Президентский ФМЛ №239

$r = 4R\sin \frac A2\sin \frac B2\sin\frac C2$

$$HI = \sqrt{2r^2-4R^2\cos A\cos B\cos C}$$

Рассмотрим треугольник $\triangle A H I$. В нем $A P=2 R \cos A, A I=4 R \sin \frac{B}{2} \sin \frac{C}{2}$ and $\angle H A I=\angle \frac{B-C}{2}$ По теореме косинусов имеем:

$\begin{align}&\small PI^2=4R^2\cos^2A+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)\\&\small=4R^2\left(\cos^2A+4\sin^2\frac{B}{2}\sin^2\frac{C}{2} -4\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)\right)\\&\small=4R^2\left(\cos^2A+4\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\cdot 2\sin\frac{B}{2}\cos\frac{B}{2}\cdot 2\sin\frac{C}{2}\cos\frac{C}{2}-4\cos A\sin^2\frac{B}{2}\sin^2\frac{C}{2}\right)\\&\small=4R^2\left(\cos^2A+8\sin^2\frac{B}{2}\sin^2\frac{C}{2}\cdot \frac 12\left(1-\cos A\right)-\cos A\sin B\sin C\right)\\&\small=4R^2\left(\cos^2A+8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\sin B\sin C\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}+\cos A(\cos A-\sin B\sin C)\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}+\cos A(\cos(180^\circ-(B+C))-\sin B\sin C)\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\cos B\cos C\right)\\&\small =2\left(4R\sin \frac A2\sin \frac B2\sin\frac C2\right)^2-4R^2\cos A\cos B\cos C\\&\small=2r^2-4R^2\cos A\cos B\cos C\end{align}$

$$HI = \sqrt{4R^2-\dfrac{a^3+b^3+c^3+abc}{2p}}.$$

Чтобы центр масс треугольника попал в инцентр $I$ вершины $A, B, C$ нужно загрузить массами $a, b, c$ соответственно.

Тогда для любой точки $X$ будет верно равенство $$\overrightarrow{XI} = \dfrac{a}{a+b+c}\overrightarrow{XA}+\dfrac{b}{a+b+c}\overrightarrow{XB}+\dfrac{c}{a+b+c}\overrightarrow{XC}.$$

Тогда можно взять вместо точки $X$ ортоцентр $H$:

$$\overrightarrow{HI} = \dfrac{1}{2p}\left(a\cdot \overrightarrow{HA}+b\cdot \overrightarrow{HB}+c\cdot \overrightarrow{HC}\right)$$

Вспомним, что $HA = a\ctg{\alpha}$. Тогда

$HI^2 = \dfrac{1}{4p^2}\left(a\cdot \overrightarrow{HA}+b\cdot \overrightarrow{HB}+c\cdot \overrightarrow{HC}\right)^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)$

$$\cos{\alpha}\cos{\beta}\cos{\gamma} = \dfrac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{2}-1 = \dfrac{a^2+b^2+c^2}{8R^2}-1$$

$a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2 = $

$= a^4\ctg^2{\alpha}+b^4\ctg^2{\beta}+c^4\ctg^2{\gamma} = $

$= 16R^4\sin^4{\alpha}\ctg^2{\alpha}+16R^4\sin^4{\beta}\ctg^2{\beta}+16R^4\sin^4{\gamma}\ctg^2{\gamma} = $

$= 16R^4(\sin^2{\alpha}\cos^2{\alpha}+\sin^2{\beta}\cos^2{\beta}+\sin^2{\gamma}\cos^2{\gamma}) = $

$= 16R^4(\sin^2{\alpha}-\sin^4{\alpha}+\sin^2{\beta}-\sin^4{\beta}+\sin^2{\gamma}-\sin^4{\gamma}) = $

$= 16R^4\left(\dfrac{a^2}{4R^2}+\dfrac{b^2}{4R^2}+\dfrac{c^2}{4R^2}-\dfrac{a^4}{16R^4}-\dfrac{b^4}{16R^4}-\dfrac{c^4}{16R^4}\right) = $

$=4R^2(a^2+b^2+c^2)-(a^4+b^4+c^4)$

$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)$

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) = $

$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^4+a^3b+a^3c+ab^3+b^4+cb^3+ac^3+bc^3+c^4+a^2bc+ab^2c+abc^2)\right)=$

$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^3(a+b+c)+b^3(a+b+c)+c^3(a+b+c)+abc(a+b+c))\right) =$

$= \dfrac{1}{4p^2}\left(16R^2p^2-(a^3+b^3+c^3+abc)2p\right) =$

$= 4R^2-\dfrac{a^3+b^3+c^3+abc}{a+b+c}$

$$HI_a=\sqrt{4R^2-\dfrac{-a^3+b^3+c^3-abc}{2(p-a)}}$$

$-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)$

Тогда

$HI^2_a = \dfrac{1}{4(p-a)^2}\left(a^2 \overrightarrow{HA}^2+b^2 \overrightarrow{HB}^2+c^2 \overrightarrow{HC}^2-2ab\overrightarrow{HA}\cdot\overrightarrow{HB}+2bc\overrightarrow{HB}\cdot\overrightarrow{HC}-2ac\overrightarrow{HA}\cdot\overrightarrow{HC}\right) =$

$=\dfrac{1}{4(p-a)^2}\left(4R^2(a^2+b^2+c^2)-(a^4+b^4+c^4)+4R^2\left(-2ab+2bc-2ac\right)-(a^2+b^2+c^2)(-ab+bc-ac)\right)$ $=\dfrac{1}{4(p-a)^2}\left(4R^2(a^2+b^2+c^2-2ab+2bc-2ac)-((a^4+b^4+c^4)+(a^2+b^2+c^2)(-ab+bc-ac))\right)$ $=\dfrac{1}{4(p-a)^2}\left(4R^2(-a+b+c)^2-(a^4 - a^3 b - a^3 c + b^4 - a b^3 + b^3 c + c^4 - a c^3 + b c^3 + a^2 b c - a b^2 c - a b c^2)\right)$

$=\dfrac{1}{4(p-a)^2}\left(4R^2(-a+b+c)^2-( - a^3(-a+b+c) + b^3(-a+b+c) + c^3(-a+b+c) - abc(-a+b+c)\right)$

$=\dfrac{1}{4(p-a)^2}\left(16R^2(p-a)^2-2(p-a)( - a^3+ b^3+ c^3- abc)\right)$

$=4R^2-\dfrac{-a^3+b^3+c^3-abc}{2(p-a)}$