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There is a point $M$ in space. Four rays $MA$, $MB$, $MC$, and $MD$. Assume $MA=3$, $MB=4$, $MC=5$, $MD=6$. What is the maximum volume of tetrahedron ABCD?

It is easy to show that M should be the orthocenter of orthocentric tetrahedron. if M B C D is set, in order to achieve the max volume, MD must be perpendicular to plane BCD. so the question becomes that given the distances from orthocenter to 4 vertices, can we calculate the volume ? I think that it should have an elegant way to do it.

for orthocentric tetrahedron, https://en.wikipedia.org/wiki/Orthocentric_tetrahedron

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  • $\begingroup$ Please give some idea about what you know about the problem and/or where you got stuck. This information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) telling you things you already know, duplicating your effort, or using techniques for which you aren't yet ready. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, you might give the source of the problem as an indication of its expected level of difficulty. Also: What formula(s) do you know for the volume of a tetrahedron? $\endgroup$ – Blue Apr 25 '19 at 3:28
  • $\begingroup$ It should be Orthocentric tetrahedron and M is its orthocenter. $\endgroup$ – Weiqing Wu Apr 25 '19 at 3:34
  • $\begingroup$ Please edit the question to add any clarifications. (Comments are easily overlooked.) Knowing that the target tetrahedron is orthocentric is helpful, but you still haven't given any indication about what you have done yourself to solve the problem. Do you know, for instance, how one might show that a tetrahedron is orthocentric? $\endgroup$ – Blue Apr 25 '19 at 3:42
  • $\begingroup$ I have an idea from linear algebra, let me see if I can work it out. $\endgroup$ – Yuval Apr 25 '19 at 3:44
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Let $a, b, c, d$ denote the four vectors $MA, MB, MC, MD$, written as column vectors. We see the tetrahedon $M-ABC$ has volume $\pm\frac 16 \text{det}[a, b, c]$, where $[a, b, c]$ is the $3\times 3$ matrix with columes $a, b, c$. We want to pick a suitable number $\mu>0$ and consider the $4\times 4$ matrix $$ U=\left[ \begin{array}{ccccc} \mu & \mu &\mu &\mu \\ a & b & c & d \end{array} \right], $$ that is, put the row $[\mu, \mu, \mu, \mu]$ on top of the $3\times 4$ matrix $[a, b, c, d]$.

Obviously we only need to consider the situation that any two among $a, b, c, d$ has an angle $>90^\circ$. Under this assumption, in the formula $$ \text{det} U= \mu \text{det}[b, c, d]- \mu \text{det}[a, c, d] + \mu \text{det}[a, b, d]- \mu \text{det}[a, b, c], $$ we see all four terms are of the same sign! You can see this by drawing $a, b, c, d$ with mutual angle $>90^\circ$ (from the point of view of orientation there are only two ways to draw this). Thus $$ \text{det} U=\pm 6\mu \text{Vol}(ABCD). $$ By Hadamard's theorem (see https://math.stackexchange.com/q/3196666 or
https://en.wikipedia.org/wiki/Hadamard%27s_inequality ), we see $|\text{det}U|\leq$ the product of the lengths of its columns, thus $$ |\text{det}U|^2\leq (\mu^2+3^2)(\mu^2+4^2)(\mu^2+5^2)(\mu^2+6^2), $$ with equal sign holds when the columns of $U$ are mutually perpendicular. Overall, $$ \text{Vol}(ABCD)\leq \frac{\sqrt{(\mu^2+3^2)(\mu^2+4^2)(\mu^2+5^2)(\mu^2+6^2)}}{6\mu}. $$

So the problem here is to decide $\mu>0$ so that there exists four $3$-vectors $a, b, c, d$ of lengths $3, 4, 5, 6$ so that the four columns of $U$ are mutually perpendicular.

We make the following construction. Let $e_1, e_2, e_3, e_4$ be the standard orthonormal basis of ${\mathbb R}^4$. Let the four columns of $U$ be $x_1e_1, x_2e_2, x_3e_3, x_4e_4$ (of course they are written in a "wrong" coordinate system so you don't see their first coordinate is $\mu$), with all $x_i>0$. Let $v=a_1e_1+a_2e_2+a_3e_3+a_4e_4$ be the unit vector in the direction of the first coordinate of the "correct" coordinate system, with all $a_i>0$. First of all, $$ a_1^2+a_2^2+a_3^2+a_4^2=1. $$ Next, $(x_1e_1)\cdot v=\mu$, and so on; thus $a_1x_1=a_2x_2=a_3x_3=a_4x_4=\mu$. So $x_i=\mu/a_i$. Now the component in $x_1e_1$ that is perpendicular to $v$ should have length $3$, thus $3^2+\mu^2=x_1^2$. Thus we see $$ 3^2+\mu^2=\frac{\mu^2}{a_1^2}, \,\,\,\, 4^2+\mu^2=\frac{\mu^2}{a_2^2}, \,\,\,\, 5^2+\mu^2=\frac{\mu^2}{a_3^2}, \,\,\,\, 6^2+\mu^2=\frac{\mu^2}{a_4^2}. \,\,\,\, $$ So $a_1^2=\frac{\mu^2}{3^2+\mu^2}$, etc. Put this into $a_1^2+...+a_4^2=1$, we get $$ \frac{\mu^2}{3^2+\mu^2}+\frac{\mu^2}{4^2+\mu^2} +\frac{\mu^2}{5^2+\mu^2}+\frac{\mu^2}{6^2+\mu^2}=1. $$ According to Wolfram-alpha we solve $\mu^2=\frac{\sqrt{6169}-43}{6}$. With this $\mu$, we see the maximum volume is (from Wolfram-alpha) $$ \frac{\sqrt{(\mu^2+3^2)(\mu^2+4^2)(\mu^2+5^2)(\mu^2+6^2)}}{6\mu} =\sqrt{\frac{482147}{486}+\frac{6169\sqrt{6169}}{486}}\approx 44.6 $$

Please let me know if there is anything incorrect.

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  • $\begingroup$ +1 clever geometric construction. $\endgroup$ – achille hui Apr 25 '19 at 5:46
  • $\begingroup$ really amazing. I have a simple question. If the problem is reduced to 2D, what happens. if MA=3, MB=4, MC=5, what is the largest area of triangle ABC ? $\endgroup$ – Weiqing Wu Apr 25 '19 at 16:22
  • $\begingroup$ @Weiqing Wu : We can use a similar linear algebra argument. Or, we can assume the angles between $MA, MB, MC$ to be $\alpha, \beta, \gamma$ and do a Lagrange multiplier argument to find $\alpha, \beta,\gamma$, but seems one has to solve a 3rd or 4th order algebraic equation, which may not be very simple (Wolfram-alpha needed?). On the other hand, $MA, MB, MC$ are the three altitudes of $ABC$ when area is maximal, but it is not clear to me how to argue along that direction. $\endgroup$ – Yuval Apr 25 '19 at 17:20
  • $\begingroup$ @Yu Ding : I found that M being orthocenter there are many triangles. if d is the distance from M to BC, the distance from M to AC is 3d/5 and M to AB is 4d/5. Is there a more elegant solution to this 2D problem ? maybe maximize (sqrt(16-dxd)+sqrt(25-dxd))(3+d) ? $\endgroup$ – Weiqing Wu Apr 25 '19 at 18:27

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