{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-similarity-01",
  "mko_id": "mko-similarity",
  "title": {
    "zh-Hant": "相似：以指定正倍率判定標記三角形相似並求面積",
    "en": "Similarity: deciding labeled triangle similarity at a specified positive scale and finding the area"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-similarity-01",
    "source": {
      "mko_id": "mko-similarity",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "d60b59d396063a61d7b04c2039c88873846f8e3486fe0fa326adf55123c006f4",
      "english_sha256": "a5cdadef7a5b49238b9ced0b42ab794c51fdffe9bd83f46239b1c56d962af72e"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "eq_congruence",
      "power_laws",
      "frac_rules",
      "zero_product",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在同一等單位正交平面中，A1=(0,0)、A2=(4,0)、A3=(0,2)，B1=(1,1)、B2=(1,7)、B3=(4,1)，對應 Ai↔Bi，倍率方向為 |BiBj|=k|AiAj|，指定 k=3/2。判定兩者在 k=3/2 下相似，並由已知面積 Area(A)=4 求 Area(B)；再說明 k=2 只被否定為見證、k=−3/2 不合資格，以及方向相反不改變正倍率。",
      "en": "In one equal-unit orthonormal plane, let A1=(0,0), A2=(4,0), A3=(0,2) and B1=(1,1), B2=(1,7), B3=(4,1), with correspondence Ai↔Bi and scale direction |BiBj|=k|AiAj|, at the specified k=3/2. Decide that they are similar at k=3/2, and from the given area Area(A)=4 find Area(B); then explain why k=2 is rejected only as a witness, why k=−3/2 is not eligible, and why opposite orientation does not change the positive scale."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：同一等單位正交歐氏平面 ℝ²，整數座標。",
          "en": "Model: one equal-unit orthonormal Euclidean plane ℝ², integer coordinates."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A1=(0,0)、A2=(4,0)、A3=(0,2)；B1=(1,1)、B2=(1,7)、B3=(4,1)；對應 Ai↔Bi；倍率方向 |BiBj|=k|AiAj|。",
          "en": "A1=(0,0), A2=(4,0), A3=(0,2); B1=(1,1), B2=(1,7), B3=(4,1); correspondence Ai↔Bi; scale direction |BiBj|=k|AiAj|."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "指定倍率 k=3/2（p=3、q=2）；已知 Area(A)=4；比較用候選倍率 k=2 與 k=−3/2。",
          "en": "Specified scale k=3/2 (p=3, q=2); given Area(A)=4; comparison candidate scales k=2 and k=−3/2."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "S(A,B)\\ \\text{at a specified scale}\\ k",
        "after_tex": "S_k(A,B)\\iff k>0\\ \\land\\ A,B\\ \\text{nondegenerate}\\ \\land\\ |B_iB_j|=k|A_iA_j|\\ (ij=12,13,23)",
        "relation": "definition",
        "rule": "mko_sim_slice",
        "conditions": [
          "A_i\\leftrightarrow B_i\\ \\text{explicit}",
          "\\text{scale direction explicit}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 固定判準：在明示對應與倍率方向下，A、B 以倍率 k 相似，當且僅當 k 為正、兩者非退化，且三組對應邊都滿足 |BiBj|=k|AiAj|。方向很重要：這裡是 B 的邊等於 k 乘 A 的邊，不是反過來（反過來的倍率是 1/k）。",
          "en": "Fix the criterion from the MKO: with the correspondence and scale direction explicit, A and B are similar at scale k exactly when k is positive, both are nondegenerate, and all three corresponding sides satisfy |BiBj|=k|AiAj|. The direction matters: here a side of B equals k times a side of A, not the other way round (that scale would be 1/k)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_sim_slice"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出判準，不推出新事實；k 是被檢查的見證，不是由圖形猜出。",
            "en": "This step only states the criterion and derives no new fact; k is a witness to be checked, not guessed from a figure."
