Mathematics: Class 7 (Part 2) — Chapter 6: Constructions and Tilings

The symmetrical 'Eyes' construction relies on finding two centers, $A$ and $B$, from which the lower and upper arcs of an eye are drawn respectively. These arcs define a supporting line segment $XY$.

• For the eye to be symmetrical, the upper and lower arcs must have the same radius: $AX = BX$ and $AY = BY$.
• Since the radii are equal, we have: $$AX = AY = BX = BY$$
• The line segment $AB$ connecting the two centers is the perpendicular bisector of the supporting line segment $XY$.
• Bisection is the division of a geometrical object into two identical parts. A line that bisects a given line segment and is perpendicular to it is called its perpendicular bisector.

Proof of Perpendicular Bisecting Property via Congruence:

• Let $O$ be the intersection of $AB$ and $XY$.
• In $\Delta ABX$ and $\Delta ABY$:
  • $AX = AY$ (given equal radii)
  • $BX = BY$ (given equal radii)
  • $AB$ is common to both triangles
  • Therefore, $\Delta ABX \cong \Delta ABY$ by SSS congruence.
  • This implies $\angle XAB = \angle YAB$ (or $\angle XAO = \angle YAO$) as corresponding parts of congruent triangles (CPCT).
• In $\Delta AOX$ and $\Delta AOY$:
  • $AX = AY$
  • $AO$ is common
  • $\angle XAO = \angle YAO$
  • Therefore, $\Delta AOX \cong \Delta AOY$ by SAS congruence.
  • This implies $OX = OY$ (so $O$ is the midpoint of $XY$) and $\angle AOX = \angle AOY$.
  • Since $\angle AOX + \angle AOY = 180^\circ$ (straight angle), $\angle AOX = \angle AOY = 90^\circ$. Thus, $AB \perp XY$.

To construct the perpendicular bisector of a line segment $XY$ using an unmarked ruler and a compass:

1. Using a fixed radius greater than half the length of $XY$, draw arcs from center $X$ and then from center $Y$ above the segment $XY$. Let these arcs intersect at point $A$.
2. Using the same radius, draw arcs from $X$ and $Y$ below the segment $XY$. Let these arcs intersect at point $B$.
3. Draw a straight line passing through $A$ and $B$. The line $AB$ is the required perpendicular bisector of $XY$.

To construct a perpendicular ($90^\circ$ angle) at a specific point $O$ on a given line:

1. Using a compass with $O$ as the center, mark two points $X$ and $Y$ on the line at equal distances on either side of $O$, making $O$ the midpoint of $XY$.
2. Using a radius larger than $OX$, draw two intersecting arcs above the line with centers $X$ and $Y$. Let the intersection point be $A$.
3. Join $A$ and $O$. The line $AO$ is perpendicular to the given line at $O$, forming a $90^\circ$ angle.

Ancient Indian mathematicians used exact geometric procedures to construct perpendiculars and altars. These methods are recorded in the Śulba-Sūtras (Vedic geometric texts, part of the Vedāṅgas).

• Instead of a modern compass, they utilized a rope to draw circles, arcs, and straight lines.
• Rope Method for Perpendicular Bisector (Kātyāyana-Śulbasūtra 1.2):
  1. Fix pegs at the endpoints $X$ and $Y$ of a line segment on the ground.
  2. Take a rope, fold it in half to find and mark its midpoint.
  3. Fasten the ends of the rope to the pegs at $X$ and $Y$.
  4. Pull the midpoint of the rope taut above $XY$ and mark this point as $A$.
  5. Pull the midpoint of the rope taut below $XY$ and mark this point as $B$.
  6. The line $AB$ forms the perpendicular bisector.

To bisect any given angle $\angle XOY$:

1. With $O$ as the center and any convenient radius, draw an arc intersecting the arms $OX$ and $OY$ at points $B$ and $A$ respectively, such that $OA = OB$.
2. With $A$ and $B$ as centers and a sufficiently large equal radius, draw arcs that intersect each other at point $C$.
3. Join $O$ and $C$. The ray $OC$ is the angle bisector of $\angle XOY$, meaning $\angle BOC = \angle AOC$.

