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如何通过三角形的顶点画一个圆？

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如何通过三角形的顶点画一个圆？

我想构造角度，但要清楚地显示圆规的弧。例如，60 度角必须用其弧来表示，并且垂直平分线必须以弧作为两边的交点

答案1

请查看是否符合要求

在此处输入图片描述

\documentclass{standalone}
\usepackage{tikz}

\usetikzlibrary{calc,positioning,angles,arrows.meta,quotes,intersections}
\usetikzlibrary{through}

\begin{document}
\begin{tikzpicture}[declare function={alpha=60; beta=75; sigma=110;}]

\path (0,0) coordinate (A)node[below]{A}  --  ++ (alpha:7cm) coordinate 
(B)node[above]{B} --
([turn]alpha:-7cm) coordinate (C)node[below]{C};
\draw(A)--(B)--(C)--cycle;

\coordinate (D) at ($(A)!(B)!(C)$);
\draw[gray!50,dashed] (B)--(D);

\coordinate (E) at ($(C)!(A)!(B)$);
\draw[gray!50,dashed] (A)--(E);

\coordinate (F) at (intersection of B--D and A--E) ;
\node[fill=red,inner sep=1pt, circle] at(F){};
\node [draw,blue,thick,name path=circle,](c) at (F) [circle through={(A)}] 
{};


\pic[ draw,<->,>=stealth,blue, "$60^0$"{fill=white},inner sep=1pt, circle,  
draw,angle 
eccentricity=1.1, angle radius = 10mm] {angle = C--A--B};

\pic[ draw,<->,>=stealth,blue, "$60^0$"{fill=white},inner sep=1pt, circle,  
draw,angle 
eccentricity=1.1, angle radius = 10mm] {angle = B--C--A};

\draw[line width=0.5pt] (0,-10mm) -- (0,2mm);
\draw[line width=0.5pt] (7cm,-10mm) -- (7cm,2mm);
\draw[<->,>=stealth,](0,-8mm) --node[midway,fill=white](){7cm}(7cm,-8mm);

\coordinate (G) at (0,5);
\path[name path=fg] (F)--(G);
\path [name intersections ={of=fg and circle,name=i}](i-1)  coordinate [];
\draw [->,>=stealth,](F)--node[pos=0.7,fill=white,sloped]{radius}(i-1);
\end{tikzpicture}
\end{document}

编辑——添加了另一个带有 tkz euclide 的选项

在此处输入图片描述

\documentclass{article} % or another class
\usepackage{xcolor} % before tikz or tkz-euclide if necessary

\usepackage{tkz-euclide} % no need to load TikZ

\usetikzlibrary{babel} %if there are problems with the active characters
\begin{document}
    \noindent\hspace{-4.5cm}
          \begin{tikzpicture}
          %equilateral triangle
          \tkzDefPoint(0,0){A}
          \tkzDefPoint(7,0){B}
          %draw intersecting circles
          \tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D}
          \tkzDrawPoints[color=red](A,B,C)
          \tkzDrawCircle[dashed, color=blue](A,B)
          \tkzDrawCircle[dashed, color=blue](B,A)
          \tkzCompass[color=blue,line width=2pt](A,C)
          \tkzCompass[color=blue,line width=2pt](B,C)
          %draw triangle
          \tkzDrawPolygon[color=red, ](A,B,C)
          \tkzMarkSegments[mark=s|,color=blue](A,C B,C)
          \tkzLabelPoints[](A,B)
          \tkzLabelPoints[above](C)
          %circumscribed circle
          \tkzDefMidPoint(A,B)\tkzGetPoint{P}
          \tkzLabelPoints[](P)
          \tkzDefLine[orthogonal =through P](A,B)\tkzGetPoint{X}
          \tkzLabelPoints(X)
          %orthogonal midpoints of two sides
          \tkzDefMidPoint(B,C)\tkzGetPoint{Q}
          \tkzLabelPoints[](Q)
          \tkzDefLine[orthogonal =through Q](B,C)\tkzGetPoint{Y}
          \tkzLabelPoints(Y)
          %mark right angles
          \tkzDrawLines[dashed,green](P,X Q,Y)
          \tkzMarkRightAngles(B,Q,Y A,P,X)
          %find center of circumscribe circle
          \tkzInterLL(P,X)(Q,Y)\tkzGetPoint{Z}
          \tkzLabelPoints(Z)
          \tkzDrawPoint[green](Z)
          %calculate radius of circum circle
          \tkzCalcLength[cm](Z,A)\tkzGetLength{rZA}
          \tkzDrawCircle[green!50!black,line width=2pt,R ](Z, \rZA cm)
          %mark all three angles of equilateral triangle
          \tkzMarkAngle[size=1cm,color=cyan,mark=||](B,A,C)
          \tkzMarkAngle[size=1cm,color=cyan,mark=||](A,C,B)
          \tkzMarkAngle[size=1cm,color=cyan,mark=||](C,B,A)
          %find the angle of eq triangle
          \tkzFindAngle(B,A,C)  
          \tkzGetAngle{angleBAC}
          \edef\angleBAC{\fpeval{round(\angleBAC)}}
          \tkzLabelAngle[pos=0.7](B,A,C){\angleBAC$^\circ$} 
          %draw radius of the circum circle
          \tkzDefShiftPoint[Z](135:\rZA){z}
          \tkzDrawSegments[arrows=-stealth](Z,z)  
          \tkzLabelLine[pos=0.7,black,rotate=-45,fill=white](Z,z){radius}   
                                                                        
