1 problem found
A plane contains \(n\) distinct given lines, no two of which are parallel, and no three of which intersect at a point. By first considering the cases \(n=1,2,3\) and \(4\), provide and justify, by induction or otherwise, a formula for the number of line segments (including the infinite segments). Prove also that the plane is divided into \(\frac{1}{2}(n^{2}+n+2)\) regions (including those extending to infinity).
Solution: With \(n=1\) line, the plane is divided in half. With \(n=2\) lines the plane is divided into four pieces. (Each of the previous pieces are split in half) With \(n=3\) lines the plane is divided into up to \(7\) pieces. (The new line crosses two lines in two places dividing \(3\) regions into \(2\), thus increasing the number of regions by \(3\)). With \(n=4\) lines the plane is divided into \(11\) pieces. (The new line crosses three lines in three places doubling the number of regions of \(4\) places). Claim: With \(n\) lines the plane is divided into \(\frac12(n^2+n+2)\) regions. Proof: (By induction) (Base case) When \(n=1\) clearly the line is divided into \(2\) regions, and \(\frac12 (1^2 + 1^2 + 2) = 2\) so the base case is true. (Inductive step) Suppose our formula is true for \(n=k\), so we have placed \(k\) lines in general position and divided the plane into \(\frac12(k^2+k+2)\) regions. When we place a new line it will meet those \(k\) lines in \(k\) places (since no lines are parallel) and there will be k+1 regions the line will run through (since no three lines meet at a point). Each of those \(k+1\) regios is now split in half, so there are \(k+1\) "new regions". Therefore there are now \(\frac12(k^2+k+2)+(k+1) = \frac12(k^2+k+1+2k+2) = \frac12 ((k+1)^2+(k+1)+1)\) regions, ie our hypothesis is true for \(n=k+1\). (Conclusion) Therefore since our statement is true for \(n=1\) and since if it is true for some \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) Proof: (Alternative). There are \(\binom{n}{2}\) places where the lines meet. Each intersection is the most extreme point (say lowest) for one region (if two are tied, rotate by a very small amount) so this is a unique mapping. There will be \(n+1\) regions which are infinite and don't have a most extreme point, hence \(\binom{n}{2} + n+1 = \frac12(n^2-n)+n+1 = \frac12(n^2+n+2)\)