Mathematicianadda

How many numbers can we select from $\{1,2, \ldots, 2016\}$ such that sum of any four of them cannot be divided by $11$

It's not hard to come up with some combinations, but the question is how to prove it's the largest set.

For example if we select numbers in forms of $11N+1,11N+4,11N+9$ will yield us $184 + 183 + 183$ numbers.

it looks like proof will be somewhat of an inequalities problem. Let $a_{i_0}, a_{i_1}, \ldots, a_{i_k} > 0$ be the count of numbers we select from each modulo class, and we want to maximize $a_{i_0} + a_{i_1} + \cdots + a_{i_k}$ But how to express the constraint is tricky.

from Hot Weekly Questions - Mathematics Stack Exchange

More specifically, how would you evaluate the below formula? $$\lim_{n\to\infty}\sum_{k=n/2}^{n}\frac{1}{k}$$ I know that the harmonic series starting at any point diverges, but when we limit it in this way, does the series diverge or converge?

If it diverges:

• How might you determine that?

• Is there some $d$ that we can replace with $2$ to make the sequence converge?

If it converges:

• What does it converge to, and how might you determine that?

• The sequence must converge for any $d>2$. Is there a formula for the series generalized for any $d$?

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Solving Word Problems Involving the Shapes Squares and Rectangles - Concept and examples with step by step solution

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Solving Word Problems Involving Age - Concept and example problems with step by step solution

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I know that if $|G| = n$, then $G$ can be embedded into $S_n$. But the group $S_n$ is very large compared to $G$, so I was wondering if there are general ways of embedding $G$ into a smaller symmetric group. (By general I don't mean that it has to work for all groups, but hopefully for large classes of groups)

Also, I was wondering embedding groups into smaller symmetric groups is common/useful, or just a curiosity?

The only approach I could think of is to let $G$ act on various things, and hope that the action is faithful. One nice thing, for example, is that if $G$ has a simple subgroup $H$ (which is itself non-normal in $G$), then the action of $G$ on the left cosets of $H$ is faithful, since according to this question, the kernel must be trivial.

Furthermore, if you let $G$ act by conjugation on a subgroup, I believe the action is sometimes faithful, sometimes not (has to do with weather the conjugates are disjoint).

from Hot Weekly Questions - Mathematics Stack Exchange

For this episode we compiled some of the key insights from our guests throughout the year, and pulling it together is our way of celebrating the important work happening in the field of math education.

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Evaluate the sum $$\frac{1}{3} + \frac{1}{3^{1+\frac{1}{2}}}+\frac{1}{3^{1+\frac{1}{2}+\frac{1}{3}}}+\cdots$$

It seems that $1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}$ approaches $\ln n$ as $n\to \infty$, but I'm not sure if this is useful. Also, $3^{\ln n} =e^{\ln n\cdot \ln 3}= n^{\ln 3}$, but I'm also not sure how this is useful.

edit: I know how to prove that it converges, but I was wondering if there was a closed form for this sum.

from Hot Weekly Questions - Mathematics Stack Exchange

It is well known that for a jointly normally distributed random variables $(X_1,...,X_n)^T,$ they are uncorrelated if and only if independent.

It is also well-known that for any random variable, independent implies uncorrelated but not the converse.

Here comes my question:

Question: Besides jointly normal random variable, what other distribution satisfies uncorrelated if and only if independent?

from Hot Weekly Questions - Mathematics Stack Exchange

Each month on the MIND blog, we share stories of how our organization, our partners, and educators across the country are advancing the mission to mathematically equip all students to solve the world’s most challenging problems. We also share resources for educators and students, and cover some of the exciting events we take part in across the country.

Welcome to the MIND Blog Rewind! December is coming to a close and along with it, 2019. I don't know about you, but I'm ready for the roaring 20's. Let's kick off the new year with gratitude and celebration for what we achieved together. Here is a recap of important news, content, and resources that came out of MIND and our community in December 2019.

Winter break is here but learning never stops! With your help, ST Math is reaching more students than ever. We are so thankful to our partners, donors, and ST Math champions for being a critical part of MIND's mission. We shared our annual digital holiday card and rounded up our favorite ways to celebrate the holidays with the Winter Break Challenge, holiday coloring sheets, and winter STEAM activities.

We may be at the end of the year, but how did we get here? We kicked off the holiday season with a gratitude tree in our Irvine office. Colleagues throughout the organization were encouraged to take a moment and reflect on what they were thankful for. I was excited to hear what my MIND family held in their hearts, but with how busy everyone was, I wasn't sure what would happen. Our gratitude tree bloomed in the most incredible way with the love and care of everyone.

What's wrong with memorizing? Jo Garrett, CFO at MIND, answered this question and shared her own experiences on the way children are being taught math today. Parents are a core part of student learning ecosystems and embracing the way that your child is learning math is a great way to support them.

What is spatial-temporal reasoning? How exactly does it work? Research shows that a visual approach to conveying math concepts can be highly effective. Ki Karou, Director of Product, ST Math, shared the science of spatial-temporal reasoning and how we can use spatial-temporal methods to teach mathematics.

Our website is chock full of great resources for parents like blog posts, posters, and an ST Math Parent Kit. Liz Neiman, VP of Engagement at MIND, is also an ST Math parent. In the past, she shared her first experience with ST Math and how it provided a challenge for her gifted son. This December, she answered many of the questions parents have when they begin using our program for the first time. Check out our new blog post called, "So Your Child is New to ST Math...Now What?"

In our latest podcast episode, EVP and Executive Director of Social Impact, Karin Wu, back to the show. Karin discussed her recent feature in STEMConnector’s Million Women Mentors: Leaders to Learn From series and recapped her recent tour across country with our donors and strategic partners, like John Deere, Phillips 66, Rockwell Automation, the One8 Foundation, and more.

