Constructions and Tilings | FIO

Question 1

When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below $XY$ ? Explore this through construction, and then justify your answer.

[Hint 1: Any point that is of the same distance from $X$ and $Y$ lies on the perpendicular bisector.

Hint 2: We can draw the whole line if any two of its points are known.]

Solution

Understand the Question

Any point equidistant from endpoints $X$ and $Y$ lies on the perpendicular bisector of segment $XY$ .
Since any two points define a unique straight line, we only need to locate two points equidistant from $X$ and $Y$ (one above and one below $XY$ ).
For a single point, the distance from $X$ must equal the distance from $Y$ . However, the radius used for the point above ( $k$ ) does not need to equal the radius used for the point below ( $k'$ ), provided each radius is strictly greater than $\dfrac{1}{2}XY$ .

Step 1 · Draw the Line Segment

Draw a straight line segment with endpoints $X$ and $Y$ . Diagram 1

Step 2 · Construct Point $A$ Above the Segment

Open the compass to a radius $k > \dfrac{1}{2}XY$ . Diagram 2

With center $X$ and radius $k$ , draw an arc above $XY$ .
With center $Y$ and the same radius $k$ , draw another arc intersecting the first arc at point $A$ .

$AX = AY = k$

Since point $A$ is equidistant from $X$ and $Y$ , $A$ lies on the perpendicular bisector of $XY$ .

Step 3 · Construct Point $B$ Below the Segment

Open the compass to a different radius $k' > \dfrac{1}{2}XY$ (where $k' \neq k$ ). Diagram 3

With center $X$ and radius $k'$ , draw an arc below $XY$ .
With center $Y$ and the same radius $k'$ , draw another arc intersecting the first arc at point $B$ .

$BX = BY = k'$

Since point $B$ is equidistant from $X$ and $Y$ , $B$ also lies on the perpendicular bisector of $XY$ .

Step 4 · Draw the Line and Justify the Result

Draw a straight line passing through points $A$ and $B$ . Diagram 4

Line $AB$ passes through two points ( $A$ and $B$ ) that both lie on the perpendicular bisector of $XY$ .
Since two points uniquely define a line, line $AB$ is the perpendicular bisector of $XY$ .
The radii $k$ and $k'$ do not need to be equal; they only need to satisfy $k, k' > \dfrac{1}{2}XY$ so the arcs intersect.

Answer

No, it is not necessary to have the same radius for the arcs above and below $XY$ . As long as the radius from $X$ equals the radius from $Y$ for each individual point and is greater than $\dfrac{1}{2}XY$ , the line connecting the two intersection points will always be the perpendicular bisector.

Common Mistakes

Unequal Radii for the Same Point: Changing the compass width between the arc from $X$ and the arc from $Y$ when constructing point $A$ or $B$ . For each intersection point individually, the distances from $X$ and $Y$ must be strictly equal.
Radius Too Small: Setting the compass radius $k \le \dfrac{1}{2}XY$ , which causes the arcs to either not meet at all or touch at only a single point along the segment.

Question 1

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