Core practice ← Back to lesson

Symmetry: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 Check the joined sides

A triangular outline joins $D(-4,0)$ , $E(4,0)$ , and $F(0,5)$ with straight segments. Test reflection across the $y$ -axis and a half-turn about the origin. Give a whole-outline reason for each decision. The labels name vertices and are not decorations.
1. Hint 1
  A successful symmetry must account for every side as well as the vertices.
2. Hint 2
  Find all three image vertices for each motion. A failed vertex rejects the motion; when all match, explain what happens to their joining segments.
Answer

The $y$ -axis reflection matches the outline; the half-turn does not.

Full solution

Reflection across the $y$ -axis swaps D and E and fixes F.
It maps DE onto itself and swaps DF with EF, so every straight side matches.

The half-turn swaps D and E but sends F to (0,-5).
That point is absent from the outline, so the entire triangle does not match.

Answer

The $y$ -axis reflection matches the outline; the half-turn does not.

Key idea

Vertex matches establish a straight-sided outline only when its joined sides match too.
Problem 2 Count without repeating identity

A figure has finite rotational symmetry about C. Its least positive matching angle is $22.5^\circ$ . Find its order, including identity once, and the number of nonidentity matching rotations. Explain how the full-turn position is counted.
1. Hint 1
  Matching turns are equally spaced by the least positive angle.
2. Hint 2
  Divide the full turn by the least angle, then distinguish the one identity mapping from the other rotations.
Answer

Order 16; 15 nonidentity rotations. The 0-degree and 360-degree positions count as the same identity.

Full solution

The number of equal least turns in a full turn is
$360\div22.5=16$

The order includes one identity.
The number of nonidentity matches is
$16-1=15$

Use 0 degrees through but not including 360 degrees to count positions.
The full-turn position repeats the starting mapping.

Answer

Order 16; 15 nonidentity rotations. The 0-degree and 360-degree positions count as the same identity.

Key idea

Finite order counts distinct rotations, with identity appearing once.
Problem 3 Read the exact letter

The exact block E outline below has the following vertices in order. Join successive vertices and close the outline.

$(-3,-3)$ , $(3,-3)$ , $(3,-2)$ , $(-2,-2)$ , $(-2,-\tfrac12)$ , $(1,-\tfrac12)$ , $(1,\tfrac12)$ , $(-2,\tfrac12)$ , $(-2,2)$ , $(3,2)$ , $(3,3)$ , $(-3,3)$ .

Test reflection across each coordinate axis and a half-turn about the origin. Give a point or a correspondence explaining each result. Judge this outline rather than another font.

Text description of this figure

The block E outline has a full-height left bar, equal top and bottom bars extending right, and a shorter centered middle bar. Its vertex coordinates are supplied in the statement.
1. Hint 1
  The top and bottom bars have the same dimensions, but the left and right parts differ.
2. Hint 2
  Pair the listed vertices under each rule. For a failed candidate, find just one image point missing from the outline.
Answer

Reflection across the $x$ -axis matches; reflection across the $y$ -axis and the half-turn fail.

Full solution

The $x$ -axis reflection pairs vertices at opposite heights with the same horizontal coordinate.
The top and bottom bars exchange and the middle bar covers itself, so all joined edges match.

The $y$ -axis reflection sends the boundary point $(-3,0)$ to $(3,0)$ , which is absent from the outline.
The candidate fails.

The half-turn also sends (-3,0) to the absent point (3,0).
Therefore there is no point symmetry about the origin.

Answer

Reflection across the $x$ -axis matches; reflection across the $y$ -axis and the half-turn fail.

Key idea

A letter name supplies no symmetry information beyond its exact drawn outline.
Problem 4 Turn around the stated center

A rectangular boundary has corners $J(1,-4)$ , $K(9,-4)$ , $L(9,2)$ , and $M(1,2)$ in that order. Its center is $C(5,-1)$ . A rectangle has four right angles.

Test a $180^\circ$ turn and a $90^\circ$ counterclockwise turn about C. Give the order and least positive matching angle about C. Explain using the complete boundary, not only C.
1. Hint 1
  Compare each corner with the stated center before turning.
2. Hint 2
  A half-turn reverses both relative coordinates. A quarter-turn exchanges horizontal and vertical relative distances with the counterclockwise sign change.
Answer

The half-turn matches; the quarter-turn fails. Order 2; least positive matching angle $180^\circ$ .

Full solution

J is four units left and three below C.
Its half-turn image is L, four right and three above C.
K and M also exchange.
The joined opposite sides pair, matching the whole boundary.