Kelsey Skaggs, our Public Relations and Communications Specialist, has lived in Texas for over two years now. Inspired by her state and the educational leaders she admires within, she compiled a helpful list of 20 Texas education leaders to follow on Twitter. From breaking news to helpful resources, these leading voices in education will inform and inspire you.

Thank you for a great 2019! Here’s to 2020.

And with that, we're signing off for the final time this year. In 2019, we made a lot of amazing things happen together with the help and support of our incredible partners, teachers, administrators, students, and ST Math parents. Thank you for everything!

Happy new year from all of us at MIND. We'll see you in 2020!

Additional Highlights:

• MIND Blog Rewind: November 2019
• MIND Blog Rewind: October 2019
• MIND Blog Rewind: September 2019

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It’s that time of year when we take a look at the UK Government’s New Years Honours list for any particularly mathematical entries. Here is the selection for this year – any more, let us know in the comments and we’ll add to the list. 

• Prof. Nick Woodhouse, Emeritus Professor of Mathematics, University of Oxford. Appointed CBE for services to Mathematics.
• Prof. Abdel Babiker, Professor of Epidemiology & Medical Statistics, UCL. Appointed OBE for services to medical research.
• Agnes Johnstone, Head of Mathematics, Oban High School. Awarded BEM for services to STEM Education and the community in Oban.

Get the full list here.

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One of my daughters is headed to middle school next year, and I had a thought about getting her started on adding and subtracting integers.  But..she doesn't like math...sigh!!! so I have to trick her and make it into a game :)

I bought some blank dice from amazon. (48 Blank Dice)  Using a sharpie, I started with using 4 of these dice.  On one of the dies I put the positive numbers from 1 to 6, on the second die, I put the negative numbers from -1 to -6, the third die had sides with addition signs and subtraction signs, and the 4th die had = signs on all of the sides.

Here are a couple examples of what the dice look like once they are rolled.  Notice that the subtraction sign is purple - I originally did all of the dice in black, but then we couldn't tell the subtraction sign from the number one.  So, I would recommend that you use a different color on the die with the operation signs.







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Solving Age Word Problems with Equations - Examples with step by step solution

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I hope the New Year is better than we deserve.

I will be spending most of the first six months in Cambridge, at a group theory programme at the Isaac Newton Institute. So hopefully I will have some time for catching up and tidying up; things have got in rather a mess, and it has been a couple of years since I had a holiday.

This also means that there might be a higher proportion of group theory in what I have to say.

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Finding the Required Fraction by Solving Word Problems - Concept and example problems with step by solution

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I'm studying abelian varieties from Milne's book, but I'm having difficulty juggling different conventions and definitions of basic concepts, like those of algebraic and projective varieties. First, let me write some terminology the way I understand it.

Classically, an affine $k$-variety is a Zariski closed (zero set of some family of polynomial), irreducible subset of $k^n$ where $k$ is any field, and an affine algebraic set is just any Zariski closed subset. A projective $k$-variety is a Zariski closed (zero set of some family of homogeneous polynomials) subset of $\mathbb P^n$.

The definition of an algebraic ($k$-)variety is somewhat more delicate. According to Milne's Algebraic Geometry notes, an affine $k$-variety is any locally ringed space isomorphic to some $(V,\mathcal{O}_V)$ where $V$ is affine algebraic and $\mathcal{O}_V(U)$ is the set of regular functions on $U$ (rational functions with a denominator that doesn't vanish on $U$). First question: how is this definition of an affine $k$-variety related to the one above?

Next, Milne defines an algebraic prevariety over $k$ as a locally ringed space admitting a finite open cover of affine $k$-varieties, and then an algebraic $k$-variety is a separated algebraic prevariety over $k$. What confuses me with this definition is that I don't recognize a scheme structure on Milne's definition of a variety. My second question is, is it true that the affine $k$-varieties $(V,\mathcal{O}_V)$ are affine (k-)schemes. In particular, to what ring $R$ is $(V,\mathcal{O}_V)$ isomorphic to $\operatorname{Spec} R$?

Finally, Qing Liu's definition of an algebraic $k$-variety is as follows. An affine $k$-variety is "the affine scheme associated to a finitely generated (reduced) $k$-aglebra". Which affine scheme is this? Is it simply $\operatorname{Spec} k[T_1,\dots,T_n]/I$ with $I$ radical? How does one make this association precise? Finally, Liu's algebraic $k$-variety is a $k$-scheme admitting an finite cover of affine $k$-varieties.

I suppose the ultimate question is this. What is the relationship between affine algebraic sets in $k^n$, sheaves of reduced, finitely-generated $k$-algebras, and the scheme $\operatorname{Spec}k[T_1,\dots,T_n]/I$ where $I$ is a radical ideal, and what does it mean for a variety to be affine or projective in these contexts?

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Have you checked out the Mathematical Association of America’s Math Values blog? The site includes posts about inclusivity, community, communication, teaching and learning, and more. Please join me on a blog tour highlighting some Math Values posts that I find … Continue reading →

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A friend and I came up with this puzzle and I'm looking for a proof.

Given an equilateral triangle of area 1, color parts of the triangle red, blue, and green such that

• Each color makes exactly one connected region strictly inside the triangle
• There is no line parallel to one of the sides that contains points of multiple colors

Let $X$ be the minimum area of the red, blue, and green regions. Find the maximum value of $X$ over all possible colorings.

