There are five copies of a book each 20mm thick. When tilted, they fit snugly inside a cubic box with a gap of 16mm at the bottom left as shown.  I struggled with this in a recent practice paper, any help appreciated.
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There are five copies of a book each 20mm thick. When tilted, they fit snugly inside a cubic box with a gap of 16mm at the bottom left as shown.   I struggled with this in a recent practice paper, any help appreciated.
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I want to stack some boxes which are 14" x 10" with some 12" records inside. This means that there will be a 2" overhang outside of each box.    I know that to have an item to balance on a edge 50% of an item needs to be supported. So I think I'm safe to load the boxes in this way.   If I were to stack boxes loaded in this way on top of each other is there a limit to how high they can go?
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Code: \documentclass{article} \usepackage[a5paper]{geometry} \usepackage{multicol} \setlength{\parindent}{0cm} \begin{document} \begin{multicols}{2}   \begin{enumerate}   \item text 1   \item text 2   \item text 3   \item text 4   \item text 5   \end{enumerate} \end{multicols} \end{document}  Gives this:  How to move &quot;text 5&quot; upwards adjacent to &quot;text 4&quot;?
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How is it that you read a mathematics book? Do you keep a notebook of definitions? What about theorems? Do you do all the exercises? Focus on or ignore the proofs?  I have been reading Munkres, Artin, Halmos, etc. but I get a bit lost usually around the middle. Also, about how fast should you be reading it? Any advice is wanted, I just reached the upper division level.
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Background  Looking to achieve a fairly complex layout from a Markdown source document. The layout resembles:    Code  \usemodule[newcolumnsets,visual]  \setuphead[chapter][   header=empty,   footer=empty,   number=no,   page=yes,   before={\blank[none]},   after={\blank[none]}, ]  \setuphead[section][   number=no,   before={\blank[small]},   after={\blank[small]}, ]  \definepapersize[BookPageSize][   width=601mm,   height=305mm, ]  \definelayout[BookPageLayout][   backspace=10mm,   cutspace=10mm,   width=585mm,   height=147mm,   topspace=10mm,   header=0mm,   footer=0mm, ]  \setuppapersize[BookPageSize] \setuplayout[BookPageLayout]  % No page number at the top. \setupheadertexts[] \setupfootertexts[]  \setupindenting[yes, small]  \definecolumnset[BookColumns][n=4,distance=0.0333\makeupwidth]  % left page (even page numbers) \setupcolumnset[BookColumns:1][width=.5\makeupwidth] \setupcolumnset[BookColumns:2][width=.125\makeupwidth] \setupcolumnset[BookColumns:3][width=.125\makeupwidth] \setupcolumnset[BookColumns:4][width=.125\makeupwidth] % right page (odd page numbers) \setupcolumnset[BookColumns:5][width=.5\makeupwidth] \setupcolumnset[BookColumns:6][width=.125\makeupwidth] \setupcolumnset[BookColumns:7][width=.125\makeupwidth] \setupcolumnset[BookColumns:8][width=.125\makeupwidth]  \definecolumnsetspan[BookIllustration][n=4]  \starttext   \startbodymatter     \startcolumnset[BookColumns]       \chapter[title={Title},reference={title}]       \subsection[title={Subsubtitle},reference={subsubtitle}]       \column        \input ward        \section[title={Subtitle},reference={subtitle}]        \input knuth       \input tufte        \section[title={Subtitle},reference={subtitle}]        \startformula  \frac{\dot{R}^2}{R^2} = \frac{8\pi G}{3}\rho - \frac{kc^2}{R^2}  \stopformula        \startformula  \frac{\ddot{R}}{R} = -\frac{4\pi G}{3} \left(\rho + \frac{3p}{c^2}\right)  \stopformula        \input tufte        \placefigure[bots]{}{\externalfigure[https://i.imgur.com/nSBuPAA.jpg][width=\paperwidth]}     \stopcolumnset   \stopbodymatter \stoptext   This code produces the following, which is pretty close:    Problems  There are a few issues:   The fourth column contains a lot of whitespace at the top. The second column contains garbled text (could be a page height issue). The image is displayed on the second page.   Questions  How can I resolve these issues without changing the body matter, but affording some tweaks to \placefigure such that:   the last column is aligned to the top; the garbled text does not appear; and the image is on the same page as the text?   Version  Using ConTeXt 2019.04.04 13:31.
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Am I the only one who feels like the logo is slightly misaligned? I think the first angled bar should be moved a little bit to the right, so that both corners are aligned. (pink)  I also found out that the padding between the container and the first bar is uneven as well. (blue)    Don't get me wrong, I love the concept of the logo, but I feel that it's a little bit off. Did anyone else notice this?
