AMC 10 Step by Step
Printable problems

Transformations & Symmetry

Rotations, reflections, folding, tessellations, symmetric figures

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 90∘90^\circ counterclockwise rotation about the origin sends (x,y)(x,y) to (−y,x)(-y,x). Example: (3,1)(3,1) goes to (−1,3)(-1,3), and both are 10\sqrt{10} from the origin.
  • Reflecting across a line. If P′P' is the reflection of PP, the line is the perpendicular bisector of PP′PP': it meets PP′PP' at a right angle, through its midpoint. Example: reflecting (1,4)(1,4) across the line y=1y = 1 gives (1,−2)(1,-2), and the midpoint (1,1)(1,1) is on the line.
  • Folding is reflecting across the crease. If a fold puts point PP on point P′P', the crease is the perpendicular bisector of PP′PP', and every length on the folded flap stays the same.
  • Symmetry of regular polygons. A regular polygon with nn sides has nn lines of symmetry, and rotational symmetry of order nn: it looks the same after each of nn turns of 360∘n\frac{360^\circ}{n} about its center. Example: a square has 44 lines of symmetry and matches itself after turns of 90∘90^\circ, 180∘180^\circ, 270∘270^\circ, 360∘360^\circ.

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 60∘60^\circ or 90∘90^\circ about one of its corners; find the area it sweeps out, or its overlap with its image.

Standard approaches

  1. For folds, call the landing point P′P', mark every length the fold keeps, and write one Pythagorean equation for the unknown crease position.
  2. When the problem says exactly which point lands on which, use coordinates and the perpendicular bisector.
  3. For swept areas, split the region into sectors centered at the center of rotation, plus triangles.

Worked example

Square ABCDABCD has side length 88. The paper is folded so that vertex AA lands on the midpoint MM of side BC‾\overline{BC}. The crease meets AB‾\overline{AB} at PP and CD‾\overline{CD} at QQ. What is the length of PQ‾\overline{PQ}?

(A) 88 (B) 454\sqrt5 (C) 99 (D) 636\sqrt3 (E) 1010

Solution. Place the square with A=(0,0)A = (0,0), B=(8,0)B = (8,0), C=(8,8)C = (8,8), D=(0,8)D = (0,8), so M=(8,4)M = (8,4).

Step 1. The fold sends AA to MM, so the crease is the perpendicular bisector of AM‾\overline{AM}. Every point on it is as far from AA as from MM, so PA=PMPA = PM and QA=QMQA = QM.

Step 2. Let PA=xPA = x. Then PB=8−xPB = 8 - x and BM=4BM = 4. Right triangle PBMPBM gives
x2=(8−x)2+42=x2−16x+80, x^2 = (8-x)^2 + 4^2 = x^2 - 16x + 80,
so 16x=8016x = 80 and x=5x = 5. That puts PP at (5,0)(5,0).

Step 3. Let DQ=yDQ = y, so QC=8−yQC = 8 - y. Right triangles QDAQDA and QCMQCM give
y2+82=(8−y)2+42,that is,y2+64=y2−16y+80, y^2 + 8^2 = (8-y)^2 + 4^2, \quad\text{that is,}\quad y^2 + 64 = y^2 - 16y + 80,
so 16y=1616y = 16 and y=1y = 1. That puts QQ at (1,8)(1,8).

Step 4. From P=(5,0)P = (5,0) to Q=(1,8)Q = (1,8) is 44 across and 88 up:
PQ=42+82=80=45. PQ = \sqrt{4^2 + 8^2} = \sqrt{80} = 4\sqrt5.
The answer is (B) 45\boxed{\textbf{(B)}\ 4\sqrt5}.

Pitfalls

  • Treating the crease as segment AMAM itself. The crease is the perpendicular bisector of AMAM.
  • Making up new unknowns for the folded flap instead of reusing the lengths the fold keeps. In the example, PMPM is just PAPA.
  • Writing (y,−x)(y,-x) for a 90∘90^\circ counterclockwise turn. That is the clockwise turn. Example: counterclockwise, (1,0)(1,0) goes up to (0,1)(0,1), not down to (0,−1)(0,-1).

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 © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.