I like this kind of problems! 1) $P \in BC$ Let $PC=x \Rightarrow PB=1-x$ $\triangle CPA: \ \ PA^2=PC^2+CA^2=x^2+1\Rightarrow \ \ PA=\sqrt{x^2+1}$ $PA \cdot PB \cdot PC = x(1-x)\sqrt{x^2+1}$ So we want to maximize the function $h(x) = x(1-x)\sqrt{x^2+1}$, when $x \in [0,1]$ 2) $P \in AB$ $PH \perp AC, H \in AC$ $CH=x \Rightarrow HA=PH=1-x$ $\triangle CPH: \ \ PC^2 = CH^2+PH^2 = x^2+(1-x)^2$ from the 45-45-90 triangle $PHA$ we have $PA = \sqrt{2}\cdot HA \Rightarrow PA = \sqrt{2}(1-x)$ similarly $PB = \sqrt{2}\cdot x$ $PA \cdot PB = 2x(1-x)$ $PA \cdot PB \cdot PC = 2x(1-x) \sqrt{x^2+(1-x)^2}$ In order to find the derivative in a shorter way, we define the functions: $f(x) = 2x(1-x)$ $g(x) = x^2+(1-x)^2$ product: $F(x) = f(x) \sqrt{g(x)}$ (of course $g(x)>0$) now, notice that $g(x)+f(x) = x^2+(1-x)^2+2x(1-x) =[x+(1-x)]^2 = 1\Rightarrow$ $g(x) = 1-f(x)$ $g'(x) = [1-f(x)]' = -f'(x)$ $F(x) = f(x) \sqrt{g(x)}$ $F'(x) = f'(x) \sqrt{g(x)} + f(x) \frac{g'(x)}{2\sqrt{g(x)}}$ $F'(x) = f'(x) \sqrt{g(x)} - f(x) \frac{f'(x)}{2\sqrt{g(x)}}$ $F'(x) = f'(x) [\sqrt{g(x)} - \frac{f(x)}{2\sqrt{g(x)}}]$ $F'(x) = f'(x) [\frac{2g(x)-f(x)}{2\sqrt{g(x)}}]$ $F'(x) = f'(x) [\frac{2[1-f(x)]-f(x)}{2\sqrt{g(x)}}]$ $F'(x) = f'(x) [\frac{2-3f(x)}{2\sqrt{g(x)}}]$ The fraction $\frac{2-3f(x)}{2\sqrt{g(x)}}$ is always positive. Indeed: the denominator is always positive also: $f(x) \leq \frac{1}{2} \Rightarrow$ $-3f(x) \geq -\frac{3}{2} \Rightarrow$ $2-3f(x) \geq 2-\frac{3}{2} \Rightarrow$ $2-3f(x) \geq \frac{1}{2}>0$ $F'(x)=0 \Leftrightarrow f'(x)=0$ $f'(x) = 2-4x$ It has a root at $x=\frac{1}{2}$ and $x<\frac{1}{2}\Rightarrow f'(x)>0$ $x>\frac{1}{2}\Rightarrow f'(x)<0$ The same happens from $F'(x)$ Hence $x=\frac{1}{2}$ is a local maximum of $F(x)$ $f(\frac{1}{2}) = \frac{1}{2} = g(\frac{1}{2})$ applying these values to $F(x) = f(x) \sqrt{g(x)}$ we get $F(\frac{1}{2}) =\frac{1}{2} \cdot \frac{\sqrt 2}{2} = \frac{\sqrt 2}{4}$ Now let's see why this maximum is greater than the maximum of $h(x) = x(1-x)\sqrt{x^2+1}$ $h(x) = x(1-x)\sqrt{x^2+1}$ $F(x) = 2x(1-x) \sqrt{x^2+(1-x)^2}$ We will show that $h(x)<F(x)$ in $(0,1)$ $h(x)<F(x) \Leftrightarrow$ $x(1-x)\sqrt{x^2+1} < 2x(1-x) \sqrt{x^2+(1-x)^2} \Leftrightarrow$ (divide with the positive $x(1-x)$) $\sqrt{x^2+1} < 2 \sqrt{x^2+(1-x)^2} \Leftrightarrow$ $x^2+1 < 4 [x^2+(1-x)^2] \Leftrightarrow$ $x^2+1 < 4x^2+4(1-x)^2 \Leftrightarrow$ $x^2+1 < 4x^2+4(1+x^2-2x) \Leftrightarrow$ $x^2+1 < 4x^2+4+4x^2-8x \Leftrightarrow$ $0 < 7x^2-8x+3,$ true because discriminant: $D = 8^2-4\cdot 7\cdot 3<0,$ so $7x^2-8x+3$ is always positive $\Rightarrow h(x)<F(x), \ \forall x\in (0,1)$ hence, if $h(x_o)$ is maximum of $h(x)$ then $h(x_o)<F(x_o)$ Finally, the maximum value of $PA \cdot PB \cdot PC$ for all possible positions of $P$ is $F(\frac{1}{2}) = \frac{\sqrt 2}{4}$ $x=\frac{1}{2}$ means that the point $P$ is the midpoint of $AB$

My solution was essentially the same as above. Would like to other solutions as well.

Nice problem

Given $\Delta ABC,$ with $AB=AC=1~\text{and}~BC=\sqrt2.$ Consider the set $\mathbb{S}=\Delta ABC\cup\text{Int}(\Delta ABC).$ Determine $$\max_{M\in\mathbb{S}}{\left(MA\cdot MB\cdot MC\right)}.$$Preluata

Old? Where

mihaig wrote: Given $\Delta ABC,$ with $AB=AC=1~\text{and}~BC=\sqrt2.$ Consider the set $\mathbb{S}=\Delta ABC\cup\text{Int}(\Delta ABC).$ Determine $$\max_{M\in\mathbb{S}}{\left(MA\cdot MB\cdot MC\right)}.$$Preluata Anyone?