Pizza Problem
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1. The document discusses the "pizza problem" of dividing a circle into slices using chords and determining if the black and white areas are equal. 2. It introduces the Tiffany Lemma, which shows that the sum of the squares of the lengths of four chords through a point is a constant equal to 4 times the radius squared. 3. By using polar coordinates and the Tiffany Lemma, the solution shows that the integral of the black and white areas over any interval of length π/2 is π/2, proving the areas are equal.
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![implies OM A ∼ CN O. But OA = CO is the radii, so the similarity is in
fact congruence (AAS), which is the desired result.
We then have the following equality as a corollary.
AE 2 + BE 2 + CE 2 + DE 2 = 4R2
where R is the radius of the circle.
The Pythagorean Theorem implies that AE 2 + BE 2 = AB 2 and CE 2 +
DE 2 = CD2 . Applying the Tiffany Lemma, we have
AE 2 + BE 2 + CE 2 + DE 2 = AB 2 + CD2
= (2ON )2 + (2CN )2
= 4(ON 2 + CN 2 )
= 4OC 2
= 4R2
Therefore, the sum AE 2 + BE 2 + CE 2 + DE 2 is a constant which equals four
times the square of the radius.
3 The Solution
Without loss of generality, suppose that the circle has radius of unit length and
that its interior contains the origin. Using polar coordinates, parametrize the
circle by r(θ), where θ ∈ R. Define the four chords by θ = kπ , where k ∈ N.
4
By the Tiffany Lemma, we have
π 3π
r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 = 4
2 2
and so the required area is
π π
1 4 π 3π 1 4
[r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 ]dθ = 4dθ
2 0 2 2 2 0
which equals, obviously, π . QED.
2
In general, the integral can be taken over any measurable subset A of the
interval [0, π ] such that m(A) = π .
2 4
2](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Anderson Emath Paper2_printed
(1) The document is an examination paper for Secondary 4/5 students in mathematics. It consists of 13 printed pages containing 11 questions testing various math concepts.
(2) Instructions are provided for candidates, including writing their name, working clearly, using calculators where appropriate, and expressing some answers to a given degree of accuracy or in terms of pi.
(3) The exam covers topics like algebra, trigonometry, geometry, calculus, statistics, and financial mathematics. Questions involve factorizing expressions, solving equations, using circle properties, graphing functions, and probability.
Inmo 2013 test_paper_solution
This document provides instructions for the 28th Indian National Mathematical Olympiad exam to be held on February 03, 2013. It states that calculators and protractors are not allowed, but rulers and compasses are. It includes 6 multi-part math problems to be solved on separate pages with clear numbering. The problems cover topics like properties of circles touching externally, positive integer solutions to equations, properties of polynomial equations, subsets with integer mean averages, relationships between areas of triangles formed by triangle centers, and inequalities relating positive real numbers.
2sol
1) The document contains 6 multi-step math problems and their solutions.
2) The problems involve concepts like trigonometry, inequalities, divisibility, and proportions in geometric figures.
3) The solutions use techniques like cyclic quadrilaterals, similarity of triangles, and algebraic manipulation of equations.
Monfort Emath Paper2_printed
The document is the question paper for a Secondary 4 mathematics examination consisting of 11 questions testing topics including algebra, geometry, trigonometry, statistics, and sequences and series. The exam has a maximum score of 100 marks and covers areas such as simplifying expressions, solving equations, calculating lengths, areas, volumes, probabilities, and interpreting graphs. Candidates are instructed to show working, use calculators appropriately, and express answers to a given degree of accuracy.
Circle 10 STB.pptx
1. The document defines various terms related to circles such as chord, diameter, secant, tangent, and tangent segment. It also states some properties of these terms like a secant containing a chord and a tangent segment joining any point on a tangent to the point of contact.
2. Several theorems related to circles are presented, such as a tangent being perpendicular to the radius at the point of contact, and the lengths of two tangent segments drawn from an external point to a circle being equal.
3. Proofs of some circle theorems are shown, for example proving that a line perpendicular to the radius at its outer end is a tangent to the circle.
Geometry s
The document contains 10 geometry problems with solutions. Problem 1 asks about the total number of sides of seven polygons given the sum of their interior angles. Problem 2 asks about the shaded area of an infinitely repeating pattern of equilateral triangles. Problem 3 asks about the ratio of the diagonal to the side of a regular pentagon.
