Constructions and Tilings | FIO

Question 2

Is it necessary to construct the pairs of arcs above and below $XY$ ? Instead, can we construct both the pairs of arcs on the same side of $XY$ ? Explore this through construction, and then justify your answer.

Solution

Understand the Question

To construct the perpendicular bisector of a line segment $XY$ , we only need to identify two distinct points that are each equidistant from $X$ and $Y$ .
It is not necessary to place these points on opposite sides. We can construct both points on the same side of $XY$ by choosing two different compass radii (both greater than $\dfrac{1}{2}XY$ ).
Connecting and extending the line through these two points yields the perpendicular bisector.

Step 1 · Steps of Construction

Draw a line segment $XY$ .
Set the compass opening to a radius $k > \dfrac{1}{2}XY$ . With $X$ and $Y$ as centers, draw two intersecting arcs on the same side of $XY$ . Let their intersection point be $A$ .
Change the compass opening to a different radius $k' > \dfrac{1}{2}XY$ (where $k' \neq k$ ). With $X$ and $Y$ as centers, draw another pair of intersecting arcs on the same side of $XY$ . Let their intersection point be $B$ .
Draw a straight line passing through points $A$ and $B$ , extending it to intersect line segment $XY$ at point $O$ . Join $AX$ , $AY$ , $BX$ , and $BY$ .

Step 2 · Justification of the Construction

In $\Delta ABX$ and $\Delta ABY$ :

$AX = AY = k$ (radii of the first pair of arcs)
$BX = BY = k'$ (radii of the second pair of arcs)
$AB = AB$ (common side)

By SSS congruence criterion: $\Delta ABX \cong \Delta ABY$

Therefore, by CPCTC: $\angle XAB = \angle YAB \implies \angle XAO = \angle YAO$

Now, in $\Delta AOX$ and $\Delta AOY$ :

$AX = AY$
$\angle XAO = \angle YAO$
$AO = AO$ (common side)

By SAS congruence criterion: $\Delta AOX \cong \Delta AOY$

Therefore, by CPCTC: $OX = OY \quad \text{and} \quad \angle AOX = \angle AOY$

Since $OX = OY$ , $O$ is the midpoint of $XY$ .

Since $\angle AOX$ and $\angle AOY$ form a linear pair on $XY$ : $\angle AOX + \angle AOY = 180^\circ$

$2\angle AOX = 180^\circ$

$\angle AOX = \dfrac{180^\circ}{2} = 90^\circ$

Thus, the line $AB$ is perpendicular to $XY$ and bisects it at $O$ . Hence, $AB$ is the perpendicular bisector of $XY$ .

Answer

No, it is not necessary to construct arcs on both sides of $XY$ . Constructing two pairs of arcs on the same side of $XY$ using two different radii ( $k \neq k' > \dfrac{1}{2}XY$ ) successfully gives the perpendicular bisector.

Common Mistakes

Using the same radius for both points: Using the same radius ( $k' = k$ ) will simply trace over the same intersection point $A$ . The two radii must be distinct ( $k \neq k'$ ) to get two different points $A$ and $B$ .
Radius too small: Choosing a radius $k \le \dfrac{1}{2}XY$ will result in arcs that do not intersect at all.

Question 2

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