14th Annual FTTS MATH CHALLENGE~2012 Easy Round -1 Find the number of diagonals in a heptagon. Solution: ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐ = = Answer: 14 diagonals ๐โ(๐−3) 2 7โ(4) 2 = 14 ๐๐๐๐๐๐๐๐๐ HS 3&4 P a g e |1 HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e |2 Easy Round -2 In a triangle ABC, the measurement of angle A is 30°, measurement of angle B is 135° and side ๐ = 4√2cm. Find the value of side ๐. Solution: Using the Law of Sine, ๐ sin ๐ด ๐ sin 30° ๐ 1 2 4√2๐๐ = sin 135° = ๐= Answer: 4cm ๐ = sin ๐ต 4√2๐๐ √2 2 4√2๐๐โ √2 2 1 2 = 4๐๐ HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e |3 Easy Round -3 Calculate cot 300 °. Solution: cot 300° = cos 300° , sin 300° cos 300° but since 300° is in the 4th quadrant so sin 300° is negative thus we have, 1 2 cot 300° = − sin 300° = − √3 =− 2 Answer: − √3 3 1 √3 =− √3 3 HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e |4 Easy Round -4 Find x in the figure. A 135ห x C B Solution: Using the Law of Cosine, we have ๐2 = ๐ 2 + ๐ 2 − 2๐๐ โ cos ๐ด (√10)2 = (√2)2 + ๐ฅ 2 − 2(√2)๐ฅ โ cos 135° 10 = 2 + ๐ฅ 2 − 2√2๐ฅ โ (− 1 ) √2 10 = 2 + ๐ฅ 2 + 2๐ฅ 0 = ๐ฅ 2 + 2๐ฅ − 8 0 = (๐ฅ + 4)(๐ฅ − 2) ๐ฅ = −4 ๐ฅ=2 Since the side of a triangle can never be negative then ๐ฅ = 2 Answer: 2 by factoring 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e |5 Easy Round -5 A rectangular piece of paper ABCD 4cm x 1cm is folded along the line MN such that vertex C coincides with vertex A, as shown in the picture. What is the area of quadrilateral ANMD’ A D B C D A C B D’ M A N B Solution: 1 A(ANMD) = 2 ๐ด(๐ด๐ต๐ถ๐ท) 1 A(ANMD) = 2·4·1 = 2๐๐2 Answer : 2 cm2 D C HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e |6 Average Round – 1 The measure of an interior angle in a regular polygon is 108°. How many sides does the polygon have? Solution: ๐๐๐ก๐๐๐๐๐ ๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐ข๐๐๐ ๐๐๐๐ฆ๐๐๐ = (๐−2)โ180° ๐ 108° = (๐−2)โ180° ๐ 108°๐ = (๐ − 2) โ 180° 108°๐ = 180°๐ − 360° 360° = 180°๐ − 108°๐ 360° = 72°๐ 5=๐ Answer: 5 sides 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e |7 Average Round – 2 A cylinder has a radius equal to its height. The total surface area of the cylinder is 100๐๐๐2 . Find its volume. Solution: ๐ก๐๐ก๐๐ ๐ ๐ข๐๐๐๐๐ ๐๐๐๐ = 2๐๐ 2 + 2๐โ โ ๐ 100๐๐๐2 = 2๐๐ 2 + 2๐๐ โ ๐ since โ = ๐ 100๐๐๐2 = 4๐๐ 2 5๐๐ = ๐ = โ ๐๐๐๐ข๐๐ ๐๐ ๐๐ฆ๐๐๐๐๐๐ = ๐๐ 2 โ โ ๐ = ๐ โ 5๐๐2 โ 5๐๐ ๐ = 125๐๐3 ๐ = (125 โ 3.14)๐๐3 ≈ 392.5๐๐3 Answer: 125๐๐๐3 ≈ 392.5๐๐3 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e |8 Average Round – 3 A three-digit number is formed by choosing elements from the set {0, 1, 3, 4, 5, 7, 8, 9}. How many numbers are divisible by four (4) without repetition? Solution: A number is divisible by four (4) if the last two digits is divisible by four. This is sequence of three-digits divisible by four are ๐04, ๐08, ๐12, ๐16, ๐20, ๐24, ๐28, ๐32, ๐36, ๐40, ….The sequence of last digit is 4, 8, and 0 since there is no 2 and 6 in the