Oblicz pole trójkąta T2, którego boki stanowią środkowe trójkąta T1, którego pole wynosi S.

△ABC:

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

AE, BF, CG – środkowe

|AE| = e; |BF| = f; |CG| = g

${\begin{matrix} e = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}} \\ f = \frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}} \\ g = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}} \end{matrix}$

TRÓJKĄT O WSKAZANCYH BOKACH

$P_{ABC} = \sqrt{(\frac{a + b + c}{2}) (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - b) (\frac{a + b + c}{2} - c)} = \sqrt{(\frac{a + b + c}{2}) (\frac{b + c - a}{2}) (\frac{a + c - b}{2}) (\frac{a + b - c}{2})} = \sqrt{(\frac{[b + c] + a}{2}) (\frac{[b + c] - a}{2}) \cdot (\frac{a + [c - b]}{2}) (\frac{a - [c - b]}{2})} = \sqrt{\frac{{(b + c)}^{2} - a^{2}}{4} \cdot \frac{a^{2} - {(c - b)}^{2}}{4}} = \frac{1}{4} \sqrt{[{(b + c)}^{2} - a^{2}] [a^{2} - {(c - b)}^{2}]} = \frac{1}{4} \sqrt{(b^{2} + 2 bc + c^{2} - a^{2}) (a^{2} - c^{2} + 2 bc - b^{2})} = \frac{1}{4} \sqrt{(2 bc + (b^{2} + c^{2} - a^{2})) (2 bc - (b^{2} + c^{2} - a^{2}))} = \frac{1}{4} \sqrt{4 b^{2} c^{2} - {(b^{2} + c^{2} - a^{2})}^{2}} = \frac{1}{4} \sqrt{4 b^{2} c^{2} - (a^{4} + b^{4} + c^{4} + 2 a^{2} c^{2} - 2 a^{2} b^{2} + 2 b^{2} c^{2})} = \frac{1}{4} \sqrt{- (a^{4} + b^{4} + c^{4} + 2 a^{2} c^{2} + 2 a^{2} b^{2} + 2 b^{2} c^{2})} = S$

TRÓJKĄT ZBUDOWANY ZE ŚRODKOWYCH:

