Crimson-boi[11] 创新高考题

[latexpage]

题目背景

总有人喜欢在高考数学中嚷嚷"仿射变换", 于是一道纯正射影几何高考"创新题"应运而生(其实还顺便测试一下QuickLaTex).

题面

\begin{tikzpicture}
[+preamble]
\usepackage{tikz}
\usepackage{pgfplots}
[/preamble]
\draw [domain=-4:0] plot(\x,{(-0-2.0146588528648297* \x)/-1.3729088238892677});
\draw (0,0) -- (0,-7);
\draw [domain=0:2.2] plot(\x,{(-0-2.5845421587489756* \x)/0.8548168263851235});
\draw [domain=-1.3729088238892677:3.0908775282059806] plot(\x,{(-5.270505022314892-0.5698833058841459* \x)/2.227725650274391});
\draw [domain=-3.0097654971040817:3.0908775282059806] plot(\x,{(-15.880699537575175--0.8643414322362726* \x)/4.1846616907620025});
\draw [domain=-1.3729088238892677:2.568596963168242] plot(\x,{(-4.311174395741616-1.1255164601094387* \x)/1.3729088238892677});
\draw [domain=-3.0097654971040817:2.568596963168242] plot(\x,{(-14.639726260916422-0.44743051972991044* \x)/3.0097654971040817});
\draw [domain=0:3.6954095336934416] plot(\x,{(-2.684274695329576--0.5556331542252928* \x)/0.8548168263851235});
\draw [domain=0:3.6954095336934416] plot(\x,{(-5.7147836523125--1.311771951966183* \x)/1.1748961936579208});
\draw [dashed,domain=2.35:3.9] plot(\x,{(--17.4898214026939-4.507772434886537* \x)/-1.1268125705251997});
\draw [fill=black] (0,0) circle (0.05) node[above right]{O};
\draw [fill=black] (-1.3729088238892677,-2.0146588528648297) circle (0.05) node[left]{A};
\draw [fill=black] (0,-3.1401753129742684) circle (0.05) node[right]{B};
\draw [fill=black] (0.8548168263851235,-2.5845421587489756) circle (0.05);
\draw (0.9424581431106462,-2.2620782226544286) node {C};
\draw [fill=black] (-3.0097654971040817,-4.416644862555649) circle (0.05) node[left]{A'};
\draw [fill=black] (0,-4.864075382285559) circle (0.05) node[below right]{B'};
\draw [fill=black] (1.1748961936579208,-3.5523034303193763) circle (0.05) node[below right]{C'};
\draw [fill=black] (3.0908775282059806,-3.1565576871217527) circle (0.05);
\draw (3.279001882554508,-2.9550223768325687) node {Q};
\draw [fill=black] (3.6954095336934416,-0.7381495315640785) circle (0.05);
\draw (3.882348565753318,-0.531233115016662) node {R};
\draw [fill=black] (2.568596963168242,-5.245921966450616) circle (0.05);
\draw (2.748472902500383,-5.045930709987063) node {P};
\end{tikzpicture}

如上图, O, A, A'共线, O, B, B'共线, O, C, C'共线, 且A, B, C不共线, A', B', C'不共线.

(1) 若ABC与A'B'C'不是同一平面, 求证AB与A'B'的交点P, AC与A'C'的交点Q, BC与B'C'的交点R共线.

(2) 若ABC与A'B'C'共平面, 试通过将其转化到(1)的方式证明结论任然成立.

(3) 称一个点集到自身的双射为自同构, 如果在此映射下原先共线的点任然共线. 例如在上图的点集中, 满足f(A)=B, f(B)=A, f(A')=B', f(B')=A', 其余点都映到自身的f就是一个自同构. 求上图点集自同构的个数.

先想再看提示哦

提示

前两小题应该都不难, 最后一题请考虑一个显然的事实: 两条相交直线确定一个平面.

愿你无需看标答

标答

先摸摸. 高考练习乱看答案小心去大专!