我的环境minipage左边有一个环境tikzpicture。在这些环境之后，我有三个equation环境。有一些额外的垂直空间，因为tikzpicture环境的显示比环境占用更多的垂直空间minipage。我如何LaTeX忽略tikzpicture环境的垂直空间？（这里有很多代码。）
\documentclass[10pt]{amsart}
\usepackage{mathtools}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
A right circular cone with a base radius and height of $R$ and $H$, respectively, circumscribes another right circular cone with the same axis. Determine the dimensions of the inscribed cone containing the maximum volume.
\noindent \begin{minipage}[t]{4in}
\noindent \textbf{Determination of the requisite dimensions} \vskip1.25mm
\noindent \raggedright{The intersection between the plane containing the axis of the \\
cones and the cones is a pair of isosceles triangles --- one in- \\
scribed in the other so that their altitudes are collinear. The \\
base radius and altitude of the circumscribing triangle are $R$ \\
and $H$, respectively. If the base radius and altitude of the \\
inscribed triangle are $r$ and $h$, respectively,}
\end{minipage}
%
\hspace{0.5cm}
%
\begin{tikzpicture}[baseline=(current bounding box.north)]
%Two isosceles triangles --- one inscribed in the other so that the bases are parallel, and the
%altitude of the smaller triangle is part of the altitude of the circumscribing triangle. If the
%altitude and base of the circumscribing isosceles triangle are H and R, respectively, the
%base radius and altitude of the inscribed triangle enclosing the biggest area is r=(2/3)R and
%h=(HR-Hr)/R, respectively. In this diagram, R=3/2 and H=4, and so r=1 and h=4/3.
%The circumscribing isosceles triangle has vertices A, B, and C, and the center of its base AB is
%called O.
\coordinate (O) at (0,0);
\draw[fill] (O) circle (1.5pt);
%
\path (-3/2,0) coordinate (A) (3/2,0) coordinate (B) (0,4) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;
%The altitude of $\triangle{ABC}$ is drawn.
\draw[dashed] (O) -- (0,4);
%The inscribed isosceles triangle has vertices E, F, and O, and the center of its base EF is
%called P.
\coordinate (P) at (0,4/3);
\draw[fill] (P) circle (1.5pt);
%
\path (-1,4/3) coordinate (E) (1,4/3) coordinate (F);
\draw (E) -- (F) -- (O) -- cycle;
%The labels for the lengths of two congruent sides and the base of the isosceles triangle are typeset.
%
\draw[|<->|] ($($(A) +(-0.45cm,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.45cm,0)$)$) -- ($($(A) +(0,4) +(-0.45cm,0)$)!-0.5cm!90:($(A) +(-0.45cm,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.55,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.55,0)$)$)!0.5!($($(A) +(0,4) +(-0.55,0)$)!-0.5cm!90:($(A) +(-0.55,0)$)$)$){$H$};
%
\draw[|<->|] ($($(A) +(-15pt,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-15pt,0)$)$) -- ($($(A) +(0,4/3) +(-15pt,0)$)!0.15cm!90:($(A) +(-15pt,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.1,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-0.1,0)$)$)!0.5!($($(A) +(0,4/3) +(-0.1,0)$)!0.15cm!90:($(A) +(-0.1,0)$)$)$){$h$};
%
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($($(A)!0.5!(O)$) +(0,-0.1)$){$R$};
%
\path node[anchor=south, inner sep=0, font=\footnotesize] at ($($(0,4/3)!0.5!(-1,4/3)$) +(0,0.1)$){$r$};
\end{tikzpicture}
\hspace{\fill}
\begin{equation*}
\frac{H - h}{r} = \frac{H}{R} .
\end{equation*}
\begin{equation*}
HR - hR = Hr .
\end{equation*}
\begin{equation*}
h = \frac{HR - Hr}{R} .
\end{equation*}
\end{document}
答案1
将其包装tikzpicture成将\raisebox图片的高度和深度设置为零的。
\raisebox{0mm}[0mm][0mm]%
{\begin{tikzpicture}
...