10 Mathematics Standard.pdf
This document provides instructions for a mathematics exam for Class X. It has the following key details:
- The exam has 4 sections (A, B, C, D) with a total of 40 questions. All questions are compulsory.
- Section A has 20 one-mark multiple choice questions. Section B has 6 two-mark questions. Section C has 8 three-mark questions. Section D has 6 four-mark questions.
- There is no overall choice but some questions provide an internal choice between alternatives. Students must attempt only one of the choices for those questions.
- Calculators are not permitted. The instructions provide details about the number and type of questions in each section and remind students
Final
The document summarizes research on finding polychromatic solutions to linear equations in an r-bounded coloring of the natural numbers. The research builds on previous work exploring rainbow analogues of classical Ramsey theory results. Key findings include:
1) Theorems were proven showing that for equations of the form ax-by=c with integers a,b,c and gcd(a,b)|c, there exists a polychromatic solution in any r-bounded coloring.
2) Explicit recurrence relations and formulas were defined to generate chains of solutions to equations with two variables.
3) The results for two variables were used to show polychromatic solutions exist for equations with three variables of the form a1
Catalan number edit
This document discusses Catalan numbers, which are a sequence of integers that appear in many counting problems. It provides:
1) An introduction and the formula for Catalan numbers in terms of binomial coefficients.
2) A definition of Catalan numbers using totally balanced sequences and an example of the first few Catalan families and numbers.
3) A proof of the formula for Catalan numbers using Euler's formula for triangulating polygons.
Prove it!
1. No matter where a point P is placed on a circle, the angle subtended at the center (α) is always 90 degrees.
2. If AB is a chord of a circle with center O, the angle subtended by the arcs APB is always equal to the angle subtended by the arcs AOB.
3. If AB and CD are chords of equal length in a circle, then the perpendicular distances of their midpoints from the center (d1 and d2) are also equal.
Cat tep0511 paper
This document contains a test with 26 multiple choice questions covering topics in mathematics, geometry, operations, and data interpretation from graphs and tables. The questions range in complexity and cover finding ratios, remainders, number of solutions to equations, rates, paths on grids, areas after operations, and interpreting data from tables and graphs.
A Proof of Twin primes and Golbach's Conjecture
The document discusses several theorems related to twin prime conjectures:
- Theorem 1 states that a prime p can be written in the form 3k+1 or 3k-1, with k being even.
- Theorem 4 characterizes twin primes as pairs where n(n+2) satisfies a modular condition.
- Theorems aim to prove there are infinitely many twin primes, relying on the ratio of primes increasing without bound as n increases without bound.
BukitPanjang GovtHigh Emath Paper2_printed
This document contains instructions and questions for a mathematics preliminary examination. It consists of 7 questions testing skills in algebra, trigonometry, geometry, statistics, and problem solving. Students are instructed to show their working, use formulas provided, and give answers to a specified degree of accuracy. A total of 100 marks are available across the exam.
IOQM_22_finalversion_for_publish.pdf
The document contains 24 math word problems. It asks to find specific values related to the configurations, equations and inequalities described in each problem. The problems cover a range of mathematical topics including geometry, number theory, sequences, permutations and combinations.
Problems and solutions inmo-2012
The document presents solutions to 6 problems:
1) Finding the maximum area of a quadrilateral inscribed in a circle.
2) Proving that 30 divides the difference of two prime number sets with differences of 8.
3) Showing polynomials with integer coefficients satisfy a recursive relation.
4) Determining the number of points dividing a triangle into equal area triangles.
5) Proving an acute triangle is equilateral given conditions on angle bisectors and altitudes.
6) Characterizing a function on integers satisfying two properties.
Peperiksaan pertengahan tahun t4 2012 (2)
This document contains 12 mathematics questions testing skills such as solving simultaneous linear equations, quadratic equations, calculating areas and perimeters of shapes, set theory, and logical reasoning. The questions cover topics like functions, sequences, proportions, geometry, and Venn diagrams. Students are required to show their work and provide answers for full marks.
Imo2012 sl
This document contains solutions to 7 problems in algebra from the 53rd International Mathematical Olympiad held in 2012. The first problem asks to find all functions f from integers to integers that satisfy a given quadratic equation, and its solution shows that the only such functions are f(x) = kx^2 for some integer k.
