Have you ever wondered whether geometry and fashion have anything in common? I mean, both ask you to walk a certain line, make bold choices, and trust that the smallest detail can change the whole look. Today, darling, that line is a Pythagorean path across a 7×7 grid. The accessories: five marked dots. The request: a single continuous path that gives lengths, in order, 1, √10, 5, √5. The question I whisper to my stilettoed brain is not so much who wore it best, but which dot goes next?
Let me set the scene. We have five points, labelled for clarity:
- A = (5,0)
- B = (6,3)
- C = (1,3)
- D = (6,2)
- E = (3,2)
Think of these like stops on a glamorous shopping trip. We must visit each exactly once, in a single stroll, with the four consecutive distances between stops equal to the chic sequence: 1, √10, 5, √5. The good news? The Pythagorean theorem is our stylist: distance between (x1,y1) and (x2,y2) is √[(x2−x1)^2 + (y2−y1)^2]. But we save time and drama by comparing squared distances: 1^2=1, (√10)^2=10, 5^2=25, (√5)^2=5. No unnecessary square-rooting while you sip your coffee.
Now the delicious, methodical tasting of possibilities. You catalogue obvious pairings — pairs that give squared distances 1, 10, 25 and 5. There is a tiny, delightful surprise: one pair gives 1 and it is unique. Look:
Compute squared distances for the key pairs used in the final path: B = (6,3), D = (6,2) BD: dx = 0, dy = 1 => 0^2 + 1^2 = 1 => distance = 1 B = (6,3), A = (5,0) BA: dx = -1, dy = -3 => 1 + 9 = 10 => distance = √10 A = (5,0), C = (1,3) AC: dx = -4, dy = 3 => 16 + 9 = 25 => distance = 5 C = (1,3), E = (3,2) CE: dx = 2, dy = -1 => 4 + 1 = 5 => distance = √5
See? BD is the only pair with squared distance 1. That is the fashion equivalent of a red shoe - instantly recognizable and impossible to miss. Because the list must begin with a segment of length 1, BD must be that first segment, or its reverse, DB. Having placed that first delicate stitch, the rest of the dress — or path — falls into place much more gracefully.
So stitch the path: D → B → A → C → E. In coordinate talk: (6,2) → (6,3) → (5,0) → (1,3) → (3,2). Each consecutive pair produces the required distances in order: 1, √10, 5, √5. We tested the arithmetic; we didn’t skip the tasting. That little Pythagorean check is the signature move that separates careful reasoning from guesswork.
Why does this feel unique? Because that one-length-1 pair pins two adjacent positions in any valid ordering. Once BD sits in the chain, the point B must also connect next to a point at distance √10. Scanning the points, only A sits exactly √10 from B. Then A must meet a point at distance 5 — and lo, C is the one. Finally, C meets E at √5. The sequence folds together without contradiction. If any of these pairings had competitors, we would have to trial many more outfits; instead the math gives us a tiny, decisive wardrobe miracle.
Hints — softly given, like a stylist with a measuring tape:
- Always compare squared distances to match required squared lengths 1, 10, 25, 5.
- Make a small table of distances between every pair of the five points before you try ordering them; it tastes each ingredient separately.
- Spot the unique pair for distance 1 — that gives you a necessary adjacency and prunes the search drastically.
- If you prefer pictures, draw the 7×7 grid and label the points, then mark possible length-vectors like (1,0), (1,3), (2,1) and their negatives.
Rubrics — Nigella-style, because beauty and precision deserve to be celebrated with a spoonful of warmth:
Proficient
- Path is correct and connects the five points matching the given lengths.
- Distances are calculated using Pythagoras; arithmetic mostly correct.
- Work is readable; student explains main choices (for example, why length 1 must come first).
- Minor omissions may be present, but the solution is complete — like a well-executed dinner with one missing garnish.
Exemplar
- Every step crisply shown: each pair checked, squares and sums displayed — nothing left to guess.
- Strategic reasoning: unique distances used to prune possibilities, symmetry noted, dead-ends explained.
- Neat diagram with labels, path drawn and each segment annotated with its length.
- Reflection included: why other orders fail, and a suggested extension to try next.
Alignment note for teachers: this task uses the distance formula (Pythagoras in coordinates), reasoning about vectors like (1,2)→√5 and (1,3)→√10, and demonstrates planning and checking — exactly the Year 9 style ACARA outcomes for geometry and reasoning.
And finally, a little parting praise, because we should always applaud precise work: you used the Pythagorean theorem like a teaspoon of salt — carefully, confidently and at just the right moments. You ordered the dots, checked each pair, and found the only path that sings. Shall we call it fashionably mathematical? Absolutely.
Faintly glamorous teacher note: The chain D (6,2) → B (6,3) → A (5,0) → C (1,3) → E (3,2) is the valid path. Show your calculations; prove uniqueness by demonstrating the unique distance-1 pair and following forced choices.
Now go practice this elegant little choreography. Draw, label, square, add and reduce — and like any great Carrie moment, ask yourself: if geometry is a stroll, where will your next pair of shoes take you?
50-word exemplar teacher comment (Tiger Mother cadence): Excellent: you solved the Pythagorean Paths accurately, showing precise coordinate calculations, systematic checking of each segment length, and clear reasoning. Continue rigorous practice: compare squared distances, reduce errors, and justify choices. Meets ACARA v9 expectations for applying the Pythagorean theorem in coordinate contexts with procedural fluency and logical argumentation consistently.