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Let $ \angle{BMA'} = \alpha_1\in (0;\frac {\pi}{2}); \ \angle{CMA'} = \alpha_2\in (0;\frac {\pi}{2}); \\ \angle {CMB'} = \beta_1 \in (0;\frac {\pi}{2}); \ \angle{AMB'} = \beta_2\in (0;\frac {\pi}{2}); \\ \angle{AMC'} = \gamma_1\in (0;\frac {\pi}{2}); \ \angle{BMC'} = \gamma_2\in (0;\frac {\pi}{2}); \\ (\alpha_1 + \alpha_2) + (\beta_1 + \beta_2) + (\gamma_1 + \gamma_2) = 2.\pi. \\ MA' = MB.cos\alpha_1 = MC.cos\alpha_2\Rightarrow MA' = \sqrt {MB.MC.cos\alpha_1.cos\alpha_2.}$ analog: $ MB' = \sqrt {MC.MA.cos\beta_1.cos\beta_2}; \ MC' = \sqrt {MA.MB.cos\gamma_1.cos\gamma_2} \\ \Rightarrow MA'.MB'.MC' = MA.MB.MC.\sqrt {cos\alpha_1.cos\alpha_2.cos\beta_1.cos\beta_2.cos\gamma_1.cos\gamma_2} \\ \Rightarrow \frac {MA'.MB'.MC'}{MA.MB.MC} = \sqrt {cos\alpha_1.cos\alpha_2.cos\beta_1.cos\beta_2.cos\gamma_1.cos\gamma_2}$ and $ f(x) = ln {cos x},$ for $ x\in (0;\frac {\pi}{2})$ is concave and acording Jensen inegality $ cos\alpha_1.cos\alpha_2.cos\beta_1.cos\beta_2.cos\gamma_1.cos\gamma_2\le cos^6 \frac {\sum\alpha_1}{6} = cos^6\frac{2.\pi}{6} = cos^2\frac{\pi}{3} = (\frac {1}{2})^6 \\ \Rightarrow max \frac {MA'.MB'.MC'}{MA.MB.MC} = (\frac {1}{2})^3 = \frac {1}{8}$ which triangle ABC echilateral and $ M\equiv O$(circumcentre), the inequality Jensen egality for $ \alpha_1 = \alpha_2 = \beta_1 = \beta_2= \gamma_1 = \gamma_2 = \frac {2.\pi}{6} = \frac {\pi}{3} \Rightarrow MA = MB = MC.$

Let $ m \angle AMB' = \alpha$, $ m \angle BMC' = \beta$, and let$ m \angle CMA' = \gamma$. $ MA = MC' \csc \alpha$, $ MB = MC' \sin \beta$, and $ MC = = MA' \csc \gamma$. Then $ \frac {MA' \cdot MB' \cdot MC}{MA \cdot MB \cdot MC} = \sin \alpha \sin \beta \sin \gamma$. Since $ \sin \alpha \sin \beta \sin \gamma$ is positive, it is sufficient to find the values of $ \alpha, \beta, \gamma$ that maximize that expression. By trig Ceva, $ \sin \alpha \sin \beta \sin \gamma = \sin (A - \alpha) \sin (B - \beta) \sin (C - \gamma)$, (where $ A = m \angle CAB$, $ B = m \angle ABC$, and $ C = m \angle BCA$) so we in fact seek to maximimize $ \sin \alpha \sin(A - \alpha) \sin \beta \sin(B - \beta) \sin \gamma \sin (C - \gamma)$. But this is equal to $ \frac {1}{8}(\cos(A - 2\alpha) - \cos A)(\cos(B - 2\beta) - \cos B)(\cos(C - 2 \gamma) - \cos C)$. Since $ 0 \leq \cos(A - 2\alpha) - \cos A \leq 1 - \cos A$, $ 0 \leq \cos(B - 2\beta) - \cos B \leq 1 - \cos B$, and $ 0 \leq \cos(C - 2 \gamma) - \cos C \leq 1 - \cos C$, we have that $ \frac {1}{8}(\cos(A - 2\alpha) - \cos A)(\cos(B - 2\beta) - \cos B)(\cos(C - 2 \gamma) - \cos C)$$ \leq \frac {(1 - \cos A)(1 - \cos B)(1 - \cos C)}{8}$, with equality holding iff $ \alpha = \frac {A}{2}$, $ \beta = \frac {B}{2}$, and $ \gamma = \frac {C}{2}$, so the point $ M$ that maximizes $ p(M)$ is the incenter. $ 8p(M)^2 = (1 - \cos A)(1 - \cos B)(1 - \cos C)$. If $ f(x) = \ln(1 - \cos x)$, $ f''(x) = \frac {1}{\cos x - 1} < 0$, so $ f$ is concave. Hence, by Jensen's inequality, $ f(A) + f(B) + f(C) \leq 3f \left (\frac {A + B + C}{3} \right)$, that is, $ (1 - \cos A)(1 - \cos B)(1 - \cos C) \leq \left (1 - \cos \frac {\pi}{3} \right)^3$, with equality holding iff $ A = B = C$. Hence, $ p(M)$ is maximized when the triangle is equilateral.

