(antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g What's the significance of this further decomposition? A tensor is a linear vector valued function defined on the set of all vectors . 1.5) are not explicitly stated because they are obvious from the context. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. The trace decomposition theory of tensor spaces, based on duality, is presented. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share â¦ Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric â¦ The symmetry-based decompositions of finite games are investigated. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Sci. This is exactly what you have done in the second line of your equation. This is an example of the Youla decomposition of a complex square matrix. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . Decomposition of tensor power of symmetric square. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. It is a real tensor, hence f Î±Î² * is also real. P i A ii D0/. Yes. In these notes, the rank of Mwill be denoted by 2n. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Sponsoring Org. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 AÄ± ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components â¦. Cartan tensor is equal to minus the structure coeï¬cients. Thus, the rank of Mmust be even. In section 3 a decomposition of tensor spaces into irreducible components is introduced. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. We begin with a special case of the definition. The result is Polon. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The bases of the symmetric subspace and those of its orthogonal complement are presented. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. : USDOE â¦ If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. MT = âM. An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This decomposition, ... ^2 indicates the antisymmetric tensor product. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. Antisymmetric and symmetric tensors. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. â What symmetry does represent?Kenta OONOIntroduction to Tensors For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. This makes many vector identities easy to prove. For N>2, they are not, however. Contents. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Symmetric tensors occur widely in engineering, physics and mathematics. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. These relations may be shown either directly, using the explicit form of f Î±Î², and f Î±Î² * or as consequences of the HamiltonâCayley equation for antisymmetric matrices f Î±Î² and f Î±Î² *; see, e.g., J. PlebaÅski, Bull Acad. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. Cl. THE INDEX NOTATION Î½, are chosen arbitrarily.The could equally well have been called Î± and Î²: vâ² Î± = n â Î²=1 AÎ±Î² vÎ² (âÎ± â N | 1 â¤ Î± â¤ n). Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Each part can reveal information that might not be easily obtained from the original tensor. There is one very important property of ijk: ijk klm = Î´ ilÎ´ jm âÎ´ imÎ´ jl. The N-way Toolbox, Tensor Toolbox, â¦ If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Vector spaces will be denoted using blackboard fonts. Antisymmetric and symmetric tensors. Ask Question Asked 2 years, 2 months ago. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 1.4) or Î± (in Eq. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. Active 1 year, 11 months ago. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. Physics 218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. A related concept is that of the antisymmetric tensor or alternating form. LetT be a second-order tensor. Decomposition. 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# What Is a Trapezoid? More on Inclusive Definitions

A month ago, I wrote about classifying shapes, discussing inclusive and exclusive definitions, and variations in different contexts. I promised to return to the subject, moving on to the specific issue of trapezoids, and some other related topics. Now is the time.

## You say trapezium, I say trapezoid

We have to start with a regional issue: The word “trapezoid” doesn’t mean the same thing in every country. In our FAQ on geometrical formulas, we head one article with two names and a footnote:

Trapezoid (American)
Trapezium (British)*
...
*From The Words of Mathematics by Steven Schwartzman (1994, Mathematical Association of America): trapezoid (noun); trapezoidal (adjective); trapezium, plural trapezia (noun): ...

Some Americans define a trapezoid as a quadrilateral with at least one pair of parallel sides. Under that definition, a parallelogram is a special kind of trapezoid. For other Americans, however, a trapezoid is a quadrilateral with one and only one pair of parallel sides, in which case a parallelogram is not a trapezoid.

The situation is further confused by the fact that in Europe a trapezoid is defined as a quadrilateral with no sides equal. Even more confusing is the existence of the similar word trapezium, which in American usage means "a quadrilateral with no sides equal," but which in European usage is a synonym of what Americans call a trapezoid. Apparently to cut down on the confusion, trapezium is not used in American textbooks.

Taking the last issue first, when we get a question about a trapezium, we generally assume it is used in the European sense (though rarely we might see it in the American sense); if it mentions parallel sides, we can go on our way with confidence, as we did here:

Cyclic Quadrilateral

For an isosceles trapezium ABCD with AB parallel to DC and AB < CD, prove that:

1) angle ADC = angle BCD
2) ABCD is a cyclic quadrilateral
3) the diagonals of ABCD are equal

There’s no question what is being asked here. Some of us (such as Doctor Floor and Doctor Anthony), are themselves European, and may use “trapezium” even when the question is about a “trapezoid”. And sometimes we just have to ask, if the question is unclear about which is meant. For example, a few years ago I started an answer with,

If you live in a country where "trapezium" means that two sides are parallel, and if you know which two they are, then ...

On the other hand, I started another answer with,

First, we need to be sure of your definition of the word "trapezium", which varies among countries. I think that you are using it to mean a general quadrilateral, with no parallel sides. Is that correct?

## Exactly, or at least?

Now let’s move on to the other issue, which tends to generate more questions, like this one from 2004:

Inclusive Definitions: Trapezoids

As far as I know, a trapezoid is defined as a quadrilateral with exactly one set of parallel sides.  Most textbooks and websites will confirm this definition.  However, a very highly regarded educator and textbook author recently argued that this definition is incorrect.  His definition of a trapezoid is that it is a quadrilateral that has at least one pair of parallel sides.  A square, therefore, would be considered a trapezoid.  He even included this definition in the glossary of a newly published textbook.  Is he correct or are thousands of books going to be published with the wrong definition?  As a teacher looking to buy new books for my school, I would really like to know.  Thanks.

I can’t vouch for the claim that most textbooks state the exclusive definition (saying that figures with a second pair of parallel sides are excluded from being trapezoids); but are they wrong, as this author reportedly says? Or is he wrong?

I started with the usual explanation of inclusive and exclusive definitions, emphasizing that both forms of definition are valid:

Both definitions are in use, so neither is wrong!  That does lead to confusion, but each author has to choose the definition that makes most sense in his context.

Quadrilateral Classification: Definition of a Trapezoid
http://mathforum.org/library/drmath/view/54901.html

Inclusive and Exclusive Definitions
http://mathforum.org/library/drmath/view/55295.html

The same sort of issue arises with other shapes, such as the rectangle.  Is a square a rectangle?  Not to a child; we tell them "This is a square, and that is a rectangle," and they learn that a rectangle is like a square but doesn't have equal sides.

Yet to a mathematician, such exclusive definitions are awkward, because everything that is true of a rectangle is true of a square, and we'd like to use one word to cover both when we write a theorem. For example, any quadrilateral with three right angles is a rectangle --why should we have to add "or a square"?  And if we prove something is true of any parallelogram, we don't want to have to add "or rhombus, or rectangle, or square."

So although even mathematicians find the exclusive definition useful when we want to point out objects (we generally use the most specific term we can, so that we wouldn't call a square a rectangle when we are trying to ask for one), for technical purposes we prefer the inclusive definition, and would prefer that it be taught in schools.

As before, inclusive definitions fit better in a formal mathematical context with theorems, while exclusive definitions fit an informal context, where we usually use the strongest description possible.

It's a little more subtle with trapezoids, because there are fewer theorems about them, so we have less commitment to an inclusive definition.  There are probably mathematicians, and certainly educators, who don't use the inclusive definition in this case.  But as you'll see in the links above, the inclusive definition makes the relationships among quadrilaterals clearer.

This may well explain the perception (and perhaps the fact) that most textbooks use the exclusive definition for the trapezoid: they are using the word not in theorems, but in relatively informal descriptions.

## Implicitly inclusive

On the other hand, it may be that they are really using the inclusive definition, but it isn’t obvious. Their wording may sound exclusive, but really be inclusive:

I should also mention that when a mathematician says "a trapezoid is a quadrilateral with two sides parallel," he probably means "at least two sides," not "exactly two sides"; that is the usual understanding of such a phrase, because we get used to speaking that way.  It may not always be clear to non-mathematicians!

We are so used to inclusive definitions that, in effect, we define “two” inclusively: If we say two sides are parallel, we are not mentioning the other sides, which may also be parallel! (In the same way, we may say that an isosceles triangle has two congruent sides, meaning that if two are congruent, it doesn’t matter if the third side is, too.) But to a non-mathematician, “two” may convey the meaning “exactly two, and no more”. If no theorems are shown where the meaning of the word is unpacked and used, you may not notice what meaning is intended.