I suspect the maximum occurs in the following arrangement of teardrop-shaped figures, which each have an area of $\frac{4}{45}$ (new bound found by Daniel Mathias). This is an awfully strange number for what seems like a nice problem, so I'm not sure if it's correct.

If you consider the triangle formed by the three points closest to the center and call $x$ the side length, $\frac{4}{45}$ can be reached when $x=\frac{4}{5\cdot 3^{3/4}}$. If $s$ is the side length of the original triangle, each region has area $x\left(\frac{s}{3}-\frac{5x\sqrt{3}}{12}\right)$. Maximizing this gives $\frac{4}{45}$.

Does anybody have an idea of a proof (or a counterexample) that this indeed gives the maximum? If it is valid, is there any intuition behind the value $\frac{4}{45}$ that makes it so special?

Also, we can look at the discrete case of this puzzle on a triangular grid with $n$ vertices on each side where we color vertices three colors. Asymptotically, this should have the same behavior as the original problem. I couldn't see a very nice pattern with small values—does anybody have a solution to this modified problem?

We tried to look for problems similar to this one; it seems like it should be well known! However, we couldn't find anything. If anybody could help us, that would be greatly appreciated.

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Solving Word Problems Based on Two Digit Number - Concept and example problems with step by step explanation

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I came across the following interesting result on the internet:

$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$

I'm looking for a (satisfying) explanations for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$. This post asks the same question but has no answers posted.

I am hoping to post my answer myself, but any input is appreciated. Thanks for your consideration.

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Solving Algera Word Problems with Two Numbers - Concept and examples with step by step solution

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Shoenfield's absoluteness theorem says that every $\Sigma^1_3$ statement is upwards absolute between models of ZF with the same ordinals. As a consequence, it implies that ZF and ZFC+V=L prove the same $\Sigma^1_3$ sentences. Meanwhile, this is sharp since "Every real is constructible" is a $\Pi^1_3$ theorem of ZFC+V=L which is independent of ZFC.

In the realm between ZFC and ZF, however, we can find significantly more absoluteness. For example, suppose $M$ is a model of ZF+DC. Then countably closed forcing adds no new reals, so when we force over $M$ with partial bijections $2^\omega\rightarrow\Theta$ with range bounded below $\Theta$ the resulting extension $M[G]$ has the same reals as $M$. But so does $L[G]^{M[G]}$, and this latter model satisfies choice since $G$ is appropriately equivalent to a set of ordinals (consider the set of pairs $\langle n,\alpha\rangle\in\omega\times\Theta^M$ such that $G^{-1}(\alpha)(n)=1$). Since $M$ and $L[G]^{M[G]}$ have the same reals, they satisfy the same projective sentences.

However, there are models of ZF without DC which do not have the same reals as any model of ZFC: for example, any model of ZF + "$2^\omega$ is a countable union of countable sets" has this property. So even ignoring the forcing details, the "model-swithcing" idea above breaks down when we try to answer the natural follow-up question:

Is ZFC conservative over ZF for projective sentences?

A natural candidate counterexample is "For every uniformly $\bf \Sigma^1_k$ sequence of countable sets of reals $(A_n)_{n\in\omega}$, there is an injection from $\bigcup_{n\in\omega}A_n$ to $\omega$" for some large enough $k$, but that would rely on building a model of ZF + "$2^\omega$ is projectively definably a countable union of countable sets which I don't see how to do. Variations on this idea hit the same problem.

On the other hand, I don't see how to make any nontrivial positive progress; in particular, I can't even resolve the following: is ZFC at least $\Pi^1_3$ conservative over ZF?

EDIT: Actually, unless I'm missing something the naive idea above does show that ZFC is conservative over ZF for $\Pi^1_3$ sentences since we still wind up building for a given model $M$ of ZF a model $N$ of ZFC with the same ordinals as $M$ and with $\mathbb{R}^M\subseteq\mathbb{R}^N$. Now for each $r\in\mathbb{R}^M$ and each $\Sigma^1_2$ formula $\varphi$ we have by Shoenfield that $M\models\varphi(r)$ iff $N\models\varphi(r)$, and this means that $\Pi^1_3$ facts holding in $N$ also hold in $N$.

• In fact, what's really going on is the following. Shoenfield immediately says that if $(*)$ is any sentence in the language of set theory with the property that for every model $M$ of ZF there is some model $N$ of ZF+$(*)$ such that $M$ and $N$ have the same ordinals and $\mathbb{R}^M\subseteq\mathbb{R}^N$, then ZF+$(*)$ and ZF have the same $\Pi^1_3$ consequences. The forcing described above shows that AC has this property (and meanwhile V=L obviously doesn't).

So - unless I've made a silly mistake here - it's at the fourth level of the projective hierarchy that things become nontrivial.

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Every odd number lies between two even numbers. Accordingly we have two categories of consecutive even number pairs; those pairs which surround primes and those pairs which surround odd composites. Some even numbers can belong to both categories as explained in the example below.

E.g: The pair $(8,10)$ will fall in the category of composite since it contains the odd composite number $9$. The pair $(10, 12)$ belongs to the category of primes since they contain the prime $11$. Hence there will be some overlap on the boundaries of primes as is the case with $10$ in this example. As primes thin out, such overlaps will also thin out accordingly.

Data: Experimental data shows that the even numbers which surround a prime have on a average about $28\%$ more divisors and $7\%$ more distinct prime factors than the even numbers which surround odd composites. For numbers up to $3.5 \times 10^7$,

1. The average number of divisors of the even pairs surrounding primes is $35.39$ while that of those which surround odd composite numbers is only $27.70$.
2. Moreover, difference between the average number of distinct prime factors of these two categories seems to converge to a value in the neighborhood of $0.27$

Question 1: How or why does the act of surrounding a prime give the two surrounding even numbers a higher number of divisors and distinct prime factors?