          }
        },
        "leaf_basis": "mko_sim_slice"
      },
      {
        "id": "S02",
        "uses": [
          "G3",
          "order_basics"
        ],
        "before_tex": "k=\\tfrac{p}{q}=\\tfrac{3}{2}",
        "after_tex": "p=3>0,\\ q=2>0\\ \\Rightarrow\\ k>0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "q>0"
        ],
        "explanation": {
          "zh-Hant": "MKO 假設 a2：k=p/q 且 q>0，倍率合格當且僅當 p>0。這裡 p=3>0、q=2>0。由 2>0 得 1/2>0（因 2·(1/2)=1>0，若 1/2≤0 則乘以正數 2 得 1≤0，矛盾），再以正數 3 乘之得 3/2>0。所以 k=3/2 是合格的正倍率。",
          "en": "MKO assumption a2: k=p/q with q>0, and the scale is eligible exactly when p>0. Here p=3>0 and q=2>0. From 2>0 we get 1/2>0 (since 2·(1/2)=1>0; if 1/2≤0, multiplying by the positive 2 would give 1≤0, a contradiction), and multiplying by the positive 3 gives 3/2>0. So k=3/2 is an eligible positive scale."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只用正數乘正數仍為正；本步不宣稱 3/2 是唯一可能的倍率（見 S14）。",
            "en": "Only 'positive times positive is positive' is used; the step does not claim that 3/2 is the only possible scale (see S14)."
          }
        },
        "leaf_basis": "order_basics + mul_inverse_nonzero"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "det2_def",
          "det_collinear"
        ],
        "before_tex": "\\det(A_2-A_1,\\,A_3-A_1)",
        "after_tex": "\\det((4,0),(0,2))=4\\cdot 2-0\\cdot 0=8\\neq 0:\\ A\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "det_collinear",
        "conditions": [
          "8\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "A2−A1=(4,0)、A3−A1=(0,2)，行列式 4·2−0·0=8≠0，所以 A1、A2、A3 相異且不共線，A 是非退化三角形（MKO 假設 a1）。",
          "en": "A2−A1=(4,0) and A3−A1=(0,2), with determinant 4·2−0·0=8≠0, so A1, A2, A3 are distinct and noncollinear and A is a nondegenerate triangle (MKO assumption a1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "det2_def",
            "det_collinear"
          ],
          "explanation": {
            "zh-Hant": "重算見 N1；只用 det≠0 ⇒ 不共線。",
            "en": "The recomputation is N1; only det≠0 ⇒ noncollinear is used."
          }
        },
        "leaf_basis": "det2_def + det_collinear + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G2",
          "det2_def",
          "det_collinear"
        ],
        "before_tex": "\\det(B_2-B_1,\\,B_3-B_1)",
        "after_tex": "\\det((0,6),(3,0))=0\\cdot 0-6\\cdot 3=-18\\neq 0:\\ B\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "det_collinear",
        "conditions": [
          "-18\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "B2−B1=(0,6)、B3−B1=(3,0)，行列式 0·0−6·3=−18≠0，所以 B 也是非退化三角形。負號在 S12 解讀為方向。",
          "en": "B2−B1=(0,6) and B3−B1=(3,0), with determinant 0·0−6·3=−18≠0, so B is also a nondegenerate triangle. The minus sign is read as orientation in S12."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "det2_def",
            "det_collinear"
          ],
          "explanation": {
            "zh-Hant": "重算見 N2；負號不影響非退化。",
            "en": "The recomputation is N2; the minus sign does not affect nondegeneracy."
          }
        },
        "leaf_basis": "det2_def + det_collinear + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "dist_formula"
        ],
        "before_tex": "|B_iB_j|=k|A_iA_j|",
        "after_tex": "|B_iB_j|^2=k^2|A_iA_j|^2",
        "relation": "equivalent",
        "rule": "L_scale_sq",
        "conditions": [
          "k>0",
          "|A_iA_j|\\ge 0",
          "|B_iB_j|\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_scale_sq（u=|AiAj|、v=|BiBj|），在 k>0 且長度非負時，長度關係與平方長度關係等價。這正是 MKO 所說「長度乘 k 等價於平方長度乘 k²」；引理在 S05.1 至 S05.3 證明。",
          "en": "By Lemma L_scale_sq (u=|AiAj|, v=|BiBj|), when k>0 and lengths are nonnegative, the length relation is equivalent to the squared-length relation. This is the MKO's 'lengths scale by k exactly when squared lengths scale by k²'; the lemma is proved in S05.1 to S05.3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_scale_sq",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都在子步證明。本稿以 k>0 與非負長度保證 ku≥0，讓平方關係推回長度關係；這是所採用的充分前提。k=0 的算術邊界也可成立，但不符合 S02 的倍率資格；負 k 配正長度時則有符號問題。",
            "en": "Both directions are proved in the sub-steps. The sufficient assumptions k>0 and nonnegative lengths give ku≥0, allowing the squared relation to imply the length relation. The arithmetic boundary k=0 can also work but is ineligible under S02; negative k with positive length causes the sign obstruction."