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I've been having a problem with these problems for a week now so I thought I'd post it on MSE. Read the text below. . I've also typed it up:  Annabel made a shape by placing identical square tiles in a frame as shown in the diagram above. The tiles are arranged in columns. Each column touches the base but no columns touch the side or top. There are no empty gaps between columns. The frame can be enlarged as needed.  These are the questions. I've also typed them up.    A. Show that it is possible to arrange $7$ tiles so that the ant walks $8,9,10,11,12,13,14,15$ tile edges.   B. Show 6 ways of arranging 7 tiles so that the ant walks a total of 9 tile edges.  C. Show it is possible to arrange 49 tiles so that the ant walks a total of less than 21 tile edges.  D. Show four arrangements of 137 tiles, each arrangement with a different maximum height, so that the ant walks a total of 34 tile edges.  For question $a$, is there a formula relating the number of tile edges exposed to the number of tiles. These are my answers for $a$    For the above picture, it's the left hand side which has the $a$ answers.  However, I'm having a bit of trouble with $b$. I've already got four answers down but where are the other two. My four current answers are in the right hand side of the last image. Also, I'll also appreciate a formula to show the relationship.  I'm also having trouble with $c$. I found out that you get 21 tile edges exposes when you make a $7*7$ square but the answer is asking for less than 21. Can anyone help me. Also, as in the previous questions, a formula to show the relationships would be great.  And for $d$, I am totally lost. I have no idea where to start, and I certainly can't find out four arrangements. So can anyone help me? And as always, a formula is nice  P.S. Please explain this in an understandable formula, as I'm only a Year 7 with the ability to understand linear algebra and parts of quadratics and trigonometry. Also, sorry about the poor imaging.
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I am using an exam package and the following code  \fillin[$P(X\le6)-P(X\le3)$][3.7in]   but there is an vertical space issue which I could not find a solution for it between the equation and the horizontal line. Please see the picture
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The question at length is exactly this:     There are five distinct computer science books, three distinct   mathematics books, and two distinct art books. In how many ways can   these books be arranged on a shelf if one of the mathematics books is   on the left of all computer science book, and another mathematics book   is on the right of all the computer science books?   I have searched a ton online for a question that's even similar to this but have yet to find one. Most examples have set groupings of books to create an 'object' but none have the kind of modular specifications of this question.  I realize that there is ambiguity about if the math books are directly flush with the computer science books, or if they can be 'anywhere to the left/right' of the computer science books. I've attempted this question under the more challenging assumption and have gotten an answer as follows:  There are 4 different placements of two math books that will allow for the computer science books to be completely between the two.  1) 8 spaces between the two math books M__________M (1 way)  2) 7 spaces between the two math books: M_______M_ and _M_______M (2 ways)  3) 6 spaces: M______M__, _M______M_, __M______M (3 ways)  4) 5 spaces: M_____M___, _M_____M__, __M_____M_, ___M_____M. (4 ways)  Then I wanted to find the permutations of each of the ways above by the Rule of Product, and add them together by the Rule of Sum to get my final answer.  (1) 8 spaces: P(3, 2) * P(8, 8)   [Delimiting math books]  * [permutations in between]  (2) 7 spaces: P (3, 2) * P(3, 1) * P(7, 7) * 2  [Delimiting math books] * [Options for book outside of delimitation] * [Permutations in-between] * [number of ways to achieve 7 spaces]  (3) 6 spaces: P (3, 2) * P(3, 2) * P(6, 5) * 3  [Delimiting math books] * [options for 2 books outside delimitation] * [permutation in-between] * [number of ways to achieve 6 spaces]  (4) 5 Spaces: P (3, 2) * P(3,3) * P(5, 5) * 4  [Delimiting math books] * [options for 3 books outside delimitation] * [permutation in-between] * [number of ways to achieve 5 spaces]  Then my final answer would be (1) + (2) + (3) + (4). My worry is that I'm missing some duplicates somehow.  Also conveniently, I believe the total for 5 spaces would satisfy the simpler question if it is actually referring to permutations with the chunk MCCCCCM.  Thanks in advance! I'd love to see specifically what I'm doing wrong (if I am indeed wrong).
