正三角形的最小整数边长
已知，在正三角形ABC中，有P，Q两个内点，这两点到三个顶点的距离分别是六个不同的正整数，求三角形ABC的最小整数边长。 这个可以尝试使用四面体体积公式：https://bbs.emath.ac.cn/thread-2515-1-1.html
让四面体a=b=c,而且要求a,d,e,f都是正整数，体积为0，看不定方程解的情况，而且至少要求a比d,e,f都大 化简后好像要求
$(-d^4 -e^4 -f^4 + 2e^2 d^2 + 2f^2 d^2 + 2f^2 e^2 ) a^2 = f^2 e^2 d^2$
所以要求$-d^4 -e^4 -f^4 + 2e^2 d^2 + 2f^2 d^2 + 2f^2 e^2 $是完全平方，而且$a|f e d$ 如同mathe所说，四面体体积为0，得到\[
\begin{vmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & a^2 & a^2 & d^2 \\
1 & a^2 & 0 & a^2 & e^2 \\
1 & a^2 & a^2 & 0 & f^2 \\
1 & d^2 & e^2 & f^2 & 0 \\
\end{vmatrix} =0\]
也就是 $a^4 -a^2(d^2 + e^2 + f^2) + (d^4 + e^4 + f^4-d^2 f^2 - e^2 f^2 - d^2 e^2) =0$,于是$a^2 = \frac{1}{2} \left(d^2+e^2+f^2 \pm \sqrt{3} \sqrt{2 d^2 e^2+2 d^2 f^2+2 e^2 f^2-d^4-e^4-f^4}\right)$
暴力美学，算得最小的解是
${5, 5,5, 3, 7, 8},{5, 5,5, 16, 19, 21}$
{5,{{3,5,7,8},{5,16,19,21}}}
{7,{{3,5,7,8},{7,8,13,15},{7,33,37,40}}}
{8,{{3,5,7,8},{7,8,13,15}}}
{9,{{9,15,21,24},{9,56,61,65}}}
{10,{{6,10,14,16},{10,32,38,42}}}
{11,{{11,24,31,35},{11,85,91,96}}}
{13,{{7,8,13,15},{13,35,43,48}}}
{14,{{6,10,14,16},{14,16,26,30},{14,66,74,80}}}
{15,{{7,8,13,15},{9,15,21,24},{15,25,35,40},{15,48,57,63}}}
{16,{{5,16,19,21},{6,10,14,16},{14,16,26,30},{16,39,49,55}}}
{19,{{5,16,19,21},{19,80,91,99}}}
{20,{{12,20,28,32},{20,64,76,84}}}
{21,{{5,16,19,21},{9,15,21,24},{21,24,39,45},{21,35,49,56}}}
{24,{{9,15,21,24},{11,24,31,35},{21,24,39,45},{24,40,56,64}}}
{26,{{14,16,26,30},{26,70,86,96}}}
{28,{{12,20,28,32},{28,32,52,60}}}
{30,{{14,16,26,30},{18,30,42,48},{30,50,70,80}}}
{32,{{10,32,38,42},{12,20,28,32},{28,32,52,60},{32,45,67,77}}}
{33,{{7,33,37,40},{33,55,77,88}}}
{35,{{11,24,31,35},{13,35,43,48},{15,25,35,40},{21,35,49,56},{35,40,65,75}}}
{39,{{16,39,49,55},{21,24,39,45}}}
{40,{{7,33,37,40},{15,25,35,40},{24,40,56,64},{35,40,65,75},{40,51,79,91}}}
{42,{{10,32,38,42},{18,30,42,48},{42,48,78,90}}}
{45,{{21,24,39,45},{27,45,63,72},{32,45,67,77}}}
{48,{{13,35,43,48},{15,48,57,63},{18,30,42,48},{22,48,62,70},{42,48,78,90}}}
{49,{{16,39,49,55},{21,35,49,56}}}
{55,{{16,39,49,55},{33,55,77,88}}}
{56,{{9,56,61,65},{21,35,49,56},{24,40,56,64}}}
{60,{{28,32,52,60},{36,60,84,96}}}
{63,{{15,48,57,63},{17,63,73,80},{27,45,63,72}}}
{64,{{20,64,76,84},{24,40,56,64}}}
{65,{{9,56,61,65},{35,40,65,75}}}
{70,{{22,48,62,70},{26,70,86,96},{30,50,70,80}}}
{77,{{32,45,67,77},{33,55,77,88}}}
{80,{{14,66,74,80},{17,63,73,80},{19,80,91,99},{30,50,70,80}}}
{84,{{20,64,76,84},{36,60,84,96}}}
{91,{{11,85,91,96},{19,80,91,99},{40,51,79,91}}}
{96,{{11,85,91,96},{26,70,86,96},{36,60,84,96}}}
wayne 发表于 2020-9-11 17:37
如同mathe所说，四面体体积为0，得到\[|\left(
\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1 \\
要求是内点,不能在边上或者外部。 wayne 发表于 2020-9-11 17:37
如同mathe所说，四面体体积为0，得到\[|\left(
\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1 \\
从中提出合乎内点要求的结果。
{{{7, 33, 37}, {15, 25, 35}}, 40}
{{{13, 35, 43}, {18, 30, 42}}, 48}
{{{14, 66, 74}, {17, 63, 73}}, 80}
{{{11, 85, 91}, {26, 70, 86}, {36, 60, 84}}, 96} {{331},111,221,280}
{{331},49,285,296}
f:=Times@@Cases,{x_,y_}->x^Floor]
Do],(a^4-a^2b^2+b^4-a^2c^2-b^2c^2+c^4-a^2#^2-b^2#^2-c^2#^2+#^4==0&&c<#<a+b)&];If],{c,3,1000},{b,c-1},{a,Select,GCD[#,b,c]==1&]}]出现两个311就可以了
直接用公式在a,b,c较大的时候效率更高些：
DoSqrt[-a^4+2a^2b^2-b^4+2a^2c^2+2b^2c^2-c^4]]/Sqrt];
If],
{c,3,1000},{b,Floor,c-1},{a,Select,GCD[#,b,c]==1&]}]
根式里的刚好是海伦公式的表达$2 a^2 b^2+2 a^2 c^2-a^4+2 b^2 c^2-b^4-c^4=(-a+b+c) (a+b-c) (a-b+c) (a+b+c) = xyz(x+y+z)$,也即是，$a=\frac{1}{2} \sqrt{x^2+y^2+z^2+y z+x y+x z\pm2 \sqrt{3} \sqrt{x y z (x+y+z)}}$
如果能得到$xyz(x+y+z) =3n^2$的通解就好了。
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