          }
        },
        "leaf_basis": "eq_congruence + power_laws + zero_product + order_basics"
      },
      {
        "id": "S05.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "k>0,\\ u\\ge 0",
        "after_tex": "ku\\ge 0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "k>0"
        ],
        "explanation": {
          "zh-Hant": "若 u=0，ku=0≥0；若 u>0，兩邊乘正數 k 保持不等號：ku>k·0=0。所以 ku≥0。",
          "en": "If u=0 then ku=0≥0; if u>0, multiplying by the positive k keeps the inequality: ku>k·0=0. So ku≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "本稿採 k>0 作充分前提，且它是相似倍率的資格條件。單就 ku≥0 而言，u≥0 時 k=0 也可用；若 k<0 且 u>0 則 ku<0。這不把零倍率納入本稿的相似資格。",
            "en": "The guide uses k>0 as a sufficient hypothesis and as an eligibility condition for similarity scales. For the arithmetic conclusion ku≥0 alone, k=0 also works when u≥0; if k<0 and u>0 then ku<0. This does not make zero scale eligible in the declared similarity model."
          }
        },
        "leaf_basis": "order_basics + mul_zero + case_analysis"
      },
      {
        "id": "S05.2",
        "uses": [
          "eq_congruence",
          "power_laws"
        ],
        "before_tex": "v=ku",
        "after_tex": "v^2=(ku)^2=k^2u^2",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊平方（同一運算作用於相等的數），再用 (ab)²=a²b²。這是引理的 ⇒ 方向。",
          "en": "Square both sides (the same operation applied to equal numbers), then use (ab)²=a²b². This is the ⇒ direction of the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "只宣稱 ⇒；反向在 S05.3 使用非負條件處理。",
            "en": "Only ⇒ is claimed; the converse uses nonnegativity in S05.3."
          }
        },
        "leaf_basis": "eq_congruence + power_laws"
      },
      {
        "id": "S05.3",
        "uses": [
          "S05.1",
          "zero_product",
          "distributive"
        ],
        "before_tex": "v^2=k^2u^2,\\ v\\ge 0,\\ ku\\ge 0",
        "after_tex": "v=ku",
        "relation": "case_split",
        "rule": "zero_product",
        "conditions": [
          "v\\ge 0",
          "ku\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "v²−(ku)²=(v−ku)(v+ku)=0（N17 核對此恆等式），依零積性質 v−ku=0 或 v+ku=0。前者即 v=ku。後者時 v=−ku≤0（因 S05.1 的 ku≥0），又 v≥0，故 v=0，進而 ku=0，仍得 v=ku。S05.1 至 S05.3 合起來證得引理 L_scale_sq。",
          "en": "v²−(ku)²=(v−ku)(v+ku)=0 (N17 checks this identity), so by the zero-product property v−ku=0 or v+ku=0. The first gives v=ku. In the second, v=−ku≤0 (as ku≥0 by S05.1) and v≥0, so v=0, hence ku=0, and again v=ku. S05.1 to S05.3 together prove Lemma L_scale_sq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05.1",
            "zero_product",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "第二情況使用 ku≥0，本稿由 k>0、u≥0 得到它。若 k=−3/2 且 u>0，ku<0，這條推理不能直接使用；若 u=0 則 ku=0。具體非退化三角形的邊長為正，仍依 S02 排除負倍率。",
            "en": "The second case uses ku≥0, supplied here by k>0 and u≥0. If k=−3/2 and u>0, then ku<0 and that reasoning cannot be used directly; if u=0 then ku=0. The concrete nondegenerate triangles have positive side lengths and S02 still excludes negative scales."
          }
        },
        "leaf_basis": "zero_product + distributive + case_analysis + order_basics"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "frac_rules"
        ],
        "before_tex": "|B_iB_j|^2=\\left(\\tfrac{p}{q}\\right)^2|A_iA_j|^2",
        "after_tex": "|B_iB_j|^2\\,q^2=|A_iA_j|^2\\,p^2",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "q\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "(p/q)²=p²/q²，兩邊同乘 q²（q≠0，所以 q²≠0）得整數交叉乘積式。這是 MKO 所說的精確檢查 B²q²=A²p²：只用整數乘法，不用小數。",
          "en": "(p/q)²=p²/q²; multiplying both sides by q² (q≠0, so q²≠0) gives the integer cross-multiplied form. This is the MKO's exact check B²q²=A²p²: only integer multiplication, no decimals."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向是兩邊同除以 q²，需要 q≠0，由 MKO 的 q>0 保證。",
            "en": "Backward, divide both sides by q², which needs q≠0, guaranteed by the MKO's q>0."
