Be Mathusay!: October 2019

Wednesday, October 9, 2019

MATHEMATICS IN THE MODERN WORLD (POWERPOINT LESSONS)

Course Syllabus:

De Lasalle MMW

CHED MMW

Week 1: Mathematics in our World

Week 2: Language in Mathematics

Week 3: Math Logic

Week 4: Problem Solving and Reasoning

Week 5: Mathematics as a Practical Tool

Sunday, October 6, 2019

Math eBooks 10

Math eBooks 10

This zip file contains the following:
1. Mathematical Olympiad Treasures - Andreescu & Enescu
2. Solved Problems in Analysis as Applied to Gamma, Beta, Legendre, Bessel Functions - Farrell O., Ross B
3. Solving Mathematical Problems - Terence Tao

Math eBooks 9

Math eBooks 9

This zip file contains the following:
1. Mathematical Olympiad in China (2007-2008)
2. Mathematical Olympiad in China, Problems and Solutions (2002-2006) - Xiong Bin & Lee Peng Yee

Math eBooks 8

Math eBooks 8

This zip file contains the following:
1. Junior Balkan Mathematical Olympiads - Branzei, Serdean & Serdean
2. Mathematical Diamonds - Ross Honsberger
3. Polynomials - E. J. Barbeau

Math eBooks 7

Math eBooks 7

This zip file contains the following:
1. Kvant Selecta (Algebra and Analysis I) - Serge Tabachnikov
2. Kvant Selecta (Combinatorics I) - Serge Tabachnikov

Math eBooks 6

Math eBooks 6

This zip file contains the following:
1. Challenging Mathematical Problems with Elementary Solutions, Volume 1 (Combinatorial Analysis & Probability Theory) - Yaglom & Yaglom
2. Challenging Mathematical Problems with Elementary Solutions, Volume 2 (Problems from Various Branches of Mathematics) - Yaglom & Yaglom
3. From Erdos to Kiev - Problems of Olympiad Caliber - Ross Honsberger

Math eBooks 5

Math eBooks 5

This zip file contains the following:
1. Contests Around the World - Titu Andreescu
2. A Primer for Mathematics Competitions - Zawaira & Hitchcock
3. Ants, Bikes, and Clocks Problem Solving for Undergraduates - William Briggs
4. Australian Mathematical Olympiads(1979-1995) - Lausch & Taylor

Source: Math Enthusiastic Quiz Group

Math eBooks 4

Math eBooks 4

This zip file contains the following:
1. GeneratingFunctionology - Herbert Wilf
2. Mathematical Olympiad Challenges 2nd ed - Titu Andreescu & Razvan Gelca
3. Problems & Solutions in Euclidean Geometry - Aref & Wernick
4. Professor Stewart's Cabinet of Mathematical Curiosities - Ian Stewart
5. The Contest Problem Book 2 - Charles Salkind
6. The IMO Compendium - A Collection of Problems Suggested for the IMO(1959 - 2004) - Djukic, Jankovic, Matic & Petrovic
7. The Complete Book of Number System
8. USA Mathematical Olympiads (1972-1986) - Murray Klamkin

Source: Math Enthusiastic Quiz Group

Math eBooks 3

This zip file contains the following:

1. Chinese Mathematics Competitions and Olympiads (1981-1993) - Andy Liu

2. Chinese Mathematics Competitions and Olympiads (1993-2001) - Andy Liu

3. Combinatorics, A Problem Oriented Approach - Daniel Marcus

4. Geometric Problems On Maxima And Minima - Andreescu, Mushkarov & Stoyanov

5. Lecture Notes on Mathematical Olympiad Courses, Volume 1 - Xu Jiagu

6. Lecture Notes on Mathematical Olympiad Courses, Volume 2 - Xu Jiagu

Math eBooks 2

Math ebooks 2

This zip file contains the following:
1. Aha! Solutions - Martin Erickson
2. Discrete Mathematics - Kevin Ferland