\end{tikzpicture}%
}%
查看LaTeX 维基百科详情请参阅\raisebox。
答案2
只需将方程式放在小页面中：
\documentclass[10pt]{amsart}
\usepackage{mathtools}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
A right circular cone with a base radius and height of $R$ and $H$,
respectively, circumscribes another right circular cone with the
same axis. Determine the dimensions of the inscribed cone containing
the maximum volume.
\medskip
\noindent
\begin{minipage}[t]{4in}
\textbf{Determination of the requisite dimensions}
\medskip
The intersection between the plane containing the axis of the
cones and the cones is a pair of isosceles triangles --- one
inscribed in the other so that their altitudes are collinear.
The base radius and altitude of the circumscribing triangle
are $R$ and $H$, respectively. If the base radius and altitude
of the inscribed triangle are $r$ and $h$, respectively,
\begin{gather*}
\frac{H - h}{r} = \frac{H}{R} . \\
HR - hR = Hr . \\
h = \frac{HR - Hr}{R} .
\end{gather*}
\end{minipage}\hfill
\begin{tikzpicture}[baseline=(current bounding box.north)]
%Two isosceles triangles --- one inscribed in the other so that the bases are parallel, and the
%altitude of the smaller triangle is part of the altitude of the circumscribing triangle. If the
%altitude and base of the circumscribing isosceles triangle are H and R, respectively, the
%base radius and altitude of the inscribed triangle enclosing the biggest area is r=(2/3)R and
%h=(HR-Hr)/R, respectively. In this diagram, R=3/2 and H=4, and so r=1 and h=4/3.
%The circumscribing isosceles triangle has vertices A, B, and C, and the center of its base AB is
%called O.
\coordinate (O) at (0,0);
\draw[fill] (O) circle (1.5pt);
%
\path (-3/2,0) coordinate (A) (3/2,0) coordinate (B) (0,4) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;
%The altitude of $\triangle{ABC}$ is drawn.
\draw[dashed] (O) -- (0,4);
%The inscribed isosceles triangle has vertices E, F, and O, and the center of its base EF is
%called P.
\coordinate (P) at (0,4/3);
\draw[fill] (P) circle (1.5pt);
%
\path (-1,4/3) coordinate (E) (1,4/3) coordinate (F);
\draw (E) -- (F) -- (O) -- cycle;
%The labels for the lengths of two congruent sides and the base of the isosceles triangle are typeset.
%
\draw[|<->|] ($($(A) +(-0.45cm,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.45cm,0)$)$) -- ($($(A) +(0,4) +(-0.45cm,0)$)!-0.5cm!90:($(A) +(-0.45cm,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.55,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.55,0)$)$)!0.5!($($(A) +(0,4) +(-0.55,0)$)!-0.5cm!90:($(A) +(-0.55,0)$)$)$){$H$};
%
\draw[|<->|] ($($(A) +(-15pt,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-15pt,0)$)$) -- ($($(A) +(0,4/3) +(-15pt,0)$)!0.15cm!90:($(A) +(-15pt,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.1,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-0.1,0)$)$)!0.5!($($(A) +(0,4/3) +(-0.1,0)$)!0.15cm!90:($(A) +(-0.1,0)$)$)$){$h$};
%
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($($(A)!0.5!(O)$) +(0,-0.1)$){$R$};
%
\path node[anchor=south, inner sep=0, font=\footnotesize] at ($($(0,4/3)!0.5!(-1,4/3)$) +(0,0.1)$){$r$};
\end{tikzpicture}\hspace*{-1in}
\bigskip
A filler. A right circular cone with a base radius and height of $R$ and $H$,
respectively, circumscribes another right circular cone with the
same axis. Determine the dimensions of the inscribed cone containing
the maximum volume.
