If the sum of the lengths of the six edges of a trirectangular tetrahedron $ PABC$ (i.e., $ \angle APB = \angle BPC = \angle CPA = 90^\circ$) is $ S$, determine its maximum volume.
Problem
Source: 1976 USAMO Problem 4
Tags: geometry, 3D geometry, tetrahedron, Pythagorean Theorem
Farenhajt
04.04.2010 20:38
$ V={ab\over 2}\cdot{c\over 3}={abc\over 6}$
\begin{eqnarray*} S &=& a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\\ &\geqslant& a+b+c+\sqrt{2ab}+\sqrt{2ac}+\sqrt{2bc}\\ &\geqslant& 3\sqrt[3]{abc}+3\sqrt[3]{2^{3/2}abc}\\ &=& 3(\sqrt{2}+1)\sqrt[3]{abc}\end{eqnarray*}
Hence $ abc\leqslant {S^3\over 27(\sqrt{2}+1)^3}={S^3(5\sqrt{2}-7)\over 27}$, yielding $ V\leqslant {S^3(5\sqrt{2}-7)\over 162}$ for $ a=b=c={S(\sqrt{2}-1)\over 3}$
Dr Sonnhard Graubner
04.04.2010 20:42
hello, let $ \overline{PB} = a,\overline{PA} = b,\overline{PC} = c$ then we get the volume of our trirectangular tetraeder in the form of $ V = \frac {1}{6}abc\le\left(\frac {a + b + c}{3}\right)^{\frac {1}{3}}$ after AM-GM. Further we can compute $ s$ with the Pythagorean theorem as $ s = a + b + c + \sqrt {a^2 + b^2} + \sqrt {a^2 + c^2} + \sqrt {b^2 + c^2}$ since $ \frac {x + y}{2}\le\sqrt {\frac {x^2 + y^2}{2}}$ we get $ s\geq(\sqrt {2} + 1)(a + b + c)$ so our $ V$ can be estimate as $ V = \frac {1}{6}abc\le\frac {1}{6}\left(\frac {a + b + c}{3}\right)^{\frac {1}{3}}\le \frac {1}{6}\left(\frac {s}{3(\sqrt {2} + 1)}\right)^{\frac {1}{3}}$. Sonnhard.
phiReKaLk6781
19.04.2011 03:00
Sorry to revive, but what exactly is a trirectangular tetrahedron? I'm having trouble visualizing this.
kacheep
19.04.2011 03:12
PhireKaLk6781 wrote: Sorry to revive, but what exactly is a trirectangular tetrahedron? I'm having trouble visualizing this. This is what google says.
phiReKaLk6781
19.04.2011 08:54
I see no rectangles. I'm pretty sure that's not right.
kacheep
20.04.2011 00:47
PhireKaLk6781 wrote: I see no rectangles. I'm pretty sure that's not right. this is another one. it's inside a rectangular prism... so...
huashiliao2020
14.04.2023 08:14
phiReKaLk6781 wrote: I see no rectangles. I'm pretty sure that's not right. Trirectangular does not say three rectangles, but rather, three right angles. First imagine a right triangle base, then imagine from the vertex that contains the right angle, extend that directly above, which makes a perpendicular segment with the plane the triangle lies on. Farenhajt wrote:
$ V={ab\over 2}\cdot{c\over 3}={abc\over 6}$
\begin{eqnarray*} S &=& a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\\ &\geqslant& a+b+c+\sqrt{2ab}+\sqrt{2ac}+\sqrt{2bc}\\ &\geqslant& 3\sqrt[3]{abc}+3\sqrt[3]{2^{3/2}abc}\\ &=& 3(\sqrt{2}+1)\sqrt[3]{abc}\end{eqnarray*}
Hence $ abc\leqslant {S^3\over 27(\sqrt{2}+1)^3}={S^3(5\sqrt{2}-7)\over 27}$, yielding $ V\leqslant {S^3(5\sqrt{2}-7)\over 162}$ for $ a=b=c={S(\sqrt{2}-1)\over 3}$
Nice sol! Yeah, it’s basically just inequalities and using basic AM-GMs on variable side lengths. I did it the same way.
AshAuktober
20.10.2024 07:21
Same as sir Sonnhard above.