Geometry - Geometrical Constructions

A map has two towns $A$ and $B$ which are $8\text{ cm}$ apart. Construct the locus of points that are exactly $5\text{ cm}$ from town $A$ and equidistant from towns $A$ and $B$. Find how many such points exist.

1. Draw a line segment $AB = 8\text{ cm}$.
2. Construct the perpendicular bisector of $AB$: Place the compass at $A$, open to $r > 4\text{ cm}$, and draw arcs above and below $AB$. Repeat from $B$. Draw a line through the arc intersections.
3. Construct the locus of points $5\text{ cm}$ from $A$: Use a compass to draw a circle with center $A$ and radius $5\text{ cm}$.
4. The intersection points of the perpendicular bisector and the circle are the required points.
5. Since the distance from the midpoint of $AB$ to $A$ is $4\text{ cm}$ and $4 < 5$, the circle will intersect the perpendicular bisector at two distinct points.

The condition 'exactly $5\text{ cm}$ from $A$' defines a circle. The condition 'equidistant from $A$ and $B$' defines the perpendicular bisector. The solution is the intersection of these two loci.

Construct a $45^{\circ}$ angle at point $P$ on line segment $PQ$.

1. At point $P$, construct a perpendicular line to $PQ$ to create a $90^{\circ}$ angle.
2. To do this, extend $QP$ to the left. Place compass at $P$, draw arcs on both sides of $P$ on the line. From these two points, draw intersecting arcs above $P$ and join to $P$.
3. Now, bisect the $90^{\circ}$ angle. Place the compass at $P$, draw an arc intersecting both the perpendicular and $PQ$.
4. From these two intersection points, draw two arcs that intersect in the middle of the angle.
5. Draw a line from $P$ through this intersection. The resulting angle is $45^{\circ}$.

A $45^{\circ}$ angle is half of a $90^{\circ}$ angle. We use the construction of a perpendicular followed by an angle bisector.

1. Draw line segment $AB = 8\text{ cm}$.
2. Using a compass set to $6\text{ cm}$, draw an arc from $A$.
3. Using a compass set to $7\text{ cm}$, draw an arc from $B$.
4. Label the intersection $C$ and join $AC$ and $BC$.
5. To find points closer to $AB$ than $AC$, construct the angle bisector of $\angle BAC$.
6. The required region is the area inside $\triangle ABC$ on the $AB$ side of the angle bisector.

The condition 'closer to $AB$ than $AC$' refers to the region bounded by the angle bisector of the angle formed by those two lines. Points on the bisector are equidistant from both lines.

1. Draw a rectangle $100\text{ mm}$ by $60\text{ mm}$ (using a scale of $1\text{ cm} : 1\text{ m}$).
2. Mark the center by intersecting the diagonals.
3. Draw a circle with radius $40\text{ mm}$ from the center.
4. The region outside this circle but inside the rectangle is the area not reached by the sprinkler.

The locus of points reached by the sprinkler is the area defined by $x^2 + y^2 \leq r^2$. The region not reached is the complement of this set within the rectangle boundaries.