The inclusive definition can sometimes be discerned, well hidden within the usage of the word. One place where the word “trapezoid” is used is in discussing the “trapezoidal approximation” in calculus. Here is a picture illustrating it; we choose points along a curve and draw (right-angled) trapezoids consisting of a chord of the curve, two vertical lines, and a piece of the x-axis:

But what if two consecutive points on the curve have the same y-coordinate, so that the chord is horizontal? Than this “trapezoid” is really a rectangle, and it we were using the exclusive definition, it would not be a trapezoid! So implicitly, when we talk about the trapezoidal rule (as opposed to the “trapezoid-or-rectangle rule”), we are defining “trapezoid” inclusively, even if we elsewhere defined it exclusively!

In my answer to Peter, I went on to refer to two random sites I had found that discuss the variation in definition among textbooks; each then states what definition they will use, and they choose differently. One of the links no longer works; the other, which agrees with me, says

The difference is that under the second definition parallelograms are trapezoids and under the first, they are not.

The advantage of the first definition is that it allows a verbal distinction between parallelograms and other quadrilaterals with some parallel sides. This seems to have been most important in earlier times. The advantage of the inclusive definition is that any theorem proved for trapezoids is automatically a theorem about parallelograms. This fits best with the nature of twentieth-century mathematics.

It is possible to function perfectly well with either definition. However, it is important to have agreement in a math class on the definition used in the class.

I concluded, in agreement,

Again, each definition has its place, and should be used in the appropriate context.  The inclusive definition fits well into the context of geometry, and I recommend it.

## The challenge of the isosceles trapezoid

Let me add one more comment: Under the inclusive definition, a parallelogram is a special kind of trapezoid. An often-unnoticed consequence is that we have to carefully define the other special kind of trapezoid, the isosceles trapezoid. This is commonly defined as a trapezoid in which the non-parallel sides are congruent. There are two problems here: there are not always any non-parallel sides; and in a parallelogram, if you pick one pair of parallel sides, the other pair will always be congruent!

We don’t want to call a parallelogram an isosceles trapezoid, because theorems about the latter typically do not apply to the former. This is because the latter has a symmetry that the former does not. Therefore, with the inclusive definition, it is best to define an isosceles trapezoid not in terms of congruent legs, but of symmetry. One way to do this is to require the base angles (angles at the ends of a side that is parallel to another) to be congruent. That is done, for example, here:

Wolfram MathWorld: Isosceles Trapezoid

An isosceles trapezoid (called an isosceles trapezium by the British; Bronshtein and Semendyayev 1997, p. 174) is trapezoid in which the base angles are equal and therefore the left and right side lengths are also equal.

Wikipedia explicitly uses symmetry for its definition:

Isosceles Trapezoid

In Euclidean geometry, an isosceles trapezoid (isosceles trapezium in British English) is a convex quadrilateral with a line of symmetry bisecting one pair of opposite sides. It is a special case of a trapezoid. In any isosceles trapezoid two opposite sides (the bases) are parallel, and the two other sides (the legs) are of equal length (properties shared with the parallelogram). The diagonals are also of equal length. The base angles of an isosceles trapezoid are equal in measure (there are in fact two pairs of equal base angles, where one base angle is the supplementary angle of a base angle at the other base).

And this site not only uses base angles in their definition, but explains why (perhaps because parents will expect otherwise):

Math Bits Notebook: Theorems Dealing With Trapezoids and Kites

Here are two other pages that touch on classification of trapezoids:

Quadrilateral Classification: Definition of a Trapezoid

Venn Diagram to Classify Quadrilaterals

The first of these includes links to discussions among mathematicians; the second provides a Venn diagram. Note that by our definitions, a rectangle is an isosceles trapezoid. Is that surprising?

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# Why does the $(n-1)!$ rule not work in all cases of circular permutations? [duplicate]

$$5$$ Boys and $$5$$ girls sit alternatively around a round table. In how many ways can this be done?

I solved it like this : $$5$$ boys can be arranged in $$(5-1)!$$ ways. After that the $$5$$ girls can be arranged in the gaps in $$(5-1)!$$ ways. So, the answer should be $$4!×4!$$ but the actual answer is $$4!×5!$$. After seeing the answer I can guess that they have considered $$5$$ instead of $$(5-1)$$ in any one of the cases. I have learnt that number of circular arrangements = $$(n-1)!$$. So, why did I get wrong answer?

EDIT

rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone 2, 4, 6, or 8 seats to their left, then we will get another seating arrangement that will be counted again.

Does that mean that if for example, $$5$$ men, $$5$$ women and $$5$$ children are to sit alternately then the answer should be $$\frac{5!×5!×5!}{5×3/3}$$ because neither can we rotate each member to the left by one place nor by two places? So does that mean that in this type of questions, we have to group the objects and then divide the result with total number of groups?

• The girls can be arranged in the gaps in $5!$ ways, not $(5-1)!$ ways. The boys' positions are already established. You have $5$ spaces to place that first girl. Jul 23, 2021 at 1:55
• The reason that you use $(n-1)!$ in these circular arrangements is that you have a rotational symmetry. Once you seat the boys, this symmetry no longer exists. Jul 23, 2021 at 2:37

The circular arrangement formula takes into account that you can rotate the positions, and nothing of substance changes. If you're putting 5 people on a circular table, you get $$5!$$ permutations, divided by $$5$$ to account for the fact that we can rotate the seating arrangement $$5$$ different ways without substantial change to the arrangement.
That is, when you count the placements as $$5!$$, you are over-counting by a factor of $$5$$, because each of the $$5!$$ placements can be rotated to $$4$$ other seating arrangements.
In your case, you seat $$5$$ girls and $$5$$ boys alternately. Without accounting for rotations, there are $$5! \times 5!$$ ways of seating the children. But then we must account for rotations. How many other equivalent seating arrangements can we form by rotating the seats?
Note that rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone $$2$$, $$4$$, $$6$$, or $$8$$ seats to their left, then we will get another seating arrangement that will be counted again. This tells us that by computing $$5! \times 5!$$, we have again over-counted by a factor of $$5$$, so the result is $$\frac{5! \times 5!}{5} = 5! \times \frac{5!}{5} = 5! \times 4!.$$
By computing $$4! \times 4!$$, you are implicitly assuming that you can rotate the boys and girls independently. This isn't really allowed in the problem, as it would produce different seating arrangements.
For example, if we had the cyclic order $$\text{matt}, \text{hannah}, \text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia},$$ then we get an equivalent order by shuffling everyone $$2$$ spots to the left: $$\text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia}, \text{matt}, \text{hannah}.$$ Everyone still has the same person to their left, and to their right. On a circular table, nobody would know the difference. But, if we were to apply the $$5 \times 5$$ independent rotations of the girls and the boys, we could rotate just the girls one spot to the left: $$\text{matt}, \text{elizabeth}, \text{charlie}, \text{jenny}, \text{warren}, \text{veronica}, \text{peter}, \text{celia}, \text{cameron}, \text{hannah},$$ and we get a totally different order. Elizabeth used to be next to Charlie and Warren, but now she's next to Matt and Charlie (and Charlie is sitting on the opposite side of her to where he previously sat). As far as the table's participants are concerned, this is a different configuration.


# Probability of pair of gloves selection

In his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen.

I am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair.

This is what I came up with, that is, the probability of success for each choice:

$$\frac{20}{20}×\frac{18}{19}×\frac{16}{18}×...×\frac{6}{13}=\frac{384}{4199}≈0.09145$$

At first, I was a little confused by the wording but I believe this seems about right.

Is there an alternative way to get the desired probability, e.g. with $$1-...$$?

Thanks in advance for any feedback.