For the quarter-turn, J has relative position (-4,-3).
The image relative position is (3,-4), giving the point (8,-5).
That point is absent from the boundary, so the quarter-turn fails.

Any matching turn must send corners to corners.
The long horizontal side at J must become a long horizontal side, so only identity and the half-turn can match this unequal-sided rectangle.
Its order is 2.

Its least positive matching angle is
$360^\circ\div2=180^\circ$

Answer

The half-turn matches; the quarter-turn fails. Order 2; least positive matching angle $180^\circ$ .

Key idea

A center staying fixed does not settle whether the rest of the figure matches.
Problem 5 Include the printed mark

A sticker design is the boundary of the square with corners $(-2,-2)$ , $(2,-2)$ , $(2,2)$ , and $(-2,2)$ , together with a single filled dot at $(1,1)$ . The dot belongs to the figure. A square has four right angles and four equal sides.

Test reflection across $y=x$ and a half-turn about the origin. Give the complete figure's rotational order about the origin, including identity.

Text description of this figure

A square boundary surrounds a single filled dot in its upper right interior. The dot is at (1,1) in the coordinates supplied in the statement; no image marks are shown.
1. Hint 1
  Treat the dot as part of the figure, not as a label.
2. Hint 2
  A motion must match both the square boundary and the unique dot. Find the image of the dot in each test.
Answer

The $y=x$ reflection matches; the half-turn fails. Rotational order 1 about the origin.

Full solution

Reflection across $y=x$ exchanges two square corners and fixes the other two, pairing all sides.
It also fixes $(1,1)$ , so the whole figure matches.

The half-turn matches the square boundary but moves the only dot to (-1,-1), where no dot is present.
Thus it fails for the design.

A nonidentity turn about the origin moves the noncentral dot to a different position.
No such turn matches the full design, giving order 1.

Answer

The $y=x$ reflection matches; the half-turn fails. Rotational order 1 about the origin.

Key idea

Included decoration can remove symmetries of an outer boundary.
Problem 6 Complete the set of marks

Draw the boundary of the square with corners $(-3,-3)$ , $(3,-3)$ , $(3,3)$ , and $(-3,3)$ on paper. A square has four right angles and four equal sides. Put a filled dot at $P(1,-2)$ .

Add the fewest distinct dots needed so that the boundary together with all dots matches a $90^\circ$ counterclockwise rotation about the origin. Give every added coordinate, explain why the whole figure then matches, and state its rotational order about the origin.
1. Hint 1
  Follow the given mark through repeated quarter-turns. The image of every dot must be another dot.
2. Hint 2
  Use the counterclockwise rule (-y,x), continuing until the starting dot returns. Check the square boundary as well.
Answer

Add $(2,1)$ , $(-1,2)$ , and $(-2,-1)$ , three dots. The completed figure has order 4 about the origin.

Full solution

The first quarter-turn sends (1,-2) to (2,1), so that second position is required.
Continuing sends (2,1) to (-1,2), then to (-2,-1), then back to (1,-2).

All four positions differ, so three additions are necessary and sufficient for the marks.
The quarter-turn cycles them and maps each square side onto the next square side, matching the whole figure.

The square boundary itself permits only quarter-turn multiples about the origin.
All four of these match the completed marks, so the order is 4.

Answer

Add $(2,1)$ , $(-1,2)$ , and $(-2,-1)$ , three dots. The completed figure has order 4 about the origin.

Key idea

To build a symmetry, include every position reached by repeating the motion.
Problem 7 Separate the circle tests

An unmarked circle has center $O(0,0)$ and radius 4 units. The figure is its boundary only.

Case A. Test reflection across $x=1$ .

Case B. Test a $13^\circ$ counterclockwise turn about O. Decide whether there is a least positive matching angle and whether the circle has point symmetry about O.
1. Hint 1
  Use the fixed-radius definition for every proposed motion.
2. Hint 2
  For the offset mirror, track the leftmost circle point. For center rotations, consider whether halving a matching angle still matches.
Answer

Case A fails. Case B matches; there is no least positive matching angle, and the circle has point symmetry about O.

Full solution

Case A.
The circle contains (-4,0).
Reflecting across x=1 sends it to (6,0), six units from O, which is not on the circle.
That failed point rejects the offset mirror.

Case B.
Every circle point stays four units from O under the center turn, and the inverse turn covers every circle position.
The entire circle matches.

