Pythagorean Path (age 13) — solve the sequence 2,2,1,10,5,5,8,1,2,2,2 with Pythagoras

Goal: Connect a sequence of 12 points so that the lengths between consecutive points (in order) are 2, 2, 1, 10, 5, 5, 8, 1, 2, 2, 2. We'll use integer coordinate moves and check each segment with the Pythagorean theorem.

Strategy — step by step

1. Understand the rule: A straight-line distance between two lattice points (points with integer coordinates) is sqrt(dx^2 + dy^2). To make that distance equal an integer L, choose integer dx, dy with dx^2 + dy^2 = L^2.
2. Work from the long distances first. Long distances (like 10 and 8) restrict how you can move and reduce options, so place them early to avoid impossible backtracking.
3. Use common Pythagorean triples: 3–4–5 (so a length 5 can be dx=3, dy=4 or dx=4, dy=3); 6–8–10 (so length 10 can be dx=6, dy=8 or 10 and 0 if you prefer axis move). Small lengths 1 and 2 can be axis-aligned (dx=1,dy=0 etc.) or other small combinations.
4. Keep coordinates simple (start at (0,0)) and check each step with dx^2 + dy^2 = L^2. Avoid crossing yourself if the puzzle requires a single continuous non-self-crossing path.

A valid worked example (one solution)

Start at P0 = (0,0). The path of points P0 → P1 → … → P11 below gives the required segment lengths in order.

• P0 = (0, 0)
• P1 = (2, 0) — segment length = sqrt((2-0)^2 + (0-0)^2) = sqrt(4) = 2
• P2 = (4, 0) — length = sqrt((4-2)^2 + 0^2) = 2
• P3 = (5, 0) — length = sqrt((5-4)^2 + 0^2) = 1
• P4 = (5, -10) — length = sqrt((5-5)^2 + (-10-0)^2) = sqrt(100) = 10
• P5 = (8, -6) — length = sqrt((8-5)^2 + (-6+10)^2) = sqrt(3^2 + 4^2) = 5
• P6 = (12, -3) — length = sqrt((12-8)^2 + (-3+6)^2) = sqrt(4^2 + 3^2) = 5
• P7 = (20, -3) — length = sqrt((20-12)^2 + 0^2) = 8
• P8 = (21, -3) — length = 1
• P9 = (23, -3) — length = 2
• P10 = (25, -3) — length = 2
• P11 = (27, -3) — length = 2

Quick checks (Pythagorean verification)

• Segments 1–3: 2, 2, 1 are axis moves — obvious by dx values (2, 2, 1).
• Segment 4 (length 10): dx=0, dy= -10 → dx^2+dy^2=0+100=100=10^2.
• Segments 5 and 6 (each length 5): we used dx=3,dy=4 and dx=4,dy=3, so dx^2+dy^2 = 9+16 = 25 = 5^2.
• Segment 7 (length 8): dx=8, dy=0 → 64=8^2.
• Remaining short steps 1,2,2,2 are axis moves with dx equal to 1 or 2, so they check trivially.

This sequence of integer-coordinate points is a correct Pythagorean Path for the given lengths. There are many possible solutions; the key is choosing integer dx,dy so dx^2+dy^2 = length^2 and arranging them so the path is continuous.

Hints you can use on future puzzles

• List the possible (dx,dy) pairs for each required integer length first — a little table of triples and axis moves is invaluable.
• Place the longest lengths early so your remaining space is feasible.
• Watch parity: dx and dy both even or both odd affects whether next moves can land on lattice points with the needed distances.
• If you get stuck, backtrack only a couple of steps and try a different triple for a long move — systematic trial is faster than random wandering.

ACARA v9 alignment — what you’ve practiced

By solving this you demonstrated key Year 8–9 skills: using the Pythagorean theorem to calculate and verify straight-line distances; working with coordinates; reasoning with integer squares and Pythagorean triples; and checking solutions carefully. These are exactly the kinds of proficiencies called for in the ACARA v9 content descriptions for applying Pythagoras and solving geometric problems.

Teacher feedback (in a Nigella Lawson cadence) — for a proficient exemplary outcome

Oh darling — the way you arranged those lengths was simply delicious. You began with the big, bold flavours — the ten and the eight — and built everything else around them with such calm confidence. The arithmetic was pristine, each dx and dy folded into place like butter into flour; your checks with dx^2 + dy^2 = length^2 were meticulous and comforting, like a perfectly glazed tart. For exemplary polish, whisper a little more about why you chose each triple — that gentle narration will make your reasoning irresistible. Keep practising your backtracking technique and listing possible (dx,dy) pairs first; do that and your solutions will be as satisfying and inevitable as the final spoonful of a good pudding.

Bravo — keep feeding your curiosity with these puzzles, and the theorem will reward you every time.

If you’d like, I can: (a) draw this path on a pictured grid, (b) produce the small-table of possible dx,dy pairs for lengths 1–10, or (c) search for a minimal-area embedding of this path that fits inside a smaller grid. Which would you prefer next?