今年奥数几何题

denglongshan · 发表于 2023-7-17 23:02

本帖最后由 denglongshan 于 2023-7-17 15:23 编辑

设ABC是一个正三角形，点A,BC在三角形ABC的内部且满足\(BA_{1}=A_{1}C,CB_{1}=B{1}A,AC_{1}=C_{1}B\),及
\(∠BA_{1}C+∠CB_{1}A+∠AC_{1}B=480\)度，
设直线\(BC_{1}\)与\(B_{1}C\)交于点\(A_{2}\),直线\(C_{1}A\)与\(A_{1}C\)交于点\(B_{2}\),直线\(A_{1}B\)与\(B_{1}A\)交于点\(C_{2}\)。
证明:若三角形\(A_{1}B_{1}C_{1}\)的三边长度两两不等,则三角形\(AA_{1}A_{2}\)、\(BB_{1}B_{2}\)和\(CC_{1}C_{2}\)的外接圆都经过两个公共点.

天山草 · 发表于 2023-7-19 18:41

本帖最后由天山草于 2023-7-20 10:27 编辑

程序代码：

Clear["Global`*"];(*u=\[ExponentialE]^(2\[ImaginaryI]\[Theta]);v=\[ExponentialE]^(2\[ImaginaryI]\[Beta]);*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = 1/2 + I Sqrt[3]/2;
\!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a;
(*u=\[ExponentialE]^(2\[ImaginaryI]\[Theta]);v=\[ExponentialE]^(2\[ImaginaryI]\[Beta])*)
\[Angle]A1BC = \[Theta]; \[Angle]C1BA = \[Beta]; \[Angle]B1CA = \[Gamma]; \[Angle]BA1C = \[Pi] - 2 \[Theta];
\[Angle]AB1C = \[Pi] - 2 \[Gamma]; \[Angle]AC1B = \[Pi] - 2 \[Beta];
(*解 Solve[{\[Angle]BA1C+ \[Angle]AB1C+\[Angle]AC1B\[Equal](480\[Pi])/180},{\[Gamma]}]得到：*)
\[Gamma] = \[Pi]/6 - \[Beta] - \[Theta];(*现在只剩二个变量 \[Beta] 和 \[Theta]*)
kBA1 = u; kBC1 = E^(2 I (\[Pi]/3))/v; kCB1 = E^(2 I ((2 \[Pi])/3 + \[Gamma])); kCB1 = E^(2 I ((5 \[Pi])/6))/(u v);
mBC = (b + c)/2; \!\(\*OverscriptBox[\(mBC\), \(_\)]\) = (\!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2; mAB = (a + b)/2; \!\(\*OverscriptBox[\(mAB\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\))/2; mAC = (a + c)/2;
\!\(\*OverscriptBox[\(mAC\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2;
K[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
(*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
a1 = Simplify@Jd[kBA1, b, -K[b, c], mBC]; \!\(\*OverscriptBox[\(a1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kBA1, b, -K[b, c], mBC]; Print["A1 = ", a1];
b1 = Simplify@Jd[kCB1, c, -K[a, c], mAC]; \!\(\*OverscriptBox[\(b1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kCB1, c, -K[a, c], mAC]; Print["B1 = ", b1];
c1 = Simplify@Jd[kBC1, b, -K[a, b], mAB]; \!\(\*OverscriptBox[\(c1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kBC1, b, -K[a, b], mAB]; Print["C1 = ", c1];
a2 = Simplify@Jd[K[b, c1], c1, K[c, b1], b1]; \!\(\*OverscriptBox[\(a2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[b, c1], c1, K[c, b1], b1];Print["A2 = ", a2];
c2 = Simplify@Jd[K[b, a1], a1, K[a, b1], b1]; \!\(\*OverscriptBox[\(c2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[b, a1], a1, K[a, b1], b1];Print["C2 = ", c2];
b2 = Simplify@Jd[K[a, c1], c1, K[c, a1], a1]; \!\(\*OverscriptBox[\(b2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[a, c1], c1, K[c, a1], a1];Print["B2 = ", b2];
Waixin[a_, b_, c_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + b \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + c \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b) )/( \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b));
\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[a_, b_, c_] := -((a \!\(\*OverscriptBox[\(a\), \(_\)]\) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\)) + b \!\(\*OverscriptBox[\(b\), \(_\)]\) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) + c \!\(\*OverscriptBox[\(c\), \(_\)]\) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) )/( \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b)));
o1 = Simplify@Waixin[a, a1, a2]; \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[a, a1, a2];
o2 = Simplify@Waixin[b, b1, b2]; \!\(\*OverscriptBox[\(o2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[b, b1, b2];
o3 = Simplify@Waixin[c, c1, c2]; \!\(\*OverscriptBox[\(o3\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[c, c1, c2];
Print["O1 = ", o1]; Print["O2 = ", o2]; Print["O3 = ", o3];
XiangjiaoxuanLianxin[o1_, a_, o2_, b_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) - b \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\) o2 + b \!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) o1 - a \!\(\*OverscriptBox[\(o1\), \(_\)]\) + \!\(\*OverscriptBox[\(o2\), \(_\)]\) o1 - o2 \!\(\*OverscriptBox[\(o1\), \(_\)]\))/(2 (\!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(o1\), \(_\)]\)));
\!\(\*OverscriptBox[\(XiangjiaoxuanLianxin\), \(_\)]\)[o1_, a_, o2_, b_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) - b \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\) o2 + b \!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) o1 - a \!\(\*OverscriptBox[\(o1\), \(_\)]\) + o2 \!\(\*OverscriptBox[\(o1\), \(_\)]\) - \!\(\*OverscriptBox[\(o2\), \(_\)]\) o1)/(2 (o2 - o1));
(*两圆连心线与公共弦的交点P，O1、O2是两圆圆心，A是O1上任一点，B是O2上任一点。如果两圆相切的，所求交点即是切点*)
m = Simplify@XiangjiaoxuanLianxin[o1, a, o2, b];
m1 = Simplify@XiangjiaoxuanLianxin[o1, a, o3, c];
SimpleFuc :=
Factor[#, Extension -> Sqrt[3]] &@(FactorTerms[Numerator[#]]/
FactorTerms[Denominator[#]]) &@Factor[#, Extension -> Sqrt[3]] &;
Print["M = ", SimpleFuc[m]];
Print["M1 = ", SimpleFuc[m1]];
Print["由于 M 和 M1 点的坐标相同，即圆O1、圆O2的根轴中心与圆O1、圆O3的根轴中心重合，因此三个圆只有两个公共交点。"];

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天山草 · 发表于 2023-7-19 09:20

本帖最后由天山草于 2023-7-19 17:29 编辑

说明：
① M 是圆 O1、圆O2 连心线与公共弦的交点， M1 是圆 O1、圆O3 连心线与公共弦的交点。如果 M 与 M1 重合，则三个圆 O1、O2、O3 交于两个点。
② 程序中倒数第2、3、4 行化简复数表达式的方法，是 creasson 大侠提供的。在此表示感谢！

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今年奥数几何题

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