system: [SYSTEM]

user: 3. 热尔岗(Gergonne)点的坐标公式 对于图1所示的四面体ABCD，设在A、B、C、D四点所对之面上的热尔岗点分别为 G 1 、 G 2 、 G 3 、 G 4 。 3.1. 三角形的热尔岗(Gergonne)点 如图2所示，对于图1所示四面体ABCD中的三角形ABC：A、B、C分别与三角形ABC的内切圆在对边上的切点 M B C 、 M A C 、 M A B 的连线 A M B C 、 B M A C 、 C M A B 三线相交的热尔岗点为 G 4 。由前面引言可知 A B = a + b ， B C = b + c ， A C = a + c . 则 G 4 的坐标为 O G 4 → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → 1 a + 1 b + 1 c = b c ⋅ O A → + a c ⋅ O B → + a b ⋅ O C → b c + a c + a b . 图2. 三角形的切心(热尔岗点) 证明：由定比分点公式，有 O M B C → = b ⋅ O C → + c ⋅ O B → b + c ， O M A C → = a ⋅ O C → + c ⋅ O A → a + c ， O M A B → = a ⋅ O B → + b ⋅ O A → a + b . 设 A G 4 G 4 M B C = λ 1 ， B G 4 G 4 M A C = λ 2 ， C G 4 G 4 M A B = λ 3 ， 则 O G 4 → 有以下三种表示： O G 4 → = O A → + λ 1 ⋅ O M B C → 1 + λ 1 = ( b + c ) ⋅ O A → + c λ 1 ⋅ O B → + b λ 1 ⋅ O C → ( b + c ) ( 1 + λ 1 ) ， O G 4 → = O B → + λ 2 ⋅ O M A C → 1 + λ 2 = ( a + c ) ⋅ O B → + a λ 2 ⋅ O C → + c λ 2 ⋅ O A → ( a + c ) ( 1 + λ 2 ) ， O G 4 → = O C → + λ 3 ⋅ O M A B → 1 + λ 3 = ( a + b ) ⋅ O C → + b λ 3 ⋅ O A → + a λ 3 ⋅ O B → ( a + b ) ( 1 + λ 3 ) . 对上面三式的 O A → 、 O B → 、 O C → 的系数分别进行比较，可得 1 1 + λ 1 = c λ 2 ( a + c ) ( 1 + λ 2 ) = b λ 3 ( a + b ) ( 1 + λ 3 ) ， ① 1 1 + λ 2 = a λ 3 ( a + b ) ( 1 + λ 3 ) = c λ 1 ( b + c ) ( 1 + λ 1 ) ， ② 1 1 + λ 3 = b λ 1 ( b + c ) ( 1 + λ 1 ) = a λ 2 ( a + c ) ( 1 + λ 2 ) ， ③ 把上面三式相乘并整理可得 λ 1 λ 2 λ 3 = ( a + c ) ( a + b ) ( b + c ) a b c ; 再分别把① ② ③中的任意两式相乘并整理可得 λ 1 λ 2 = ( a + c ) ( b + c ) c 2 ， λ 1 λ 3 = ( a + b ) ( b + c ) b 2 ， λ 2 λ 3 = ( a + b ) ( a + c ) a 2 . 解得 λ 1 = a ( b + c ) b c ， λ 2 = b ( a + c ) a c ， λ 3 = c ( a + b ) a b . 因此 O G 4 → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → 1 a + 1 b + 1 c = b c ⋅ O A → + a c ⋅ O B → + a b ⋅ O C → b c + a c + a b . 3.2. 四面体的热尔岗(Gergonne)点 对于图1所示的四面体ABCD，由文献 [ 5 ] 可知， A G 1 、 B G 2 、 C G 3 、 D G 4 这四条直线在四面体ABCD内交于一点，设该点为G即四面体ABCD的热尔岗(Gergonne)点。 