          }
        },
        "leaf_basis": "frac_rules + power_laws + mul_inverse_nonzero + eq_congruence"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "dist_formula"
        ],
        "before_tex": "|A_1A_2|^2,\\ |A_1A_3|^2,\\ |A_2A_3|^2",
        "after_tex": "16,\\ 4,\\ 20",
        "relation": "equivalent",
        "rule": "dist_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "按 12、13、23 順序：(4−0)²+(0−0)²=16；(0−0)²+(2−0)²=4；(0−4)²+(2−0)²=16+4=20。",
          "en": "In the order 12, 13, 23: (4−0)²+(0−0)²=16; (0−0)²+(2−0)²=4; (0−4)²+(2−0)²=16+4=20."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dist_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "重算見 N3、N4、N5；兩邊是同一個數。",
            "en": "The recomputations are N3, N4, N5; both sides name the same number."
          }
        },
        "leaf_basis": "dist_formula + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G2",
          "dist_formula"
        ],
        "before_tex": "|B_1B_2|^2,\\ |B_1B_3|^2,\\ |B_2B_3|^2",
        "after_tex": "36,\\ 9,\\ 45",
        "relation": "equivalent",
        "rule": "dist_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一順序：(1−1)²+(7−1)²=36；(4−1)²+(1−1)²=9；(4−1)²+(1−7)²=9+36=45。",
          "en": "Same order: (1−1)²+(7−1)²=36; (4−1)²+(1−1)²=9; (4−1)²+(1−7)²=9+36=45."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dist_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "重算見 N6、N7、N8。",
            "en": "The recomputations are N6, N7, N8."
          }
        },
        "leaf_basis": "dist_formula + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S06",
          "S07",
          "S08",
          "G3"
        ],
        "before_tex": "|B_1B_2|^2q^2\\ \\text{vs}\\ |A_1A_2|^2p^2",
        "after_tex": "36\\cdot 2^2=144=16\\cdot 3^2",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 12 組：左邊 36·4=144，右邊 16·9=144，相等，所以由 S06 得 |B1B2|²=(3/2)²|A1A2|²。",
          "en": "Pair 12: the left side is 36·4=144 and the right side 16·9=144, equal, so by S06 |B1B2|²=(3/2)²|A1A2|²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S08",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數相等（N9）；只處理第 12 組。",
            "en": "An exact integer equality (N9); only pair 12 is handled."
          }
        },
        "leaf_basis": "numeric_evaluation + eq_refl_symm_trans"
      },
      {
        "id": "S10",
        "uses": [
          "S06",
          "S07",
          "S08",
          "G3"
        ],
        "before_tex": "\\text{pairs }13,\\ 23",
        "after_tex": "9\\cdot 2^2=36=4\\cdot 3^2,\\quad 45\\cdot 2^2=180=20\\cdot 3^2",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 13 組：9·4=36=4·9；第 23 組：45·4=180=20·9。三組都通過交叉乘積檢查，沒有首個不等見證。",
          "en": "Pair 13: 9·4=36=4·9; pair 23: 45·4=180=20·9. All three pairs pass the cross-multiplication check, and there is no first unequal witness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S08",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數相等（N10、N11）。",
            "en": "Exact integer equalities (N10, N11)."
          }
        },
        "leaf_basis": "numeric_evaluation + eq_refl_symm_trans"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S04",
          "S05",
          "S09",
          "S10"
        ],
        "before_tex": "|B_iB_j|^2=\\tfrac{9}{4}|A_iA_j|^2\\ (ij=12,13,23)",
        "after_tex": "S_{3/2}(A,B):\\ 6=\\tfrac32\\cdot 4,\\ 3=\\tfrac32\\cdot 2,\\ 3\\sqrt5=\\tfrac32\\cdot 2\\sqrt5",
        "relation": "implies",
        "rule": "mko_sim_slice",
        "conditions": [
          "k=\\tfrac32>0",
          "A,B\\ \\text{nondegenerate}"
        ],
        "explanation": {
          "zh-Hant": "由 S09、S10 與 S06，三組平方長度都乘 9/4；由 S05（引理 ⇐ 方向，k=3/2>0），三組長度都乘 3/2：6=(3/2)·4、3=(3/2)·2、√45=3√5=(3/2)·√20=(3/2)·2√5。加上 S02 至 S04 的資格，S01 的判準全部成立，所以 A、B 在 k=3/2 下相似。",
          "en": "By S09, S10 and S06, all three squared lengths scale by 9/4; by S05 (the ⇐ direction of the lemma, with k=3/2>0), all three lengths scale by 3/2: 6=(3/2)·4, 3=(3/2)·2, √45=3√5=(3/2)·√20=(3/2)·2√5. Together with the eligibility in S02 to S04, every clause of the criterion in S01 holds, so A and B are similar at k=3/2."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S04",