Note: This question was motivated by the following question on twin primes in MSE.

Code

n = 3
pa = pb = ca = cb = 0
ip = ic = 0
target = step = 10^6
while true:
    if is_prime(n) == True:
        ip = ip + 1
        pb = pb + len(divisors(n-1))
        pa = pa + len(divisors(n+1))
    else:
        ic = ic + 1
        cb = cb + len(divisors(n-1))
        ca = ca + len(divisors(n+1))
    if n > target:
        print n, ip, pb, pa, ir, cb, ca, pb/ip.n(), (pb/ip)/(cb/ic).n(), pb/ip.n() - cb/ic.n()
        target = target + step
    n = n + 2

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As discussed at Triangular numbers ($\text{mod } 2^n$) as a permutation of $\{0,1,2,\dots,2^n-1\}$ and What is the set of triangular numbers mod $n$?, mapping the integer $n$ for $0\le n\lt2^k$ to the residue of the corresponding triangular number $\frac12n(n+1)$ modulo $2^k$ yields a permutation. For example, for $k=3$:

$$ 01234567\\ 01362754 $$

I noticed that up to $k=5$, all elements except for $0$ and $1$ (which are always mapped to themselves) form a single cycle of length $2^k-2$. The probability for a uniformly random permutation of length $n$ to consist of a single cycle is $\frac1n$, so if these permutations (excluding $0$ and $1$) could be considered uniformly random, the probability for this to happen would be only $\frac12\cdot\frac16\cdot\frac1{14}\cdot\frac1{30}=\frac1{5040}$. That was reason enough to check whether the pattern continues for greater $k$.

It turns out that it doesn't, as for $k=6$ there is a $3$-cycle: $(4,10,55)$. Nevertheless, at first unusually large cycle lengths persist: For all $k$ from $2$ to $12$, except for $k=7$, the largest cycle contains more than half the elements, whereas the probability for this to happen in a random permutation is roughly $\ln 2$. In fact, in $9$ of these $11$ cases (all except $k=6$ and $k=7$), the largest cycle contains more than $\frac45$ of the elements; the probability for that is roughly $\ln\frac54\approx0.223$ per case, so the probability for it to happen at least $9$ times out of $11$ is only $\sum_{k=9}^{11}\binom{11}k\left(\ln\frac54\right)^k\left(1-\ln\frac54\right)^{11-k}\approx5\cdot10^{-5}$.

However, this pattern, too, doesn't continue: For $k$ from $2$ to $30$, there are $21$ cases with cycles of more than half the elements, which is about the expected number $29\ln2\approx20.1$; and for $k$ from $13$ to $30$ there are only $4$ cases with cycles of more than $\frac45$ of the elements, which is almost exactly the expected number $18\ln\frac54\approx4.0$.

My question is: Is there an explanation for this initial tendency to form long cycles? Or should we put it down to coincidence?

For your convenience, here's the code I used to find the cycle lengths, and here are the results up to $k=30$:

4 : 2
8 : 6
16 : 14
32 : 30
64 : 40, 19, 3
128 : 55, 48, 14, 6, 3
256 : 247, 4, 3
512 : 488, 7, 6, 6, 3
1024 : 818, 157, 47
2048 : 1652, 371, 23
4096 : 4060, 25, 9
8192 : 3754, 3609, 412, 321, 79, 12, 3
16384 : 15748, 292, 190, 71, 24, 22, 13, 13, 9
32768 : 20161, 11349, 333, 305, 281, 218, 72, 44, 3
65536 : 20128, 17231, 16759, 8072, 2377, 579, 295, 60, 33
131072 : 85861, 26603, 9389, 3887, 3365, 682, 594, 488, 118, 49, 23, 6, 5
262144 : 159827, 89991, 5749, 5465, 592, 231, 118, 100, 42, 24, 3
524288 : 211265, 176243, 59029, 35639, 28496, 6122, 4245, 1239, 713, 632, 244, 146, 133, 59, 39, 36, 6
1048576 : 620076, 216520, 131454, 68118, 7535, 2111, 1235, 1028, 225, 213, 36, 20, 3
2097152 : 993084, 583840, 394263, 87941, 31835, 3389, 1648, 459, 306, 273, 45, 35, 14, 10, 8
4194304 : 1487646, 1119526, 942359, 429054, 118037, 64446, 28806, 3238, 323, 291, 186, 126, 118, 102, 12, 11, 10, 7, 4
8388608 : 2542051, 2462220, 2040680, 1138236, 93072, 45880, 19664, 16473, 14243, 6319, 2917, 2598, 2160, 1414, 514, 118, 23, 19, 5
16777216 : 12137774, 4004239, 271250, 253890, 43860, 33597, 25495, 4132, 2575, 157, 116, 67, 35, 9, 8, 6, 4
33554432 : 28169497, 2552414, 1401622, 1019221, 356682, 21006, 14735, 10242, 8223, 566, 135, 45, 21, 15, 6
67108864 : 32223531, 29360424, 3530597, 932310, 809707, 99109, 83093, 67418, 1612, 364, 248, 248, 166, 21, 14
134217728 : 87591110, 34361487, 3360928, 3343185, 3291274, 1345478, 353498, 323522, 158252, 47767, 17776, 11159, 5927, 2681, 2343, 530, 235, 208, 162, 84, 59, 31, 30
268435456 : 232647749, 24918738, 5559122, 3742461, 525140, 384941, 278834, 197080, 62977, 48736, 21684, 16632, 13525, 8993, 3073, 2721, 1625, 1262, 153, 5, 3
536870912 : 379598603, 129063661, 26279056, 665648, 483286, 222289, 137686, 106713, 94323, 80276, 59199, 41767, 15498, 10615, 5066, 2816, 2699, 1579, 113, 10, 7
1073741824 : 877039442, 181409872, 7571387, 6549459, 921247, 240525, 3924, 3416, 1602, 894, 54

from Hot Weekly Questions - Mathematics Stack Exchange

Proportions

Like an iceberg
of which only an eighth is visible --
of death we show only
life . . . 
Read more »

from Intersections -- Poetry with Mathematics

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Problem:

Evaluate the following triple integral: $$ \int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx $$ Answer:
The problem is the bound $\min(8 - x - y,3)$. I would like to write it as the sum of two triple integrals with simple bounds. I could try something like: $$ \int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx = \int_1^2 \int_1^2 \int_1^{3} 2 \, dz \, dy \, dx + \int_2^3 \int_2^3 \int_1^{8 - x - y} 2 \, dz \, dy \, dx $$ but I know that is wrong. What is the right approach to evaluate this integral?

Based upon comments I received, I am updating my post. Using Wolfram, I find: $$\int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx = \frac{47}{3} $$

Using Wolfram, I find: $$ \int_2^3 \int_1^{5-x} \int_1^3 2\,dz\,dy\,dx = 2 $$

Using Wolfram, I find: $$ \int_2^3 \int_{5-x}^3 \int_1^{8-x-y} 2\,dz\,dy\,dx = 4 $$

Since $4 + 2 = 6$ not $\frac{47}{3}$ I am questioning the correctness of the first answer I received.

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Look at the composites between twin primes (A014574):

$$ 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, 138, 150, 180, 192, 198, 228, \\ 240, 270, 282, 312, 348, 420, 432, 462, 522, 570, 600, 618, \ldots \;. $$

Is there anything special about their distribution of factors, number of divisors, or other number-theoretical properties? Or are these twin-prime averages totally "normal" numbers, as far as we know?

from Hot Weekly Questions - Mathematics Stack Exchange

Let be the set of all times, and let be the proposition “She is a woman to me at time .” Translate the logical statement . This matches a line from “She’s Always a Woman” by Billy Joel. Context: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our […]

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We define recursive sequences $a_{n+1}=1+\frac 1{1+a_n}$, $a_1=1$ and $b_{n+1}=\frac{b_n^2+2}{2b_n}$, $b_1=1$. I wish to show that $b_{n+1}=a_{2^n}$.

This can be proven using closed forms expressions related to continued fractions. I know that $a_n$ can be expressed as $$a_n=\sqrt2\cdot \frac{(1+\sqrt 2)^n +(1-\sqrt 2)^n}{(1+\sqrt 2)^n - (1-\sqrt 2)^n}$$

On the other hand, we can prove inductively that $$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2}=\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n}$$

So the relation $a_{2^n}=b_{n+1}$ can be deduced by expanding the fractions. However the computation is rather tedious, I am looking for a proof that does not involve expanding everything into closed form expressions. Thanks.

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Suppose you have an xy plane with two circles both with a radius of one, centered at (-2,2) and (2,2). You have a line segment with one endpoint at (0,0) and forms an angle (∠a) with the x-axis. The line segment continues until it intersects one of the circles and then “bounces off” -that is it forms a new line segment with an endpoint at the intersection, forming an equivalent angle with the tangent of the circle at that point to the angle formed by the original segment and that tangent (as if the segments show the path of light and the circles are mirrors). The new segment extends until it intersects a circle and “bounces off” to form a new segment and so on. For what measure of ∠a are there infinitely many intersections? (The light never stops bouncing).

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See my question at the bottom of this post. The recurrence $P(n) x_{n+2} = Q(n)x_{n+1} - R(n)x_n$, where $P(n), Q(n), R(n)$ are polynomials of degree $1$, sometimes leads to interesting results. Probably the most basic cases are:

For $\log\alpha$:

$$P(n) = \alpha (n+2), Q(n) = (2\alpha-1)(n+1)+\alpha, R(n)=(\alpha-1)(n+1)$$ $$\mbox{with } x_1=\frac{\alpha-1}{\alpha}, x_2 = \frac{(\alpha-1) (3\alpha-1)}{2\alpha^2}$$

We have $\lim_{n\rightarrow\infty} x_n = \log\alpha$. The convergence is fastest when $\alpha$ is close to $1$. The related recurrence $$P(n) = 1, Q(n) = (2\alpha-1)(n+1)+\alpha, R(n)=(\alpha-1)\alpha(n+1)^2$$ $$\mbox{with } x_1=\alpha-1, x_2=(\alpha-1)(3\alpha-1)$$ yields $$\lim_{n\rightarrow\infty} \frac{x_n}{\alpha^n n!} = \log\alpha$$ and in addition $x_n$ is an integer if $\alpha>0$ is an integer.

For $\exp \alpha$:

$$P(n) = n+2, Q(n) = n+2+\alpha, R(n)=\alpha$$ $$\mbox{with } x_0=1, x_1 = 1+\alpha$$

We have $\lim_{n\rightarrow\infty} x_n = \exp\alpha$. The related recurrence $$P(n) = 1, Q(n) = n+2+\alpha, R(n)=\alpha(n+1)$$ $$\mbox{with } x_0=1, x_1=1+\alpha$$ yields $$\lim_{n\rightarrow\infty} \frac{x_n}{n!} = \exp\alpha$$ and in addition $x_n$ is an integer if $\alpha$ is an integer.