There is a more general formula for this.
Here you are asked that no pair is selected, but this formula will take care of any number of pairs selected

With $$10$$ be the number of pairs, and $$k$$ the number of pairs selected from $$8$$ gloves, the formula is

$$\dfrac{\dbinom{10}{k}\dbinom{10-k}{8-2k}\cdot2^{8-2k}}{\dbinom{20}{8}}$$

For the particular case for $$k=0$$, it simplifies to

$$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$$

$$= \dfrac{\dbinom{10}{0}\dbinom{10}8 \cdot2^8}{\dbinom{20}{8}}$$

• Explaining briefly for OP: this can be justified by an extension of my argument where we first choose $k$ pairs from the original $10,$ then from the remaining $10 - k$ pairs we pick the remaining $8 - 2k$ gloves and choose whether each is the left or right glove. We can actually generalize this for all numbers of original pairs of gloves and gloves selected as well, but we need to be a bit careful regarding bounds. (for instance, notice that this formula is clearly only valid when $0 \leq k \leq 4$: if $k = 5$ then the probability is $0$ because we'd need to pick at least $10$ gloves) May 16 at 7:28
• I was about to point out that $\dbinom{20}{16}$ as a denominator would seem impossible as it yields $2.3777...$, but seems like it's already corrected. Anyways, this formula is exactly what I was looking and I feel like this should be the answer. I would have definitely marked it as an answer if you would have responded 10 seconds earlier @trueblueanil May 16 at 7:29
• @nimen55290: No matter, the important thing is that I have been of help ! May 16 at 8:05
• I really appreciate it @trueblueanil May 16 at 8:16

We can use a combinatoric argument if you like: there are $$20 \choose 8$$ ways we could possibly choose $$8$$ gloves from the $$20,$$ neglecting order.

To see how many of these will involve us choosing no pairs, we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick from each pair. There are $$10$$ pairs so we have $$10 \choose 8$$ ways to choose our pairs, and then for each set of pairs there are $$2^8$$ ways that we can choose to take the left or right glove from each.

So, if all possible sets of gloves are equally likely to be taken, the probability of taking no pairs of gloves should be $$\frac{{10 \choose 8} \cdot 2^8}{20 \choose 8} = \frac{\frac{10!}{2! 8!} \cdot 2^8}{\frac{20!}{12!8!}} = \frac{(10 \cdot 9 \cdot \ldots \cdot 3) \cdot 2^8}{20 \cdot 19 \cdot \ldots \cdot 13} = \frac{20 \cdot 18 \cdot \ldots \cdot 6}{20 \cdot 19 \cdot \ldots \cdot 13}$$

There are $$\binom{20}{8}$$ ways to select the $$8$$ gloves, all of which we assume are equally likely. Let's say a selection has "Property $$i$$" if it includes both gloves of pair $$i$$, for $$1 \le i \le 10$$, and let $$S_j$$ be the total probability (with over-counting) of the selections with $$j$$ of the properties, for $$1 \le j \le 4$$. So $$S_j = \frac{\binom{10}{j} \binom{20-2j}{8-2j}}{\binom{20}{8}}$$
By inclusion-exclusion, the probability of a selection with none of the properties, i.e. with no pair of matching gloves, is $$1-S_1+S_2-S_3+S_4 = 0.0914503$$


Different ways of evaluating $\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx$

My friend showed me this integral (and a neat way he evaluated it) and I am interested in seeing a few ways of evaluating it, especially if they are "often" used tricks.

I can't quite recall his way, but it had something to do with an identity for phase shifting sine or cosine, like noting that $\cos(x+\pi/2)=-\sin(x)$ we get: $$I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx=\int_{\pi/2}^{\pi}\frac{-\sin(x)}{-\sin(x)+\cos(x)}dx\\$$ Except for as I have tried, my signs don't work out well. The end result was finding $$2I=\int_{0}^{\pi/2}dx\Rightarrow I=\pi/4$$ Any help is appreciated! Thanks.

• It seems better to make the change of variable $x \to \pi/2-x$. Jun 18 '16 at 20:52
• @OlivierOloa oops right, I'll fix it. Have any other ideas for how to integrate the above? Jun 18 '16 at 20:53
• @OlivierOloa yes, thank you! I will try to use that for mine and provide an answer Jun 18 '16 at 20:57
• Also related: math.stackexchange.com/questions/180744/… Jun 18 '16 at 21:02

$$\int_{0}^{a}{\frac{f(x)}{f(x)+f(a-x)}}dx=\frac{a}{2}$$ let $f(x)=\sin x$ and $a=\frac\pi2$

• Is there a name for this formula? Where can I find more? Jun 18 '16 at 21:06
• No there is not Jun 18 '16 at 21:08
• @qbert The formula arises from another one: $\int_{0}^{a}\mathrm{F}\left(x\right)\,\mathrm{d}x = \int_{0}^{a}\mathrm{F}\left(a - x\right)\,\mathrm{d}x$ which sometimes we say "by reflection in the mirror" especially by people that computes MonteCarlo integrals. By adding ( to the original one ) and dividing by two we usually arrives to an integrand which is somehow smooth which the MC-people like a lot because it can reduce computing time. Jun 19 '16 at 2:31

Let

$$I=\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\ dx$$

and

$$J=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\ dx$$

then

$$I+J=\frac{\pi}{2}\tag1$$

and

\begin{align} I-J&=\int_0^{\pi/2}\frac{\cos x-\sin x}{\sin x+\cos x}\ dx\\[10pt] &=\int_0^{\pi/2}\frac{1}{\sin x+\cos x}\ d(\sin x+\cos x)\\[10pt] &=0\tag2 \end{align}

Hence, $$I=J=\frac\pi4$$by linear combinations $(1)$ and $(2)$.

• You might also be interested in seeing the general method Jun 18 '16 at 22:54
• I came back to this question, and you're answer, and the link you provided is really fantastic, and quite general Jul 7 '16 at 2:40

Hint: Substitute $2i\sin(x)=e^{ix}-e^{-ix}$ and $2\cos(x)=e^{ix}+e^{-ix}$

$$I=\int_{0}^{\pi/2}\frac{2\cos(x)}{2\sin(x)+2\cos(x)}dx=\int_{0}^{\pi/2}\frac{e^{ix}+e^{-ix}}{\frac{e^{ix}-e^{-ix}}{i}+e^{ix}+e^{-ix}}dx$$

Now substitute $u=e^{ix} \implies du = ie^{ix}dx=iudx$

$$I=\int\frac{u+1/u}{\frac{u-1/u}{i}+u+1/u}\frac{du}{iu}$$

Alternative method: $$I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx=\int_{0}^{\pi/2}\frac{1}{\tan(x)+1}dx$$

Substitute: $u=\tan(x) \implies du=(1+\tan^2(x))dx=(1+u^2)dx$

$$I=\int\frac{1}{u+1}\frac{du}{1+u^2}$$

• Ah I love it when complex makes integration easy. Thanks! Jun 18 '16 at 21:01
• Added second method. Jun 18 '16 at 21:24

You're integrating on $[0, \pi/2]$ so replacing $x$ by $\pi/2 -x$ we see that $$I=\int_0^{\pi/2} \frac{\sin{x}}{\cos{x}+\sin{x}}dx.$$ Now sum this integral with the initial expression and notice that $2I=\pi/2$ hence...


# The probability that A hits a target is $\frac14$ and that of B is $\frac13$. If they fire at once and one hits the target, find $P(\text{A hits})$

The probability that A hits a target is 1/4 and the probability that B hits a target 1/3. They each fire once at the target.

If the target is hit by only one of them, what is the probability that A hits the target?

I know that this is an independent event. If I do P(A hitting) * P(B not hitting) then (1/4)(2/3) = 1/6 But when I look at the back of my book the answer is 2/5? My book is known to give wrong answers because it is quite old; therefore, I am left with self doubt. Can anyone tell me if I have the correct answer or if I am actually making a mistake?

• This is a conditional probability. Letting $A$ be the event that player $A$ hit the target (in a single shot) and $B$ the event that $B$ hit the target (in a single shot), then what you calculated was $Pr(A\cap B^c)$. What you were told to calculate was $Pr(A\mid (A\cap B^c)\cup (A^c\cap B))$, i.e. the probability that $A$ hit the target given that exactly one of them hit the target. – JMoravitz Sep 26 '17 at 4:07
• You might like this explanation of why the formulas being posted here work: arbital.com/p/bayes_rule/?l=1zq – Davislor Sep 26 '17 at 14:51
• Would also add the comment that the book is likely not full of mistakes because it is old, but that because it is old the mistakes have been found. New books are not necessarily more correct, they just haven't been around long enough for the mistakes to be as well known. Old does not imply bad. – Jared Smith Sep 26 '17 at 15:39
• The probability of A hitting and B not hitting is 1/6 when it is also possible that they both hit or both miss. But in this case, it is not possible for them both to hit or both to miss, so the probability must be greater than 1/6. (If you don't see why, imagine rolling a die. The probability of 1 coming up is 1/6, but only because there are 5 other possibilities. If you roll a die and the result is not a 3, 4, 5, or 6, now the probability it's a 1 is 1/2.) – David Schwartz Sep 26 '17 at 18:42

\begin{align} P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\ &= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\ &= \frac{5}{12} \end{align}

So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$

• I would change the LHS of the last line to "$P(\text{A hitting}|\text{target is hit once})$, to be more explicit that we're using conditional probability here. – JiK Sep 26 '17 at 8:18

Your answer is not correct because you did not account for the case where only B hits, which has probability $\frac13×\frac34=\frac14$. Then the required probability is $$\frac{\frac16}{\frac14+\frac16}=\frac25$$ as the book gives.