Math eBooks 1

Math Ebooks1

This zip file contains the following:
1. 101 Problems in Algebra - Titu Andreescu & Zuming Feng
2. 102 Combinatorial Problems from the Training of the USA IMO Team - Andreescu, Feng
3. 103 Trigonometry Problems from the Training of the USA IMO Team - Titu Andreescu & Zuming Feng
4. 104 Number Theory Problems. From the Training of the USA IMO Team - Andreescu, Andrica & Feng
5. A Tour of Triangle Geometry - Paul Yiu
6. An Introduction to Diophantine Equations - Andreescu, Andrica, Cucurezeanu
7. Challenges in Geometry for Mathematical Olympians Past and Present - Christopher Bradley
8. Combinatorics, A Problem Oriented Approach - Daniel Marcus

Saturday, October 5, 2019

Compilation of MTAP Reviewers (Grade 1-Grade 10)

Grade1-10:
2019 MMC with answer key

Grade 1
MTAP 2014-2017
2018 Division Finals-Team Orals

Grade 2:
2017 MMC Divisions
2018 Division Finals-Team Orals

Grade 3:
MTAP 2014-2017
2018 Division Finals-Team Orals

Grade 4:
MTAP 2016-2017
2018 Division Finals-Team Orals

Grade 5:
MTAP 2016-2017
2019 Eliminations
2018 Division Finals-Team Orals

Grade 6:
2015 Division Team Orals
MTAP 2016-2017
2017 Division Team Orals
2018 Division Eliminations
2018 Division Finals-Team Orals
2019 Eliminations
2019 National Individual Orals
2019 National Team Orals
MMC 2018 Math Solutions

Grade 7:
Division Orals
2015 Division Eliminations
2017 Division Eliminations CatB
2018 MMC
2019 Eliminations

Grade 8:
Team Orals
2015 MTAP Elimination
2015 Division Eliminations
System of Linear Equations and Inequalities
2017 Division Eliminations CatB
2018 MMC
2019 Eliminations

Grade 9:
2015 Division Eliminations
2017 Division Eliminations CatB
2018 MMC
2019 Eliminations

Grade 10:
2015 Division Elimination and Team Orals
2016 Division Eliminations
2016 Team Orals
2017 MMC CatA
2017 MMC CatB
2017 Team Orals
2017 MMC Regional Individuals
MTAP Solutions
2 018 Division Eliminations
2018 National- Individual Orals
2018 National-Team Orals
2019 Eliminations
2019 National Individual Orals
2019 National Team Orals

Grade7-Grade10:
2016 MMC Regionals B
2017 MMC Sectorals
Sipnayan 2017
Training Module 1
Training Module 2
2018 National Finals

For more MTAP Reviewers, click here.

MTAP Review for Grade 7

MTAP Grade 9

MTAP Grade 8

MTAP Grade 7

Grade 10 MTAP

MTAP Playlist Review

https://m.youtube.com/playlist?list=PLCVVUSE8iFbC7o-6goz-6BMgvL-nSqRr_

MTAP Grade 3- Grade 10 Video Tutorials by Engr. Maling

MTAP Division Orals Video Tutorials by Engr. Maling

2014 MTAP Division Orals Grade 7

2016 MTAP Division Orals Grade 10
j

2018 MTAP Grade 10 Division Orals

2018 MTAP Grade 10 Division Orals Solutions

2018 GRADE 10 DIVISION ORALS SOLUTIONS:

15-second question

1.) Difference = 4-1 = 3. Thus, 4 + 2(3) = 10

2.) x² + x - 12 = 0. (x+4)(x-3) = 0. Thus, roots are -4 and 3

3.) 4! = (4)(3)(2)(1) = 24

4.) 10! / 7! = (10)(9)(8) = 720

5.) 2/4 = 3/x. Thus, x = 6

6.) √(2² + 3²) = √13

7.) 3/15 or 1/5

8.) sin²A + cos²A = 1
9/49 + cos²A = 1
cos²A = 40/49
cosA = 2√10 / 7

9.) x-1 = 0, x = 1
(1)² + 1 + 1 = 3

10.) Ratio = cbrt(24/3) = cbrt(8) = 2
3 * 2² = 12

11.) Discriminant should be zero
b² - 4ac = 0
81 - 4k = 0
k = 81/4 or 20.25

30-second question

1.) Ratio = 7/3 / 7/2 = 2/3
Sum = a_1 / (1-r)
Sum = 7/2 / (1/3) = (7/2)(3) = 21/2 or 10.5

