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		<summary type="html">&lt;p&gt;by freem.ai...&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&lt;br /&gt;
{{DISPLAYTITLE:Tetrahedron volume calculation}}&lt;br /&gt;
&lt;br /&gt;
&amp;#039;&amp;#039;This conversation was exported from ChatGPT&amp;#039;&amp;#039;&amp;lt;ref&amp;gt;{{cite web|title=Tetrahedron volume calculation|url=https://chatgpt.com/share/69440b7b-41f8-8010-9331-140a6dd5eed2|publisher=ChatGPT|access-date=2025-12-18}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== gpt-5-2-thinking ==&lt;br /&gt;
=== User: Think well to give a correct answer, and if possible use two methods to make sure the result is correct. ===&lt;br /&gt;
Think well to give a correct answer, and if possible use two methods to make sure the result is correct. Just to make sure you correctly understand the requirements:&lt;br /&gt;
* the sides are integer numbers&lt;br /&gt;
* the volume is an integer number&lt;br /&gt;
* the areas of the sides are irrational&lt;br /&gt;
* for (1) find the volume of the tetrahedron with minimum a+b+c+d+e+f&lt;br /&gt;
* for (2) find the volume of the tetrahedron that has volume = a+b+c+d+e+f&lt;br /&gt;
&lt;br /&gt;
=== Assistant: I’ll use the standard labeling with vertices A,B,C,DA,B,C,DA,B,C,D and edge-lengths ===&lt;br /&gt;
I’ll use the standard labeling with vertices A,B,C,DA,B,C,DA,B,C,D and edge-lengths&lt;br /&gt;
&lt;br /&gt;
AB=a, AC=b, AD=c, BC=d, BD=e, CD=fAB=a,\ AC=b,\ AD=c,\ BC=d,\ BD=e,\ CD=fAB=a, AC=b, AD=c, BC=d, BD=e, CD=f&lt;br /&gt;
Face-triangles are: (a,b,d),(a,c,e),(b,c,f),(d,e,f)(a,b,d), (a,c,e), (b,c,f), (d,e,f)(a,b,d),(a,c,e),(b,c,f),(d,e,f).&lt;br /&gt;
&lt;br /&gt;
A face area is rational iff&lt;br /&gt;
&lt;br /&gt;
16 Δ2=(x+y+z)(−x+y+z)(x−y+z)(x+y−z)16\,\Delta^2=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)16Δ2=(x+y+z)(−x+y+z)(x−y+z)(x+y−z)&lt;br /&gt;
is a perfect square integer (then Δ=K/4\Delta=\sqrt{K}/4Δ=K​/4 is rational). So “irrational face area” ⇔ that KKK is not a square.&lt;br /&gt;
&lt;br /&gt;
Volume check (two independent exact methods):&lt;br /&gt;
&lt;br /&gt;
Method 1 (Gram 3×3):&lt;br /&gt;
Let u=AB→,v=AC→,w=AD→u=\overrightarrow{AB}, v=\overrightarrow{AC}, w=\overrightarrow{AD}u=AB,v=AC,w=AD. Then&lt;br /&gt;
&lt;br /&gt;
u ⁣⋅ ⁣v=a2+b2−d22,u ⁣⋅ ⁣w=a2+c2−e22,v ⁣⋅ ⁣w=b2+c2−f22u\!\cdot\! v=\frac{a^2+b^2-d^2}{2},\quad&lt;br /&gt;
u\!\cdot\! w=\frac{a^2+c^2-e^2}{2},\quad&lt;br /&gt;
v\!\cdot\! w=\frac{b^2+c^2-f^2}{2}u⋅v=2a2+b2−d2​,u⋅w=2a2+c2−e2​,v⋅w=2b2+c2−f2​&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
V2=det⁡(Gram(u,v,w))36.V^2=\frac{\det(\text{Gram}(u,v,w))}{36}.V2=36det(Gram(u,v,w))​.&lt;br /&gt;
Method 2 (Cayley–Menger 5×5):&lt;br /&gt;
&lt;br /&gt;
288V2=det⁡(0111110a2b2c21a20d2e21b2d20f21c2e2f20).288V^2=\det&lt;br /&gt;
\begin{pmatrix}&lt;br /&gt;
0&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1\\&lt;br /&gt;
1&amp;amp;0&amp;amp;a^2&amp;amp;b^2&amp;amp;c^2\\&lt;br /&gt;
1&amp;amp;a^2&amp;amp;0&amp;amp;d^2&amp;amp;e^2\\&lt;br /&gt;
