In the attached image i have shown a square whose area comes out as 64. Now that square is cutted to make a rectangle and the area of rectangle comes out as 65. How is it possible when same material is their and no more material is added ? Thanks for your time .
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how is this possible? I know there is some trick, should someone please explain?!
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We have a $100\times100$ board divided into $10^4$ unit squares. These squares are coloured with four colours so that every row and every column has $25$ squares   of each colour. Prove that there are $2$ rows and $2$ columns such that their $4$ intersections are painted with different colors.   My attempt:  I first counted number of pairs $\mathcal P$ $(r_1,r_2)$ such that $r_1,r_2$ belongs to same row, but different in colour. This can be done in $$\binom 42\times25\times 25$$ ways. So, in $100$ rows, this number is $$100\times\binom 42\times25\times 25$$ Now, we can find a pair of columns in $$\binom{100}2$$ ways.  So, by Pigeonhole Principle, we can find at least a pair of columns containing $$\left\lceil\frac{100\times\binom 42\times25\times 25}{\binom{100}2}\right\rceil\\=\left\lceil\frac{100\times75\times 50}{50\times99}\right\rceil\\=76$$ of those pairs $\mathcal P$.    But unable to proceed anymore. And this approach too seems misleading, because in every pair of columns, we can always find $\binom42\times25\times25$ pairs from $\mathcal P$. So, this approach makes no sense, I think.  What should be correct approach? And also, "if we pour $nk+1$ objects into $n$ urns, then there is an urn with $k+1$ objects", can this concept be plugged in this question?
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Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where.
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can we do this without integration ( breaking the area into triangles &amp; rectangles etc ) ?
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I was looking at a 10x10 multiplication table, and I decided to count the unique products. There are 42 out of a possible 100 numbers represented. I had to wonder, why 42? I counted the 58 non-listed numbers -- most of them are either primes >10 or multiples of those primes. I couldn't figure out a satisfying reason like that for 75, 84, 96 and 98. I feel like the number 42 should be calculable from the natural log of 100 and 10, or something, but I can't really figure it out.  Also, I counted the frequency of all the numbers represented. They all have either 1, 2, 3, or 4 instances. The six numbers that have only 1 instance are all squares -- those of 1, 5, 7, 8, 9, and 10. This makes sense -- squares don't fall into that 4x5 = 5x4 redundancy. The 23 numbers with 2 instances are kind of the default I guess. The four with 3 are the remaining squares -- 4, 9, 16, and 36. 4 and 9 make sense to me, as they are squares within the basic range of the times table, so they are hit by themselves, their square root, and 1. I'm having trouble accounting for why 16 and 36 have three representatives. I'm also having trouble seeing the pattern (or patterns) in the numbers with four representatives.  I ultimately want to find a way to, given a times table of any size, calculate whether a given number will appear, and how many times.  Also, I'm pretty sure that for times tables 7x7 and below, more than half of the possible numbers are represented on the table, but for times tables above 7x7, less than half are represented. Why is 7 the turning point?
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A 5x5 board has $25$ cells. The numbers $\{1,2,3,4,5\}$ are written on every row,every column and the two main diagonals without any repetition. If the sum of the numbers of the diagonal below the leading diagonal be represented as $A$. Show that $A\ne 20$. Also find the maximum possible value of $A$.    I could prove that $A\ne 20$ by showing that each of the cells of the diagonal below the leading diagonal cannot be $20$ as if such arrangement exists, then the board cannot have $5$ five times, which actually should be by the given condition of the question.    For the second part, I can imagine the maxiumum value to be $17$. For which I proved that $5$ can occur only $2$ times in the four cells and probably $4$ cannot occur twice. But I got no formal proof. Please help!
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Probably as easy question, I know. But how do I draw something like this:     I know how to draw a 5x5 grid, but I'm not sure how to remove squares from it. Thanks!
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Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.