            "S05",
            "S09",
            "S10"
          ],
          "explanation": {
            "zh-Hant": "目標 G：S_{3/2}(A,B)。足夠條件 H：k>0、兩者非退化、三組邊長都乘 k；H⇒G 由判準 S01。H 由 S02 至 S10 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: S_{3/2}(A,B). Sufficient condition H: k>0, both nondegenerate, all three sides scaled by k; H⇒G by the criterion S01. H is proved forwards in S02 to S10; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_sim_slice + L_scale_sq + and_intro_elim"
      },
      {
        "id": "S12",
        "uses": [
          "S03",
          "S04",
          "S11"
        ],
        "before_tex": "\\det_A=8>0,\\ \\det_B=-18<0",
        "after_tex": "\\text{opposite orientations; scale still}\\ k=\\tfrac32>0",
        "relation": "implies",
        "rule": "mko_sim_slice",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個行列式異號，B 的方向與 A 相反：B 可由 A 先關於 y=x 鏡射、再均勻放大 3/2、再平移 (1,1) 得到，即 Bi=(1+(3/2)y, 1+(3/2)x)，其中 Ai=(x,y)（N18、N19 核對 B2、B3）。MKO 指出鏡射不改變正長度倍率；倍率是長度之比，長度非負，所以倍率仍是 3/2，不是 −3/2。",
          "en": "The determinants have opposite signs, so B has the opposite orientation: B is obtained from A by reflection across y=x, uniform enlargement by 3/2 and translation by (1,1), that is Bi=(1+(3/2)y, 1+(3/2)x) for Ai=(x,y) (N18 and N19 check B2 and B3). The MKO says reflection does not change the positive length scale; the scale is a ratio of nonnegative lengths, so it stays 3/2, not −3/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "S11",
            "mko_sim_slice"
          ],
          "explanation": {
            "zh-Hant": "方向只由行列式正負讀出，相似判定（S11）沒有用到方向；映射公式只是交叉核對。",
            "en": "Orientation is read only from the signs of the determinants, and the decision in S11 did not use orientation; the map formula is only a cross-check."
          }
        },
        "leaf_basis": "mko_sim_slice + order_basics + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "G3"
        ],
        "before_tex": "S_{3/2}(A,B),\\ \\mathrm{Area}(A)=4",
        "after_tex": "\\mathrm{Area}(B)=\\left(\\tfrac32\\right)^2\\cdot 4=9",
        "relation": "theorem_application",
        "rule": "mko_sim_slice",
        "conditions": [
          "S_{3/2}(A,B)",
          "\\mathrm{Area}(A)=4>0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 的面積條款：已確立以正倍率 k 相似，且給定 A 的正面積，則 B 的面積是 k² 倍。k²=9/4，所以 Area(B)=(9/4)·4=9。不是 k·4=6：面積是二維量，乘的是 k²。",
          "en": "By the MKO's area clause: once similarity at a positive scale k is established and a positive area of A is given, the area of B is k² times it. k²=9/4, so Area(B)=(9/4)·4=9. It is not k·4=6: area is two-dimensional and scales by k²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "G3",
            "mko_sim_slice"
          ],
          "explanation": {
            "zh-Hant": "需要 S11 的相似結論與已給的正面積；本步不從座標推導面積理論。",
            "en": "It needs the similarity of S11 and the given positive area; the step does not derive any area theory from coordinates."
          }
        },
        "leaf_basis": "mko_sim_slice + power_laws + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S07",
          "S08"
        ],
        "before_tex": "k=2=\\tfrac{2}{1}",
        "after_tex": "36\\cdot 1^2=36\\neq 64=16\\cdot 2^2:\\ k=2\\ \\text{rejected as a witness}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以 p=2、q=1 做第 12 組交叉乘積：36·1=36，16·4=64，不等，所以第 12 組就是首個不等見證，k=2 不是相似倍率見證。MKO 假設 a3：這只否定 k=2 這個見證，不代表 A、B 不相似（S11 已證 k=3/2 可行）。事實上在此對應下倍率被第 12 組決定：若 |B1B2|=k'|A1A2|，則 6=4k'，k'=3/2。",
          "en": "Cross-multiply pair 12 with p=2, q=1: 36·1=36 and 16·4=64 differ, so pair 12 is the first unequal witness and k=2 is not a similarity witness. MKO assumption a3: this rejects only the witness k=2, not similarity itself (S11 shows k=3/2 works). In fact, under this correspondence pair 12 pins the scale: if |B1B2|=k'|A1A2| then 6=4k', so k'=3/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S08",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "用到 S06 的等價（不等的平方式推出不等的長度關係）；不宣稱任何其他對應下的結論。",
            "en": "It uses the equivalence of S06 (an unequal squared form gives an unequal length relation); nothing is claimed about any other correspondence."