$P_{PQR} = \frac{1}{4} \sqrt{[{(f + g)}^{2} - e^{2}] [e^{2} - {(g - f)}^{2}]} = \frac{1}{4} \sqrt{[f^{2} + 2 fg + g^{2} - e^{2}] [e^{2} - g^{2} + 2 fg - f^{2}]}$ $e^{2} = {(\frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}})}^{2} = \frac{1}{4} (2 b^{2} + 2 c^{2} - a^{2})$ $f^{2} = {(\frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}})}^{2} = \frac{1}{4} (2 a^{2} + 2 c^{2} - b^{2})$ $g^{2} = {(\frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}})}^{2} = \frac{1}{4} (2 a^{2} + 2 b^{2} - c^{2})$ $2 fg = 2 \cdot \frac{1}{4} (2 a^{2} + 2 c^{2} - b^{2}) \cdot \frac{1}{4} (2 a^{2} + 2 b^{2} - c^{2}) = \frac{1}{8} (2 a^{2} + 2 c^{2} - b^{2}) (2 a^{2} + 2 b^{2} - c^{2})$ $[f^{2} + 2 fg + g^{2} - e^{2}] = \frac{a^{2}}{2} + \frac{c^{2}}{2} - \frac{b^{2}}{4} + 2 fg + \frac{a^{2}}{2} + \frac{b^{2}}{2} - \frac{c^{2}}{4} - \frac{b^{2}}{2} - \frac{c^{2}}{2} + \frac{a^{2}}{4} = \frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4} + 2 fg$ $[e^{2} - g^{2} + 2 fg - f^{2}] = \frac{b^{2}}{2} + \frac{c^{2}}{2} - \frac{a^{2}}{4} - \frac{a^{2}}{2} - \frac{b^{2}}{2} + \frac{c^{2}}{4} + 2 fg - \frac{a^{2}}{2} - \frac{c^{2}}{2} + \frac{b^{2}}{4} = \frac{- {5 a}^{2}}{4} + \frac{b^{2}}{4} + \frac{c^{2}}{4} + 2 fg$ $P_{PQR} = \frac{1}{4} \sqrt{(\frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4} + 2 fg) (\frac{- {5 a}^{2}}{4} + \frac{b^{2}}{4} + \frac{c^{2}}{4} + 2 fg)} = \frac{1}{4} \sqrt{(2 fg + (\frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4})) (2 fg - (\frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4}))} = \frac{1}{4} \sqrt{{(2 fg)}^{2} - {(\frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4})}^{2}} = \frac{1}{4} \sqrt{4 f^{2} g^{2} - {(\frac{{5 a}^{2}}{4} - \frac{b^{2}}{4} - \frac{c^{2}}{4})}^{2}} = \frac{1}{4} \sqrt{4 \cdot \frac{1}{4} (2 a^{2} + 2 c^{2} - b^{2}) \cdot \frac{1}{4} (2 a^{2} + 2 b^{2} - c^{2}) - {(\frac{1}{4} ({5 a}^{2} - b^{2} - c^{2}))}^{2}} = \frac{1}{4} \sqrt{\frac{1}{4} (2 a^{2} + 2 c^{2} - b^{2}) (2 a^{2} + 2 b^{2} - c^{2}) - \frac{1}{16} {({5 a}^{2} - b^{2} - c^{2})}^{2}}$ $(2 a^{2} + 2 c^{2} - b^{2}) (2 a^{2} + 2 b^{2} - c^{2}) = 4 a^{4} + 4 a^{2} b^{2} - 2 a^{2} c^{2} + 4 a^{2} c^{2} + 4 b^{2} c^{2} - 2 c^{4} - 2 a^{2} b^{2} - 2 b^{4} + b^{2} c^{2} = 4 a^{4} - 2 b^{4} - 2 c^{4} + 2 a^{2} b^{2} + 2 a^{2} c^{2} + 5 b^{2} c^{2}$ ${({5 a}^{2} - b^{2} - c^{2})}^{2} = ({5 a}^{2} - b^{2} - c^{2}) ({5 a}^{2} - b^{2} - c^{2}) = 25 a^{4} - 5 a^{2} b^{2} - 5 a^{2} c^{2} - 5 a^{2} b^{2} + b^{4} + b^{2} c^{2} - 5 a^{2} c^{2} + b^{2} c^{2} - c^{4} = 25 a^{4} + b^{4} - c^{4} - 10 a^{2} b^{2} - 10 a^{2} c^{2} + 2 b^{2} c^{2}$ $P_{PQR} = \frac{1}{4} \sqrt{\frac{1}{16} [4 (2 a^{2} + 2 c^{2} - b^{2}) (2 a^{2} + 2 b^{2} - c^{2}) - {({5 a}^{2} - b^{2} - c^{2})}^{2}]} = \frac{1}{16} \sqrt{4 (4 a^{4} - 2 b^{4} - 2 c^{4} + 2 a^{2} b^{2} + 2 a^{2} c^{2} + 5 b^{2} c^{2}) - (25 a^{4} + b^{4} - c^{4} - 10 a^{2} b^{2} - 10 a^{2} c^{2} + 2 b^{2} c^{2})} = \frac{1}{16} \sqrt{- 9 a^{4} - 9 b^{4} - 9 c^{4} + 18 a^{2} b^{2} + 18 a^{2} c^{2} + 18 b^{2} c^{2}} = \frac{1}{16} \sqrt{- 9 (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} - 2 b^{2} c^{2})} = \frac{3}{16} \sqrt{- (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} - 2 b^{2} c^{2})}$ $\frac{P_{PQR}}{P_{ABC}} = \frac{\frac{3}{16} \sqrt{- (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} - 2 b^{2} c^{2})}}{\frac{1}{4} \sqrt{- (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} - 2 b^{2} c^{2})}} = \frac{3}{16} : \frac{1}{4} = \frac{3}{16} \cdot 4 = \frac{3}{4}$ $⇓$ $P_{PQR} = \frac{3}{4} P_{ABC} \Leftrightarrow P_{PQR} = \frac{3}{4} S$

Skorzystaj z twierdzenia Cantora, które wiąże długość środkowej d opuszczonej na bok długości c: $d = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}}$ . 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: $(a - b) (a + b) = a^{2} - b^{2}, {(a \pm b)}^{2} = a^{2} \pm 2 ab + b^{2}$ .