Wyznacz pole powierzchni trójkąta, w którym środkowe wynoszą odpowiednio 5, 12 i 13.

△ABC:

|AB| = c; |BC| = a; |AC| = b

AE, BF, CG – środkowe

|AE| = 5; |BF| = 12; |CG| = 13

${\begin{matrix} 5 = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}} \\ 12 = \frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}} \\ 13 = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}} \end{matrix}$ ${\begin{matrix} 25 = \frac{1}{4} (2 b^{2} + 2 c^{2} - a^{2}) \\ 144 = \frac{1}{4} (2 a^{2} + 2 c^{2} - b^{2}) \\ 169 = \frac{1}{4} (2 a^{2} + 2 b^{2} - c^{2}) \end{matrix}$ ${\begin{matrix} 100 = 2 b^{2} + 2 c^{2} - a^{2} \\ 576 = 2 a^{2} + 2 c^{2} - b^{2} \\ 676 = 2 a^{2} + 2 b^{2} - c^{2} \end{matrix}$ $b^{2} = t; 2 c^{2} = u; a^{2} = v (t, u, v > 0)$ ${\begin{matrix} 100 = 2 u + 2 v - t \Rightarrow t = 2 u + 2 v - 100 \\ 576 = 2 t + 2 v - u \\ 676 = 2 t + 2 u - v \end{matrix}$ ${\begin{matrix} 576 = 2 (2 u + 2 v - 100) + 2 v - u \\ 676 = 2 (2 u + 2 v - 100) + 2 u - v \end{matrix}$ ${\begin{matrix} 576 = 4 u + 4 v - 200 + 2 v - u \\ 676 = 4 u + 4 v - 200 + 2 u - v \end{matrix}$ ${\begin{matrix} 776 = 3 u + 6 v \\ 876 = 6 u + 3 v \end{matrix}$ ${\begin{matrix} 1552 = 6 u + 12 v \\ 876 = 6 u + 3 v \end{matrix} \lor -$ $676 = 9 v$ $v = \frac{676}{9}$ $6 u + 3 \cdot \frac{676}{9} = 876$ $6 u = \frac{1952}{3}$ $u = \frac{976}{9}$ $t = 2 \cdot \frac{976}{9} + 2 \cdot \frac{676}{9} - 100 = \frac{2404}{9}$ ${\begin{matrix} a = \sqrt{t} = \sqrt{\frac{2404}{9}} \\ b = \sqrt{u} = \sqrt{\frac{976}{9}} \\ c = \sqrt{v} = \sqrt{\frac{676}{9}} \end{matrix}$ ${\begin{matrix} a = \frac{2 \sqrt{601}}{3} \\ b = \frac{4 \sqrt{61}}{3} \\ c = \frac{26}{3} \end{matrix}$ $P_{ABC} = \sqrt{(\frac{a + b + c}{2}) (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - b) (\frac{a + b + c}{2} - c)} = \sqrt{(\frac{a + b + c}{2}) (\frac{b + c - a}{2}) (\frac{a + c - b}{2}) (\frac{a + b - c}{2})} = \sqrt{(\frac{[b + c] + a}{2}) (\frac{[b + c] - a}{2}) \cdot (\frac{a + [c - b]}{2}) (\frac{a - [c - b]}{2})} = \sqrt{\frac{{(b + c)}^{2} - a^{2}}{4} \cdot \frac{a^{2} - {(c - b)}^{2}}{4}} = \sqrt{\frac{{(\frac{4 \sqrt{61}}{3} + \frac{26}{3})}^{2} - {(\frac{2 \sqrt{601}}{3})}^{2}}{4} \cdot \frac{{(\frac{2 \sqrt{601}}{3})}^{2} - {(\frac{26}{3} - \frac{4 \sqrt{61}}{3})}^{2}}{4}} = \sqrt{\frac{{(\frac{4 \sqrt{61} + 26}{3})}^{2} - \frac{2404}{9}}{4} \cdot \frac{\frac{2404}{9} - {(\frac{26 - 4 \sqrt{61}}{3})}^{2}}{4}} = \sqrt{\frac{\frac{976}{9} + \frac{208 \sqrt{61}}{9} + \frac{676}{9} - \frac{2404}{9}}{4} \cdot \frac{\frac{2404}{9} - \frac{976}{9} + \frac{208 \sqrt{61}}{9} - \frac{676}{9}}{4}} = \sqrt{\frac{\frac{208 \sqrt{61}}{9} - \frac{752}{9}}{4} \cdot \frac{\frac{752}{9} + \frac{208 \sqrt{61}}{9}}{4}} = \sqrt{\frac{(\frac{208 \sqrt{61}}{9} - \frac{752}{9}) (\frac{208 \sqrt{61}}{9} + \frac{752}{9})}{16}} = \sqrt{\frac{{(\frac{208 \sqrt{61}}{9})}^{2} - {(\frac{752}{9})}^{2}}{16}} = \sqrt{\frac{\frac{2639104}{81} - \frac{565504}{81}}{16}} = \sqrt{\frac{\frac{2073600}{81}}{16}} = \frac{\frac{1440}{9}}{4} = \frac{1440}{36} = 40$

Skorzystaj z twierdzenia Cantora, które wiąże długość środkowej d opuszczonej na bok długości c: $d = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}}$ . Z układu równań wyznacz długości boków trójkąta (a, b, c). Aby obliczyć pole tego trójkąta skorzystaj ze wzoru Herona: $P = \sqrt{(\frac{a + b + c}{2}) (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - b) (\frac{a + b + c}{2} - c)}$ . Skorzystaj z WSM: $(a - b) (a + b) = a^{2} - b^{2}, {(a \pm b)}^{2} = a^{2} \pm 2 ab + b^{2}$ .