given sequence If the last digit is four (4), we have โ 6 โโ 1โโ 1=6 {1,3,5,7,8,9} 0 4 โ 5 {1,3,5,7,9} โโ 1โโ 1=5 8 4 If the last digit is eight (8), we have โ 6 โโ 1โโ 1=6 {1,3,4,5,7,9} 0 8 โ 5 {1,3,5,7,9} โโ 1โโ 1=5 4 If the last digit is zero (0), we have 6โ โ 2 โโ 1 = 12 {4,8} 0 Adding all the results, we have 6 + 5 + 6 + 5 + 12 = 34 Answer: 34 ways 8 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 Average Round – 4 How many integer solution does the inequality x3 < 64 < x2 have ? Solution: By solving the inequality separately ๐ฅ 3 <64 64<๐ฅ 2 ๐ฅ 3 -64<0 64-๐ฅ 2 <0 (x-4)(๐ฅ 2 +4x+16)<0 (8-x)(8+x)<0 x<4 -8<x<8 Since the discriminant of ๐ฅ 2 +4x+16 is less than 0, it does not have real solution. We get x<4 and -8<x<8. By combining the two inequalities, we will have -8<x<4. Therefore, the integers from -7 to 3 are solutions. There are 11 integers. Answer : 11 P a g e |9 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 10 Average Round – 5 The square ABCE has side length 4 cm and the same area as the triangle ECD. What is the distance from the point D to the line g? 4 4 Solution: 1 โ ๐ธ๐ถ โ โ { 2 ๐ด(๐ด๐ต๐ถ๐ธ) = 4 · 4 = 16 ๐ด(๐ธ๐ถ๐ท) = 1 โ4โโ =4โ4 2 h = 8cm The distance from D to the line g is 8cm+4cm = 12cm Answer: 12 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 11 Difficult Round – 1 ๐โ๐ =6 Given { ๐ โ ๐ = 15 where ๐, ๐ ๐๐๐ ๐ ∈ Ζ+ . Find the value of a, b, and c. ๐ โ ๐ = 10 Solution: Multiplying the given equations, we have ๐ โ ๐ โ ๐ โ ๐ โ ๐ โ ๐ = 6 โ 15 โ 10 ๐2 โ ๐ 2 โ ๐ 2 = 3 โ 2 โ 3 โ 5 โ 2 โ 5 ๐2 โ ๐ 2 โ ๐ 2 = 22 โ 32 โ 52 we will just take the positive since ๐, ๐ ๐๐๐ ๐ ∈ Ζ+ ๐โ๐โ๐ =2โ3โ5 ๐= ๐๐๐ ๐๐ = 2โ3โ5 3โ5 =2 ๐= ๐๐๐ ๐๐ = 2โ3โ5 2โ5 =3 ๐= ๐๐๐ ๐๐ = 2โ3โ5 2โ3 =5 Answer: ๐ = 2, ๐ = 3, ๐๐๐ ๐ = 5 HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e | 12 Difficult Round – 2 1 Find the constant term in the expansion of (๐ฅ + ๐ฅ 2 )12. Solution: 1 12 Let (๐−1 )(๐ฅ)12−๐+1 (๐ฅ2 )๐−1 be the constant term. 12 (๐−1 ) ๐ฅ 12−๐+1 ๐ฅ 2๐−2 12 12 = (๐−1 )๐ฅ 12−๐+1−2๐+2 = (๐−1 )๐ฅ 15−3๐ If this is the constant term then, the power of ๐ฅ must be zero (0). This gives us 15 − 3๐ = 0 and ๐ = 5. In other words, the constant term is the fifth term. Substituting ๐ = 5 in the expression, we have 1 1 12 (5−1 )(๐ฅ)12−5+1 (๐ฅ2 )5−1 = (12 )(๐ฅ)8 (๐ฅ 2 )4 = (12 ) 4 4 12! (12 ) = 8!โ4! = 4 Answer: 495 12โ11โ10โ9โ8! 8!โ4โ3โ2 = 495 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 13 Difficult Round – 3 Ma’am Wee picks out two numbers a and b from the set {1, 2, 3, …, 26}. The product ab is equal to the sum of the remaining 24 numbers. What is the value of |a-b|? Solution: ๐โ๐ = 26 โ 27 −๐−๐ 2 ๐๐ + ๐ + ๐ = 351 (๐ + 1)(๐ + 1) − 1 = 351 (๐ + 1)(๐ + 1) = 352 We know that a and b must be less than or equal to 26. The only factors of 352 that satisfies the condition are 22 and 16. Let’s assume that, a+1 = 22 and b+1 = 16. So, a is 21 and b is 15. The value of|๐ − ๐| is |21 − 15| = 6 Answer : 6 