          }
        },
        "leaf_basis": "numeric_evaluation + contrapositive"
      },
      {
        "id": "S15",
        "uses": [
          "G3",
          "S01",
          "S05"
        ],
        "before_tex": "k=-\\tfrac32=\\tfrac{-3}{2}",
        "after_tex": "p=-3\\le 0:\\ k\\ \\text{ineligible}",
        "relation": "implies",
        "rule": "mko_sim_slice",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a2：資格題允許 p≤0，並答否。這裡 p=−3≤0，所以 k=−3/2 不合格。注意 (−3/2)²=9/4，平方交叉乘積會全部通過，所以符號必須另外檢查；直接看長度也一樣：|B1B2|=6，而 (−3/2)·4=−6，長度不可能是負數。",
          "en": "MKO assumption a2: eligibility questions allow p≤0 and answer no. Here p=−3≤0, so k=−3/2 is ineligible. Note that (−3/2)²=9/4, so every squared cross-multiplication would pass; the sign must be checked separately. Looking at lengths directly gives the same: |B1B2|=6 while (−3/2)·4=−6, and a length cannot be negative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "S01",
            "S05",
            "mko_sim_slice"
          ],
          "explanation": {
            "zh-Hant": "S05 的 ⇐ 方向需要 k>0，所以平方檢查不能代替符號檢查；本步不宣稱負倍率的任何幾何意義。",
            "en": "The ⇐ direction of S05 needs k>0, so a squared check cannot replace the sign check; the step claims no geometric meaning for a negative scale."
          }
        },
        "leaf_basis": "mko_sim_slice + order_basics + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "A、B 都非退化（行列式 8 與 −18）。在對應 Ai↔Bi 下三組平方邊長 16、4、20 與 36、9、45 都滿足 B²·2²=A²·3²，所以 A、B 以 k=3/2 相似，平方長度乘 9/4；由 Area(A)=4 得 Area(B)=9。B 的方向與 A 相反，但倍率仍是正的 3/2。k=2 在第 12 組失敗，只否定該見證；k=−3/2 因 p≤0 不合格。",
      "en": "A and B are nondegenerate (determinants 8 and −18). Under Ai↔Bi the squared sides 16, 4, 20 and 36, 9, 45 all satisfy B²·2²=A²·3², so A and B are similar at k=3/2, with squared lengths scaled by 9/4; from Area(A)=4 we get Area(B)=9. B has the opposite orientation, but the scale is still the positive 3/2. k=2 fails at pair 12, which rejects only that witness; k=−3/2 is ineligible because p≤0."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "Area(B)=k·Area(A)=(3/2)·4=6",
          "en": "Area(B)=k·Area(A)=(3/2)·4=6"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "面積乘 k²：(9/4)·4=9。",
          "en": "Area scales by k²: (9/4)·4=9."
        }
      },
      {
        "claim": {
          "zh-Hant": "平方長度也只乘 k，所以 |B1B2|²=(3/2)·16=24",
          "en": "Squared lengths also scale by k, so |B1B2|²=(3/2)·16=24"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "|B1B2|²=36=(9/4)·16；平方長度乘 k²。",
          "en": "|B1B2|²=36=(9/4)·16; squared lengths scale by k²."
        }
      },
      {
        "claim": {
          "zh-Hant": "k=2 在第 12 組失敗，所以 A 與 B 不相似",
          "en": "k=2 fails at pair 12, so A and B are not similar"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "只否定 k=2 這個見證；k=3/2 三組都通過（S11）。",
          "en": "Only the witness k=2 is rejected; k=3/2 passes all three pairs (S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "k=−3/2 的平方交叉乘積全部通過，所以 −3/2 也是合格倍率",
          "en": "With k=−3/2 every squared cross-multiplication passes, so −3/2 is also an eligible scale"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "p=−3≤0 不合格；平方檢查無法分辨 k 與 −k，而 (−3/2)·4=−6 不可能等於長度 6。",
          "en": "p=−3≤0 is ineligible; a squared check cannot tell k from −k, and (−3/2)·4=−6 cannot equal the length 6."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "mko_sim_slice",
        "kind": "cited",
        "source": {
          "mko_id": "mko-similarity",
          "sha256": "d60b59d396063a61d7b04c2039c88873846f8e3486fe0fa326adf55123c006f4",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對明示 Ai↔Bi 的兩個非退化歐氏平面三角形，正倍率 k 的 SSS 相似切片要求每組對應邊長 BiBj=k·AiAj；等價地三組平方長度都乘 k²；若給定 A 的正面積值，B 的面積為 k² 倍；鏡射不改變正長度倍率。",
          "en": "For two nondegenerate Euclidean plane triangles with explicit correspondence Ai↔Bi, the SSS similarity slice at a specified positive scale k requires |BiBj|=k|AiAj| for every corresponding side; equivalently, all three squared side lengths scale by k². If a positive area of A is given, the area of B is k² times that value. Reflection does not change the positive length scale."