The answer is indeed 2/5 I believe.

\begin{align} \mathbb{P}[\text{A hit | only one hit}] &= \frac{\mathbb{P}[\text{A hit} \,\cap\, \text{only one hit}]}{\mathbb{P}[\text{only one hit}]} \\ &= \frac{\mathbb{P}[\text{A hit}\,\cap\,\text{B didn't hit}]}{\mathbb{P}[\text{A hit}\,\cap\, \text{B didn't hit}] + \mathbb{P}[\text{A didn't hit}\,\cap\, \text{B hit}]} \\ &= \frac{1/4 \cdot 2/3}{1/4 \cdot 2/3 + 3/4 \cdot 1/3} \\ &=\frac{2}{5} \end{align}

Without using the conditional probability formula:

There are four cases:

1. Both miss
2. A hits and B misses
3. B hits and A misses
4. Both hit

We're only interested in (2) and (3). (2) has probability $\frac{1}{4}*\frac{2}{3} = \frac{1}{6}$. (3) has probability $\frac{1}{3}*\frac{3}{4}=\frac{1}{4}$. And we need $\frac{(2)}{(2) + (3)}$.

• What's the last formula if not the conditional probability formula? – JiK Sep 26 '17 at 16:48
• @JiK common sense – MattPutnam Sep 26 '17 at 17:42

The probability that only one person hits the target is $$1/4 * 2/3 + 1/3 * 3/4 = 5/12$$ The first event occurs when A hits and B misses, and the second when B hits and A misses. So if only one hit occurs, A hits 2/5 of the time and B 3/5 of the time.

This is an application of Bayes's law. You have a theory: A hit the target. You have data: there's only one hit. What is the probability your theory is true, given the data? 2/5. If you saw two bullet holes, then your theory would be true with probability 1 because A had to hit the target, given those data.

• We don't consider $P(\text{only one hit}|\text{A hit the target})$ or $P(\text{only one hit}|\text{A didn't hit the target})$ explicitly, so I don't see how this is an application of Bayes's law. – JiK Sep 26 '17 at 8:20


# How to plot slices of a surface of an iterative function parametrized by the iterator k?

I am trying to plot a surface of

$$z=\sin^{(k)}(x),\text{where (k) means nesting the function k times}$$

to visualise the fixed points and their neighbourhood to visually analyse their behaviour.

Currently, the following (adapted from this link) give me a contour version of the above:

f[x_] := Sin[x]
Show[Table[Plot[Nest[f, x, i], {x, -π, π},
PlotRange -> {-1, 1}, PlotStyle -> ColorData["Rainbow", 0.5 + i/10]], {i, 1, 10}]]

However, I want to space out the contours along the $k$ axis so that e.g. $\sin(x)$ corresponds to $k=1$, $\sin(\sin(x))$ corresponds to $k=2$ and so on...

Below is my most recent attempt at doing it:

 f[x_] := Sin[x]
data[x_] := Table[{Nest[f, x, i], i}, {i, 0, 10}]
ListPlot3D[data[x], {x, -π, π}]

which gives me an error

 SetDelayed::write: Tag List in {{x,0},{Sin[x],1},{Sin[Sin[x]],2},
{Sin[Sin[Sin[x]]],3},{Sin[Sin[Sin[Sin[x]]]],4},{Sin[Sin[Sin[<<1>>]]],5},
{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],6},
{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],7},
{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],8},
{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],9},
{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],10}}[x_] is Protected. >>

Strangely the data behind seemed to be interpreted correctly

ListPlot3D[{{x, 0}, {Sin[x], 1}, {Sin[Sin[x]], 2}, {Sin[Sin[Sin[x]]],
3}, {Sin[Sin[Sin[Sin[x]]]], 4}, {Sin[Sin[Sin[Sin[Sin[x]]]]],
5}, {Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]],
6}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]],
7}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]],
8}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]],
9}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]]], 10}}[x], {x, -\[Pi], \[Pi]}]

I was suspecting that ListPlot3D cannot read my input is probably because I have mixed data type. In details

$$z\in \mathbb{R}$$ $$x \in [-\pi,\pi]$$ but $$k \in \{0,1,2,3,4,5,6,7,8,9,10\}$$

From browsing the documentation, I am not aware of any examples of plots made from a mix of discrete and continuous variables as plotting arguments, thus I am not sure how to plot the surface I want.

I am not sure how to circumvent/cheat it without taking too much computation time since if my set of points $x$ is too sparse, it will fail to display the sinusoidal feature (which will be a problem because I am planning to apply this code on other iterative functions, such as the logistic map), but if my sampling is too dense, it will probably took too much computation time

Any ideas on what I can do?

P.S. To give an idea on what I am trying to achieve, refer to the below sketch:

which after interpolation along $k$, will give a nice surface.

• Related: (1413). – march Sep 14 '15 at 17:07

Use Interpolation if you want a regular function. Just for the plot you can also use ListPlot3D.

fun = Interpolation[
Flatten[Table[{x, k, Nest[Sin, x, k]}, {x, -Pi, Pi, .1}, {k, 1, 10,1}], 1]];

Plot the continuous function and those $k$-mesh lines!

Plot3D[fun[x, k], {x, -Pi, Pi}, {k, 1, 10}, MeshFunctions -> {#2 &},
Mesh -> 10, PerformanceGoal -> "Quality", MeshStyle -> {{Black, Thin}}]

If you only want discrete lines you can use ParametricPlot3D in combination with Map or Table embedded in a Show.

Below the Blend function is used to add a variable color (optional). Black is Sin[x] and Red is the curve nested ten times.

Show[
Map[
ParametricPlot3D[{u, #, Nest[Sin, u, #]}, {u, -\[Pi], \[Pi]},
PlotStyle -> Blend[{Black, Red}, #/10],
PlotRange -> {{-\[Pi], \[Pi]}, {0, 10}, {-1, 1}}
] &,
Range[10]
]
]

This is, I think, a dupe of Plotting several functions, except that that thread displayed only the contours.

An approach simpler than the other posted answers proceeds like so:

Plot3D[Nest[Sin, x, Round[k]], {x, -π, π}, {k, 1, 10},
MeshFunctions -> {#2 &}, Mesh -> 10]


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# Probability that a random tetrahedron over a sphere contains its center

I got interested in this problem watching the YouTube channel 3Blue1Brown, by Grant Sanderson, where he explains a way to tackle the problem that is just … elegant!

I can't emphasize enough how much I like this channel. For example, his approach to linear algebra in Essence of linear algebra is really good. I mention it, just in case you don't know it.

## The problem

Let's talk business now. The problem was originally part of the 53rd Putnam competition on 1992 and was stated as

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is in- dependently chosen relative to a uniform distribution on the sphere.)

As shown in the mentioned video, the probability is $1/8$. Let's come with an algorithm to obtain this result —approximately, at least.

## The proposed approach

The approach that we are going to use is pretty straightforward. We are going to obtain a sample of (independent) random sets, with four points each, and check how many of them satisfy the condition of being inside the tetrahedron with the points as vertices.

For this approach to work, we need two things:

1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need to do much about it.

2. A way to check if a point is inside a tetrahedron.

### Checking that a point is inside a tetrahedron

To find if a point is inside a tetrahedron, we could compute the barycentric coordinates for that point and check that all of them are have the same sign. Equivalently, as proposed here, we can check that the determinants of the matrices

\begin{equation*} M_0 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_1 = \begin{bmatrix} x &y &z &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_2 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x &y &z &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_3 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x &y &z &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_4 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x &y &z &1 \end{bmatrix}\, , \end{equation*}

have the same sign. In this case, $(x, y, z)$ is the point of interest and $(x_i, y_i, z_i)$ are the coordinates of each vertex.

