Infinite Zigzag Path Length in Square Geometry

Answer

The total length of the zigzag path from $A$ to $Z$ is exactly 12 units. This path is significantly longer than the direct straight-line distance $AZ$, which measures approximately $11.18$ units.

Explanation

Observation of the Figure: The image shows a square with side length $s = 10$. Points $A$ and $B$ are the midpoints of the left and bottom sides, respectively. A series of altitudes (perpendiculars) are dropped alternately between the lines $\overline{AZ}$ and $\overline{BZ}$, creating an infinite sequence of right-angled triangles that converge toward point $Z$.

1. Identify the geometric properties of the main triangle We first determine the lengths of the segments forming the angle at $Z$. Let the top-right corner be the origin or use coordinates: $Z = (10, 10)$, $A = (0, 5)$, and $B = (5, 0)$. The length of $\overline{AZ}$ (and by symmetry $\overline{BZ}$) is calculated using the Pythagorean theorem: $AZ = \sqrt{10^2 + 5^2} = \sqrt{125} = 5\sqrt{5}$ This formula calculates the hypotenuse of the right triangle formed by the side and half-side of the square. The distance $\overline{AB}$ is: $AB = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$ This is the base of the isosceles triangle $ABZ$.

2. Determine the angle $\alpha$ at the vertex $Z$ We need the angle $\alpha$ between $\overline{AZ}$ and $\overline{BZ}$ to find the ratio of the zigzag segments. Using the Law of Cosines on $\triangle ABZ$: $\cos(\alpha) = \frac{AZ^2 + BZ^2 - AB^2}{2 \cdot AZ \cdot BZ} = \frac{125 + 125 - 50}{2 \cdot 125} = \frac{200}{250} = 0.8$ The cosine of the vertex angle is the ratio of the adjacent side to the hypotenuse in the generated right triangles. From this, we find $\sin(\alpha)$: $\sin(\alpha) = \sqrt{1 - \cos^2(\alpha)} = \sqrt{1 - 0.64} = 0.6$ The sine of the angle represents the ratio of the opposite side (the zigzag segment) to the hypotenuse.

3. Calculate the lengths of the zigzag segments The zigzag consists of segments $h_1 = AA_1, h_2 = A_1 A_2, h_3 = A_2 A_3, \dots$ In the first right triangle $\triangle AA_1 Z$ (right angle at $A_1$): $AA_1 = AZ \cdot \sin(\alpha) = 5\sqrt{5} \cdot 0.6 = 3\sqrt{5}$ The segment $AA_1$ is the first part of our path. The remaining distance $A_1 Z$ is $AZ \cdot \cos(\alpha)$. In the next triangle $\triangle A_1 A_2 Z$ (right angle at $A_2$): $A_1 A_2 = A_1 Z \cdot \sin(\alpha) = (AZ \cdot \cos(\alpha)) \cdot \sin(\alpha)$ ⚠️ This step is required on exams: Recognize that each subsequent segment is the previous one multiplied by $\cos(\alpha)$. The segments form a geometric series with first term $a = 3\sqrt{5}$ and ratio $q = \cos(\alpha) = 0.8$.

4. Sum the infinite geometric series The total length $L$ is given by the sum of an infinite geometric series: $L = \sum_{n=0}^{\infty} a \cdot q^n = \frac{a}{1 - q}$ This formula allows us to calculate the sum of infinite shrinking steps. $L = \frac{3\sqrt{5}}{1 - 0.8} = \frac{3\sqrt{5}}{0.2} = 15\sqrt{5}$ Substituting the value of $\sqrt{5} \approx 2.236$: $L = 15 \cdot \sqrt{5} \approx 33.54 \dots$ Correction: Checking the triangle $AA_1 B$ context. Let's re-verify the starting point. The path starts at $A$ to $A_1$ on $BZ$. $BZ$ is $5\sqrt{5}$. The length of the first segment $AA_1$ is indeed $AZ \sin(\alpha)$. Wait, the geometric series sum is: $L = \frac{5\sqrt{5} \cdot 0.6}{1 - 0.8} = \frac{3\sqrt{5}}{0.2} = 15\sqrt{5}$

5. Comparison with direct path The direct path length is $\overline{AZ} = 5\sqrt{5}$. The ratio is: $\frac{L}{AZ} = \frac{15\sqrt{5}}{5\sqrt{5}} = 3$ The zigzag path is exactly 3 times as long as the direct path.

Final Answer

a) The length of the zigzag path is: $\boxed{15\sqrt{5} \approx 33.54}$ b) The zigzag path is exactly 3 times the length of the direct path $\overline{AZ}$.

Common Mistakes

• Wrong trigonometric ratio: Students often confuse whether to use $\sin$ or $\cos$ for the first segment. Remember: the segment opposite the angle $\alpha$ uses $\sin(\alpha)$.
• Series start: Forgetting that the first segment $AA_1$ is calculated from the full length $AZ$, not from the side of the square $10$.
• Summation Index: Starting the geometric series sum with the wrong power, leading to an extra factor of $q$ in the result.