由前面引言可知 A B = a + b ， B C = b + c ， A C = a + c ， A D = a + d ， B D = b + d ， C D = c + d . 则G的坐标为 O G → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → + 1 d ⋅ O D → 1 a + 1 b + 1 c + 1 d = b c d ⋅ O A → + a c d ⋅ O B → + a b d ⋅ O C → + a b c ⋅ O D → b c d + a c d + a b d + a b c . 证明：由于 O G 4 → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → 1 a + 1 b + 1 c ; 同理 O G 1 → = 1 b ⋅ O B → + 1 c ⋅ O C → + 1 d ⋅ O D → 1 b + 1 c + 1 d ， O G 2 → = 1 a ⋅ O A → + 1 c ⋅ O C → + 1 d ⋅ O D → 1 a + 1 c + 1 d ， O G 3 → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 d ⋅ O D → 1 a + 1 b + 1 d . 因为点G为 A G 1 、 B G 2 、 C G 3 、 D G 4 这四条直线的交点，设 A G G G 1 = λ 1 ， B G G G 2 = λ 2 ， C G G G 3 = λ 3 ， D G G G 4 = λ 4 . 则 O G → 有以下四种表示： O G → = O A → + λ 1 ⋅ O G 1 → 1 + λ 1 = ( 1 b + 1 c + 1 d ) ⋅ O A → + λ 1 ⋅ ( 1 b ⋅ O B → + 1 c ⋅ O C → + 1 d ⋅ O D → ) ( 1 b + 1 c + 1 d ) ( 1 + λ 1 ) ， O G → = O B → + λ 2 ⋅ O G 2 → 1 + λ 2 = ( 1 a + 1 c + 1 d ) ⋅ O B → + λ 2 ⋅ ( 1 a ⋅ O A → + 1 c ⋅ O C → + 1 d ⋅ O D → ) ( 1 a + 1 c + 1 d ) ( 1 + λ 2 ) ， O G → = O C → + λ 3 ⋅ O G 3 → 1 + λ 3 = ( 1 a + 1 b + 1 d ) ⋅ O C → + λ 3 ⋅ ( 1 a ⋅ O A → + 1 b ⋅ O B → + 1 d ⋅ O D → ) ( 1 a + 1 b + 1 d ) ( 1 + λ 3 ) ， O G → = O D → + λ 4 ⋅ O G 4 → 1 + λ 4 = ( 1 a + 1 b + 1 c ) ⋅ O D → + λ 4 ⋅ ( 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → ) ( 1 a + 1 b + 1 c ) ( 1 + λ 4 ) . 对上面四式的 O A → 、 O B → 、 O C → 、 O D → 的系数分别进行比较，可得 1 1 + λ 1 = 1 a λ 2 ( 1 a + 1 c + 1 d ) ( 1 + λ 2 ) = 1 a λ 3 ( 1 a + 1 b + 1 d ) ( 1 + λ 3 ) = 1 a λ 4 ( 1 a + 1 b + 1 c ) ( 1 + λ 4 ) ， 1 1 + λ 2 = 1 b λ 3 ( 1 a + 1 b + 1 d ) ( 1 + λ 3 ) = 1 b λ 4 ( 1 a + 1 b + 1 c ) ( 1 + λ 4 ) = 1 b λ 1 ( 1 b + 1 c + 1 d ) ( 1 + λ 1 ) ， 1 1 + λ 3 = 1 c λ 4 ( 1 a + 1 b + 1 c ) ( 1 + λ 4 ) = 1 c λ 1 ( 1 b + 1 c + 1 d ) ( 1 + λ 1 ) = 1 c λ 2 ( 1 a + 1 c + 1 d ) ( 1 + λ 2 ) ， 1 1 + λ 4 = 1 d λ 1 ( 1 b + 1 c + 1 d ) ( 1 + λ 1 ) = 1 d λ 2 ( 1 a + 1 c + 1 d ) ( 1 + λ 2 ) = 1 d λ 3 ( 1 a + 1 b + 1 d ) ( 1 + λ 3 ) . 