## The algorithm

Below is a Python implementation of the approach discussed before

from __future__ import division, print_function
from numpy import (random, pi, cos, sin, sign, hstack,
column_stack, logspace)
from numpy.linalg import det
import matplotlib.pyplot as plt

def in_tet(x, y, z, pt):
"""
Determine if the point pt is inside the
tetrahedron with vertices coordinates x, y, z
"""
mat0 = column_stack((x, y, z, [1, 1, 1, 1]))
det0 = det(mat0)
for cont in range(4):
mat = mat0.copy()
mat[cont] = hstack((pt, 1))
if sign(det(mat)*det0) < 0:
inside = False
break
else:
inside = True
return inside

#%% Computation
prob = []
random.seed(seed=2)
N_min = 1
N_max = 5
N_vals = logspace(N_min, N_max, 100, dtype=int)
for N in N_vals:
inside_cont = 0
for cont_pts in range(N):
phi = random.uniform(low=0.0, high=2*pi, size=4)
theta = random.uniform(low=0.0, high=pi, size=4)
x = sin(theta)*cos(phi)
y = sin(theta)*sin(phi)
z = cos(theta)
if in_tet(x, y, z, [0, 0, 0]):
inside_cont += 1

prob.append(inside_cont/N)

#%% Plotting
plt.figure(figsize=(4, 3))
plt.hlines(0.125, 10**N_min, 10**N_max, color="#3f3f3f")
plt.semilogx(N_vals, prob, "o", alpha=0.5)
plt.xlabel("Number of trials")
plt.ylabel("Computed probability")
plt.tight_layout()
plt.show()

As expected, when the number of samples is sufficiently large, the estimated probability is close to the theoretical value: 0.125. This can be seen in the following figure.


SEARCH HOME
 Math Central Quandaries & Queries
 Question from r.m, a student: question from calculus exam: what is the figure obtained having eqn.r=10cos(t) in cylindrical coordinates? i know it is a cylinder with center (5,0) ,but can't the equation represent two cylinders, one with center (5,0) and the other with center (-5,0). thanks for any help.

Hi,

I want to use the cartesian graph of $y = \cos(x)$ for reference.

Now let's plot $r = 10 \cos(\theta)$ in polar coordinates for $0 \le \theta \le 2 \pi.$

$\cos(0) = 1$ and hence the graph starts at $(r, \theta_1) = (10,0)$ which is the point $P_1$ in my diagram.

Now let $0 < \theta_2 < \frac{\pi}{2}$ then $\cos(\theta_2)$ is positive and resulting point $P_2$ is on the upper half of the circle with center $(5, 0)$ and radius 10 as in my diagram. When $\theta_3 = \frac{\pi}{2}$ then $\cos(\theta_3) = 0$ and the resulting point on the graph is $P_3$.

For $\frac{\pi}{2} < \theta le \pi$ as $\theta_4$ in my diagram, $\cos(\theta) < 0$ and the resulting point (for example $P_4$) is on the bottom half of the circle with center $(5, 0)$. When $\theta = \pi$ then $\cos(\theta) = -1$ and we are back at $P_1$.

For $\pi < \theta \le \frac{3 \pi}{2}$ as $\theta_5$ in my diagram, $\cos(\theta)$ is still negative and the resulting point (for example $P_5$) is on the top half of the circle with center $(5, 0)$. When $\theta = \frac{3 \pi}{2}$ then $\cos(\theta) = 0$ and we are back at $P_3$.

Finally for $\frac{3 \pi}{2} < \theta \le 2 \pi, \cos(\theta)$ is positive and the resulting point, as $P_5$ in my diagram is on the bottom half of the circle and when $\theta = 2 \pi$ we are back at $P_1$.

Hence as $\theta$ moves from $0$ to $2 \pi$ the point defined by $r = 10 \cos(\theta)$ moves twice around the circle with center $(r \theta) = (5, 0)$ and radius 10.

Penny

r.m replied

sir,
in reply to your answer for r=10cost in cylinderical coordinates. if we follow the same method as you explained to sketch r=sin(t/2),i expected graph to be in only first two quadrants, but the graph was covering all four quadrants ?
If $0 < \theta <2 \pi$ then $0< \large \frac{\theta}{2} \normalsize < \pi$ and $\sin \left(\large \frac{\theta}{2}\right) >0.$ Thus, for example if $\theta = \large \frac{3 \pi}{2}$ then
$r = \sin \left(\frac{\theta}{2}\right) = \sin \left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt 2} = 0.70711$


Is the blue area greater than the red area?

Problem:

A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.

Which area is greater?

Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.

Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.

But how can it be proven?

I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.

Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$

According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}

Assertion:

To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.

Is this true? If so, is there another way of proving the assertion?

Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.

This question is very similar to this post.

Source:

The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$.

• i think you can tile the red area 4 times to get the entire square – gt6989b May 11 '18 at 4:25
• Hint: the sum of the two red sides that don't touch the center is $1$. – dxiv May 11 '18 at 4:29
• @user477343 Glad the hint helped. You can make that into a full-fledged answer, and I'll +1 it. – dxiv May 11 '18 at 4:31
• Is this a problem from "Brilliant" – Rohan Shinde May 11 '18 at 4:32
• Note that purely from exam technique alone, the answer is likely to be "they are the same size". Indeed, the problem has not told you by how much the rotation occurs, and why privilege a rotation of $0$ over a rotation of some greater angle? This is not a proof; but the phrasing of the question has told you what answer to look for. (This is a more general point than your "it works this way for 45 degrees": this is a demonstration that no mathematical reasoning at all is required to exam-technique that the answer is "they're the same".) – Patrick Stevens May 11 '18 at 20:24

The four numbered areas are congruent.

[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

• This answer also deserves the tick for artistic reasons. – BenM May 11 '18 at 6:09
• A great example of "proof by picture" that actually works. – Bristol May 11 '18 at 14:46
• This is not the same as the answer by Ross and Zoltan. I like this one better. Theirs was the first that came to my mind, too. – Carsten S May 11 '18 at 23:09
• Can Wolfram Alpha draw that? – Willtech May 12 '18 at 2:44
• @FrankShmrank The original poster asked within the question how it can be proven that the red area equals 1/4 (which would settle the title question). My answer makes it clear (without a formal proof, but in proof-by-picture, that’s par for the course) that the red area is one of four congruent areas that partition the unit square, so its area is 1/4. I agree my answer is less than a complete proof of the original question, but I think (and I guess many upvoters think) that it’s convincing. There are other excellent answers that are more traditionally proof-like, so upvote your favorite(s). – Steve Kass May 19 '18 at 17:14

I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.

• This method is similar to @RossMillikan 's answer above, but not quite the same :) I have to wait $9$ hours before I can upvote as I have reached my daily limit... but when I can, $$(+1)$$ – Mr Pie May 11 '18 at 15:00
• It's not only similar, now that I read that solution, it's actually the exact same idea. Unfortunatly that answer didn't contain any images and I just looked at the images before posting my own answer. :) – Zoltan May 11 '18 at 15:05
• Well congratulations on your first answer on the MSE! Yours is still a good answer :)) – Mr Pie May 11 '18 at 15:08
• This is the clearest image to understand. +1 – qwr May 12 '18 at 20:36
• @qwr Indeed! If only I could grab this answer and drag it below the accepted answer. That way, nobody would have to scroll all the way down to see this. It is my own answer that should probably be at the very bottom :) – Mr Pie May 13 '18 at 1:52

Note that for equal angles $\angle A'OB' = \angle AOB = 90^\circ$, when we subtract a common part $\angle A'OB$ from both sides, we have $\angle AOA' = \angle BOB'$, so the red and cyan triangles are congruent: $\triangle AOA' \sim \triangle BOB'$.

That implies their areas are equal, and when we add a common part $\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle.

• Shouldn't be $\angle AOA' = \angle BOB'$? – Pedro May 13 '18 at 3:59
• @Petro Right, thank you. – CiaPan May 13 '18 at 7:10
• Do you mean to say that $\Delta AOA' \color{red}{\cong} \Delta BOB'$? – Mr Pie May 14 '18 at 3:08
• This is the way I saw it – MichaelChirico May 15 '18 at 2:59

The two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.