          
                \end{tikzpicture}
          
    
\end{document}

答案2

我假设 c、alpha 和 beta 已知；然后：

新的更简单的答案：

在此处输入图片描述

\documentclass[margin=5pt, tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections}
\usetikzlibrary{backgrounds}
\usetikzlibrary{patterns}
\usetikzlibrary{positioning}
\usetikzlibrary{angles, quotes, babel}
\usepackage{amsmath, amssymb}

\begin{document}
% Gegebene Größen
\pgfmathsetmacro{\c}{7} 
\pgfmathsetmacro{\Alpha}{60} 
\pgfmathsetmacro{\Beta}{60} 

% Seitenlängen
\pgfmathsetmacro{\Gamma}{180-\Alpha-\Beta} 
\pgfmathsetmacro{\a}{\c*sin(\Alpha)/sin(\Gamma)} %  
\pgfmathsetmacro{\b}{sqrt(\a*\a +\c*\c -2*\a*\c*cos(\Beta))} %  

\pgfmathsetmacro{\R}{\a/(2*sin(\Alpha))} %  
\pgfmathsetmacro{\McU}{\R*abs(cos(\Gamma))} % 

\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
]

% Dreieckskonstruktion
%\pgfmathsetmacro{\Alpha}{acos((\b^2+\c^2-\a^2)/(2*\b*\c))} % 
\coordinate[label=below:$A$] (A) at (0,0); 
\coordinate[label=below:$B$] (B) at (\c,0); 
\coordinate[label=$C$] (C) at (\Alpha:\b); 
\draw[local bounding box=dreieck] (A) -- (B) -- (C) --cycle; 

% Umkreis
\draw[red] ($(A)!0.5!(B)$) coordinate[label=-90:$M_c$] (Mc) -- +(90:\McU) coordinate[label=110:$U$](U) node[midway, right] {$|M_cU|$};
\draw[densely dashed, red] (U) -- (A) node[midway, above] {$R$};;
\draw[] (U) circle[radius=\R];
\draw pic [angle radius=3mm, %angle eccentricity=1.2, 
draw,   "$\cdot$", red
] {angle =U--Mc--A};

% Annotationen - Dreieck
\draw[thick] (A) -- (B)  node[pos=0.25, below]{$c/2$} node[pos=0.75, below]{$c/2$};
\draw pic [angle radius=6mm, %angle eccentricity=1.2, 
draw,   "$\alpha$", thick
] {angle =B--A--C};
\draw pic [angle radius=6mm, %angle eccentricity=1.2, 
draw,   "$\beta$", thick
] {angle =C--B--A};

\draw[-latex] (U) -- +(44:\R) node[near end, above]{$R$};

% Annotationen - Aufgabe
\pgfmathsetmacro{\x}{max(\a, \b,\c)} %
\begin{scope}[shift={($(dreieck.north west)+(-\x cm-3mm,0)$)}]
% Strecken
\foreach[count=\y from 0] \s/\S in {c/c}{%%
\draw[|-|, yshift=-\y*5mm, local bounding box=strecken] (0,0) -- (\csname \s \endcsname,0) node[midway, above]{$\S$ %= \csname \s \endcsname cm
};}%%
\end{scope}
% Winkel
\pgfmathsetmacro{\Winkel}{\Alpha}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={}, 
"$\alpha$", 
] {angle =R--Q--P};
% Winkel 2
\pgfmathsetmacro{\Winkel}{\Beta}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(15mm+\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={}, 
"$\beta$", 
] {angle =R--Q--P};