This construction works because $\Delta OBC \cong \Delta OAC$ by SSS congruence ($OA = OB$, $BC = AC$, and $OC$ is common).

To construct an exact copy of a given angle $\angle A$ at a new vertex $X$:

1. Draw an arc with center $A$ intersecting the arms of the angle at points $B$ and $C$. This forms an isosceles triangle $\Delta ABC$.
2. With the same radius, draw an arc from the new vertex $X$ intersecting the base line at point $Z$.
3. Measure the distance $BC$ using a compass.
4. With $Z$ as the center and radius equal to $BC$, draw an arc intersecting the previous arc at point $Y$.
5. Draw a ray from $X$ through $Y$. The angle $\angle YXZ$ is equal to $\angle BAC$ (by SSS congruence, $\Delta ABC \cong \Delta XYZ$).

To construct a line parallel to a given line $m$ through a point $B$ not on $m$:

1. Draw a transversal line $l$ passing through $B$ and intersecting line $m$ at point $A$.
2. At point $B$, construct an angle equal to the corresponding angle (or alternate interior angle) formed at $A$ by copying the angle.
3. Draw a line $n$ through $B$ along the copied angle. Since the corresponding angles are equal, line $n$ is parallel to line $m$ ($m \parallel n$).

• A regular hexagon can be decomposed into six congruent equilateral triangles meeting at a central point.
• Since the sum of angles around a point is $360^\circ$, six $60^\circ$ angles fit perfectly around the center without gaps or overlaps ($6 \times 60^\circ = 360^\circ$).
• The interior angle of a regular hexagon is $60^\circ + 60^\circ = 120^\circ$.
• Constructing a 60° Angle:
  1. Draw a line segment $AX$.
  2. With $A$ as the center, draw an arc intersecting $AX$ at $B$.
  3. With the same radius and $B$ as the center, draw an arc intersecting the first arc at $C$.
  4. Join $AC$. The angle $\angle CAX = 60^\circ$ because $\Delta ABC$ is an equilateral triangle ($AB = AC = BC$).

• Tiling (or tessellation) is the covering of a flat region using a set of shapes without any gaps or overlaps.
• Tangrams: A traditional Chinese puzzle consisting of 7 geometric pieces obtained by dividing a square: 5 triangles (of various sizes), 1 square, and 1 parallelogram. These pieces can be rearranged to form diverse shapes (e.g., animals, figures, letters).

Analyzing whether a grid of size $m \times n$ can be tiled using $2 \times 1$ domino tiles:

• A $2 \times 1$ tile covers exactly 2 unit squares. Therefore, any grid tileable by $2 \times 1$ tiles must contain an even total number of unit squares.
• If both $m$ and $n$ are even, or if one is even and the other is odd, the total area $m \times n$ is even, and a tiling strategy exists (e.g., laying tiles systematically along columns or rows).
• If both $m$ and $n$ are odd, the total area $m \times n$ is odd, making tiling impossible.
• The Black-and-White Coloring Proof (Parity):
  • Color the grid like a chessboard so that adjacent squares have different colors.
  • Every $2 \times 1$ tile must cover exactly one black square and one white square.
  • Therefore, any region that can be tiled with $2 \times 1$ tiles must contain an equal number of black and white squares.
  • For example, if a $5 \times 3$ grid (15 squares) has one corner square removed, it leaves 14 squares. However, depending on which square is removed, the remaining grid may have 8 white and 6 black squares. Because the counts are unequal, this modified grid cannot be tiled, even though its total area (14) is even.

A plane can be tiled infinitely using regular polygons if the interior angles of the polygons meeting at any vertex add up to exactly $360^\circ$:

• Squares: Four squares meet at each vertex ($4 \times 90^\circ = 360^\circ$).
• Equilateral Triangles: Six triangles meet at each vertex ($6 \times 60^\circ = 360^\circ$).
• Regular Hexagons: Three hexagons meet at each vertex ($3 \times 120^\circ = 360^\circ$).
• A plane can also be tiled using combinations of multiple shapes, non-regular polygons, or artistic organic shapes (such as those created by the artist M.C. Escher).