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I found the following question in a paper I was trying to solve:  The following figure shows a $3^2 \times 3^2$ grid divided into $3^2$ subgrids of size $3 \times 3$. This grid has $81$ cells, $9$ in each subgrid.    Now consider an $n^2 \times n^2$ grid divided into $n^2$ subgrids of size $n \times n$. Find the number of ways in which you can select $n^2$ cells from this grid such that there is exactly one cell coming from each subgrid, one from each row and one from each column.  My try:  Since we have $n^2$ rows, $n^2$ columns and $n^2$ subgrids in total, we have to choose one and only one cell from each of them. Let's choose them one at a time. We can choose the first cell in $n^4$ many ways. Then, we'll have to avoid that subgrid, that column and that row that we've chosen the first one from when choosing the second cell. So, we have $n^4-n^2-2n(n-1)$ choices. We can continue this to get the total number of possible ways. But, I think there's a hole. Say, we've chosen the first cell from the subgrid of the up-left corner and the second from the subgrid just right to it so that it doesn't violate any rules. Then, when finding the number of ways we can choose the third cell, we would have substracted some of the cells twice. I think you get it. Please, if anyone can help me solving this problem, it'd be greatly appreciated.
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Let   consider   a  square $10\times 10$   and  write  in  the   every   unit  square    the   numbers   from  $1$  to  $100$   such  that   every  two   consecutive  numbers  are  in   squares   which  has  a  common edge.  Then  there  are  two  perfect  squares   on  the  same line  or  column. Can  you   give  me  an  hint? How  to  start?
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Question:  Consider the following square sheet of metal (grey) with the side s = 10 cm. From it, we want to cut the parts of a cylindrical can with a lid and a bottom (bblack). Determine the radius and height of the can when the volume is maximal.   The rectangle and circles have to be cut in entire, singular pieces, so no trickery to minimize waste and no diagonal placement of the rectangle.  Solution has to use optimization / differentiation.  Attempted solution:  In order for this to work out the circumference of the circles and one side of the rectangle has to be identical. Otherwise they will not fit together in a cylindrical can.  This leads us to two cases that needs to be treated separately.  Case 1: The rectangle is standing upright.  Case 2: The rectangle is lying on its longest side.  There are two cases that needs to be considered separately.  Case 1:  The right-hand side of the above image must be the circumference of a circle, so  $$s_1 = 2r\pi$$  The bottom-side must be the height of the cylinder plus two radii of the circles:  $$s_1 = h + 2r$$  Putting these together becomes:  $$2r\pi = h + 2r = 10$$  Solving for h:  $$h = 10 - 2r$$  The volume of the can:  $$V = r^2\pi h$$  Adding in the earlier formula to get rid of h:  $$V = r^2\pi h = r^2\pi (10 - 2r) = 10r^2\pi - 2r^3\pi$$  Taking the derivative:  $$V' = 20\pi r - 6\pi r^2$$  Setting derivative equal to zero:  $$V' = 0 \Rightarrow r = \frac{10}{3}$$  (neglecting trivial r = 0 solution)  Height is then:  $$h = 10 - 2r = 10 - \frac{20}{3} = \frac{10}{3}$$  But this is not really possible given the side of the metal sheet cannot be more than 10 cm and $$2r\pi &gt; 10$$.  Case 2:  The bottom becomes:  $$2r + 2r\pi = 10 \Rightarrow r= \frac{5}{1+\pi}$$  The right side becomes:  $$h = 10$$  However, none of these answers match the solution:  $$r = \frac{5}{\pi}$$  $$h = 10(1-\frac{1}{\pi})$$  Did I get the starting point for the possible cases right? Where did I go wrong? How come it appears that no calculus is needed in the second step?
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Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?   I can see it has something to do with Heron's formula but I just can't put t together.