2.) x+4 = 0, x = -4
2(-4)³ + 6(-4)² + 5(-4) + 2
-128 + 96 - 20 + 2 = -50

3.) 7C3 * 5C2
(7! / 4!3!)(5! / 3!2!) = (35)(10) = 350

4.) radius² = 87 + (14/2)² + (16/2)²
radius² = 87+49+64 = 200
radius = 10√2

5.) Arc AD + Arc CB = 2 * Angle AED
55° + Arc CB = 2(40°)
Arc CB = 25°

6.) Difference = 2.5
Sum = (a_1 + a_n)(n/2)
a_n = a_1 + (n-1)d
Sum = (a_1 + a_1 + 2.5(n-1))(n/2)
374 = (2 + 2 + 2.5(n-1))(n/2)
748 = 2.5n² + 1.5n
5n² + 3n - 1496 = 0
(5n+88)(n-17) = 0
Thus, answer is 17

1-minute question

1.) Distance of (6,3) and (12,-7) is √136, which is also the diameter
Midpoint is (9,-2), which is also the center
Note that new circle has twice the radius of the old, thus, the diameter of the old is the radius of the new.
Thus, (x-9)² + (y+2)² = 136

2.) By synthetic division trial and error, roots are -1, ±2, 3

3.) Let x be the first term of arithmetic
Let x+3 be the second term of arithmetic
Let x+6 be the third term of arithmetic
Let x-2 be the first term of geometric
Let 2x+6 be the second term of geometric
Let 5x+30 be the third term of geometric

(2x+6) / (x-2) = (5x+30) / (2x+6)
(2x+6)² = (5x+30)(x-2)
4x² + 24x + 36 = 5x² + 20x - 60
x² - 4x - 96 = 0
(x-12)(x+8) = 0, x = 12
Thus, 12,15,18

4.) Since Arc BC = 120°, meaning, angle BDC and angle CAB are 60° , and since arc AD is 60°, meaning, angle ACD and angle ABD is 30°. We will notice that the center are 90°, and this forms 4 right triangles with hypotenuse as the sides of the quadrilateral.
Let E be the point of intersection, meaning, by 30-60-90 triangle, BE is 6√3, AE is 6, CE is 9√3, and DE is 9.
BD = BE + DE = 9 + 6√3
AC = AE + CE = 6 + 9√3
By observation, we will notice that 9 + 6√3 is shorter.

5.) Maximum sum is 4*6 = 24. Question is "at least 23", thus, 23 and 24.
To get a sum of 23, the dice must be (6,6,6,5). Thus, it can be arranged in 4!/3! ways or 4 ways
To get a sum of 24, there is only 1 possible way, and that is if all dices show 6.
Thus, (4+1) / 6⁴ = 5 / 1296

6.) a = 1 and c = -8
-8 can be 4 ways:
(8)(-1) , (-8)(1) , (4)(-2) , (-4)(2).
If (8)(-1) , b = 7
If (-8)(1) , b = -7
If (4)(-2) , b = 2
If (-4)(2) , b = -2
By trial and error, b = 2

Clincher question

1.) (4² + 1) - (3² + 1) = 7

2.) 2(4x - 10) = 6x + 10
4x - 10 = 3x + 5
x = 15

3.) Since P(1) = P(2) = P(3) = P(4) = 0, then, they are zeros of the function. Thus, P(x) = a(x-1)(x-2)(x-3)(x-4)
Solve for a by substituting 5
P(5) = a(4)(3)(2)(1)
120 = 24a , thus, a = 5
Substitute 6 to get P(6)
P(6) = 5(5)(4)(3)(2)(1)
P(6) = (5)(5!) = (5)(120) = 600

Do-or-Die question

Notice that (9)(2018)² = (3²)(2018²) = (3*2018)² = 6054²
Last number of 6054² is 6. Thus, the integer must be ending in 4 or 6. By trial and error, you will get 6056

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