Similar to Pizza Problem (20)
More from HC Wong
忘盡背後 一世不離棄
對很多人來說，畢業是人生中一個很重要的里程碑。當子良老師提出要在畢業晚宴上演出時，我便著意將《忘盡背後》重新編曲，以鼓勵在場的每一位同學。盼望大家在畢業後能忘記背後，努力面前的，為著上主所賜的目標而努力。
在這首樂曲中，我刻意地引用了德伏扎克 E 小調第 9 交響曲的第 2 樂章，並重新譜上中文歌詞。這樣的編排有著雙重的意義：其一、這是敝校的校鐘，是多年來天天伴隨著我們成長的集體回憶；其二、喻意同學們畢業後將踏進「新的世界」，或是在本地升學，或是到外地進修，都不忘在這裡所學習到的一切。作為 12 年級的班主任，能看見同學們的成長、更新與改變，實在感恩。但願大家都能一生信靠主，一世不離棄。
Uiit itce portfolio submission sheet
1. This document is a portfolio submission sheet for an ITCE (Information Technology for Continuous Education) program submitted by Wong Hang Chi, a student in the PGDE(S) Mathematics program at HKIEd with a student number of 09451331 and expected graduation date of July 2010.
2. The sheet includes a declaration that the three required ITCE portfolio items have been completed at the Upper Intermediate Level, including a project-based learning proposal, reflection on changes to teaching philosophy/subject due to IT, and a conclusion on the pros and cons of integrating ICT in education in Hong Kong.
3. Wong Hang Chi signs the sheet to declare the work as his own and the instructor,
Uiit ai3
1) ICT tools can enhance the effectiveness of teaching by allowing teachers to demonstrate concepts like graphs and functions in new ways. However, it also increases the pressure on teachers who must acquire new skills and ensure appropriate use of technologies.
2) While ICT helps students learn independently, it risks exacerbating inequality if not all students have equal access. Weaker students especially benefit from tools like graphing calculators, but quality of access may differ by individual backgrounds.
3) Schools must be aware of both the advantages of ICT in modernizing education, as well as the disadvantages like potential increases in differences between strong and weak students due to unequal access, in order to develop policies that mitigate risks.
Uiit ai2
The document discusses two approaches to teaching students about the area of a circle - using ICT tools or a traditional non-ICT method. Using ICT tools like Mathematica Player allows the teacher to demonstrate incrementally increasing the number of sectors in a circle to show the limiting case of a rectangle, which helps students understand that the area is πr2. The traditional method involves students physically cutting a paper circle into sectors but is more time consuming and limited. While ICT tools save time and help demonstrate the limiting concept, the traditional approach allows students to physically manipulate the materials.
Uiit ai1
The document proposes a project-based learning activity to teach students about Fermat's problem for Torricelli. Students will form groups and use dynamic geometry software to investigate Viviani's theorem. They will then perform an experiment involving drilling holes in acrylic boards and attaching strings to masses to measure angles. Finally, students will research the Fermat point online and present their findings to the class. The teacher's role is to facilitate learning through questioning rather than simply providing answers. ICT tools like dynamic geometry software and videos are used to enhance collaboration and make the activities more interactive.
10 Logical Reasoning
This document discusses logical reasoning and provides 3 examples of problems solved through logical reasoning. Logical reasoning involves breaking down problems into logical steps and using techniques like deduction and contradiction to arrive at solutions. The first example asks if 44 silver dollars can be distributed into 10 pockets with each pocket containing a different number of dollars. The second example asks if a notched checkerboard with squares removed can still be covered with the remaining dominoes. The third example asks if the equation x2 + y2 = 3(u2 + v2) has any integral solutions other than (0, 0, 0, 0). The document concludes by listing references on problem solving strategies and mathematical methods.
Epa5075
The document discusses the trial scheme on school drug testing in Tai Po District from three sociological perspectives - structural functionalism, social conflict theory, and symbolic interactionism. From the structural functionalist view, the scheme aims to fulfill the social functions of the education system and narcotics division by deterring drug use and motivating abusers to seek help. However, it also creates social dysfunctions like increased public tension and labeling effects. The social conflict perspective sees the scheme as imposing government authority and creating social inequality between authorities and students. Finally, the symbolic interactionism perspective analyzes how the government constructs positive perceptions of the scheme through explanations, media coverage, and protecting students' self-concepts and esteem.
教大理想
This document is a 3-part choral work titled "Hong Kong Youth" arranged for SATB voices with piano accompaniment. It was written by Chong Wing Fat, Kai Yi Ni, and Chan King Ho, and arranged by Wong Hang Chi. The choral work has an allegro tempo of 120 beats per minute and is in 3/4 time. It features the soprano, alto, tenor and bass vocal parts, with the piano providing harmonic and rhythmic support.
