Verify that the program segment $$ \begin{array}{c}{x :=2} \\ {z :=x+y} \\ {\text { if } y>0 \text { then }} \\\ {z :=z+1} \\ {\text { else }} \\ {z :=0}\end{array} $$ is correct with respect to the initial assertion \(y=3\) and the final assertion \(z=6\)

Use the principle of mathematical induction to show that \(P(n)\) is true for \(n=b, b+1, b+2, \ldots,\) where \(b\) is an integer, if \(P(b)\) is true and the conditional statement \(P(k) \rightarrow\) \(P(k+1)\) is true for all integers \(k\) with \(k \geq b\) .

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Pick's theorem says that the area of a simple polygon \(P\) in the plane with vertices that are all lattice points (that is, points with integer coordinates) equals \(I(P)+B(P) / 2-1\) where \(I(P)\) and \(B(P)\) are the number of lattice points in the interior of \(P\) and on the boundary of \(P,\) respectively. Use strong induction on the number of vertices of \(P\) to prove Pick's theorem. [Hint: For the basis step, first prove the theorem for rectangles, then for right triangles, and finally for all triangles by noting that the area of a tri- angle is the area of a larger rectangle containing it with the areas of at most three triangles subtracted. For the inductive step, take advantage of Lemma \(1 . ]\)

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