For $\sqrt{2}$:

$$P(n) = 4(n+2), Q(n) = 6n+11, R(n)=2n+3$$ $$\mbox{with } x_0=1, x_1 = \frac{5}{4}$$

We have $\lim_{n\rightarrow\infty} x_n = \sqrt{2}$. The related recurrence $$P(n) = 2(n+2), Q(n) = 4(6n+11), R(n)=32(2n+3)$$ $$\mbox{with } x_0=1, x_1=10$$ yields $$\lim_{n\rightarrow\infty} \frac{x_n}{8^n} = \sqrt{2}$$ and in addition $x_n$ is an integer.

Comment

These formulas (and tons of other similar formulas) are easy to obtain, yet I could not find any reference in the literature. It would be interesting to see if one is available for $\gamma$ (the Euler Mascheroni constant), but I don't think so. Also, what happens when you change the initial conditions? What if you replace the recurrence by its equivalent differential equation, for instance $$(x+2) f(x) - (x+2+\alpha) f'(x) + \alpha f''(x) =0$$ corresponding to the case $\exp\alpha$?

How to obtain these recursions?

The case $\sqrt{2}$ can be derived from this other question. To me, it is the most interesting case as it allows you to study the digits of $\sqrt{2}$ in base 2. Some of these recursions can be computed with WolframAlpha, see here for the exponential case, and here for $\sqrt{2}$. Numerous other recurrences, with much faster convergence, can be derived from combinatorial sums featured in this WA article.

My question

I am looking for some literature on these linear, non-homogeneous second order recurrences involving polynomials of degree $1$. Also, I will accept any answer for a recurrence that yields $\pi$. Should be easy, using formulas (37) or (38) in this article as a starting point.

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So I am preparing to go to this olympiad. I received previous years problems and the toughest problem in the definite integral section was this

$$\int_1^a \sqrt[5]{x^5-1}\ dx \ +\ \int_0^b \sqrt[5]{x^5+1}\ dx$$ $$a^5-b^5 = 1$$

I tried substituting the whole root sign in the respective integrals but that led to nowhere. I don't see how trigonometric substitution could be used, dummy variables or the DI method. I am really at a loss here.

Any ideas?

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I had two expression that simplified to the ones in the title.

Obviously, I can't use a calculator.

We didn't learn the double angle (or half angle) formulas so Ill have to find a different way, maybe with the unit circle.

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Let $p_n$ the probability that a random matrix $M\in\mathcal{M}_n(\mathbb{R})$ such that its entries $(a_{i,j})_{1\leqslant i,j\leqslant n}$ are independant and following an uniform distribution over $[-1,1]$, is diagonalizable. I was wondering how to calculate $p_n$ and maybe how to find its limit or an equivalent.

Diagonalization in $\mathbb{C}$ : I proved that $p_n=1$ for all $n\in\mathbb{N}$ if we talk about diagonalization in $\mathbb{C}$ :

Let $$ \Phi_n : \left|\begin{aligned} &\ \ \ \ \ \ \ _ \ \ \ \ \mathbb{R}_{=n}[X] &\longrightarrow &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbb{C} \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P &\longmapsto &\prod_{1\leqslant i<j\leqslant n}\left(\lambda_i(P)-\lambda_j(P)\right) \end{aligned}\right. $$ where $(\lambda_i(P))_{1\leqslant i\leqslant n}$ are the roots of $P$ ordered in the lexicographical order. For any $P\in\mathbb{R}_{=n}[X]$, $\Phi_n(P)$ is, by a factor, the discriminant of $P$ and thus is a polynomial function of the coefficients of $P$. Moreover, $\Phi_n(P)=0$ if and only if $P$ has a multiple root so that $$ p_n \geqslant \mathbb{P}(\Phi_n(\chi_M)\neq 0)=1-\mathbb{P}(\Phi_n(\chi_M)=0) $$ because $M$ is diagonalizable in $\mathbb{C}$ if $\chi_M$ has no multiple root.

Moreover, if we denote $\lambda_n$ the Lebesgue measure on $\mathbb{R}^n$, one can show that for any non-constant $P\in\mathbb{R}[X_1,\ldots,X_n]$, if $$ \zeta(P) := \{ x\in\mathbb{R}^n\ |\ P(x)=0 \} $$ then $\lambda_n(\zeta(P))=0$. We show it by induction on $n$ : if $n=1$ then $\zeta(P)$ is finite so that $\lambda_1(\zeta(P))=0$. If $n\geqslant 2$, we write $$ \zeta(P)=\bigcup_{t\in\mathbb{R}}\zeta(P(\cdot,t)) $$ where $P(\cdot,t):(x_1,\ldots,x_{n-1})\mapsto P(x_1,\ldots,x_{n-1},t)$. By hypothesis $\lambda_{n-1}(\zeta(P(\cdot,t)))=0$ for all $t\in\mathbb{R}$, thus using Fubini's theorem we have $$ \lambda_n(\zeta(P))=\int_{-\infty}^{+\infty}\lambda_{n-1}(\zeta(P(\cdot,t)))dt=0 $$

Finally, since $M\mapsto\Phi_n(\chi_M)$ is a polynomial function of the coefficients of $M$ (because $\Phi_n$ and $M\mapsto\chi_M$ are), the measure of the set $$ \{M\in\mathcal{M}_n(\mathbb{R})\ |\ \Phi_n(\chi_M)=0\} $$ is $0$ and $\mathbb{P}(\Phi_n(\chi_M)=0)=0$ and thus $p_n=1$.

Diagonalization in $\mathbb{R}$ : Because of what said above, for any $M\in\mathcal{M}_n(\mathbb{R})$, $\chi_M$ has no multiple root almost surely so that $$ p_n=\mathbb{P}(\text{Sp}(M)\subset\mathbb{R}) $$ I believe that $\lim\limits_{n\rightarrow +\infty}p_n=0$ but I don't know how to prove it, and even less how to find an equivalent of $p_n$.

