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Centrally symmetric figures
DEFINITION: A planar (or spatial) figure is centrally symmetric (point-symmetric) if there is a point in the plane (space) the figure reflected in which is an invariant figure.
Point O is the centre of symmetry of the figure if the figure is an invariant figure when reflected in this point.
The circle and the sphere are centrally symmetric in their centre. (Figure 28)
From the definition and the properties of point reflection it can be shown that if a planar polygon is centrally symmetric, then it has even number of vertices. Hence there is no centrally symmetric triangle.
Figure 28
\latex{ O }
\latex{ P' }
\latex{ P }
\latex{ P }
\latex{ P' }
\latex{ O }
Centrally symmetric quadrilateral
Two and two vertices of a centrally symmetric quadrilateral are each other's mirror images when reflected in the centre of symmetry. The centrally symmetric quadrilaterals are parallelograms. (Figure 29)
The central symmetry (or point symmetry) implies the following about a parallelogram:
Figure 29
\latex{ D=B' }
\latex{ C=A' }
\latex{ B=D' }
\latex{ A=C' }
- its opposite sides are of equal length;
- its opposite sides are parallel;
- its opposite angles are of equal measure;
- the sum of any two adjacent angles is \latex{ 180º };
- its diagonals bisect each other; their intersection point is the centre of symmetry;
- two opposite sides are parallel and of equal length.
It can be proven that any of the listed properties implies all the rest and the central symmetry. It means that any of the properties by itself is suitable for defining the parallelogram.
Example 1
Let us construct a parallelogram if the length of the two diagonals and the angle included between the diagonals are given.
\latex{ \varphi }
Figure 30
\latex{ A }
\latex{ B }
\latex{ D }
\latex{ C }
\latex{ f }
\latex{ e }
\latex{ M }
Solution
During the construction we use that the diagonals of the parallelogram bisect each other. By using the notations of figure 30 the process of construction is as follows:
- Taking diagonal AC = e and constructing its midpoint M.
- Measuring \latex{ \varphi } on AC at M.
- Marking off the distance \latex{\frac{BD}{2} = \frac{f}{2}} on the straight line of the resulting arm of angle in both directions from M.
- Connecting the corresponding vertices.
Example 2
What can we say about the below quadrilaterals regarding central symmetry?
- trapezium
- square
- rhombus
- kite
- rectangle
Solution
Squares, rhombi and rectangles are special parallelograms; therefore these quadrilaterals are centrally symmetric (Figure 31). Kites and trapezia in general are not centrally symmetric.
Figure 31
\latex{ O }
\latex{ O }
\latex{ O }
Example 3
Let us prove that if a planar figure is axially symmetric about two perpendicular axes, then it is also centrally symmetric.
Solution
Let the axes of symmetry be\latex{ t_{1} } and \latex{ t_{2} } . (Figure 32)
Let us reflect the figure first about axis \latex{ t_{1} } and then about axis \latex{ t_{2} } . The figure is invariant when reflected about any of the two axes, so it is also invariant when doing the transformation resulting by doing the two reflections in succession. We saw it earlier that this transformation is the point reflection in the intersection point of the axes of symmetry. Is the converse of the statement true?
Figure 32
\latex{ l }
\latex{ t_{1} }
\latex{ t_{2} }
Example 4
Let us prove that regular polygons with even number of sides are centrally symmetric.
Solution
We know that a regular polygon with n sides has n axes of symmetry. If n is even, then these axes pass through the opposite vertices or the midpoints of the opposite sides.
If we can prove that in the case of an even n there are two perpendicular ones among these axes, then based on example 3 we will have proven the statement.
Let n = 2k, where k is a whole number not less than \latex{ 2 }.
The axes intersect each other at one point and they divide the \latex{ 360º } angle around the common point into 2n pieces of angle of equal measure. (Figure 33)
The measure of the angle included between two adjacent axes is
\latex{\frac{360°}{2n} =\frac{180°}{n} =\frac{180°}{2k}=\frac{90°}{k}}
It means that when starting from any fixed axis into whichever direction the kth axis will be perpendicular to the chosen fixed axis. We have found two perpendicular axes, and thus we have shown the statement.
Figure 33
\latex{\frac{360°}{2n} =\frac{90°}{k}}
Exercises
{{exercise_number}}. Decide which of the below statements are true and which are false.
- There is a centrally symmetric triangle.
- There is a centrally symmetric quadrilateral.
- Every quadrilateral is centrally symmetric.
- If a quadrilateral is centrally symmetric, then its diagonals bisect each other.
- If the diagonals of a quadrilateral bisect each other, then it is centrally symmetric.
- If two opposite sides of a quadrilateral are parallel and of equal length, then it is centrally symmetric.
- Every regular polygon is centrally symmetric.
- There is a centrally symmetric polygon which is regular.
- If the number of sides of a regular polygon is even, then it has a centre of symmetry.
{{exercise_number}}. Construct a parallelogram if the intersection point of its diagonals and two adjacent vertices are given.
{{exercise_number}}. In the Cartesian coordinate system two adjacent vertices of a parallelogram are: A(–2; 5), B(–4; –2). Determine the coordinates of the other two vertices, if we know that the diagonals of the parallelogram intersect each other at the origin.
{{exercise_number}}. Reflect a triangle in the midpoint of one of its sides. What planar figure is defined by the union of the original triangle and the image triangle? (Justify your answer.)
{{exercise_number}}. Take a convex angle and point O in the angular domain. Construct a parallelogram so that one of its vertices is the vertex of the angle, two adjacent sides lie on the two arms of the angle, and the intersection point of its diagonals is O.
{{exercise_number}}. Give the angles of the parallelogram if the ratio of two of its interior angles is:
- \latex{2:3};
- \latex{4:5};
- \latex{3:7};
- \latex{p:q}.
{{exercise_number}}. Justify that the interior angle bisectors belonging to opposite vertices of the parallelogram are parallel. What can we state about the parallelogram if the mentioned two angle bisectors are collinear? Justify your statement.
{{exercise_number}}. Is the following statement true? “If a hexagon is centrally symmetric, then it is regular.” (Justify your answer.)
Puzzle
Is there a point set which has infinitely many centres of symmetry?