• I see what you mean. There was no need to describe the result when drawing the vertical and horizontal lines that define the blue area on the diagram of the red area $-$ it was clear as day that they would be equal after looking at the newly formed triangles! I like your method of showing they were equal :) $$(+1)$$ – Mr Pie May 11 '18 at 5:29

By pinning a square's vertex to the center of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry of n=4. Since the rotation number is an integer multiple of the slice number, the area is invariant of rotation. You can apply this generally as well. A 120 degree slice of an equilateral triangle will be invariant. A 60 degree slice of a uniform hexagon will too. 120 degrees will work for the hexagon as well since that's 3 slices on a rotation number of 6.

• FWIW I like this answer the best. It is a simple, brief proof that uses clear logic instead of math. – Bohemian May 12 '18 at 14:28
• @Bohemian, the reasoning is of course maths. – Carsten S May 12 '18 at 23:24
• @carsten but it’s basic geometry, without any calculations, arithmetic or formulae, such that someone without any mathematical know-how could follow. It’s only barely maths (and I’m not in the mood to play semantics) – Bohemian May 13 '18 at 8:37
• @Bohemian: Whatever mood you are in, this is very much a mathematical answer. Looking for ideas like this will help you find solutions when manipulating formulæ gets you stuck. – PJTraill May 13 '18 at 21:39
• @Bohemian, I may be a bit touchy on this subject, I hope I did not come across as rude. It is just that a recognize a misconception of what is mathematical in this, even though you may not hold it. It reminds me of beginners asking questions on how they can make their perfectly fine argument "more mathematical", by which they mean that they feel that they should use formulas. – Carsten S May 14 '18 at 15:37

Let $f(\alpha)$ be the length of the segment from the center of the square to the outside of the square on the line at an angle of $\alpha$ degrees from the horizontal line pointing right.

Suppose that the first side of the square (in counterclockwise order) makes an angle of $\alpha$, then area you want is $\int\limits_{\alpha}^{\alpha+\frac{\pi}{2}} \frac{f(x)^2}{2} dx$ and since $f$ is periodic with period $\frac{\pi}{2}$ this is independent of $\alpha$.

• This is much too advanced for my skill level. $$(+1)$$ – Mr Pie May 11 '18 at 4:35
• in hindsight the other approach is better, but looking at it from the calculus point of view probably wont hurt :) – Jorge Fernández Hidalgo May 11 '18 at 4:36
• I am a high school student who is familiar with integrals and radians... but the statement, "$f$ is periodic," I don't know what that means. Is it ok if you could explain to me? Other than that, your answer is great! Thanks :) – Mr Pie May 11 '18 at 4:38
• @AHB yeah I do. In my opinion, it is an act of kindness, especially when one has at least $-3$ downvotes or lower. However, I let the user know what I believe is (or might be) wrong with their question as if I did put a downvote. Also, I think there are some badges earnt when using all the upvotes in one day or something like that, idk for sure. I have only ever downvoted $1$ post, only to earn a badge of my first donwvote. – Mr Pie May 12 '18 at 10:03
• @user477343 Yup, two actually, both bronze, 'Suffrage: 30 votes in a day', 'Vox Populi: all 40'. – Artemis Fowl May 19 '18 at 13:19

$\hspace{5cm}$

$$b^2+b^2=(a-c)^2+c^2 \Rightarrow \frac{b^2}{2}=\frac{(a-c)^2+c^2}{4}\\ S=\frac{b^2}{2}+\frac{(a-c)c}{2}=\frac{(a-c)^2+c^2+2(a-c)c}{4}=\frac{a^2}{4}.$$

• The number of ways one can work this out is amazing!! Also, your answer is pure math(s)! However, I have to wait $1$ hour before I can upvote as I have reached my daily voting limit. $$(+1)$$ – Mr Pie May 16 '18 at 22:47
• Could you elaborate on how you can assert that the 2 bs are actually equal to each other? – Frank Shmrank May 19 '18 at 14:52
• @FrankShmrank, such problems help intuitive thinking and imagination. If the lower square turns clockwise, its top two sides will turn to the same angle with respect to their original positions and the sides $b$-$b$ will increase equally. – farruhota May 19 '18 at 16:21
• @FrankShmrank I had the same problem, and then I deleted my comment and put up a new one (namely, my current one above) because I found out that by looking at the accepted answer, if the four triangles are congruent, then the two $b$s are equal to each other :) – Mr Pie May 20 '18 at 8:03

Solution:

Although the red area is not a triangle, the sum of its sides that do not touch the centre is equal to $1$. This can only mean that no matter how many degrees the square is rotated, no area will be greater; the red area will always be equal to the blue area, i.e. $$\frac 14$$

Credit to @dxiv who pointed this out as a hint in a comment!

• This is similar to Captain Morgan’s answer,but I find it less clearly expressed than that. – PJTraill May 13 '18 at 21:42
• @PJTraill Yes, you are correct $-$ Captain Morgan has a much better answer :) – Mr Pie May 13 '18 at 21:49

If we use $$\overline{FB}$$ for the base of $$\triangle FEB$$, then its altitude is $$\frac 12s$$. If we use $$\overline{BG}$$ for the length of the base of $$\triangle BEG$$, then its altitude is $$\frac 12s$$.

So the area of $$\square FBGE$$ is $$\frac 12(\frac 12s)(s-x) + \frac 12(\frac 12s)(x) = \frac 14s^2$$. Which is one-fourth of the area of the square.

The advantage of this method is that it allows you to break the square into $$n$$ pieces of equal area quite easily. In fact, the same method applies to an regular polygon.

• Thank you for letting me know. The diagram was very clear, and this method is very much applicable to the problem. $$(+1)$$ (in at least $19$ hours $-$ I have reached my daily voting limit). Also, how did you construct the picture? – Mr Pie May 16 '18 at 4:03
• I used GeoGebra. – steven gregory May 16 '18 at 4:48
• Thank you, again, for telling me :) – Mr Pie May 16 '18 at 6:55
• This is by far the best answer as it doesn't assert anything that isn't given. All the other answers assert things that aren't necessarily known. It's sad that the chosen answer was chosen because that user had the lowest rep. – Frank Shmrank May 19 '18 at 14:55
• @FrankShmrank all the answers are great, imho. – Mr Pie May 20 '18 at 8:02

The two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.


# Is it possible to cover a $8 \times8$ board with $2 \times 1$ pieces?

We have a $8\times 8$ board, colored with two colors like a typical chessboard. Now, we remove two squares of different colour. Is it possible to cover the new board with two-color pieces (i.e. domino pieces)?

I think we can, as after the removal of the two squares, we are left with $64-2=62$ squares with $31$ squares of each colour, and - since the domino piece covers two colours - we can cover the new board with domino pieces.

But how should one justify it mathematically?

• It's possible to remove four squares, two of each color, and not cover the remaining with dominos, because the new set is disconnected. So you need to know more than that there are 31 of each color. Oct 15 '16 at 16:49
• Your constraints are valid. But there must be more constraints. Such that the colours alternate on the board, that dominos are made of two different colours, next to each other. Then the shape of the board.
– mvw
Oct 15 '16 at 17:05
• @mvw "like a typical chessboard" should cover all of that. Oct 15 '16 at 17:11
• @JMoravitz That was not my point. The problem is clear. It is about the proof attempt. hetajr just cared about a subset of all constraints that play a role here.
– mvw
Oct 15 '16 at 17:13
• @ThomasAndrews Ah cutting off two corners, leaving single corner stones.
– mvw
Oct 15 '16 at 17:23

Assume without loss of generality that the two squares to be removed are in different rows. (Otherwise turn the board 90°).

First cover the board in horizontal dominoes, and connect the two squares by a zig-zag line like this:

which follows the rule that if the line goes through one end of a domino, it immediately connects to another end. (The requirement that the two squares have different colors ensure that this will be true of the end of the path if only we start out in the right direction for this to hold at the beginning). Now you can flip dominoes along the zig-zag line to produce a covering that avoids the two squares.

With a bit of (easy) special-casing for the same-row case, this strategy can be extended to any size board as long as one of the side lengths is even and the other is $\ge 2$.