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For anyone unfamiliar with multifactorial notation, I will give a quick rundown of it (at least, to the best of my understanding) for non-negative integer values of $n$:

$$n!=n(n-1)(n-2)(n-3)...(n-a), (n-a) > 0$$ $$n!!=n(n-2)(n-4)(n-6)...(n-a), 2 \geq (n-a) > 0$$ $$n!!!=n(n-3)(n-6)(n-9)...(n-a), 3 \geq (n-a) > 0$$

A more generalised expression can be given like so, where $k$ represents the number of factorial symbols:

$$n!^{k}=\left( \begin{cases} 1 & n=0\\ n & 1\leq n\leq k \\ n(n-k)(n-2k)(n-3k)...(n-a) & n>k \end{cases} \right), k \geq (n-a) > 0$$

Now, this is great for when you're working with (primarily) positive integers, but I'm curious how you'd go about extending the definition such that it will be valid for all real and complex numbers. While a general definition for any multifactorial would be amazing, I am primarily just looking for for the definition regarding the triple factorial.

I already know it's possible to do so for the double factorial; $z!!=2^{(1+2z-\cos(\pi z))/4}\pi^{(\cos(\pi z)-1)/4}\Gamma(z/2+1), z \in \mathbb{C}$, so it has me hopeful that it's also possible to define $z!!!$ in a similar such manner.

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Solving Word Problems Involving Two Numbers - Example problems with step by step solution

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Solving Word Problems Involving Consecutive Integers - Concept and examples problems with step by step solution

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Prove without actually evaluating the integrals that:$$2\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx \ \Leftrightarrow \ 2\mathcal I=\mathcal J\tag{*}$$

I am already aware how to evaluate the integrals as we have: $$\mathcal I= \int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx\overset{ x\to \tan \frac{x}{2}}=-2\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-\frac{\pi^3}{8}$$ And the latter integral is evaluated in many ways here so if you have other approaches, please add them there.

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Here's how I came up with $(*)$.
I knew from here that: $$I\left(\frac{3\pi}{2}\right)=\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx=-\frac{3\pi^2}{8}$$ And since this result is very similar to the one from above, I tried to show that $\mathcal I=\frac{\pi}{3} I\left(\frac{3\pi}{2}\right)$.
I also noticed that we have:

$$\mathcal J=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\overset{\pi-x\to x}=\int_0^\frac{\pi}{2}\frac{(\pi-x)\ln(1-\sin x)}{\sin x}dx$$ $$\Rightarrow \mathcal I+\mathcal J=\int_0^\pi \frac{x\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)=-\frac{3\pi^3}{8}$$ Of course now it's trivial to deduce that $2\mathcal I=\mathcal J$ as we know the result for $\mathcal I$, but I'm interest to show that relationship without making use of the result or by calculating any of the integrals. If possible showing $(*)$ using only integral manipulation (elementary tools such as substitution/integration by parts). I hope there's a nice slick way to do it as it will give an easy evaluation of the main integral.

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The holidays are a perfect time to unwind, reflect, and spend time with loved ones. For me, it is also a great time to browse the internet for fun activities to do. In this post, I highlight some of the … Continue reading →

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I've done proofs in discrete mathematics, but I'm still at the stage where proofs with more than a few steps make me uncomfortable.

From Apostol's Mathematical Analysis [2nd Ed.] on page 5, we have

Theorem 1.6. Every pair of integers $a$ and $b$ has a common divisor $d$ of the form $$ d = ax + by $$ where $x$ and $y$ are integers. Moreover, every common divisor of $a$ and $b$ divides this $d$.

The proof (with my questions throughout) goes as follows:

Proof. First assume that $a \geq 0, b \geq 0$ and use induction on $n = a + b$. If $n = 0$ then $a = b = 0$, and we can take $d = 0$ with $x = y = 0$. Assume, then, that the theorem has been proved for $0, 1, 2, ..., n - 1$.

I am a little confused about taking $n$ to be $a + b$, since it's not obvious that all pairs $\{a, b\}$ would be covered by induction for all combinations of $a, b \in \mathbb{Z}$.

By symmetry, we can assume $a \geq b$. If $b = 0$ take $d = a, x = 1, y = 0$.

OK.

If $b \geq 1$ we can apply the induction hypothesis to $a - b$ and $b$, since their sum is $a = n - b \leq n - 1$. Hence there is a common divisor $d$ of $a - b$ and $b$ of the form $d = (a - b)x + by$.

I'm going to let $a' = a - b$, let $b' = b$ and let $d' = a'x + b'y$. (I wish Apostol did something like this to make his proofs clearer.)

I don't understand this logical step. Why does the fact that $a' + b' \leq n - 1$ imply that $d'$ exists and is a common divisor of $a'$ and $b'$? This seems like a huge leap.

This $d$ also divides $(a - b) + b = a$, so $d$ is a common divisor of $a$ and $b$ and we have $d = ax + (y-x)b$, a linear combination of $a$ and $b$.

At this point I am clueless. Why does $d$ divide $a$ and why does this imply it also divides $b$? And where does Apostol get $y-x$ from??

To complete the proof we need to show that every common divisor divides $d$. Since a common divisor divides $a$ and $b$, it also divides the linear combination $ax + (y-x)b = d$. This completes the proof if $a \geq 0$ and $b \geq 0$. If one or both of $a$ and $b$ is negative, apply the result just proved to $|a|$ and $|b|$.

Why not just do the entire proof with absolute values from the beginning?

