Eufisky - The lost book

## 谢之题解：一道综合性的解几题

$\overrightarrow {{n_i}} \cdot \overrightarrow {{n_j}} = {a^2}{x_i}{x_j} + {b^2}{y_i}{y_j} + {c^2}{z_i}{z_j} = 0\left( {i \ne j} \right).$

$\frac{1}{{\sqrt {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} }}\left( {i = 1,2,3} \right).$

${x^2} + {y^2} + {z^2} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} .$

$1 = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right) = {t_i}\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right).$

\begin{align*}&{x^2} + {y^2} + {z^2} = \left( {\begin{array}{*{20}{c}}x&y&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{t_1}}&{{t_2}}&{{t_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{b{y_1}}&{c{z_1}}\\{a{x_2}}&{b{y_2}}&{c{z_2}}\\{a{x_3}}&{b{y_3}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)\\=& \sum\limits_{i = 1}^3 {{t_i}^2\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} = \sum\limits_{i = 1}^3 {\frac{{a{x_i}^2 + b{y_i}^2 + c{z_i}^2}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} \\=& a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right).\end{align*}

\begin{align*}a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{a},\\b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{b},\\c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{c}.\end{align*}

$\frac{a^2x_1^2}{d_1^2}+\frac{a^2x_2^2}{d_2^2}+\frac{a^3x_3^2}{d_3^2}=1\Rightarrow \frac{ax_1^2}{d_1^2}+\frac{ax_2^2}{d_2^2}+\frac{ax_3^2}{d_3^2}=\frac{1}{a}.$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ax_1^2+by_1^2+cz_1^2}{d_1^2}+\frac{ax_2^2+by_2^2+cz_2^2}{d_2^2}+\frac{ax_3^2+by_3^2+cz_3^2}{d_3^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.$

## 来自西哥的一道解几题

(1)求$\triangle ABF_2$内心的轨迹；

(2)当$\triangle ABF_2$内切圆面积最大时，求直线$AB$的方程.
(1)根据题意有$F_1(-3,0),F_2(3,0)$，不妨设$A(x_1,y_1),B(x_2,y_2)$，且
$\left\{ \begin{array}{l}{x_1} = 5\cos \alpha \\{y_1} = 4\sin \alpha\end{array} \right.\text{及}\left\{ \begin{array}{l}{x_2} = 5\cos \beta \\{y_2} = 4\sin \beta\end{array} \right.$

$\frac{{4\sin \beta - 4\sin \alpha }}{{5\cos \beta - 5\cos \alpha }} = \frac{{4\sin \beta }}{{5\cos \beta + 3}} \Leftrightarrow 3\left( {\sin \beta - \sin \alpha } \right) = 5\sin \left( {\alpha - \beta } \right).$
$5\cos \frac{{\alpha - \beta }}{2} = - 3\cos \frac{{\alpha + \beta }}{2}\Leftrightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2} = - 4.$

\begin{align*}\left| {A{F_2}} \right| &= 5 - 3\cos \alpha \\\left| {B{F_2}} \right| &= 5 - 3\cos \beta \\\left| {AB} \right| &= \left| {A{F_1}} \right| + \left| {A{F_1}} \right| = 3\left( {\cos \beta  + \cos \alpha }\right) + 10.\end{align*}

$\left\{ \begin{array}{l}x = \frac{{\left| {B{F_2}} \right|{x_1} + \left| {A{F_2}} \right|{x_2} + 3\left| {AB} \right|}}{{20}}\\y = \frac{{\left| {B{F_2}} \right|{y_1} + \left| {A{F_2}} \right|{y_2}}}{{20}}\end{array} \right.$