1&amp;amp;b^2&amp;amp;d^2&amp;amp;0&amp;amp;f^2\\&lt;br /&gt;
1&amp;amp;c^2&amp;amp;e^2&amp;amp;f^2&amp;amp;0&lt;br /&gt;
\end{pmatrix}.288V2=det​01111​10a2b2c2​1a20d2e2​1b2d20f2​1c2e2f20​​.&lt;br /&gt;
&lt;br /&gt;
==== (1) Minimum ====&lt;br /&gt;
𝑎&lt;br /&gt;
+&lt;br /&gt;
𝑏&lt;br /&gt;
+&lt;br /&gt;
𝑐&lt;br /&gt;
+&lt;br /&gt;
𝑑&lt;br /&gt;
+&lt;br /&gt;
𝑒&lt;br /&gt;
+&lt;br /&gt;
𝑓&lt;br /&gt;
a+b+c+d+e+f ⇒ report the volume&lt;br /&gt;
&lt;br /&gt;
The minimum total edge length is 28, and one such tetrahedron is:&lt;br /&gt;
&lt;br /&gt;
(a,b,c,d,e,f)=(2,4,5,4,6,7),a+b+c+d+e+f=28.(a,b,c,d,e,f)=(2,4,5,4,6,7),\quad a+b+c+d+e+f=28.(a,b,c,d,e,f)=(2,4,5,4,6,7),a+b+c+d+e+f=28.&lt;br /&gt;
Volume (both methods agree):&lt;br /&gt;
* Cayley–Menger determinant =10368=288⋅62⇒V=6=10368=288\cdot 6^2\Rightarrow V=6=10368=288⋅62⇒V=6.&lt;br /&gt;
* Gram method gives the same V=6V=6V=6.&lt;br /&gt;
&lt;br /&gt;
All four face areas are irrational (each K=16Δ2K=16\Delta^2K=16Δ2 is non-square):&lt;br /&gt;
* (a,b,d)=(2,4,4): K=240(a,b,d)=(2,4,4):\ K=240(a,b,d)=(2,4,4): K=240 (not a square)&lt;br /&gt;
* (a,c,e)=(2,5,6): K=351(a,c,e)=(2,5,6):\ K=351(a,c,e)=(2,5,6): K=351&lt;br /&gt;
* (b,c,f)=(4,5,7): K=1536(b,c,f)=(4,5,7):\ K=1536(b,c,f)=(4,5,7): K=1536&lt;br /&gt;
* (d,e,f)=(4,6,7): K=2295(d,e,f)=(4,6,7):\ K=2295(d,e,f)=(4,6,7): K=2295&lt;br /&gt;
&lt;br /&gt;
✅ Answer (1): V=6V=6V=6 (at the minimum possible total edge-length 282828).&lt;br /&gt;
&lt;br /&gt;
==== (2) Find integer-edge tetrahedron with ====&lt;br /&gt;
𝑉&lt;br /&gt;
=&lt;br /&gt;
𝑎&lt;br /&gt;
+&lt;br /&gt;
𝑏&lt;br /&gt;
+&lt;br /&gt;
𝑐&lt;br /&gt;
+&lt;br /&gt;
𝑑&lt;br /&gt;
+&lt;br /&gt;
𝑒&lt;br /&gt;
+&lt;br /&gt;
𝑓&lt;br /&gt;
V=a+b+c+d+e+f&lt;br /&gt;
&lt;br /&gt;
The smallest solution (by total edge sum, hence by VVV too) I find is:&lt;br /&gt;
&lt;br /&gt;
(a,b,c,d,e,f)=(4,10,17,12,19,22)(a,b,c,d,e,f)=(4,10,17,12,19,22)(a,b,c,d,e,f)=(4,10,17,12,19,22)&lt;br /&gt;
a+b+c+d+e+f=84.a+b+c+d+e+f = 84.a+b+c+d+e+f=84.&lt;br /&gt;
Volume (both methods agree):&lt;br /&gt;
* Cayley–Menger determinant =2032128=288⋅842⇒V=84=2032128=288\cdot 84^2\Rightarrow V=84=2032128=288⋅842⇒V=84.&lt;br /&gt;
* Gram method also gives V=84V=84V=84.&lt;br /&gt;
&lt;br /&gt;
All face areas are irrational (non-square KKK’s):&lt;br /&gt;
* (4,10,12): K=5616(4,10,12):\ K=5616(4,10,12): K=5616&lt;br /&gt;
* (4,17,19): K=15360(4,17,19):\ K=15360(4,17,19): K=15360&lt;br /&gt;
* (10,17,22): K=106575(10,17,22):\ K=106575(10,17,22): K=106575&lt;br /&gt;
* (12,19,22): K=207495(12,19,22):\ K=207495(12,19,22): K=207495&lt;br /&gt;
&lt;br /&gt;
✅ Answer (2): V=84V=84V=84 (with total edge-length 848484).&lt;br /&gt;
&lt;br /&gt;
If you want, I can also list all (1)-type solutions at sum 282828 (they’re just relabelings) and/or additional equable solutions beyond the minimum.&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:ChatGPT Shares]]&lt;br /&gt;
[[Category:openai]]&lt;/div&gt;</summary>
		<author><name>Lukegao</name></author>
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