複習（二元一次方程及百分法）
The document contains several math equations, including two systems of linear equations and two other equations involving variables a and x and y. It also includes some music and financial examples, such as discussing the Well Tempered Clavier, frequency of notes, interest rates calculations on $4000 and $68,000 amounts. Overall it seems to be presenting various examples from math, music and finance to illustrate concepts.
複習（特殊四邊形）
This document discusses a proof in geometry about angles in a triangle. It shows that the angle between a line parallel to one side of a triangle and another side is equal to the opposite angle in the triangle. It provides two steps to prove this using properties of parallel lines and the angle sum of a triangle being 180 degrees.
與平行四邊形有關的簡單證明
El documento presenta una demostración geométrica de que si AB = CD y AD//BC, entonces BD = CA. Primero establece que si AB = CD y AD//BC, entonces EBC ~ EAD y BE = CE. Luego, usando propiedades de triángulos similares y congruentes, deduce que BD = CA.
平行四邊形的判定條件
El documento describe un triángulo ABCD donde se dan los ángulos A = 63° y D = 117° y las longitudes BC = AD. Se deduce que la suma de los ángulos internos es 180° y que los lados AB y CD son paralelos, por lo que ABCD es un paralelogramo.
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We will explore the capabilities of AI in understanding XML markup languages and autonomously creating structured XML content. Additionally, we will examine the capacity of AI to enrich plain text with appropriate XML markup. Practical examples and methodological guidelines will be provided to elucidate how AI can be effectively prompted to interpret and generate accurate XML markup.
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Monitoring and observability aren’t traditionally found in software curriculums and many of us cobble this knowledge together from whatever vendor or ecosystem we were first introduced to and whatever is a part of your current company’s observability stack.
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Pizza Problem
- 1. Pizza Problem Wong Hang Chi January 19, 2010 1 The Problem Draw three chords through an arbitrary point inside a circle so that they make six 60◦ angles at the point and color the resulting “pizza slices” alternately black and white. Is the black area necessarily equal to the white area? What if four chords are used, making eight 45◦ angles? 2 Tiffany Lemma In the figure, AC and BD are two chords of the circle such that AC ⊥ BD. M and N are the midpoints of AB and CD respectively. Show that 2ON = AB C . . N − . . . − B ... . .. D . . E .. . . .. .. . . . .. = . . . . . . .. ... . . O ... . .. . M . ... . . = .. ... .. .. .. A It suffices to show that OM A ∼ CN O. = Firstly, the congruence OM A ∼ OM B is established by OA = OB, = OM = OM and AM = BM (SSS). Similarly, we have CN O ∼ DN O. Let = x = AOM = BOM . Since AOB = 2x is an angle at center and ADB is an angle at circumference, we have 2x = 2 ADB, and so ADB = x. Secondly, OM A ∼ DEA because ADB = x = AOM and OM A = DEA = 90◦ (equiangular). Similarly, we have CN O ∼ DEA, which 1
- 2. implies OM A ∼ CN O. But OA = CO is the radii, so the similarity is in fact congruence (AAS), which is the desired result. We then have the following equality as a corollary. AE 2 + BE 2 + CE 2 + DE 2 = 4R2 where R is the radius of the circle. The Pythagorean Theorem implies that AE 2 + BE 2 = AB 2 and CE 2 + DE 2 = CD2 . Applying the Tiffany Lemma, we have AE 2 + BE 2 + CE 2 + DE 2 = AB 2 + CD2 = (2ON )2 + (2CN )2 = 4(ON 2 + CN 2 ) = 4OC 2 = 4R2 Therefore, the sum AE 2 + BE 2 + CE 2 + DE 2 is a constant which equals four times the square of the radius. 3 The Solution Without loss of generality, suppose that the circle has radius of unit length and that its interior contains the origin. Using polar coordinates, parametrize the circle by r(θ), where θ ∈ R. Define the four chords by θ = kπ , where k ∈ N. 4 By the Tiffany Lemma, we have π 3π r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 = 4 2 2 and so the required area is π π 1 4 π 3π 1 4 [r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 ]dθ = 4dθ 2 0 2 2 2 0 which equals, obviously, π . QED. 2 In general, the integral can be taken over any measurable subset A of the interval [0, π ] such that m(A) = π . 2 4 2