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Soft question: is it normal for authors to be very terse and not explain or give motivation for any steps? How do you go about trying to understand proofs that require a higher level of intuition than you currently have?

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We are 3 friends; just checked into an airbnb flat with 3 rooms. And to make things truly random we decided to use 1 coin to assign each person a room. What's the best way to do this? (Assuming coin flip is random.) (One way to define 'best' is minimum number of flips)

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Let $ABCD$ be a regular tetrahedron with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-3\overrightarrow{MO}.$ Prove or disprove that $$NA+NB+NC+ND\geq MA+MB+MC+MD$$

I tried to use CS in the Euclidean space $E_3$, but it does not help, because the minoration is too wide.

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I'm looking to evaluate $$\int_0^1 x^{1/x} dx$$ So far, I have that $$ \int_0^1 x^\frac{1}{x} dx \Rightarrow \int_0^1 e^{\frac{lnx}{x}} dx \Rightarrow \int_0^1 \sum_{n \geq 0} \frac{ln^nx}{x^nn!} dx $$ However, I know that $ \int_0^1 \frac{lnx}{x} dx $ diverges, so I can't see a justification for switching the integral and sum. Furthermore, substituting $x = e^\frac{-u}{n+1}$ yields that the above is equivalent to (assuming my calculations were right) $$ \sum_{n\geq0} \frac{(-1)^n}{(1-n)^{1+n}} $$ which is nonsensical. This thought process can be used to solve the "sophomore's dream" integral, which is $\int_0^1 x^{-x} dx = \sum_{n\geq0} n^{-n}$, to which I have two questions:

1. Why can't the same process be used? What changes between the two expressions? I suspect is has something to do with switching the integral and sum, but being a power series for $e^x$ I believe it should be justifiable.
2. If the above is true, what is another approach to take for this integral?

For some context, I have barely cracked into analysis, so if the answer is rudimentary, this is why. However, its very clear that the original integral will have a finite value, and being a cousin of the sophomore's dream, I suspect it could have a solution of the same form.

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I've been fiddling with the recursive sequence defined as follows:

$$\begin{equation} f_n=\begin{cases} a, & n=1.\\ b, & n=2.\\ c, & n=3.\\ f_{n-1}f_{n-2}f_{n-3} \mod[f_{n-1}+f_{n-2}+f_{n-3}], & n>3. \end{cases} \end{equation}$$

And no matter my initial choices of positive integers $a,b,c$, it seems $ \{ f_n \}$ always terminates (three consecutive zeros) or enters a cycle. For instance, if $a=12,b=12,c=9$, then the sequence becomes $12,12,9, 9,12,$ $12\dots$

My question: can we prove (or disprove) that for any positive integers $a,b,c$, the sequence $\{ f_n\}$ will always terminate (three consecutive zeros) or enter a cycle?

Remark: it seems my conjecture is true for the simpler recursive sequence

$$\begin{equation} f_n=\begin{cases} a, & n=1.\\ b, & n=2.\\ f_{n-1}f_{n-2} \mod[f_{n-1}+f_{n-2}], & n>2. \end{cases} \end{equation}$$

Perhaps this would be a better starting point.

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For $c>0$, sample repeatedly and independently from $(0, 1)$ until the sum of the samples exceeds $c$. Let $\mu_c$ be the expected size of the final sample.

For which $c$ is $\mu_c$ maximised?

It is clear that as $c$ tends to $0$, $\mu_c$ tends to $\frac{1}{2}$ and this is its minimum value.

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An exercise in a text I am reading says to find the smallest $n$ such that $\frac{\phi(n)}{n}<\frac{1}{10}$. I checked the first 10 million integers and found none with this property.

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Finding factors of a number - easy way to find factors with examples

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This seems super intuitive, but I can't seem to prove it. Is the set of equivalence classes of totally ordered sets totally ordered?

More precisely, given two totally ordered sets, $F$ and $G$, does there always exist an order preserving injection from one into the other?

I would think there is some adaptation of Zermelo's theorem that can fix the problem, but again, I can't seem to find it.

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A 40 X 40 white square is divided into 1 X 1 squares by lines parallel to its sides. Some of these 1 X 1 squares are coloured red so that each of the 1 X 1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square (not counting itself). What is the largest possible number of red squares?

What I did is as following(R is red, W is white), There are only 400 red squares. The answer should be more. (Sorry, the previous diagram was wrong, I missed every empty White lines. Now I fixed it. ):
RRWWRRWWRRWW...RRWW
WWWWWWWWWWWW...WWWW
WWRRWWRRWWRR...WWRR
WWWWWWWWWWWW...WWWW
RRWWRRWWRRWW...RRWW
WWWWWWWWWWWW...WWWW
WWRRWWRRWWRR...WWRR
...................
RRWWRRWWRRWW...RRWW
WWWWWWWWWWWW...WWWW
WWRRWWRRWWRR...WWRR
WWWWWWWWWWWW...WWWW

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Examples of Adding and Subtracting Rational Expressions - Concept and example problems with step by step explanation

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Length of arc worksheets - Concept formulas and example problems with step by step solution

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Let A be a point on a fixed semicircle with diameter BC. MNPQ is a square such that $M \in AB, N \in AC, P \in BC, Q \in BC$. Let D be the intersection of BN and CM and E be the center of the square. Prove that as A varies, DE always passes through a fixed point.

The fixed point is the midpoint of the semicircle. Any suggestion ?

Edit : I have proved that AE is the angle bisector of angle BAC and AD passes through the midpoint of BC.

I think that DE also passes through the feet of the altitude to BC. May be harmonic bundle is useful. (D,H,E,J) = -1 ?

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