$\left\{ \begin{array}{l}x = \frac{{34\left( {\cos \beta + \cos \alpha } \right) - 30\cos \alpha \cos \beta + 30}}{{20}}\\y = \frac{{5\left( {\sin \beta + \sin \alpha } \right) - 3\sin \left( {\alpha + \beta } \right)}}{5}\end{array} \right.$

\begin{align*}\sin \beta  + \sin \alpha  &= 2\sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2} =  - \frac{6}{5}\sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  + \beta }}{2}\\&=  - \frac{3}{5}\sin \left( {\alpha  + \beta } \right)\\\cos \beta  + \cos \alpha  &= 2\cos \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2} =  - \frac{6}{5}{\cos ^2}\frac{{\alpha  + \beta }}{2}\\&=  - \frac{3}{5}\cos \left( {\alpha  + \beta } \right) - \frac{3}{5}\\\cos \alpha \cos \beta  &= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) + \frac{1}{2}\cos \left( {\alpha  - \beta } \right)\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) + \frac{1}{2}\left( {2{{\cos }^2}\frac{{\alpha  - \beta }}{2} - 1} \right)\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}{\cos ^2}\frac{{\alpha  + \beta }}{2}\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}\frac{{\cos \left( {\alpha  + \beta } \right) + 1}}{2}\\&= \frac{{17}}{{25}}\cos \left( {\alpha  + \beta } \right) - \frac{8}{{25}}.\end{align*}

$\left\{ \begin{array}{l}x = - \frac{{51}}{{25}}\cos \left( {\alpha + \beta } \right) + \frac{{24}}{{25}}\\y = - \frac{6}{5}\sin \left( {\alpha + \beta } \right)\end{array} \right. \Rightarrow {\left( {\frac{{25x - 24}}{{51}}} \right)^2} + \frac{{25}}{{36}}{y^2} = 1\left( {x = \frac{{27}}{{25}},y = 0} \right).$
(2)由题意
$S = \frac{1}{2} \times 20r = \frac{1}{2} \times \left| {{F_1}{F_2}} \right| \times 4\left| {\sin \beta - \sin \alpha } \right| = \frac{1}{2} \times 24\left| {\sin \beta - \sin \alpha } \right|.$
\begin{align*}\Rightarrow r &= \frac{6}{5}\left| {\sin \beta  - \sin \alpha } \right| = 2\left| {\sin \left( {\alpha  - \beta } \right)} \right|\\&= 4\left| {\frac{{\sin \frac{{\alpha  - \beta }}{2}\cos \frac{{\alpha  - \beta }}{2}}}{{{{\sin }^2}\frac{{\alpha  -\beta }}{2} + {{\cos }^2}\frac{{\alpha  - \beta }}{2}}}} \right|\\&= 4\left| {\frac{{\tan \frac{{\alpha  - \beta }}{2}}}{{{{\tan }^2}\frac{{\alpha  - \beta }}{2} + 1}}} \right|.\end{align*}

$r = \frac{4}{{\left| {t + \frac{1}{t}} \right|}} = \frac{4}{{\left| t \right| + \frac{1}{{\left| t \right|}}}} \le \frac{4}{{\frac{4}{3} + \frac{3}{4}}} = \frac{{48}}{{25}}.$

$\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = - \frac{{16}}{5}\end{array} \right.\text{或}\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = \frac{{16}}{5}\end{array} \right.$

则$$x_{1}+x_{2}=-\dfrac{2a^2ck^2}{b^2+a^2k^2},x_{1}x_{2}=\dfrac{a^2c^2k^2-a^2b^2}{b^2+a^2k^2}$$

