# 问题及解答

## 设圆 $O$ 是 $\triangle ABC$ 的外接圆, $E$, $F$ 分别为边 $AC$, $AB$ 上的点, 使得 $BCEF$ 四点共圆, 圆心为 $J$. $BE$, $CF$ 交于点 $S$, 直线 $JS$ 交圆 $O$ 于点 $T$. 证明: $\angle ATJ=90^\circ$ .

Posted by haifeng on 2021-10-20 14:56:43 last update 2021-10-20 15:19:38 | Edit | Answers (1)

1

Posted by haifeng on 2021-10-24 17:43:41

$\begin{split} &(a-0)^2+(b-1-t)^2=(c-0)^2+(0-1-t)^2\\ \Rightarrow\ &a^2+b^2-2b(1+t)+(1+t)^2=c^2+(1+t)^2\\ \Rightarrow\ &2b(1+t)=a^2+b^2-c^2. \end{split}$

\begin{aligned} \ell_{AC}:\quad& y-0=\frac{b-0}{a-c}(x-c),\\ \ell_{AB}:\quad& y-0=\frac{b-0}{a+c}(x+c).\\ \end{aligned}

$(y_1-t)^2+x_1^2=t^2+c^2=(y_2-t)^2+x_2^2.$

$\begin{cases} y_1^2-2y_1 t+x_1^2=c^2,\\ y_2^2-2y_2 t+x_2^2=c^2. \end{cases}$

$\begin{cases} \frac{b^2}{(a-c)^2}(x_1-c)-\frac{2b}{a-c}t+x_1+c=0,\\ \frac{b^2}{(a+c)^2}(x_2+c)-\frac{2b}{a+c}t+x_2-c=0. \end{cases}$

$\begin{cases} x_1=\dfrac{2b(a-c)t+b^2 c-c(a-c)^2}{b^2+(a-c)^2},\quad (1)\\ x_2=\dfrac{2b(a+c)t-b^2 c+c(a+c)^2}{b^2+(a+c)^2}.\quad (2)\\ \end{cases}$

$\begin{cases} y_1=b\cdot\dfrac{2bt-2c(a-c)}{b^2+(a-c)^2},\quad(3)\\ y_2=b\cdot\dfrac{2bt+2c(a+c)}{b^2+(a+c)^2}.\quad(4) \end{cases}$

\begin{aligned} \ell_{BE}:\quad y-0=\frac{y_1-0}{x_1-(-c)}(x-(-c))\quad\Rightarrow\ y=\frac{y_1}{x_1+c}(x+c)\\ \ell_{CF}:\quad y-0=\frac{y_2-0}{x_1-c}(x-c)\quad\Rightarrow\ y=\frac{y_2}{x_2-c}(x-c) \end{aligned}

$\begin{split} &\frac{y_1}{x_1+c}(x+c)=\frac{y_2}{x_2-c}(x-c)\\ \Rightarrow\ & x_S=\dfrac{(\frac{y_1}{x_1+c}+\frac{y_2}{x_2-c})\cdot c}{\frac{y_2}{x_2-c}-\frac{y_1}{x_1+c}}\quad(5) \end{split}$

\begin{aligned} \frac{y_1}{x_1+c}&=\frac{bt-c(a-c)}{(a-c)t+bc},\\ \frac{y_2}{x_2-c}&=\frac{bt+c(a+c)}{(a+c)t-bc}. \end{aligned}

$x_S=\dfrac{ab(t^2+c^2)}{-bt^2+(b^2+a^2-c^2)t+bc^2}=\dfrac{a(t^2+c^2)}{t^2+2t+c^2}.$

$\begin{split} y_S=\frac{y_1}{x_1+c}(x_S+c)&=\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\biggl(\dfrac{a(t^2+c^2)}{t^2+2t+c^2}+c\biggr)\\ &=\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\frac{(a+c)(t^2+c^2)+2ct}{t^2+2t+c^2} \end{split}$

$k_{JD}=\frac{t-(2(1+t)-b)}{0-(-a)}=\frac{b-t-2}{a}.$

$\begin{split} k_{JS}&=\frac{y_S-y_J}{x_S-x_J}=\frac{y_S-t}{x_S}\\ &=\dfrac{\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\frac{(a+c)(t^2+c^2)+2ct}{t^2+2t+c^2}-t}{\frac{a(t^2+c^2)}{t^2+2t+c^2}}\\ &=\dfrac{\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\Bigl((a+c)(t^2+c^2)+2ct\Bigr)-t(t^2+2t+c^2)}{a(t^2+c^2)}\\ &=\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\biggl[\frac{(a+c)(t^2+c^2)+2ct}{a(t^2+c^2)}\biggr]-\frac{t}{a}\cdot\frac{t^2+2t+c^2}{t^2+c^2}\\ &=\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\biggl[\frac{a+c}{a}+\frac{2ct}{a(t^2+c^2)}\biggr]-\frac{t}{a}\cdot\biggl[1+\frac{2t}{t^2+c^2}\biggr]\\ &=\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\frac{a+c}{a}+\frac{bt-c(a-c)}{(a-c)t+bc}\cdot\frac{2ct}{a(t^2+c^2)}-\frac{t}{a}-\frac{2t^2}{a(t^2+c^2)}\\ &=\frac{1}{a}\cdot\biggl[\frac{bt-c(a-c)}{(a-c)t+bc}\cdot(a+c)-t\biggr]+\frac{2t}{a(t^2+c^2)}\biggl[\frac{bt-c(a-c)}{(a-c)t+bc}\cdot c-t\biggr]\\ &=\frac{1}{a}\cdot\frac{b(a+c)t-c(a^2-c^2)-(a-c)t^2-bct}{(a-c)t+bc}+\frac{2t}{a(t^2+c^2)}\cdot\frac{bct-c^2(a-c)-(a-c)t^2-bct}{(a-c)t+bc}\\ &=\frac{1}{a}\cdot\frac{bat-c(a^2-c^2)-(a-c)t^2}{(a-c)t+bc}+\frac{2t}{a}\cdot\frac{-(a-c)}{(a-c)t+bc}\\ &=\frac{-(a-c)t^2+\bigl(ba-2(a-c)\bigr)t-c(a^2-c^2)}{a\bigl[(a-c)t+bc\bigr]}. \end{split}$

$\begin{split} &k_{JS}=k_{JD}\\ \Rightarrow &\ \frac{-(a-c)t^2+\bigl(ba-2(a-c)\bigr)t-c(a^2-c^2)}{a\bigl[(a-c)t+bc\bigr]}=\frac{b-t-2}{a}\\ \Rightarrow &\ -(a-c)t^2+\bigl[ba-2(a-c)\bigr]t-c(a^2-c^2)=\bigl[(a-c)t+bc\bigr]\cdot(b-t-2)\\ \Rightarrow &\ \bigl[ba-2(a-c)\bigr]t-c(a^2-c^2)=b(a-c)t-2(a-c)t+b^2c-bct-2bc\\ \Rightarrow &\ bat-2(a-c)t-c(a^2-c^2)=bat-2bct-2(a-c)t+b^2 c-2bc\\ \Rightarrow &\ -c(a^2-c^2)=-2bct+b^2 c-2bc\\ \Rightarrow &\ 2bt=a^2-c^2+b^2-2b\\ \Rightarrow &\ 2b(1+t)=a^2+b^2-c^2. \end{split}$