欧拉的四面体问题 历史上欧拉提出了这样一个问题:如何用四面体的棱长去表示它的体积?这 个问题可以用矢量代数的基本知识来解决 设A,B,C三点的坐标分别为(a2,b,c)(i=1,2,3),并设四面体O-ABC 的六条棱长分别为l,m,n,p,q,r.由立体几何知识知道,该四面体的体积V等于 以矢量OA.OB,OC为棱的平行六面体的体积的而由空间解析几何可知,当 OA,OB,OC组成右手系时,以它们为棱的平行六面体的体积 V6=(OAx OB).OC b b2 于是得 6V b 将上式平方,得 b2 b3 C3 由于行列式转置后其值不变,将第二个进行转置后再相乘,得 b1 C1a1 a2 a b2c2|·b1b2b3 a12+b12+c12a1a2+b1b2+c1c2a1a3+b1b3+c1e3 a1a2+b1b2+c12a2+b2+c2a2a3+b2b3+c2c3 a1a3+b1b3+c1c3a2a3+b2b3+c23a32+b32+c32 根据矢量的数量积的坐标表示,有 0A 0A=012+b2+C12, 0A.0B=0102+b1b2+Cc? OA.OC=0103+bb3+C1C3, OBOB=a2+b2+c2 OB·OC=a2a3+b2b3 于是
✁✂✄☎✆✝✞ ✟✠✡☛☞✌✍✎✏✑✒✓✔✕✖✗✘✙✚✛✜✢✣✤✥✦✧★✢✜✩ ✪✏ ✓✔✕✫✬✙✭✮✯✰✢✱✲✳✴✵✶✷✸ ✹ A, B, C ✺✻✢✼✽✾✿❀ (ai , bi , ci)(i = 1, 2, 3), ❁ ✹✚✛✜ O − ABC ✢❂❃✣✤✾✿❀ l, m, n, p, q, r. ❄ ❅✜❆✘✳✴✳❇❈❉✚✛✜✢✜✩ V ❊❋ ✬✭✮ −→OA, −−→OB, −→OC ❀✣✢●❍❂✛✜✢✜✩✢ 1 6 , ■❄❏❑✶▲❆✘✫✳❈▼ −→OA, −−→OB, −→OC ◆❖P◗❘❙❈✬★❚❀✣✢●❍❂✛✜✢✜✩ V6 = (−→OA × −−→OB) · −→OC = a1 b1 c1 a2 b2 c2 a3 b3 c3 ❋❯❱ 6V = a1 b1 c1 a2 b2 c2 a3 b3 c3 ❲✡❳●❨❈❱ 36V 2 = a1 b1 c1 a2 b2 c2 a3 b3 c3 · a1 b1 c1 a2 b2 c2 a3 b3 c3 ❄❋ ❍❩❳❬❭❪❫❴❵❛❈❲❜❝✓❞❍❬❭❪❡❢❣❈ ❱ 36V 2 = a1 b1 c1 a2 b2 c2 a3 b3 c3 · a1 a2 a3 b1 b2 b3 c1 c2 c3 = a1 2 + b1 2 + c1 2 a1a2 + b1b2 + c1c2 a1a3 + b1b3 + c1c3 a1a2 + b1b2 + c1c2 a2 2 + b2 2 + c2 2 a2a3 + b2b3 + c2c3 a1a3 + b1b3 + c1c3 a2a3 + b2b3 + c2c3 a3 2 + b3 2 + c3 2 . ❤✐✭✮✢✰✮✩✢✼✽✦✧❈❥ −→OA · −→OA = a1 2 + b1 2 + c1 2 , −→OA · −−→OB = a1a2 + b1b2 + c1c2 −→OA · −→OC = a1a3 + b1b3 + c1c3, −−→OB · −−→OB = a2 2 + b2 2 + c2 2 −−→OB · −→OC = a2a3 + b2b3 + c2c3, −→OC · −→OC = a3 2 + b3 2 + c3 2 ❋❯ 1
OA. OAOA.OB OA OA. OC OB OC 由矢量的数量积的定义,又有 OA 0A=OA 0=p 同理 - q 再由余弦定理,可得 OA.OB=pq·cos6-p2+q2 同理 OB OC 2 将以上各式代入(1),得 这就是欧拉的四面体求积公式 2
36V 2 = −→OA · −→OA −→OA · −−→OB −→OA · −→OC −→OA · −−→OB −−→OB · −−→OB −−→OB · −→OC −→OA · −→OC −−→OB · −→OC −→OC · −→OC (1) ❄ ✭✮✢✰✮✩✢❦❧❈♠❥ −→OA · −→OA = | −→OA| 2 cos θ = p 2 ♥♦ −→OA · −−→OB = q 2 , −→OC · −→OC = r 2 ❡ ❄ ♣q❦♦❈✫❱ −→OA · −−→OB = p · q · cos θ = p 2 + q 2 − n 2 2 ♥♦ −→OA · −→OC = p 2 + r 2 − m2 2 , −−→OB · −→OC = q 2 + r 2 − l 2 2 ❲✬✡r❳✯s (1), ❱ 36V 2 = p 2 p 2+q 2−n 2 2 p 2+r 2−m2 2 p 2+q 2−n 2 2 q 2 q 2+r 2−l 2 2 p 2+r 2−m2 2 q 2+r 2−l 2 2 r 2 ✏t❯ ☛☞✢✚✛✜✉✩✈❳✸ 2