A smaller matching angle is
$13^\circ\div2=6.5^\circ$

The same halving works for any proposed positive least angle.
Thus none exists.
A half-turn also preserves every radius and matches the circle, giving point symmetry.

Answer

Case A fails. Case B matches; there is no least positive matching angle, and the circle has point symmetry about O.

Key idea

The finite-order formula cannot replace the circle definition in the infinite case.
Problem 8 Use two established matches

A whole figure is known to match reflection across line m and also reflection across line n. The two lines are perpendicular and meet at O. No particular shape is assumed.

Must the figure have point symmetry about O? Explain what happens to the whole figure after first reflecting in m and then in n.
1. Hint 1
  A composition of two matching motions still matches the figure.
2. Hint 2
  Use the earlier result for reflections in intersecting mirrors; their perpendicular angle determines the resulting turn.
Answer

Yes; the two successive reflections give a half-turn about O, matching the whole figure.

Full solution

The first reflection carries the figure exactly onto itself by the given symmetry.
The second therefore acts on that same figure and again carries every part onto itself.
Their composition matches the entire figure.

The mirrors meet at a right angle.
Two successive reflections turn the plane about their intersection through twice that angle:
$2\times90^\circ=180^\circ$

The matching half-turn about O is precisely point symmetry.
No claim that every point is fixed is needed.

Answer

Yes; the two successive reflections give a half-turn about O, matching the whole figure.

Key idea

Composing whole-figure symmetries produces another whole-figure symmetry.
Problem 9 When does finite order give a half-turn?

A figure has finite rotational order n about O, including identity once. Explain why it has point symmetry about O exactly when n is even. Include the order-one case.
1. Hint 1
  Think of the matching angles as n equally spaced steps around one full turn.
2. Hint 2
  A half-turn is halfway around. Decide when halfway corresponds to a whole number of the least-angle steps.
Answer

Point symmetry occurs exactly for even n. Order 1 has no point symmetry about O.

Full solution

Let the least positive matching angle be $\alpha$ .
The n equal steps complete one turn, so
$n\alpha=360^\circ.$
Halfway around requires $n/2$ steps.

If n is even, n/2 is a whole number.
Repeating the least matching turn that many times gives a matching half-turn, hence point symmetry.

If n is odd, n/2 is not a whole number.
The half-turn lies between consecutive matching angles, so it is not a symmetry.
In particular, order 1 has identity only and no nontrivial half-turn.

Answer

Point symmetry occurs exactly for even n. Order 1 has no point symmetry about O.

Key idea

For finite order about the same center, a half-turn is present exactly when the count is even.
Problem 10 Test a candidate set on the whole figure

The figure consists of the boundary of the square with corners $(-3,-3)$ , $(3,-3)$ , $(3,3)$ , and $(-3,3)$ , together with the complete diagonal segment from $(-3,-3)$ to $(3,3)$ . A square has four right angles and four equal sides. There is no other diagonal or decoration.

Decide separately whether each candidate below is a symmetry of this entire figure. Give a whole-figure match or a specific failure for each.

Candidate A. Reflection across the $x$ -axis.

Candidate B. Reflection across $y=x$ .

Candidate C. A $90^\circ$ counterclockwise turn about the origin.

Candidate D. A $180^\circ$ turn about the origin.

Candidate E. A $360^\circ$ turn about the origin.

Text description of this figure

The figure shows one square boundary and only the diagonal from its lower left corner to its upper right corner. The other diagonal is absent.
1. Hint 1
  The square boundary and the included diagonal must both pass each test.
2. Hint 2
  A point inside the diagonal, such as (1,1), can reject a candidate even when the square corners match. For successful candidates, track the complete diagonal as well as all sides.
Answer

A fails; B matches; C fails; D matches; E matches as identity.

Full solution

Candidate A.
The diagonal point (1,1) reflects to (1,-1), absent from both the boundary and the included diagonal.
The candidate fails.

Candidate B.
Reflection across $y=x$ pairs the square sides and fixes every point of the included diagonal.
The entire figure matches.

Candidate C.
The point (1,1) turns to (-1,1), absent from the figure.
The square boundary matches, but the included diagonal turns to the other diagonal, which is missing.

Candidate D.
The half-turn pairs opposite square sides.
It exchanges the diagonal endpoints and maps the whole segment between them onto itself.
Every part matches.

Candidate E.
The full turn fixes every point.
It is the identity symmetry and is counted once if distinct rotations are counted.

Answer

A fails; B matches; C fails; D matches; E matches as identity.

Key idea

A symmetry of a component must still be tested on every other component of the figure.

← Back to lesson