% Annotationen - Rechnung
\tikzset{PosUnten/.style={below=5mm of dreieck, anchor=north,}}
\tikzset{PosLinks/.style={shift={($(dreieck.north)+(-44mm,-17mm)$)}, anchor=north east,}}
\node[yshift=-0mm, draw, align=left, fill=lightgray!50,
%PosUnten,
PosLinks,
] (Rechnung){
$\begin{array}{l l}
c = \c \text{ cm}  &  \\
\alpha = \Alpha^\circ  &  \\
\beta = \Beta^\circ    &  \\   \hline
\gamma = 180^\circ-\alpha-\beta  &=\Gamma^\circ   \\
a = c\cdot\dfrac{\sin(\alpha)}{\sin(\gamma)}  & =\a \text{ cm}  \\[1em]
b = \sqrt{a^2+c^2-2ac\cos(\beta)}  & =\b \text{ cm}   \\
R = \dfrac{a}{2\sin(\alpha)}   &=\R \text{ cm}    \\[1em]
|M_cU| = R|\cos(\gamma)|  &=\McU \text{ cm}    \\ 
\end{array}$
};

\node[anchor=north west, yshift=-3mm, inner sep=0pt, draw=none] at (Rechnung.south west){
$\begin{array}{l l}
\text{Hint: } |M_cU|^2 \hspace{-3mm}& =R^2-\left(\dfrac{c}{2}\right)^2 
= R^2 - \bigl( R   \sin (\gamma)\bigr)^2 \\[0.75em]
&= R^2 \bigl(1-\sin^2(\gamma) \bigr) =R^2\cos^2(\gamma)\\[1em]
\multicolumn{2}{l}{\Rightarrow |M_cU| = R|\cos(\gamma)|} 
\end{array}$
};

%% Punkte
\foreach \P in {U, Mc}
\draw[fill=black!1, draw=red] (\P) circle (1.75pt);
\end{tikzpicture}
\end{document}

老旧，太复杂，答案：

在此处输入图片描述

暗示：从边长和角点到外接圆中心点（U）的矢量计算方式

\pgfmathsetmacro{\Da}{\a^2*(\b^2+\c^2-\a^2)} %  
\pgfmathsetmacro{\Db}{\b^2*(\a^2+\c^2-\b^2)} %  
\pgfmathsetmacro{\Dc}{\c^2*(\a^2+\b^2-\c^2)} %  
\pgfmathsetmacro{\D}{\Da+\Db+\Dc} %  
\pgfmathsetmacro{\au}{\Da/\D} %  
\pgfmathsetmacro{\bu}{\Db/\D} %  
\pgfmathsetmacro{\cu}{\Dc/\D} %    

\coordinate[] (U) at ($\au*(A)+\bu*(B)+\cu*(C)$);

位于维基百科。

平均能量损失：

\documentclass[margin=5pt, tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections}
\usetikzlibrary{backgrounds}
\usetikzlibrary{patterns}
\usetikzlibrary{positioning}
\usetikzlibrary{angles, quotes, babel}
\usepackage{amsmath, amssymb}

\begin{document}

% Gegebene Größen
\pgfmathsetmacro{\c}{7} 
\pgfmathsetmacro{\Alpha}{60} 
\pgfmathsetmacro{\Beta}{60} 

% Seitenlängen
\pgfmathsetmacro{\Gamma}{180-\Alpha-\Beta} 
\pgfmathsetmacro{\a}{\c*sin(\Alpha)/sin(\Gamma)} %  
\pgfmathsetmacro{\b}{sqrt(\a*\a +\c*\c -2*\a*\c*cos(\Beta))} %  

\pgfkeys{/tikz/savevalue/.code 2 args={\global\edef#1{#2}}}

\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
Punkt/.style 2 args={  label={[#1]:$#2$}   },  
Dreieck/.style={thick}, 
]

% Dreieckskonstruktion
%\pgfmathsetmacro{\Alpha}{acos((\b^2+\c^2-\a^2)/(2*\b*\c))} % 
\coordinate[Punkt={below}{A}] (A) at (0,0); 
\coordinate[Punkt={below}{B}] (B) at (\c,0); 
\coordinate[Punkt={above}{C}] (C) at (\Alpha:\b); 
\draw[local bounding box=dreieck] (A) -- (B) -- (C) --cycle; % Dreieck zeichnen