\end{document}
我猜您想让图片留在边缘处。
或者，使用wrapfig；检查包文档以了解参数的含义\begin{wrapfigure}。
\documentclass[10pt]{amsart}
\usepackage{mathtools}
\usepackage{wrapfig}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
A right circular cone with a base radius and height of $R$ and $H$,
respectively, circumscribes another right circular cone with the
same axis. Determine the dimensions of the inscribed cone containing
the maximum volume.
\begin{wrapfigure}[10]{r}{1in}
\begin{tikzpicture}[baseline=(current bounding box.north)]
%Two isosceles triangles --- one inscribed in the other so that the bases are parallel, and the
%altitude of the smaller triangle is part of the altitude of the circumscribing triangle. If the
%altitude and base of the circumscribing isosceles triangle are H and R, respectively, the
%base radius and altitude of the inscribed triangle enclosing the biggest area is r=(2/3)R and
%h=(HR-Hr)/R, respectively. In this diagram, R=3/2 and H=4, and so r=1 and h=4/3.
%The circumscribing isosceles triangle has vertices A, B, and C, and the center of its base AB is
%called O.
\coordinate (O) at (0,0);
\draw[fill] (O) circle (1.5pt);
%
\path (-3/2,0) coordinate (A) (3/2,0) coordinate (B) (0,4) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;
%The altitude of $\triangle{ABC}$ is drawn.
\draw[dashed] (O) -- (0,4);
%The inscribed isosceles triangle has vertices E, F, and O, and the center of its base EF is
%called P.
\coordinate (P) at (0,4/3);
\draw[fill] (P) circle (1.5pt);
%
\path (-1,4/3) coordinate (E) (1,4/3) coordinate (F);
\draw (E) -- (F) -- (O) -- cycle;
%The labels for the lengths of two congruent sides and the base of the isosceles triangle are typeset.
%
\draw[|<->|] ($($(A) +(-0.45cm,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.45cm,0)$)$) -- ($($(A) +(0,4) +(-0.45cm,0)$)!-0.5cm!90:($(A) +(-0.45cm,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.55,0)$)!-0.5cm!-90:($(A) +(0,4) +(-0.55,0)$)$)!0.5!($($(A) +(0,4) +(-0.55,0)$)!-0.5cm!90:($(A) +(-0.55,0)$)$)$){$H$};
%
\draw[|<->|] ($($(A) +(-15pt,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-15pt,0)$)$) -- ($($(A) +(0,4/3) +(-15pt,0)$)!0.15cm!90:($(A) +(-15pt,0)$)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($($($(A) +(-0.1,0)$)!0.15cm!-90:($(A) +(0,4/3) +(-0.1,0)$)$)!0.5!($($(A) +(0,4/3) +(-0.1,0)$)!0.15cm!90:($(A) +(-0.1,0)$)$)$){$h$};
%
\path node[anchor=north, inner sep=0, font=\footnotesize] at ($($(A)!0.5!(O)$) +(0,-0.1)$){$R$};
%
\path node[anchor=south, inner sep=0, font=\footnotesize] at ($($(0,4/3)!0.5!(-1,4/3)$) +(0,0.1)$){$r$};
\end{tikzpicture}\hspace*{-2in}
\end{wrapfigure}
\noindent\textbf{Determination of the requisite dimensions}
\medskip
\noindent
The intersection between the plane containing the axis of the
cones and the cones is a pair of isosceles triangles --- one
inscribed in the other so that their altitudes are collinear.
The base radius and altitude of the circumscribing triangle
are $R$ and $H$, respectively. If the base radius and altitude
of the inscribed triangle are $r$ and $h$, respectively,
\begin{gather*}
\frac{H - h}{r} = \frac{H}{R} . \\
HR - hR = Hr . \\
h = \frac{HR - Hr}{R} .
\end{gather*}
A filler. A right circular cone with a base radius and height of $R$ and $H$,
respectively, circumscribes another right circular cone with the
same axis. Determine the dimensions of the inscribed cone containing
the maximum volume.
\end{document}
