Pythagorean Paths solution & strict teacher feedback for a 14‑year‑old — Beast Academy Practice 5D (sequence 2,2,1,10,5,5,8,1,2,2,2)

What the puzzle asks

You must place a continuous path through grid points so the distances between consecutive points are, in order: 2, 2, 1, 10, 5, 5, 8, 1, 2, 2, 2. We will build one valid integer‑grid solution and explain the reasoning so you can understand (and reproduce) the method.

Key idea / allowed steps

• All points lie on integer lattice points (x,y) with unit spacing.
• A move of length L must correspond to a vector (dx,dy) with dx^2 + dy^2 = L^2.
• Small lengths like 1 and 2 are axis moves: 1 → (±1,0) or (0,±1). 2 → (±2,0) or (0,±2).
• Length 5 can be either axis 5 or the 3–4–5 triple: (±3,±4) or (±4,±3). Length 10 can be (±6,±8) or axis 10. Length 8 must be axis 8 because there is no non‑zero integer pair (a,b) with a^2+b^2=64 except (8,0) or (0,8).

One valid path (start at (0,0))

We choose step vectors that match the required lengths and produce a simple path. The list below shows each point (p0 through p11). Verify each consecutive distance equals the required length.

p0 = (0, 0)
1) length 2  → move ( +2,  0)  => p1 = ( 2,  0)
2) length 2  → move (  0, +2)  => p2 = ( 2,  2)
3) length 1  → move ( +1,  0)  => p3 = ( 3,  2)
4) length 10 → move ( +6, +8)  => p4 = ( 9, 10)
5) length 5  → move ( -3, +4)  => p5 = ( 6, 14)
6) length 5  → move ( -3, -4)  => p6 = ( 3, 10)
7) length 8  → move ( -8,  0)  => p7 = ( -5, 10)
8) length 1  → move (  0, +1)  => p8 = ( -5, 11)
9) length 2  → move (  0, +2)  => p9 = ( -5, 13)
10) length 2 → move ( +2,  0)  => p10= ( -3, 13)
11) length 2 → move ( +2,  0)  => p11= ( -1, 13)

Verification (quick)

• p0→p1: dx=2,dy=0 → length = sqrt(4)=2
• p1→p2: dx=0,dy=2 → length = 2
• p2→p3: dx=1,dy=0 → length = 1
• p3→p4: dx=6,dy=8 → length = sqrt(36+64)=sqrt(100)=10
• p4→p5: dx=-3,dy=4 → length = 5 (3–4–5 triple)
• p5→p6: dx=-3,dy=-4 → length = 5
• p6→p7: dx=-8,dy=0 → length = 8
• p7→p8: dx=0,dy=1 → length = 1
• p8→p9: dx=0,dy=2 → length = 2
• p9→p10: dx=2,dy=0 → length = 2
• p10→p11: dx=2,dy=0 → length = 2

This is one correct path. There are other valid placements: you can reflect or rotate segments, or swap the 5‑length (3,4) direction with a vertical 5. The important part is checking dx^2 + dy^2 = L^2 for every step.

How I decided each move — a reproducible strategy

1. List the possible integer vectors for each required length. For example: 1 → (±1,0),(0,±1). 2 → (±2,0),(0,±2). 5 → (±3,±4),(±4,±3),(±5,0),(0,±5). 10 → (±6,±8),(±8,±6),(±10,0),(0,±10). 8 → only axis (±8,0) or (0,±8).
2. Pick an easy starting point (0,0) and try to keep large steps from running off to impossible areas. For long lengths (8 and 10) prefer Pythagorean triples that make the path fold back toward the remaining unused area so you can still place the last small steps.
3. After choosing each step, check that the remaining room can still accommodate the remaining long steps. If you get stuck later, undo the last long step and try its other orientation (flip signs, swap dx and dy).
4. Always verify by squaring dx and dy; mistakes usually come from forgetting negative signs or swapping dx/dy incorrectly.

ACARA v9 alignment

This exercise practices these ACARA v9 ideas: using the Pythagorean theorem to calculate distances in the Cartesian plane, representing points with coordinates, reasoning through multi‑step problems and planning. It builds procedural fluency (computing lengths) and reasoning/problem solving (choosing good vectors and planning so later moves remain possible).

Teacher feedback — Amy Chua Tiger Mother cadence (for a student aiming for exemplary / proficient outcome)

Listen carefully. Good effort is not enough. You must show every calculation, check every distance, and explain why each long move (8 and 10) was chosen. If your written work omits the vector for a move—if you just draw it—then you have failed to demonstrate understanding. We will not accept guessing. We will accept a clear plan, the chosen vectors, and a final verification that dx^2 + dy^2 = L^2 for each step. That is the standard of excellence.

If you produced exactly the coordinate list above and wrote out the checks, exemplary. If you tried different vectors but could not finish the path, that’s fine — but show your trials and where the path became impossible. Show why a given choice blocks later moves. That level of reflection is required for a proficient to exemplary grade.

To improve: redo the puzzle and hand in

1. a neat coordinate list (like p0=(0,0), p1=(2,0), ...),
2. a one‑line justification for each move (e.g. p3→p4: dx=6,dy=8 so length=10),
3. a short paragraph explaining your planning for the long steps (where you avoided running off the grid),
4. a final comment: could the path be reflected or rotated? show one alternate variation or explain why not.

Do that and you will demonstrate the knowledge, the reasoning, and the attention to detail that ACARA v9 expects. Anything less — sloppiness, missing checks, or no explanation — is insufficient. I expect precision. Now finish it properly.

Quick tips for future Pythagorean Paths

• Always write possible dx,dy for each required length before drawing.
• Place long steps early but oriented so they don’t isolate small leftover regions where you can’t finish the last steps.
• When stuck, reflect a segment across an axis or swap the order of a symmetric pair of moves (e.g. use (±3,±4) as (±4,±3) instead).

If you want, upload a picture of your attempted path and I will give exact corrections in the same strict style, pointing out any arithmetic or planning errors.