对上面四式联立可得 a 1 + λ 1 = b 1 + λ 2 = c 1 + λ 3 = d 1 + λ 4 = λ 1 ( 1 b + 1 c + 1 d ) ( 1 + λ 1 ) = λ 2 ( 1 a + 1 c + 1 d ) ( 1 + λ 2 ) = λ 3 ( 1 a + 1 b + 1 d ) ( 1 + λ 3 ) = λ 4 ( 1 a + 1 b + 1 c ) ( 1 + λ 4 ) . 解得 λ 1 = a ( 1 b + 1 c + 1 d ) ， λ 2 = b ( 1 a + 1 c + 1 d ) ， λ 3 = c ( 1 a + 1 b + 1 d ) ， λ 4 = d ( 1 a + 1 b + 1 c ) . 因此 O G → = 1 a ⋅ O A → + 1 b ⋅ O B → + 1 c ⋅ O C → + 1 d ⋅ O D → 1 a + 1 b + 1 c + 1 d = b c d ⋅ O A → + a c d ⋅ O B → + a b d ⋅ O C → + a b c ⋅ O D → b c d + a c d + a b d + a b c . 3.3. N维单形的热尔岗(Gergonne)点 若n维单形 A 0 A 1 ⋯ A n 存在棱切超球，且各棱长满足： A i A j = a i + a j ， ( 0 ≤ i ≠ j ≤ n ， a i ， a j ∈ R + ). 设 A 0 、 A 1 、 ⋯ 、 A n 各点在其所对之面上的热尔岗点分别为 G 0 、 G 1 、 ⋯ 、 G n 。则诸直线 A 0 G 0 、 A 1 G 1 、 ⋯ 、 A n G n 在n维单形 A 0 A 1 ⋯ A n 内交于一点，设该点为G即n维单形 A 0 A 1 ⋯ A n 的热尔岗点，其坐标为 O G → = ∑ i = 0 n 1 a i ⋅ O A i → ∑ i = 0 n 1 a i . 证明：前面已经证明了当 n = 3 时命题成立。假设当 n = k 时命题同样成立 ( k ≥ 3 ， k ∈ Z ) ，现使用数学归纳法证明当 n = k + 1 时命题依然成立。由归纳假设可得 O G i → = ∑ j = 0 k + 1 1 a j ⋅ O A j → − 1 a i ⋅ O A i → ∑ j = 0 k + 1 1 a j − 1 a i ， 0 ≤ i ≤ k + 1 . 若点G为 A 0 G 0 、 A 1 G 1 、 ⋯ 、 A k + 1 G k + 1 这 k + 2 条直线的交点，设 A i G G G i = λ i . 则 O G → = O A i → + λ i ⋅ O G i → 1 + λ i = ( ∑ j = 0 k + 1 1 a j − 1 a i ) ⋅ O A i → + λ i ⋅ ( ∑ j = 0 k + 1 1 a j ⋅ O A j → − 1 a i ⋅ O A i → ) ( ∑ j = 0 k + 1 1 a j − 1 a i ) ( 1 + λ i ) . 对上式中的i分别取0到 k + 1 之间的各个整数值，并对 O A i → 的系数分别进行比较，可得 1 1 + λ i = λ j ⋅ 1 a i ( ∑ i = 0 k + 1 1 a i − 1 a j ) ( 1 + λ j ) ， ( 0 ≤ i ≠ j ≤ k + 1 ) . 对上式中的i、j取遍各种符合条件的取值并联立可得 a i 1 + λ i = λ i ( ∑ j = 0 k + 1 1 a j − 1 a i ) ( 1 + λ i ) . 解得 λ i = a i ( ∑ j = 0 k + 1 1 a j − 1 a i ) . 因此 O G → = ∑ i = 0 k + 1 1 a i ⋅ O A i → ∑ i = 0 k + 1 1 a i . 故 n ≥ 3 时命题均成立。