Alex stole my solution. =P

My solution is actually a mixture of the above 2 and I wonder if another exists Let $\angle BAM= \angle 1$, $\angle CAM= \angle 2$, $\angle ACM= \angle 3$, $\angle BCM= \angle 4$, $\angle CBM= \angle 5$, $\angle ABM= \angle 6$ Quite simply, $p(M)^2=\prod _{i=1}^{6} \sin (\angle i) \le (\sin (\frac{\sum \angle i}{6})^6 = \frac{1}{2^6}$ $p(M)=\frac{1}{8}$ holds when the triangle is equilateral and $M$ is circumcentre.

Note that $p(M)=\frac{MC'}{MA}\cdot\frac{MA'}{MB}\cdot\frac{MB'}{MC}=\sin\angle MAB\sin\angle MBC\sin\angle MCA$. By Trig Ceva, $\sin\angle MAB\sin\angle MBC\sin\angle MCA=\sin\angle MAC\sin\angle MBA\sin\angle MCB$, so $p(M)^2=\prod\sin\angle MAB\sin\angle MAC=\prod\frac12(\cos(\angle MAB-\angle MAC)-\cos\angle A)$ by Product to Sum. Since $|\angle MAB-\angle MAC|\leq\angle A$, we have that the expression is maximized when $\angle MAB=\angle MAC$. Similar equalities hold true for the other vertices, so $p(M)^2$ and hence $p(M)$ is maximized when $M$ is the incenter. Now $p(M)=\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$, and by AM-GM and Jensen this is maximized when $\angle A=\angle B=\angle C$, or when $ABC$ is equilateral.

Nice trigonometric problem

Let $\angle CAB=\alpha$, $\angle ABC=\beta$, $\angle BCA = \gamma$. Furthermore, let $\angle BAM=\alpha_1$, $\angle MAC=\alpha_2$, $\angle ACM=\gamma_1$, $\angle MCB=\gamma_2$, $\angle CBM=\beta_1$, and $\angle MBA=\beta_2$. Note that \[p(M)=\sin(\alpha_1)\sin(\beta_1)\sin(\gamma_1)=\sin(\alpha_2)\sin(\beta_2)\sin(\gamma_2)=\sqrt{\sin(\alpha_1)\sin(\alpha_2)\sin(\beta_1)\sin(\beta_2)\sin(\gamma_1)\sin(\gamma_2)}.\]We'll maximize $\sin(\alpha_1)\sin(\alpha_2)$ given fixed $\alpha_1+\alpha_2.$ Note that \begin{align*} \sin(\alpha_1)\sin(\alpha_2) &= \tfrac12 (\cos(\alpha_1-\alpha_2)-\cos(\alpha_1+\alpha_2)) \\ & \le \tfrac12 (\cos(0)-\cos(\alpha_1+\alpha_2)) \end{align*}Thus, $p(M)$ is minimized when $\alpha_1=\alpha_2$, $\beta_1=\beta_2$, and $\gamma_1=\gamma_2$, precisely when $M$ is the incenter, and the value of \[\mu(M)=\sqrt{\tfrac{1-\cos(\alpha)}{2}}\cdot \sqrt{\tfrac{1-\cos(\beta)}{2}}\cdot \sqrt{\tfrac{1-\cos(\gamma)}{2}}=\sin\left(\tfrac{\alpha}{2}\right)\sin\left(\tfrac{\beta}{2}\right)\sin\left(\tfrac{\gamma}{2}\right)\]Let $f(x)=\ln(\sin(x))$. What we see is that on the $f'(x)=\cot(x)$, so on the interval $\left[0,\tfrac{\pi}{2}\right)$, $f$ is concave. Indeed, \[\ln\left(\sin\left(\tfrac{\alpha}{2}\right)\right)+\ln\left(\sin\left(\tfrac{\beta}{2}\right)\right)+\ln\left(\sin\left(\tfrac{\gamma}{2}\right)\right)\le 3\ln(\sin(30^\circ))\implies \sin\left(\tfrac{\alpha}{2}\right)\sin\left(\tfrac{\beta}{2}\right)\sin\left(\tfrac{\gamma}{2}\right)\le \tfrac18\]and the optimal triangle is the equilateral.