% Umkreis
\pgfmathsetmacro{\s}{0.5*(\a+\b+\c)} % 
\pgfmathsetmacro{\F}{sqrt(\s*(\s-\a)*(\s-\b)*(\s-\c))} % 

\pgfmathsetmacro{\Da}{\a^2*(\b^2+\c^2-\a^2)} %  
\pgfmathsetmacro{\Db}{\b^2*(\a^2+\c^2-\b^2)} %  
\pgfmathsetmacro{\Dc}{\c^2*(\a^2+\b^2-\c^2)} %  
\pgfmathsetmacro{\D}{\Da+\Db+\Dc} %  
\pgfmathsetmacro{\au}{\Da/\D} %  
\pgfmathsetmacro{\bu}{\Db/\D} %  
\pgfmathsetmacro{\cu}{\Dc/\D} %    
 
\coordinate[Punkt={below}{U}] (U) at ($\au*(A)+\bu*(B)+\cu*(C)$); 
 
\pgfmathsetmacro{\R}{(\a*\b*\c)/(4*\F)} %  
\draw[] (U) circle[radius=\R];

% Annotationen - Dreieck
\draw[thick] (A) -- (B)  node[midway, below]{$c$};
\draw pic [angle radius=6mm, %angle eccentricity=1.2, 
draw,   "$\alpha$", thick
] {angle =B--A--C};
\draw pic [angle radius=6mm, %angle eccentricity=1.2, 
draw,   "$\beta$", thick
] {angle =C--B--A};

\draw[-latex] (U) -- +(33:\R) node[near end, above]{$R$};


% Annotationen - Aufgabe
\pgfmathsetmacro{\x}{max(\a, \b,\c)} %
\begin{scope}[shift={($(dreieck.north west)+(-\x cm-3mm,0)$)}]
% Strecken
\foreach[count=\y from 0] \s/\S in {c/c}{%%
\draw[|-|, yshift=-\y*5mm, local bounding box=strecken] (0,0) -- (\csname \s \endcsname,0) node[midway, above]{$\S$ %= \csname \s \endcsname cm
};}%%
\end{scope}
% Winkel
\pgfmathsetmacro{\Winkel}{\Alpha}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={}, 
"$\alpha$", 
] {angle =R--Q--P};
% Winkel 2
\pgfmathsetmacro{\Winkel}{\Beta}
\pgfmathsetmacro{\WinkelXShift}{\Winkel > 90 ? -cos(\Winkel) : 0} %
\draw[shift={($(strecken.south west)+(15mm+\WinkelXShift,50mm)$)}] (\Winkel:1)  coordinate(P) -- (0,0) coordinate(Q) -- (1,0) coordinate(R);
\draw pic [draw, angle radius=7mm, %angle eccentricity=1.3,
% pic text={$\Winkel$}, pic text options={}, 
"$\beta$", 
] {angle =R--Q--P};

% Annotationen - Rechnung
\tikzset{PosUnten/.style={below=5mm of dreieck, anchor=north,}}
\tikzset{PosLinks/.style={shift={($(dreieck.north)+(-44mm,-20mm)$)}, anchor=north east,}}
\node[yshift=-0mm, draw, align=left, fill=lightgray!50,
%PosUnten,
PosLinks,
]{
$\begin{array}{l l}
c = \c \text{ cm}  &  \\
\alpha = \Alpha^\circ  &  \\
\beta = \Beta^\circ    &  \\   \hline
\gamma = 180^\circ-\alpha-\beta  &=\Gamma^\circ   \\
a = c\cdot\dfrac{\sin(\alpha)}{\sin(\gamma)}  & =\a \text{ cm}  \\[1em]
b = \sqrt{a^2+c^2-2ac\cos(\beta)}  & =\b \text{ cm}   \\
R = \dfrac{a}{2\sin(\alpha)}   &=\R \text{ cm}    \\ 
%\beta = \Beta^\circ    & (5) \\
%\gamma = \Gamma^\circ    & (2) \\
%\multicolumn{2}{l}{s_{a, \text{max}} = \saMax  \text{ cm}} \\ 
\end{array}$
};

%% Punkte
\foreach \P in {U}
\draw[fill=black!1, draw=red] (\P) circle (1.75pt);

\end{tikzpicture}
\end{document}

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