因为$$|AF_{1}|=a+ex_{1},|AF_{2}|=a-ex_{1},|BF_{1}|=a+ex_{2},|BF_{2}|=a-ex_{2}$$

\begin{align*}y&=\dfrac{(a-ex_{2})y_{1}+(a-ex_{1})y_{2}}{4a}=\dfrac{a(y_{1}+y_{2})-e(x_{1}y_{2}+x_{2}y_{1})}{4a}\\&=\dfrac{a\cdot\dfrac{2b^2ck}{b^2+a^2k^2}-\dfrac{c}{a}\left(-\dfrac{2a^2b^2k}{b^2+a^2k^2}\right)}{4a}\\&=\dfrac{b^2ck}{b^2+a^2k^2}\end{align*}

\begin{align*}y^2&=\dfrac{b^4c^2k^2}{(b^2+a^2k^2)^2}=\dfrac{b^2c^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}}{\left(b^2+a^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}\right)^2}=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{(a^2x+c^3)b^2+a^2(b^2c-b^2x)}{a^2x+c^3}\right]^2}\\&=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{b^2c(a^2+c^2)}{a^2x+c^3}\right]^2}=\dfrac{b^2(c-x)(a^2x+c^3)}{(a^2+c^2)^2}\end{align*}

$$(a^2+c^2)^2y^2=(b^2c-b^2x)(a^2x+c^3)=b^2(cb^2x+c^4-a^2x)$$
$$a^2b^2x^2+(a^2+c^2)^2y^2-cb^4x-b^2c^4=0$$

(2)设直线$AB$方程为$x=ky-3$与椭圆联立得$$(16k^2+25)y^2-96ky-256=0$$

## 解析几何竞赛题

(1)上述$L$被双曲线$x^2-y^2=2(x>0)$所截线段的中点的轨迹为双曲线；

(2)$L$总是(1)中轨迹曲线的切线.

\begin{align}A &= \int_{{y_1}}^{{y_2}} {\left( {my + l - \sqrt {{y^2} + 2} } \right)dy}  = \left[ {\frac{{m{y^2}}}{2} + ly - \left( {\frac{{y\sqrt {{y^2} + 2} }}{2} + \ln \left( {y + \sqrt {{y^2} + 2} } \right)} \right)} \right]_{{y_1}}^{{y_2}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{y_2}\sqrt {y_2^2 + 2} }}{2} - \frac{{{y_1}\sqrt {y_1^2 + 2} }}{2}} \right) - \ln \frac{{{y_2} + \sqrt {y_2^2 + 2} }}{{{y_1} + \sqrt {y_1^2 + 2} }}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{x_2}{y_2}}}{2} - \frac{{{x_1}{y_1}}}{2}} \right) - \ln \frac{{{x_2} + {y_2}}}{{{x_1} + {y_1}}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left[ {\frac{m}{2}\left( {y_2^2 - y_1^2} \right) + \frac{l}{2}\left( {{y_2} - {y_1}} \right)} \right] - \ln \frac{{{x_1}{x_2} - {y_1}{y_2} + {x_1}{y_2} - {x_2}{y_1}}}{2}\\&= \frac{l}{2}\left( {{y_2} - {y_1}} \right) - \ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2}\end{align}
\begin{align}&\ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2} = \ln \frac{{{l^2} - 2 + \frac{{2{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{2l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} + {l^2}}}{2}\\&= \ln \left( {{l^2} - 1 + \frac{{{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right) = \ln \left( {\frac{{{m^2} + {l^2} - 1}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right).\end{align}

$A = \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} - \ln \frac{{{m^2} + {l^2} - 1 + l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.$

$\left\{ \begin{array}{l}x = \frac{{{x_1} + {x_2}}}{2} = \frac{{{y_1} + {y_2}}}{2}m + l = \frac{l}{{1 - {m^2}}}\\y = \frac{{{y_1} + {y_2}}}{2} = \frac{{ml}}{{1 - {m^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}m = \frac{y}{x}\\l = \frac{{{x^2} - {y^2}}}{x}\end{array} \right..$

$A = \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} - \ln \left[ {\left( {{x^2} - {y^2}} \right) + \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} - 1} \right].$

$A = t - \ln \left( {t + \sqrt {{t^2} + 1} } \right).\tag{1}$

(2)再之，由$2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y} = \frac{1}{m} = {k_L}$及直线$L$经过点$M$可知，直线$L$总为$M$的轨迹曲线的切线.