Transformations & Symmetry
Rotations, reflections, folding, tessellations, symmetric figures
Transformations & SymmetryAMC 10 Step by Step
Name: Date: October 28, 2026
Geometry · 15 problems have this as their main idea
15
primary-topic problems (1.2% of all)
25
more as a secondary topic
Where it appears
4
P1-10
2
P11-15
5
P16-20
4
P21-25
What you need to know
- Moves that keep size and shape. Rotations (turns), reflections (flips), and translations (slides) do not change lengths, angles, or areas. A counterclockwise rotation about the origin sends to . Example: goes to , and both are from the origin.
- Reflecting across a line. If is the reflection of , the line is the perpendicular bisector of : it meets at a right angle, through its midpoint. Example: reflecting across the line gives , and the midpoint is on the line.
- Folding is reflecting across the crease. If a fold puts point on point , the crease is the perpendicular bisector of , and every length on the folded flap stays the same.
- Symmetry of regular polygons. A regular polygon with sides has lines of symmetry, and rotational symmetry of order : it looks the same after each of turns of about its center. Example: a square has lines of symmetry and matches itself after turns of , , , .
How AMC 10 tests it
- Problems 12–18: fold a rectangle or square so a corner lands on an edge, a midpoint, or the opposite corner; find the crease length or a folded segment.
- "Which figures have a line of symmetry, or rotational symmetry?" or counting the colorings of a grid that are symmetric.
- Turn a segment or triangle or about one of its corners; find the area it sweeps out, or its overlap with its image.
Standard approaches
- For folds, call the landing point , mark every length the fold keeps, and write one Pythagorean equation for the unknown crease position.
- When the problem says exactly which point lands on which, use coordinates and the perpendicular bisector.
- For swept areas, split the region into sectors centered at the center of rotation, plus triangles.
Worked example
Square has side length . The paper is folded so that vertex lands on the midpoint of side . The crease meets at and at . What is the length of ?
(A) (B) (C) (D) (E)
Solution. Place the square with , , , , so .
Step 1. The fold sends to , so the crease is the perpendicular bisector of . Every point on it is as far from as from , so and .
Step 2. Let . Then and . Right triangle gives
so and . That puts at .
Step 3. Let , so . Right triangles and give
so and . That puts at .
Step 4. From to is across and up:
The answer is .
Pitfalls
- Treating the crease as segment itself. The crease is the perpendicular bisector of .
- Making up new unknowns for the folded flap instead of reusing the lengths the fold keeps. In the example, is just .
- Writing for a counterclockwise turn. That is the clockwise turn. Example: counterclockwise, goes up to , not down to .
Traps that recur
- Tracing the bounces one at a time, losing the thread after four or five reflections and concluding from the absence of an early corner that the path runs forever, which is (E). (2025 AMC 10B #25)
- Checking only even n (which first works at n = 720) and missing that an odd number of transformations, ending on a reflection, already returns the point at n = 359. (2022 AMC 10A #18)
- Assuming AE = EC' or that the crease passes through E; E lies on the folded image of edge CB, not on the crease. (2021 AMC 10B #21)
- Forgetting that a rotation sandwiched between two reflections is reversed, or allowing sequences with one or three reflections. (2020 AMC 10A #23)
Problems, easiest first
Open any of them to try it here; the worked solution is on the problem's own page.