Различия

Показаны различия между двумя версиями страницы.

--- math-public:otrtocentr [2019/05/26 16:56] – [Доказательство] labreslav
+++ math-public:otrtocentr [2019/05/27 07:58] (текущий) – labreslav
@@ Строка 406: / Строка 406: @@
 ===== Теорема (расстояние между ортоцентром и инцентром) =====
-$$HI = \sqrt{4r^2-\dfrac{a^3+b^3+c^3+abc}{2p}}.$$
+$$HI = \sqrt{4R^2-\dfrac{a^3+b^3+c^3+abc}{2p}}.$$
 ==== Доказательство ====
@@ Строка 423: / Строка 423: @@
-$ = \dfrac{1}{4p^2}\left(a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2+2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}+2ac\overrightarrow{HA}\cdot\overrightarrow{HC}\right) = $
+$ = \dfrac{1}{4p^2}\left(a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2+2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}+2ac\overrightarrow{HA}\cdot\overrightarrow{HC}\right)$
+==== Лемма 0 ====
+$$\cos{\alpha}\cos{\beta}\cos{\gamma} = \dfrac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{2}-1 = \dfrac{a^2+b^2+c^2}{8R^2}-1$$
+====Лемма 1====
 $a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2 = $
@@ Строка 443: / Строка 448: @@
+====Лемма 2====
 $2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}+2ac\overrightarrow{HA}\cdot\overrightarrow{HC} = $
@@ Строка 456: / Строка 461: @@
 $ = -2\left(\dfrac{a^2+b^2+c^2}{8R^2}-1\right)4R^2\left(ab+bc+ac\right) = $
-$ = 4R^2\left(2ab+2bc+2ac\right)-(a^2+b^2+c^2)(ab+bc+ac))$
+$ = 4R^2\left(2ab+2bc+2ac\right)-(a^2+b^2+c^2)(ab+bc+ac)$
+===продолжение доказательства===
+Тогда
+$HI^2 = \dfrac{1}{4p^2}\left(a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2+2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}+2ac\overrightarrow{HA}\cdot\overrightarrow{HC}\right) =$
+$= \dfrac{1}{4p^2}\left(4R^2(a^2+b^2+c^2)-(a^4+b^4+c^4)+4R^2\left(2ab+2bc+2ac\right)-(a^2+b^2+c^2)(ab+bc+ac)\right) = $
+$= \dfrac{1}{4p^2}\left(4R^2(a^2+b^2+c^2+2ab+2bc+2ac)-(a^4+b^4+c^4+(a^2+b^2+c^2)(ab+bc+ac))\right) = $
+$= \dfrac{1}{4p^2}\left(4R^2(a+b+c)^2-(a^4+b^4+c^4+a^3b+a^2bc+a^3c+ab^3+b^3c+b^2ac+abc^2+bc^3+ac^3)\right) = $
+$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^4+a^3b+a^3c+ab^3+b^4+cb^3+ac^3+bc^3+c^4+a^2bc+ab^2c+abc^2)\right)=$
+$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^3(a+b+c)+b^3(a+b+c)+c^3(a+b+c)+abc(a+b+c))\right) =$
+$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^3+b^3+c^3+abc)2p\right) =$
+$= 4R^2-\dfrac{a^3+b^3+c^3+abc}{a+b+c}$
+===== Теорема =====
+$$HI_a=\sqrt{4R^2-\dfrac{-a^3+b^3+c^3-abc}{2(p-a)}}$$
+====Доказательство====
+=====Лемма 3=====
+$-2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}-2ac\overrightarrow{HA}\cdot\overrightarrow{HC} = $
+$ = -2a^2b^2\ctg^2{\alpha}\ctg^2{\beta}\cos{(180^\circ-\gamma)} + 2b^2c^2\ctg^2{\beta}\ctg^2{\gamma}\cos{(180^\circ-\alpha)} - 2a^2c^2\ctg^2{\alpha}\ctg^2{\gamma}\cos{(180^\circ-\beta)} = $
+$ = 2a^2b^2\ctg^2{\alpha}\ctg^2{\beta}\cos{\gamma} - 2b^2c^2\ctg^2{\beta}\ctg^2{\gamma}\cos{\alpha} 2a^2c^2\ctg^2{\alpha}\ctg^2{\gamma}\cos{\beta} = $
+$ = -2\cos{\alpha}\cos{\beta}\cos{\gamma}\left(-\dfrac{a^2b^2}{\sin{\alpha}\sin{\beta}}+\dfrac{b^2c^2}{\sin{\beta}\sin{\gamma}}-\dfrac{a^2c^2}{\sin{\alpha}\sin{\gamma}}\right) = $
+$ = -2\cos{\alpha}\cos{\beta}\cos{\gamma}\left(-4R^2ab+4R^2bc-4R^2ac\right) = $
+$ = -2\left(\dfrac{a^2+b^2+c^2}{8R^2}-1\right)4R^2\left(-ab+bc-ac\right) = $
+$ = 4R^2\left(-2ab+2bc-2ac\right)-(a^2+b^2+c^2)(-ab+bc-ac)$
+===продолжение доказательства===
+Тогда
+$HI^2_a = \dfrac{1}{4(p-a)^2}\left(a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2-2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}-2ac\overrightarrow{HA}\cdot\overrightarrow{HC}\right) =$
+$=\dfrac{1}{4(p-a)^2}\left(4R^2(a^2+b^2+c^2)-(a^4+b^4+c^4)+4R^2\left(-2ab+2bc-2ac\right)-(a^2+b^2+c^2)(-ab+bc-ac)\right)$
+$=\dfrac{1}{4(p-a)^2}\left(4R^2(a^2+b^2+c^2-2ab+2bc-2ac)-((a^4+b^4+c^4)+(a^2+b^2+c^2)(-ab+bc-ac))\right)$
+$=\dfrac{1}{4(p-a)^2}\left(4R^2(-a+b+c)^2-(a^4 - a^3 b - a^3 c
++ b^4  - a b^3  + b^3 c
++ c^4 - a c^3   + b c^3
++ a^2 b c - a b^2 c - a b c^2)\right)$
+$=\dfrac{1}{4(p-a)^2}\left(4R^2(-a+b+c)^2-(
+- a^3(-a+b+c)
++ b^3(-a+b+c)
++ c^3(-a+b+c)
+- abc(-a+b+c)\right)$
+$=\dfrac{1}{4(p-a)^2}\left(16R^2(p-a)^2-2(p-a)(
+- a^3+ b^3+ c^3- abc)\right)$
+$=4R^2-\dfrac{-a^3+b^3+c^3-abc}{2(p-a)}$
-===== Лемма =====
-$$\cos{\alpha}\cos{\beta}\cos{\gamma} = \dfrac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{2}-1 = \dfrac{a^2+b^2+c^2}{8R^2}-1$$