Square Shaded Fraction E14488

1. **Problem Statement:** We have a square whose two adjacent sides are divided into $n$ equal parts. Lines are drawn connecting these division points on one side to the division points on the adjacent side, creating a pattern that shades a certain fraction of the square. We want to find: - (b) The value of $n$ such that the shaded area is exactly $\frac{1}{5}$ of the entire square. - (c) A function $f(n)$ that gives the fraction of the square shaded for any $n$, and the limit of $f(n)$ as $n$ approaches infinity. --- 2. **Understanding the Pattern:** - The square has side length 1 (we can assume this for simplicity). - The two adjacent sides are divided into $n$ equal segments, so each segment length is $\frac{1}{n}$. - Lines are drawn from each division point on one side to the corresponding division point on the adjacent side, forming a series of triangles and trapezoids. - The shaded region is the area below the curve formed by these lines. --- 3. **Finding the Fraction of the Square Shaded, $f(n)$:** - The shaded area consists of the sum of areas of smaller right triangles formed by these lines. - Each small triangle has vertices at $(0, \frac{i}{n})$, $(\frac{i}{n}, 0)$, and $(0, \frac{i-1}{n})$ or $(\frac{i-1}{n}, 0)$ for $i=1,2,...,n$. - The area of the $i$-th small triangle is: $$ A_i = \frac{1}{2} \times \frac{1}{n} \times \frac{1}{n} = \frac{1}{2n^2} $$ - But the shaded area is the sum of these triangles for $i=1$ to $n$. - Actually, the shaded area is the sum of the areas of triangles formed by connecting points $(0, \frac{i}{n})$ to $(\frac{i}{n}, 0)$ for $i=1$ to $n$. - The total shaded area is: $$ f(n) = \sum_{i=1}^n \frac{1}{2} \times \frac{i}{n} \times \frac{i}{n} = \frac{1}{2n^2} \sum_{i=1}^n i^2 $$ - Using the formula for the sum of squares: $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ - Substitute this back: $$ f(n) = \frac{1}{2n^2} \times \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{12n} $$ --- 4. **Answer to part (b): Find $n$ such that $f(n) = \frac{1}{5}$:** Set the equation: $$ \frac{(n+1)(2n+1)}{12n} = \frac{1}{5} $$ Multiply both sides by $12n$: $$ (n+1)(2n+1) = \frac{12n}{5} $$ Expand the left side: $$ 2n^2 + n + 2n + 1 = \frac{12n}{5} $$ Simplify: $$ 2n^2 + 3n + 1 = \frac{12n}{5} $$ Multiply both sides by 5 to clear denominator: $$ 10n^2 + 15n + 5 = 12n $$ Bring all terms to one side: $$ 10n^2 + 15n + 5 - 12n = 0 $$ $$ 10n^2 + 3n + 5 = 0 $$ This quadratic has no real positive roots (discriminant $= 3^2 - 4 \times 10 \times 5 = 9 - 200 = -191 < 0$), so no integer $n$ satisfies exactly $\frac{1}{5}$. Check for approximate $n$ by testing values: - For $n=1$, $f(1) = \frac{(1+1)(2+1)}{12 \times 1} = \frac{2 \times 3}{12} = \frac{6}{12} = 0.5$ (too big) - For $n=5$, $f(5) = \frac{6 \times 11}{12 \times 5} = \frac{66}{60} = 1.1$ (too big, so re-check formula or problem interpretation) Since the problem states examples for $n=5$ and $n=8$, and the shaded area is less than half, re-examine the formula: Actually, the shaded area is the sum of areas of triangles with base and height $\frac{1}{n}$ but the height changes with $i$. Recalculate $f(n)$: Each triangle area: $$ A_i = \frac{1}{2} \times \frac{1}{n} \times \frac{n - i + 1}{n} $$ Sum over $i=1$ to $n$: $$ f(n) = \sum_{i=1}^n \frac{1}{2} \times \frac{1}{n} \times \frac{n - i + 1}{n} = \frac{1}{2n^2} \sum_{i=1}^n (n - i + 1) $$ Sum inside: $$ \sum_{i=1}^n (n - i + 1) = \sum_{k=1}^n k = \frac{n(n+1)}{2} $$ So: $$ f(n) = \frac{1}{2n^2} \times \frac{n(n+1)}{2} = \frac{n+1}{4n} $$ Now solve for $f(n) = \frac{1}{5}$: $$ \frac{n+1}{4n} = \frac{1}{5} $$ Multiply both sides by $4n$: $$ n + 1 = \frac{4n}{5} $$ Multiply both sides by 5: $$ 5n + 5 = 4n $$ $$ 5n + 5 - 4n = 0 $$ $$ n + 5 = 0 $$ $$ n = -5 $$ Negative $n$ is invalid, so no integer $n$ satisfies exactly $\frac{1}{5}$. Try $n=4$: $$ f(4) = \frac{4+1}{4 \times 4} = \frac{5}{16} = 0.3125 $$ Try $n=5$: $$ f(5) = \frac{6}{20} = 0.3 $$ Try $n=6$: $$ f(6) = \frac{7}{24} \approx 0.2917 $$ Try $n=3$: $$ f(3) = \frac{4}{12} = \frac{1}{3} = 0.3333 $$ Try $n=2$: $$ f(2) = \frac{3}{8} = 0.375 $$ Try $n=1$: $$ f(1) = \frac{2}{4} = 0.5 $$ The function decreases as $n$ increases, so $f(n)$ never equals exactly $\frac{1}{5} = 0.2$ for positive integer $n$. --- 5. **Answer to part (c): Find $f(n)$ and $\lim_{n \to \infty} f(n)$:** We have: $$ f(n) = \frac{n+1}{4n} = \frac{1}{4} + \frac{1}{4n} $$ As $n \to \infty$: $$ \lim_{n \to \infty} f(n) = \frac{1}{4} + 0 = \frac{1}{4} $$ So the shaded fraction approaches $\frac{1}{4}$ of the square as $n$ becomes very large. --- 6. **Summary:** - The function for the shaded fraction is $f(n) = \frac{n+1}{4n}$. - There is no integer $n$ such that $f(n) = \frac{1}{5}$ exactly. - The shaded area approaches $\frac{1}{4}$ of the square as $n$ grows large. This matches the intuition that the shaded region is roughly a quarter of the square in the limit.