        }
      },
      {
        "id": "dist_formula",
        "kind": "cited",
        "source": {
          "mko_id": "mko-distance-formula",
          "sha256": "153d0d0feabadaf33e08081677be0f81709df28ffa5bffaf0176a0f7875e5904",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一等單位正交座標系，P=(x1,y1)、Q=(x2,y2) 的距離為 d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0；距離平方是兩個座標差平方之和。",
          "en": "In one equal-unit orthonormal frame, P=(x1,y1) and Q=(x2,y2) have distance d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0; its square is the sum of the squared coordinate differences."
        }
      },
      {
        "id": "det2_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-determinant",
          "sha256": "d49050d70b58e0d9d2cab2f5ddc64133813f5a99418865b5941af4f1ec240093",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實二階方陣 M=[[a,b],[c,d]] 的行列式為 det(M)=ad−bc；本稿把向量 (a,b)、(c,d) 作為兩行（row）。",
          "en": "A real 2×2 matrix M=[[a,b],[c,d]] has determinant det(M)=ad−bc; here the vectors (a,b) and (c,d) are its rows."
        }
      },
      {
        "id": "det_collinear",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "平面上三點 P、Q、R 共線當且僅當 det(Q−P, R−P)=0；所以 det≠0 推出三點相異且不共線。",
          "en": "Three points P, Q, R of the plane are collinear exactly when det(Q−P, R−P)=0; so det≠0 gives three distinct noncollinear points."
        }
      },
      {
        "id": "L_scale_sq",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2",
          "S05.3"
        ],
        "statement": {
          "zh-Hant": "對實數 k>0、u≥0、v≥0：v=ku ⇔ v²=k²u²。",
          "en": "For reals k>0, u≥0 and v≥0: v=ku ⇔ v²=k²u²."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "4*2-0*0",
        "rhs": "8"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "0*0-6*3",
        "rhs": "-18"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "(4-0)**2+(0-0)**2",
        "rhs": "16"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "(0-0)**2+(2-0)**2",
        "rhs": "4"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "(0-4)**2+(2-0)**2",
        "rhs": "20"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "(1-1)**2+(7-1)**2",
        "rhs": "36"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "(4-1)**2+(1-1)**2",
        "rhs": "9"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "(4-1)**2+(1-7)**2",
        "rhs": "45"
      },
      {
        "id": "N9",
        "step": "S09",
        "lhs": "36*2**2",
        "rhs": "16*3**2"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "9*2**2",
        "rhs": "4*3**2"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "45*2**2",
        "rhs": "20*3**2"
      },
      {
        "id": "N12",
        "step": "S13",
        "lhs": "(3/2)**2*4",
        "rhs": "9"
      },
      {
        "id": "N13",
        "step": "S13",
        "lhs": "(3/2)*4",
        "relation": "!=",
        "rhs": "9"
      },
      {
        "id": "N14",
        "step": "S14",
        "lhs": "36*1**2",
        "relation": "!=",
        "rhs": "16*2**2"
      },
      {
        "id": "N15",
        "step": "S14",
        "lhs": "6/4",
        "rhs": "3/2"
      },
      {
        "id": "N16",
        "step": "S15",
        "lhs": "(-3/2)**2",
        "rhs": "(3/2)**2"
      },
      {
        "id": "N17",
        "step": "S05.3",
        "vars": [
          "v",
          "k",
          "u"
        ],
        "degree": 2,
        "lhs": "(v-k*u)*(v+k*u)",
        "rhs": "v**2-k**2*u**2"
      },
      {
        "id": "N18",
        "step": "S12",
        "lhs": "1+(3/2)*4",
        "rhs": "7"
      },
      {
        "id": "N19",
        "step": "S12",
        "lhs": "1+(3/2)*2",
        "rhs": "4"
      },
      {
        "id": "N20",
        "lhs": "abs(0*0-6*3)/2",
        "rhs": "(3/2)**2*(abs(4*2-0*0)/2)"
      },
      {