• I do not get the flipping. But it seems clear the length of the zig zag line is a multiple of two, as is the reduced line, so it can be covered.
– mvw
Oct 15 '16 at 17:31
• @mvw: Initially every other of the square boundaries crossed by the red line was covered by a domino. By "flipping" I mean to instead cover the other half of the boundaries by dominos. Oct 15 '16 at 17:36
• weren't you lucky that the top and bottom part had an even number of squares ? Oct 16 '16 at 13:23
• @mercio: If the top endpoint had been at an even column, I would just have started going left instead of right. This means that the zig-zag line would always enter each of the original dominoes at the end that has the same color as the top endpoint, so the bottom endpoint will always be reached at the end of a domino. Oct 16 '16 at 13:33

From a graph theory point of view, this can be seen as a "matching problem" on a bipartite graph.

The nodes of the graph are the squares remaining.

Two nodes have an edge in the graph if the squares are neighbors - that is, if the two squares can be covered by a single domino.

Obviously, in any edge, one square is white, the other black, hence the "bipartite" nature of this graph.

So you are seeking to show there is a perfect matching for any such graph.

There is a general theorem about when there is a perfect matching for a bipartite graph, called Hall's Theorem or Hall's Marriage Theorem. It is possibly overkill for this question - induction is likely the better approach.

Per the discussion on Henning's answer, it is actually possible to prove your theorem directly using a "Hamilton cycle" on the chess board.

Consider the loop path on the board:

$$\begin{matrix}1&2&3&4&5&6&7&8\\ 64&15&14&13&12&11&10&9\\ 63&16&17&18&19&20&21&22\\ 62&29&28&27&26&25&24&23\\ 61&30&31&32&33&34&35&36\\ 60&43&42&41&40&39&38&37\\ 59&44&45&46&47&48&49&50\\ 58&57&56&55&54&53&52&51 \end{matrix}$$

So we have walked in a circle, and, if the upper left is black, then we have that odd numbers on black and the even numbers on white.

If we unwind this, and consider it 64 beads in a circle, alternating black and white, then if we remove/cut away one black and one white bead, we are either left with one string of 62 beads alternating black/white, which lets us cover those with dominos, or two seperate strings.

With two strings, here is the key: because we cut away one black and one white bead, those two strands are of even length.

For example, if we removed the square at 12 and the square at 23, then we could get the domino placement: $$(13,14),(15,16),(17,18),(19,20),(21,22), \\(24,25),\dots,(62,63),(64,1),(2,3),(4,5),(6,7),(8,9),(10,11).$$

This can be generalized as: "If a bipartite graph has a Hamiltonian cycle, then if you remove one node from each of the parts, you can still get a perfect matching."

In particular, this argument works for any $2n\times m$ board, because we can get a "King's cycle tour" on any such board.

For example, a $3\times 4$ board:

$$\begin{matrix} 1 & 2& 3&4\\ 12& 9& 8&5\\ 11&10& 7&6 \end{matrix}$$

• (+1) Interesting approach. In this framework, we just need to check that the hypothesis of Hall's theorem are met, i.e. that for any subset of white (or black) squares in the truncated chessboard, their neighbourhood has a larger cardinality, and that is clearly true. Oct 15 '16 at 17:46
• (on the other hand, induction is just the usual way for proving Hall's theorem) Oct 15 '16 at 17:54
• Hmm, yes, I suppose you could prove that for a subset of $k$ white squares on the uncut board, with $0<k<32$, the set of neighboring black squares has at least $k+1$ elements, then you can apply Hall's theorem to the cut board. @JackD'Aurizio Oct 15 '16 at 18:31
• And that can indeed be proved. (Posted as a separate answer so I can show diagrams). @JackD'Aurizio too. Oct 15 '16 at 20:50

Hint: A promising strategy is to prove that the claim

If we remove two opposite-colored squares from a $2m\times 2m$ chessboard,
we may tile the remaining part with $2\times 1$ dominoes.

by induction on $m$. The case $m=1$ is trivial. Assume that the claim holds for some $m\geq 1$ and consider a $(2m+2)\times (2m+2)$ chessboard. If both the removed squares do not lie on the boundary of the chessboard, there is nothing to prove. Hence we may assume that at least one of the removed squares lies on the boundary. And we may also start tiling by following a spiral, starting next to the removed square on the boundary:

Another interesting idea is just to place $31$ non-overlapping dominoes on a $8\times 8$ chessboard and start playing Sokoban with the placed dominoes, in order to free the wanted squares.

• I like this one the best so far. Still thinking about it.
– mvw
Oct 15 '16 at 17:19
• You might be able to prove this form $2m\times n$ for any $n>1$, which makes it easier to do the induction. Oct 15 '16 at 17:42
• @ThomasAndrews: true. Interesting remark, thanks. Oct 15 '16 at 17:53

Completing an approach suggested by Thomas Andrews, if we can show that on the complete chessboard any proper subset of the white squares have more black neighbors than it has members, then Hall's marriage theorem will apply to the chessboard with two squares erased.

Suppose therefore, that a proper subset of the white squares are given. Since the red and green lines in the following diagram connect all the white squares, there will be at least one red or green line that goes from a square in the subset to a square outside of it:

Assume without loss of generality that there is a red line joining a square in the subset to a square outside the subset. (Otherwise mirror everything around the white diagonal).

Now pair up each white square with the black neighbor it is connected to by a blue line in this diagram:

This gives one neighboring black square for each white square. However the white square with a diagonal-partner that is not selected is additionally neighbor to the non-selected square's black partner which is not otherwise used.

So, as desired, our set of white squares has more black neighbors in total than there are white squares in the set.

• Very nice! I couldn't quite find that zig-zag argument while trying to find the "extra" black square. Oct 15 '16 at 21:39
• @ThomasAndrews: Actually this is, on further thought, still more complicated than it needs to be. Just fix a Hamiltonian circuit on the board (necessarily with alternating white and black squares) and then just count neighbors along that circuit for each run of selected white squares. Oct 15 '16 at 22:47
• of course, once you have a Hamiltonian circuit, you actually don't need Hall's theorem. Just remove your two nodes, and you get one or two paths, both of even lengths. Oct 15 '16 at 23:21
• @ThomasAndrews: Indeed. I would update my main answer, except I would have to make new diagrams ... Oct 15 '16 at 23:22
• Okay, I updated my answer with the non-Hall Hamiltonian cycle answer. Oct 16 '16 at 12:57

Yes.

Consider the board initially covered with dominoes. After the two squares have been removed, the board has two holes.

If the two holes were in the same row, slide the dominoes between the holes until one of the holes is filled. This will leave two adjacent holes that can be covered by a domino.

If the holes are in different rows, slide dominoes from the lower row to the upper row up until one of the holes is filled. That leaves the lower row with two holes which can be filled as described above.

If the two rows are adjacent, slide the upper row left or right until its hole is filled and the hole has moved above the hole in the next row. This can be filled by a domino.

• How does this answer make use of the information that the holes are on squares of different colour?
– HTFB
Oct 15 '16 at 17:02
• "Slide dominoes" won't work if the holes are in column 2 and 3 of the same row. Oct 15 '16 at 17:11
• I think this answer is wrong because if the holes are of the same colour, then it should not work. Oct 15 '16 at 17:17


As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand.

Example 11.5.1 Does $\ds\sum_{n=2}^\infty {1\over n^2\ln n}$ converge?

The obvious first approach, based on what we know, is the integral test. Unfortunately, we can't compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a $p$-series, that is, $${1\over n^2\ln n}< {1\over n^2},$$ when $n\ge3$. Since adding up the terms $\ds 1/n^2$ doesn't get "too big'', the new series "should'' also converge. Let's make this more precise.

The series $\ds\sum_{n=2}^\infty {1\over n^2\ln n}$ converges if and only if $\ds\sum_{n=3}^\infty {1\over n^2\ln n}$ converges—all we've done is dropped the initial term. We know that $\ds\sum_{n=3}^\infty {1\over n^2}$ converges. Looking at two typical partial sums: $$s_n={1\over 3^2\ln 3}+{1\over 4^2\ln 4}+{1\over 5^2\ln 5}+\cdots+ {1\over n^2\ln n} < {1\over 3^2}+{1\over 4^2}+ {1\over 5^2}+\cdots+{1\over n^2}=t_n.$$ Since the $p$-series converges, say to $L$, and since the terms are positive, $\ds t_n< L$. Since the terms of the new series are positive, the $\ds s_n$ form an increasing sequence and $\ds s_n< t_n< L$ for all $n$. Hence the sequence $\ds \{s_n\}$ is bounded and so converges. $\square$

Sometimes, even when the integral test applies, comparison to a known series is easier, so it's generally a good idea to think about doing a comparison before doing the integral test.

