2
\begin{proof}
We are given that there exists a quadrilateral $ABCD$ on a sphere. Let us create a line segment $AC$ that divides our quadrilateral into two triangles. Thus, $\angle{A}$ and $\angle{C}$ have been divided by our line segment $AC$. Let the portion of $\angle{A}$ and $\angle{C}$ on the same side of $AC$ as $B$ be called $\angle{A_1}$ and $\angle{C_1}$. Also, Let the portion of $\angle{A}$ and $\angle{C}$ on the same side of $AC$ as $D$ be called $\angle{A_2}$ and $\angle{C_2}$. Therefore, 

\begin{align*}
\left|ABCD|\right &=\left|\Delta{ABC}|+|\Delta{ACD}|\right\\
\ &= \left p^2(\angle{A_1}+\angle{B}+\angle{C_1}-\pi)+p^2(\angle{A_2}+\angle{C_2}+\angle{D}-\pi)\right\\
\ &= \left p^2(\angle{A_1}+\angle{B}+\angle{C_1}+\angle{A_2}+\angle{C_2}+\angle{D}-2\pi)\right.
\end{align*}

However, \[\angle{A_1}+\angle{A_2}=\angle{A}\] and \[\angle{C_1}+\angle{C_2}=\angle{C}\] \[\left|ABCD|\right=\left p^2(\angle{A}+\angle{B}+\angle{C}+\angle{D}-2\pi)}\right\].
To generalize furthur the area of any $n$-sided polygon on the sphere is as follows, \[\left{Area of an N-sided Polygon}\right=\left(sum of the angles-(n-2)\pi\right\]

\end{proof}

This is code that is producing the missing delimiter error, it says the problem is occurring somewhere with the align environment. Anyone have suggestions?

1
  • Should be \right| and not |\right.
    – Sigur
    Commented Nov 20, 2015 at 14:17

2 Answers 2

4

the problem here is how the \left and \right commands are used. each of these commands always requires either an actual delimiter -- parenthesis, bracket, etc. -- or a "placeholder", namely a period.

in the first line of the alignment, the \right is placed after the |; it should be before.

so at the beginning of the second and third lines (and perhaps elsewhere, i didn't check), \left is not followed by any delimiter, it's not clear what delimiter is to be "matched". perhaps the opening parenthesis after \p^2?

here is the reformulated display:

\begin{align*}
\left|ABCD\right| &=\left|\Delta{ABC}\right|+\left|\Delta{ACD}\right|\\
\ &= p^2\left(\angle{A_1}+\angle{B}+\angle{C_1}-\pi)+p^2(\angle{A_2}
   +\angle{C_2}+\angle{D}-\pi\right)\\
\ &= p^2\left(\angle{A_1}+\angle{B}+\angle{C_1}+\angle{A_2}
   +\angle{C_2}+\angle{D}-2\pi\right).
\end{align*}

actually, since most of the symbols between the delimiters aren't taller than the normal text, \left and \right aren't really needed.

in the paragraph following the align*, i think you're confusing \[ and \] with the actual square brackets. the "escaped" forms indicate the beginning and end of a one-line math display. no backslashes should be used if an actual bracket is intended. however, a typeset brace must be entered preceded by a backslash. and within a math environment, normal text (the "Area of ...") needs to be so indicated, by \text{...}, and within that string, any small math expressions need to be returned to math mode by $...$.

here is that paragraph reformulated:

However, [\angle{A_1}+\angle{A_2}=\angle{A}] and
[\angle{C_1}+\angle{C_2}=\angle{C}]
[\left|ABCD\right|= p^2\left(\angle{A}+\angle{B}+\angle{C}
 +\angle{D}-2\pi\right)}].
To generalize further the area of any $n$-sided polygon on the sphere is as follows,
\[\left\{\text{Area of an $N$-sided Polygon}\right\}
 =\left\{\text{sum of the angles $-(n-2)\pi$}\right\}\]

i may have misunderstood your intent in the paragraph following the display; if you really did intend these bracketed expressions to be displayed, then instead of coding each one as a separate display, you should use the gather* environment. you don't say what document class or theorem package you are using. however, in any event, you shouldn't ever leave a blank line before \end{proof} -- that will guarantee that the "tombstone" is always set on a line by itself.

also, if you are using amsthm, you can insert \qedhere before the closing \] to place the "tombstone" on the last line of the display.

1

Two important facts about \left and \right:

  1. each is used immediately before the delimiter, so you would call \left( and not \left p^2(, and \right|, rather than |\right; and
  2. if you use a \left to open something, you need to use a \right to close it.

After fixing these points, your align* environment looks like this (you don't need to manually add space before the &=):

\begin{align*}
\left|ABCD\right| &= \left|\Delta{ABC}\right| + \left|\Delta{ACD}\right|\\
&= p^2\left(\angle{A_1} + \angle{B} + \angle{C_1} \pi\right)
   + p^2\left(\angle{A_2} + \angle{C_2} + \angle{D} - \pi\right)\\
&= p^2\left(\angle{A_1} + \angle{B} + \angle{C_1} + \angle{A_2} +
   \angle{C_2} + \angle{D} - 2\pi\right).
\end{align*}

The \DeclarePairedDelimiter command in the mathtools package may also be worth checking out: you could put code like this in your preamble, for example.

\usepackage{mathtools}
\DeclarePairedDelimiter{\paren}{(}{)}
% note: \lvert and \rvert look like | and |, but are spaced as
% delimiters,so they'll look a little nicer
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}

Then, your code can be rewritten like this.

\begin{align*}
\abs{ABCD} &= \abs{\Delta{ABC}} + \abs{\Delta{ACD}}\\
&= p^2\paren{\angle{A_1} + \angle{B} + \angle{C_1} \pi} + p^2\paren{\angle{A_2} +
   \angle{C_2} + \angle{D} - \pi}\\
&= p^2\paren{\angle{A_1} + \angle{B} + \angle{C_1} + \angle{A_2} +
   \angle{C_2} + \angle{D} - 2\pi}.
\end{align*}

In my experience, this is a little easier

1
  • I would not recommend using \paren if just for adding ()'s, it does not make the math code easier to understand
    – daleif
    Commented Nov 20, 2015 at 14:52

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