math-public:rasstojanija_zi_zia
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| — | math-public:rasstojanija_zi_zia [2019/05/27 15:15] (текущий) – создано labreslav | ||
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| + | ===== Теорема ===== | ||
| + | |||
| + | Пусть $I$ -- инцентр, | ||
| + | $$IZ^2 = \dfrac19\sqrt{9r^2-3p^2+2 (a^2+b^2+c^2)}$$ | ||
| + | |||
| + | ====Доказательство==== | ||
| + | По теореме Лейбница: | ||
| + | $IA^2+IB^2+IC^2 = 3IZ^2+(AZ^2+BZ^2+CZ^2)$ | ||
| + | |||
| + | $IA^2+IB^2+IC^2 = 3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $r^2+(p-a)^2+r^2+(p-b)^2+r^2+(p-c)^2=3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r^2+p^2-2ap+a^2+p^2-2pb+b^2+p^2-2pc+c^2=3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r^2+3p^2-2ap-2pb-2pc+(a^2+b^2+c^2)=3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r^2+3p^2-2p(a+b+c)+(a^2+b^2+c^2)=3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r^2+3p^2-2p\cdot 2p+(a^2+b^2+c^2)=3IZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r^2-p^2+\dfrac23 (a^2+b^2+c^2)=3IZ^2$ | ||
| + | |||
| + | $r^2-\dfrac13 p^2+\dfrac29 (a^2+b^2+c^2)=IZ^2$ | ||
| + | |||
| + | $IZ^2 = \sqrt{r^2-\dfrac13 p^2+\dfrac29 (a^2+b^2+c^2)}$ | ||
| + | |||
| + | $IZ^2 = \dfrac19\sqrt{9r^2-3p^2+2 (a^2+b^2+c^2)}$ | ||
| + | |||
| + | |||
| + | |||
| + | ===== Теорема ===== | ||
| + | |||
| + | Пусть $I_a$ -- эксцентр, | ||
| + | |||
| + | ====Доказательство==== | ||
| + | По теореме Лейбница: | ||
| + | $I_aA^2+I_aB^2+I_aC^2 = 3I_aZ^2+(AZ^2+BZ^2+CZ^2)$ | ||
| + | |||
| + | $I_aA^2+I_aB^2+I_aC^2 = 3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $r_a^2+p^2+r_a^2+(p-b)^2+r_a^2+(p-c)^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+p^2+p^2-2pb+b^2+p^2-2pc+c^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+3p^2-2pb-2pc+(b^2+c^2)=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+3p^2-2p(b+c)+(b^2+c^2)=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+3p^2-2p(a+b+c)+2pa+(a^2+b^2+c^2)-a^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+3p^2-4p^2+(a^2+b^2+c^2)+2pa-a^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+(a^2+b^2+c^2)-p^2+2pa-a^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2+(a^2+b^2+c^2)-(p-a)^2=3I_aZ^2+\dfrac{a^2+b^2+c^2}{3}$ | ||
| + | |||
| + | $3r_a^2-(p-a)^2+\dfrac{2}{3}(a^2+b^2+c^2)=3I_aZ^2$ | ||
| + | |||
| + | $I_aZ^2 = \sqrt{r_a^2-\dfrac13 (p-a)^2+\dfrac29 (a^2+b^2+c^2)}$ | ||
| + | |||
| + | $I_aZ^2 = \dfrac19\sqrt{9r_a^2-3(p-a)^2+2 (a^2+b^2+c^2)}$ | ||
math-public/rasstojanija_zi_zia.txt · Последнее изменение: 2019/05/27 15:15 — labreslav