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I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
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I posted the same question in . I am not able to get the figure with two nodes the same height (10 cm), when they are rectangles split in different number of parts. It appears when the object is split in more parts the total height, apparently, increases. How to fix this?  The reproducible example and the actual output is on the .  \documentclass[a4paper]{scrbook} \usepackage[hmargin=2.5cm,vmargin=2.5cm]{geometry} \usepackage[utf8x]{inputenc}  \usepackage{caption} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsmath} \usepackage{subcaption}   \usepackage{calc} \usepackage{tikz} \usetikzlibrary{shapes.geometric, arrows,positioning,shapes,calc,arrows}  \newcommand\Textbox[2]{%     \parbox[c][#1][c]{2.3cm}{\centering#2}}  \begin{document}  \begin{figure}     \begin{subfigure}[b]{0.32\textwidth}         \centering         \resizebox{\linewidth}{!}{          \begin{tikzpicture}         \node (h1) {\textbf{Text I}};         \node[draw,         align=center,         text width=2.5cm,         %minimum height=7cm,         draw=red!50,         rectangle split,          rectangle split parts=3, below of= h1,yshift=.5cm,         rectangle split part fill={gray!30, blue!20, green!15},         anchor=north east]{\textbf{Tit A}             \nodepart{two}\Textbox{6.7cm}{Cir 1\\67\% }              \nodepart{three}\Textbox{3.3cm}{Cir 2\\33\%} };         \node[draw,         align=center,         text width=2.5cm,         %minimum height=7cm,         draw=red!50,         rectangle split,          rectangle split parts=4, below of=h1,yshift=0.5cm,         rectangle split part fill={gray!30, blue!15, green!15,green!30},         anchor=north west]{\textbf{Tit B}             \nodepart{two}\Textbox{4.6cm}{Cir 1\\ 46\%}               \nodepart{three}\Textbox{3cm}{Met \\30\%}               \nodepart{four}\Textbox{2.4cm}{Let\\24\%} };         \end{tikzpicture}         }     \end{subfigure}     \begin{subfigure}[b]{0.32\textwidth}     \centering     \resizebox{\linewidth}{!}{         \begin{tikzpicture}         \node (h1) {\textbf{Text II}};         \node[draw,         align=center,         text width=2.5cm,         %minimum height=7cm,         draw=red!50,         rectangle split,          rectangle split parts=3, below of= h1,yshift=0.5cm,         rectangle split part fill={gray!30, blue!20, green!15},         anchor=north east]{\textbf{Tit A}             \nodepart{two}\Textbox{3.3cm}{Part 1\\33\% }              \nodepart{three}\Textbox{6.7cm}{Part 2\\67\%} };         \node[draw,         align=center,         text width=2.5cm,         %minimum height=7cm,         draw=red!50,         rectangle split,          rectangle split parts=5, below of=h1,yshift=0.5cm,         rectangle split part fill={gray!30, blue!15, green!15,green!30,yellow!15},         anchor=north west]{\textbf{Tit B}             \nodepart{two}\Textbox{3.3cm}{Cr1 \\ 33\%}               \nodepart{three}\Textbox{2.5cm}{Cr2\\25\%}               \nodepart{four}\Textbox{2.2cm}{Cr3\\22\%}             \nodepart{five}\Textbox{2.0cm}{Cr4\\20\%} };         \end{tikzpicture}     }     \end{subfigure}  \end{figure}  \end{document}
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By placing 27 squares with side $1$ in a circle with radius $2$, such that each square is entirely inside the circle, prove that there exists a point that 3 squares share (inside of their area).    I thought about dividing the circle into 9 equal areas and using the pigeonhole principle, but got stuck there. Any hint will help.
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I found  in the old ESRI forums with the same type of problem I am facing today. The solution to the problem is explained in little detail and since I am not an expert, I was wondering if somebody could describe it to me in little more detail.  We are doing a mental map analysis. We have let's say 50 polygons of a certain city square which are all different in shape, size and location. Our goal is to create a single polygon which would represent the answers from our whole input data. So basically, we need a method for generating a single polygon out of these 50 which would be an "average" polygon in size, shape and location.  Can somebody explain the process?
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Consider a square of sidelength $n$ and $ (n+1)^{2} $ interior points. Prove that we can choose 3 of these points so that they determine a triangle (eventually degenerated) of area at most 1/2.  I couldn't get started on this. Any ideas?