Example 11.5.2 Does $\ds\sum_{n=1}^\infty {|\sin n|\over n^2}$ converge?

We can't apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however, $${|\sin n|\over n^2}\le {1\over n^2},$$ because $|\sin n|\le 1$. Once again the partial sums are non-decreasing and bounded above by $\ds \sum 1/n^2=L$, so the new series converges. $\square$

Like the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will confirm whichever is the case. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly.

Example 11.5.3 Does $\ds\sum_{n=2}^\infty {1\over\sqrt{n^2-3}}$ converge?

We observe that the $-3$ should have little effect compared to the $\ds n^2$ inside the square root, and therefore guess that the terms are enough like $\ds 1/\sqrt{n^2}=1/n$ that the series should diverge. We attempt to show this by comparison to the harmonic series. We note that $${1\over\sqrt{n^2-3}} > {1\over\sqrt{n^2}} = {1\over n},$$ so that $$s_n={1\over\sqrt{2^2-3}}+{1\over\sqrt{3^2-3}}+\cdots+ {1\over\sqrt{n^2-3}} > {1\over 2} + {1\over3}+\cdots+{1\over n}=t_n,$$ where $\ds t_n$ is 1 less than the corresponding partial sum of the harmonic series (because we start at $n=2$ instead of $n=1$). Since $\ds\lim_{n\to\infty}t_n=\infty$, $\ds\lim_{n\to\infty}s_n=\infty$ as well. $\square$

So the general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series.

Example 11.5.4 Does $\ds\sum_{n=1}^\infty {1\over\sqrt{n^2+3}}$ converge?

Just as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately, $${1\over\sqrt{n^2+3}} < {1\over n},$$ so we can't compare the series directly to the harmonic series. A little thought leads us to $${1\over\sqrt{n^2+3}} > {1\over\sqrt{n^2+3n^2}} = {1\over2n},$$ so if $\sum 1/(2n)$ diverges then the given series diverges. But since $\sum 1/(2n)=(1/2)\sum 1/n$, theorem 11.2.2 implies that it does indeed diverge. $\square$

For reference we summarize the comparison test in a theorem.

Theorem 11.5.5 Suppose that $\ds a_n$ and $\ds b_n$ are non-negative for all $n$ and that $\ds a_n\le b_n$ when $n\ge N$, for some $N$.

If $\ds\sum_{n=0}^\infty b_n$ converges, so does $\ds\sum_{n=0}^\infty a_n$.

If $\ds\sum_{n=0}^\infty a_n$ diverges, so does $\ds\sum_{n=0}^\infty b_n$.

$\qed$

## Exercises 11.5

Determine whether the series converge or diverge.

Ex 11.5.1 $\ds\sum_{n=1}^\infty {1\over 2n^2+3n+5}$ (answer)

Ex 11.5.2 $\ds\sum_{n=2}^\infty {1\over 2n^2+3n-5}$ (answer)

Ex 11.5.3 $\ds\sum_{n=1}^\infty {1\over 2n^2-3n-5}$ (answer)

Ex 11.5.4 $\ds\sum_{n=1}^\infty {3n+4\over 2n^2+3n+5}$ (answer)

Ex 11.5.5 $\ds\sum_{n=1}^\infty {3n^2+4\over 2n^2+3n+5}$ (answer)

Ex 11.5.6 $\ds\sum_{n=1}^\infty {\ln n\over n}$ (answer)

Ex 11.5.7 $\ds\sum_{n=1}^\infty {\ln n\over n^3}$ (answer)

Ex 11.5.8 $\ds\sum_{n=2}^\infty {1\over \ln n}$ (answer)

Ex 11.5.9 $\ds\sum_{n=1}^\infty {3^n\over 2^n+5^n}$ (answer)

Ex 11.5.10 $\ds\sum_{n=1}^\infty {3^n\over 2^n+3^n}$ (answer)


# Finding sides on giant wooden cube

• May 25th 2008, 01:49 PM
annie3993
Finding sides on giant wooden cube
A giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces?

i got 72, but I don't think its right...

THANX!!
• May 25th 2008, 03:10 PM
TheEmptySet
Quote:

Originally Posted by annie3993
A giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces?

i got 72, but I don't think its right...

THANX!!

Maybe this diagram will help.

P.S. Always try to draw a picture it really helps you see what is going on.

Attachment 6507

It looks like four groups of 3 on the top

four groups of 3 in the middle

and four groups of 3 on the bottom.

$4(3)+4(3)+4(3)=4(9)=36$

I hope this helps.
• May 25th 2008, 03:22 PM
Soroban
Hello, annie3993!

Quote:

A giant wooden cube is painted green on all 6 sides and then cut into 125 identical smaller cubes.
How many of these smaller cubes are painted on exactly two faces?

This is a 5 × 5 × 5 cube.

A cube has 6 faces, 12 edges, and 8 corners (vertices).

Let's look at one face.
Code:

      * - * - * - * - * - *       | 3 | 2 | 2 | 2 | 3 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 3 | 2 | 2 | 2 | 3 |       * - * - * - * - * - *
The nine cubes in the center have one face painted green.
The four cubes in the corners have three faces painted green.

On each edge, there are three cubes with two green faces.

Since there are 12 edges, there are: $12 \times 3 \:=\:\boxed{36}$ cubes with two green faces.

• May 25th 2008, 03:23 PM
galactus
See this thread. You may find it interesting.

http://www.mathhelpforum.com/math-he...-question.html

If yu go to the bottom you will see the general formula of 12(n-2). In your case, 12(5-2)=36


# In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels

In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels

I solved and got answer as $90720$. But other sites are giving different answers. Please help to understand which is the right answer and why I am going wrong.

My Solution

Arrange 6 consonants $\dfrac{6!}{2!}$
Chose 2 slots from 7 positions $\dbinom{7}{2}$
Chose 1 slot for placing the 2 vowel group $\dbinom{2}{1}$
Arrange the vowels $3!$

Required number of ways:
$\dfrac{6!}{2!}\times \dbinom{7}{2}\times \dbinom{2}{1}\times 3!=90720$

Solution taken from http://www.sosmath.com/CBB/viewtopic.php?t=6126)

Solution taken from http://myassignmentpartners.com/2015/06/20/supplementary-3/

• Can you explain your working. Just putting down your calculation doesn't tell us why you chose to do them. – Ian Miller Nov 20 '16 at 14:30
• @sorry, edited the calculation and added the details. pl help. – Kiran Nov 20 '16 at 14:31
• I will point out that the solution in the excerpt solves a different problem. Your problem asks for "exactly two consecutive vowels", the excerpt's solution allows 3 consecutive vowels as well. As it says at the end "with at least two adjacent vowel" – ReverseFlow Nov 20 '16 at 14:36
• @Kiran You answer is right and their answer is wrong. I have added my explanation below. – user940 Nov 20 '16 at 15:12
• Checked with Python, the answer is indeed $90720$, deleted mine. – barak manos Nov 20 '16 at 15:13

The number of arrangements with 3 consecutive vowels is correctly explained in the original post: the number is $15120$.
To find the number of arrangements with at least two consecutive vowels, we duct tape two of them together (as in the original post) and arrive at $120960$.
The problem with this calculation is that every arrangement with 3 consecutive vowels was double counted: once as $\overline{VV}V$ and again as $V\overline{VV}$. To compensate for this we must subtract $15120$. The correct number of arrangements with at least two consecutive vowels is $120960-15120=105840.$
Therefore, correct number of arrangements with exactly two consecutive vowels is $105840-15120=90720.$
The total number of ways of arranging the letters is $\frac{9!}{2!} = 181440$. Of these, let us count the cases where no two vowels are together. This is $$\frac{6!}{2!} \times \binom{7}{3}\times 3! = 75600$$ Again, the number of ways in which all vowels are together is 15120. Thus the number of ways in which exactly two vowels are together is $$181440 - 75600 - 15120 = 90720$$