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[See  for more on this topic, as well as better illustrations of the final product.]  I have a document with a lot of short excerpts from another book scattered through it.  Putting frames around them looks a little too 'clean'. How can I get a 'torn page' type effect, to make it look as if the excerpt has been physically torn out of a book?  (I've occasionally seen this done in printed books, although I can't find an example right now.)  I'm not wedded to a very specific look here --  gives plenty of examples. But just to give this some frame of reference, here are some of the nicer ones:          (The strong colouration in this second example is a probably a little too distracting for actual use, although it might be interesting as an option. It's more the edge effect I'm interested in.)    Edit (again): I tried typesetting my text according to the updated versions of the answers below. Here are the results:    Or with interpolated shading:
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how I can fix the fourth row?    \begin{table}[!h]\label{Table}   \caption{random symmetrized instances.}   \begin{tabularx}{\textwidth}{c&gt;{\centering}Xc}   \toprule       Name  &amp;   The generators of the symmetry group &amp; Dimension  \\   \midrule   NS1     &amp;   $(1,7,12,16,19,21,6)$, $(2,8,13,17,20,5,11)$, $(3,9,14,18,4,10,15)$,   $(4,6,5)$, $(9,11,10)$, $(13,15,14)$, $(16,18,17)$, $(19,20,21)$  &amp;  21 \\   \midrule   NS2   &amp;   $ ( 1, 2, 7)( 3,11,27)( 4,14,31)( 5,18,32)( 6,20,36)( 8,24,39)( 9,25,28)     (10,26,42)(12,15,16)(13,30,40)(17,19,21)(22,35,44)(23,33,29)(34,43,37)     (38,45,41), ( 1, 3, 5, 6, 7,22,13,23)( 2, 8, 9,10)( 4,15,16,17,14,21,19,12      )(11,28,29,38,44,25,20,37)(18,33,34,24,40,36,41,39)(26,43,35,32,42,45,27,      30), ( 1, 4)( 3,12)( 5,19)( 6,21)( 7,14)( 8,10)(11,20)(13,16)(15,23)     (17,22)(18,33)(24,41)(25,28)(26,43)(27,32)(29,44)(30,35)(34,39)(36,40)     (42,45) $  &amp; 5\\   \bottomrule   \end{tabularx} \end{table}
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I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.  A non-graphical calculation will be very helpful too.  Thank you so much in advance!
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I often used \boxed{...} from the AMS math package to place a box around important equations.  However, this approach often produces somewhat awkward looking output.  Consider  \documentclass{article} \pagestyle{empty} \usepackage{amsmath} \begin{document}  \[  \boxed{c_i = \sum_jA_{ij}}  \]  \[  \boxed{c_i = \langle\psi|\phi\rangle}  \] \end{document}     Observe that in the first case the  box looks unbalanced while in the second case could do with more padding on both the top and bottom.  What is the best way to go about this?  I know that I could use a \phantom but am wondering if more elegant solutions exist.
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I am trying to prepare printer labels for the spine of ring binders. The text is designed to be printed on label paper sheets with A4 format, having top and bottom margins of 26.5mm and left/right margins of 9mm.  The format of the labels is 61mm by 192mm, there are 4 of them on the sheet, without any separation between them.  I use the tcolorbox environment as a 'frame' for the labels and  and perhaps my idea is to naive, but I just wrote four of such environments in a row with nobeforeafter setting for each (well, using \tcbset command).  Even with \parskip set to 0pt there is small spacing between the boxes of about 1mm, which would lead to a shift of the content consecutively, and the 4th box appears on the next page. Reducing the bottom margin is not an option, of course. I believe, that \parskip is of no use here anyway.  I am pretty sure, that the content of the boxes has no influence on the spacing, as long as the height is not too large, which does not seem to be the case here.  How can I reduce the space between those colorboxes to a effectively 'zero' value, resulting in no shift when printed?   Note  I checked the margins etc. after print out, they are correct!   Even if there might be other solution for such label printing, I would like to have some solution to reduce the spacing between the boxes (Ok, it might be a very trivial solution.)    This is the code I tried so far  \documentclass[12pt,paper=a4]{article}  \usepackage[lmargin=0.9cm,rmargin=0.9cm,tmargin=26.5mm,bmargin=26.5mm,showframe=true,a4paper]{geometry} \usepackage{tcolorbox} \usepackage{lmodern}   \begin{document} \setlength{\parindent}{0pt}% \setlength{\parskip}{0pt}% \pagestyle{empty}%  \fontsize{80}{96}\selectfont  \tcbset{height=61mm,nobeforeafter,top=2cm} \begin{tcolorbox}% \textbf{\LaTeX}% \end{tcolorbox} \begin{tcolorbox}% \textbf{is}% \end{tcolorbox} \begin{tcolorbox}% \textbf{very}% \end{tcolorbox} \begin{tcolorbox}% \textbf{sophisticated}% \end{tcolorbox}  \end{document}   I looked into other 'similar' questions here, none of them seems to be connected to this problem          Note  Here is the result with some features of the tcolorbox package and the solutions by Gonzalo Medina and egreg together with the comment by Thomas F. Sturm.