        "id": "N21",
        "lhs": "(3/2)*16",
        "relation": "!=",
        "rhs": "36"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 的 SSS 相似切片內工作：明示對應、明示倍率方向、k=p/q 且 q>0、整數座標、以交叉相乘精確比較。面積只用 MKO 給出的「面積為 k² 倍」條款，從已給的正面積計算；不發展一般面積理論，不搜尋其他對應。不改動 canonical MKO 或英文 edition。",
      "en": "Works only inside the MKO's SSS similarity slice: explicit correspondence, explicit scale direction, k=p/q with q>0, integer coordinates, exact comparison by cross-multiplication. The area uses only the MKO's clause that the area scales by k², starting from the given positive area; no general area theory is developed and no other correspondence is searched. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "k=\\tfrac{p}{q}",
        "指定的長度倍率，q>0",
        "the specified length scale, q>0"
      ],
      [
        "|A_iA_j|",
        "A 的第 ij 組邊長",
        "the side length ij of A"
      ],
      [
        "S_k(A,B)",
        "在對應 Ai↔Bi 下以倍率 k 相似",
        "similar at scale k under Ai↔Bi"
      ],
      [
        "\\mathrm{Area}(A)",
        "A 的已知正面積",
        "the given positive area of A"
      ],
      [
        "\\det(u,v)",
        "以 u、v 為兩行（row）的二階行列式",
        "the 2×2 determinant with rows u and v"
      ]
    ],
    "md_error_controls": [
      {
        "title": "長度乘 k，面積乘 k² / Lengths scale by k, area by k²",
        "zh": "MKO 的誤解之一是長度和面積都只乘 k，或把平方倍率當長度倍率。本例長度乘 3/2（6=(3/2)·4），平方長度乘 9/4（36=(9/4)·16），面積乘 9/4（4 變成 9）。若面積只乘 k 會得到 6；若把平方長度乘 k 會得到 24 而不是 36。",
        "en": "One misconception the MKO lists is scaling both lengths and area by k, or taking the squared scale for the length scale. Here lengths scale by 3/2 (6=(3/2)·4), squared lengths by 9/4 (36=(9/4)·16), and area by 9/4 (4 becomes 9). Scaling area by k would give 6; scaling squared lengths by k would give 24 instead of 36."
      },
      {
        "title": "錯誤的 k 只否定該見證 / A wrong k rejects only that witness",
        "zh": "k=2 在第 12 組失敗（36≠64），但這只說明 k=2 不是見證；以 k=3/2 檢查則三組都通過。MKO 假設 a3 明言：指定倍率不符合只否定那個倍率見證，不代表不存在其他相似倍率。",
        "en": "k=2 fails at pair 12 (36≠64), but this only shows that k=2 is not a witness; with k=3/2 all three pairs pass. MKO assumption a3 says explicitly that failure at the specified k rejects only that scale witness, not every possible similarity scale."
      },
      {
        "title": "負座標係數與負長度倍率不同 / A negative coordinate coefficient is not a negative length scale",
        "zh": "方向相反（行列式異號）或座標乘負數，都不會讓長度倍率變負。MKO 指出對座標做係數 −2 的均勻縮放，長度倍率是正的 2。本例的 B 含鏡射，但倍率仍是 3/2；k=−3/2 因 p≤0 不合格，雖然它的平方與 3/2 的平方相同。",
        "en": "Opposite orientation (determinants of opposite signs) or multiplying coordinates by a negative number never makes the length scale negative. The MKO notes that uniform scaling of coordinates by −2 has positive length scale 2. Here B involves a reflection, yet the scale is 3/2; k=−3/2 is ineligible because p≤0, even though its square equals that of 3/2."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "面積結論只用 MKO 的 k² 面積條款，從已給的 Area(A)=4 計算。N20 以 |det|/2 做獨立數值交叉核對（18/2=9），該面積公式未在本稿證明，論證不依賴它。",
        "en": "The area conclusion uses only the MKO's k² area clause, starting from the given Area(A)=4. N20 is an independent numerical cross-check by |det|/2 (18/2=9); that area formula is not proved here and the argument does not depend on it."
      },
      {
        "zh-Hant": "det_collinear 是本稿的前提，未在本稿中證明；本稿只檢查題目給定的對應。",
        "en": "det_collinear is a premise of this guide and is not proved here; only the given correspondence is checked."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "a3e5da3d8b98c14fda10b000ea7223cdb705cfc1824ae646dd1becfcce543568",
    "original_source_sha256": "58669e6075a82ec69759d174d0aa3477b52d7ff3ba2964ddac1c24f126d02aed",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M021"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}