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[几何] 三角形的内切锥线、和两边相切的锥线交点

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hbghlyj 发表于 2023-3-18 00:30 |阅读模式
三角形 $A B C$,圆锥曲线 $\mathcal{G}$ 内切于 $A B C$ 且圆锥曲线 $\mathcal{K}$ 与直线 $AB, A C$ 在点 $B, C$ 相切。设圆锥曲线 $\mathcal{G}$ 和 $\mathcal{K}$ 相交于两点 $P, Q$。证明从点 $P, Q$ 到圆锥曲线 $\mathcal{G}$ 的切线的交点在圆锥曲线 $\mathcal{K}$ 上。
size(400); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-2.2076452464988225,xmax=2.0960426841273088,ymin=-1.4037065211271376,ymax=2.104001510104192;
pair B=(-1.1470114060788423,0.688716925339064), C=(0.3677950608792688,0.7640186551456793), A=(-0.9828516794917416,1.8323840534571243), P=(-0.9863058558381781,1.0131534750865794), Q=(-0.12356570917458554,1.056040686659606);
draw(shift((0.,0.))*rotate(-45.)*xscale(1.4142135623730951)*yscale(0.816496580927726)*unitcircle); draw((xmin,6.966794791237958*xmin+8.679710014699669)--(xmax,6.966794791237958*xmax+8.679710014699669)); draw((xmin,-0.7910028332189766*xmin+1.0549455903451272)--(xmax,-0.7910028332189766*xmax+1.0549455903451272));
draw(shift(-0.58735600823044, 1.095039877980622)*rotate(-26.7342909968379)*scale(0.51211108274715,0.3231995806602)*unitcircle);
draw((xmin,-1.3499556751554866*xmin-0.31831571244125834)--(xmax,-1.3499556751554866*xmax-0.31831571244125834)); draw((xmin,-2.9765532812147604*xmin+0.6882407695703645)--(xmax,-2.9765532812147604*xmax+0.6882407695703645));
label("$\mathcal K$",(-0.6224310400769718,1.18),NE*lsf);
label("$\mathcal G$",(-0.6224310400769718,.77),NE*lsf); dot(B,ds); label("$B$",(-1.3442094234264954,0.7413918518181756),NE*lsf); dot(C,ds); label("$C$",(0.3961534074536643,0.8290845525989587),NE*lsf); dot(A,linewidth(4.pt)+ds); label("$A$",(-0.9529650660968471,1.8881425543361101),NE*lsf); dot(P,linewidth(4.pt)+ds); label("$P$",(-0.9597106584645996,1.0651802854702983),NE*lsf); dot(Q,linewidth(4.pt)+ds); label("$Q$",(-0.27166023695383884,0.9100316610119894),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);

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 楼主| hbghlyj 发表于 2024-11-6 06:01

$\mathbf{TG}_△$

On theorem generators in plane geometry

Problem 3. Let given triangle $A B C$, let conic $\mathcal{G}$ is inscribed in $A B C$ and conic $\mathcal{K}$ tangent to lines $A B, A C$ at points $B, C$. Let conics $\mathcal{G}$ and $\mathcal{K}$ intersects at two points $P, Q$. We prove that tangents to conic $\mathcal{G}$ from points $P, Q$ intersects on conic $\mathcal{K}$.

Proof. Consider algebraical group $G:=\operatorname{PGL}\left(\mathbb{C}^2\right)$ as group of projective transformations of plane $\mathbb{C}^2$. Consider object $\mathcal{X}$ as sets of conic $\mathcal{K}$ and lines $l, x, y$ which are tangent to it and where $x \| y, x$ is perpendicular to $l$. Consider morphism $\pi: \mathcal{X} \times G \rightarrow$ $\mathcal{Y}$, where $\mathcal{Y}$ is sets : conic and three lines which are tangent to it, $\pi((\mathcal{K}, l, x, y), g):=$ $(g(\mathcal{K}), g(l), g(x), g(y)) \in \mathcal{Y}$. Easy to check that $\pi$ is dominating. If for any set $\left(\mathcal{K}, l_1, l_2, l_3\right) \in$ $\mathcal{Y}$ we construct conic $\mathcal{C}$ which is tangent to lines $l_2, l_3$ at points $l_1 \cap l_2, l_1 \cap l_3$, and consider intersection point of tangents from points $P, Q:=\mathcal{K} \cap \mathcal{C}$ to conic $\mathcal{K}$, and also consider conic $\mathcal{C}$, then we get morphism $\psi: \mathcal{Y} \rightarrow \mathcal{Z}$, where $\mathcal{Z}$ is sets : point on plane and conic. Easy to check that if $\phi:=\pi \circ(\psi, e)$, then next diagram is commutative:\begin{xy}
\xymatrix {
&\mathcal X \times G \ar[dl]_\pi \ar[dr]^\phi\\
\mathcal Y \ar[rr]^\psi &&\mathcal Z
}
\end{xy}From density of $π$ we get that $\operatorname{Im}ψ ⊆ \operatorname{Im}φ$ and from problem 2 we know that $\operatorname{Im}(φ)$ is conics with points on them.

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 楼主| hbghlyj 发表于 2024-11-6 06:02

不幸的是,1 楼的图无法再加载,因此暂时手动粘贴 SVG

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 楼主| hbghlyj 发表于 2024-11-6 06:12
2# 中的证明很难理解!有人能提出更简单的证明吗?

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 楼主| hbghlyj 发表于 2024-11-7 18:34
1 楼的图无法再加载, 什么原因造成的

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