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If there are 5 points on the surface of a sphere, then there is a closed half sphere, containing at least 4 of them.  It's in a pigeonhole list of problems. But, I think I have to use rotations in more than 1 dimension.  Regards
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If I want to shuffle the question to prepare four different questions papers having same questions but with the different question numbers, what should I do?       \documentclass[12pt,a4paper]{exam}     \usepackage{amsmath,amsthm,amsfonts,amssymb,dsfont}     \setlength\parindent{0pt}         %usage \choice{ }{ }{ }{ }         %(A)(B)(C)(D)         \newcommand{\fourch}[4]{         \par                 \begin{tabular}{*{4}{@{}p{0.23\textwidth}}}                 (a)~#1 &amp; (b)~#2 &amp; (c)~#3 &amp; (d)~#4                 \end{tabular}         }          %(A)(B)         %(C)(D)         \newcommand{\twoch}[4]{                  \begin{tabular}{*{2}{@{}p{0.46\textwidth}}}                 (a)~#1 &amp; (b)~#2                 \end{tabular}         \par                 \begin{tabular}{*{2}{@{}p{0.46\textwidth}}}                 (c)~#3 &amp; (d)~#4                 \end{tabular}         }          %(A)         %(B)         %(C)         %(D)         \newcommand{\onech}[4]{         \par               (a)~#1 \par (b)~#2 \par (c)~#3 \par (d)~#4         }          \newlength\widthcha         \newlength\widthchb         \newlength\widthchc         \newlength\widthchd         \newlength\widthch         \newlength\tabmaxwidth          \setlength\tabmaxwidth{0.96\textwidth}         \newlength\fourthtabwidth         \setlength\fourthtabwidth{0.25\textwidth}         \newlength\halftabwidth         \setlength\halftabwidth{0.5\textwidth}        \newcommand{\choice}[4]{%       \settowidth\widthcha{AM.#1}\setlength{\widthch}{\widthcha}%       \settowidth\widthchb{BM.#2}%       \ifdim\widthch&lt;\widthchb\relax\setlength{\widthch}{\widthchb}\fi%       \settowidth\widthchb{CM.#3}%       \ifdim\widthch&lt;\widthchb\relax\setlength{\widthch}{\widthchb}\fi%       \settowidth\widthchb{DM.#4}%       \ifdim\widthch&lt;\widthchb\relax\setlength{\widthch}{\widthchb}\fi%       \ifdim\widthch&lt;\fourthtabwidth         \fourch{#1}{#2}{#3}{#4}       \else\ifdim\widthch&lt;\halftabwidth         \ifdim\widthch&gt;\fourthtabwidth           \twoch{#1}{#2}{#3}{#4}         \else           \onech{#1}{#2}{#3}{#4}         \fi       \fi\fi     }     \begin{document}      \begin{questions}     \question If $a = 3 + i$ and $z = 2 - 3i$ then the points on the Argand diagram     representing az, 3az and - az are     \choice{Vertices of a right angled triangle}{ Vertices of an equilateral      triangle}{Vertices of an isosceles triangle}{Collinear}     \question If z represents a complex number then $\arg (z) + \arg\left(\bar z\right)$ is      \choice{$\dfrac{\pi}{4}$}{$\dfrac{\pi}{2}$}{0}{$\dfrac{\pi}{6}$}     \question If the amplitude of a complex number is $\dfrac{\pi}{2}$ then the number is     \choice{ purely imaginary}{purely real}{0}{neither real nor imaginary}     \question The value of $i + i^{22} + i^{23} + i^{24} + i^{25}$ is     \choice{i}{-i}{1}{-1}     \question The volume generated by      rotating the triangle with vertices at     (0, 0), (3, 0) and (3, 3) about x-axis is     \choice{$18\pi$}{$2\pi$}{$36\pi$}{$9\pi$}\end{questions}     \end{document}      \end{document}   That is I need to mark the question paper given above as Code&nbsp;A. And I need to print another question paper which marked as Code&nbsp;B and in this question paper the first question Code&nbsp;A should be in question number 5 (say) and similarly others also should have different positions...
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I'm stuck on what how to continue. I know I'm missing a few steps but this is what I have so far. Thank you in advance!   ¬Cube(b) → Small(b) Small(c) → (Small(d) ∨ Small(e))  Small(d) → ¬Small(c) Cube(b) → ¬Small(e)   Small(c)   ¬Small(b)   ¬Cube(b) Small(b) (→ Elim 1,7) ⊥ (⊥intro 6, 8)    10.Cube(b) (¬Intro 7-9)   ¬Small(e) (→ Elim 4,10)     Goal: Small(c) → Small(b)