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 Difficult Round – 4 The last non-zero digit of the number K = 259ฮ34ฮ553 is Solution: K = 259 · 553 โ 34 = (2 โ 5)53 · 26 โ 34 = 1053 โ 64 โ 81 = 5184·1053 Therefore, the first non-zero digit from unit digit of K is 4. Answer: 4 P a g e | 14 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 15 Difficult Round – 5 ` Two sides of a quadrilateral are equal to 1 and 4. One of the diagonals, which is 2 in length, divides it into two isosceles triangles. Then the perimeter of the quadrilateral is equal to: Solution : A 4 D 1 2 B y x C Let ABCD be a quadrilateral. In ΔABC, by triangle inequality, we have 2–1<x<2+1 1<x<3 The length of x must be 2. In ΔACD, 4 – 2 < y < 4 + 2 2 < y <6. The length of y must be 4. So, the perimeter of the quadrilateral ABCD is 4 + 4 + 1 + 2 = 11. Answer: 11 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 16 Tie Breaker – 1 The altitude to the hypotenuse of a right triangle separates the hypotenuse into two segments of 9m and 3m. Find the length of the altitude. Solution: Using the Euclidean Theorem, we have ๐๐๐ก๐๐ก๐ข๐๐ 2 = 9๐ โ 3๐ ๐๐๐ก 2 = 27๐2 ๐๐๐ก๐๐ก๐ข๐๐ = 3√3๐ Answer: 3√3 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 17 Tie Breaker – 2 The sum of the digits of a nine-digit number is 8. What is the product of these digits? Solution: The number must have 0 as its digit(s). Thus, the product of the digit of this number will be zero. Answer: 0 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 Tie Breaker – 3 M and N are the midpoints of the equal sides of an isosceles triangle. The area of the missing (?) quadrilateral piece is: Solution: Since M and N are the midpoint, the triangle is divided into two equal parts. 3+?=3+6 3+?=9 ? =6 Answer : 6 P a g e | 18 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 19 Tie Breaker – 4 The numbers 144 and 220 when divided by the positive integer number x both give a remainder of 11. Find x. Solution: 144 = ๐๐ฅ + 11 { 220 = ๐๐ฅ + 11 ax = 133 = 19·7 bx = 209 = 19·11 The value of x must me 19. Answer : 19 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 Tie Breaker - 5 The maximum natural value n, for which n200<5300 is equal to Solution: ๐200 < 5300 ๐2·100 < 53โ100 ๐ 2 < 53 ๐2 < 125 The maximum natural number n must be 11. Answer : 11 P a g e | 20 HS 3&4 14th Annual FTTS MATH CHALLENGE~2012 P a g e | 21 Solution: DoD – 1 Find the area of the figure. 2 cm 2 cm 2 cm 2 cm 2 cm 2 cm 2 cm 2 cm 2 cm Solution: 2 cm A1 2 cm 2 cm 2 cm A2 2 cm A3 A4 ๐ด1 + ๐ด2 + ๐ด3 + ๐ด4 = ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ข๐๐ 2 cm ๐ด1 = 2๐๐ โ 2๐๐ = 4๐๐2 ๐ด2 = 2๐๐ โ 2๐๐ = 4๐๐2 2 cm ๐ด3 = 10๐๐ โ 2๐๐ = 20๐๐2 ๐ด4 = 4๐๐ โ 2๐๐ = 8๐๐2 ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ข๐๐ = 4๐๐2 + 4๐๐2 + 20๐๐2 + 8๐๐2 ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ข๐๐ = 36๐๐2 Answer: 36๐๐2 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 22 DoD - 2 What is the unit digit of 20122012 ? Solution: 2012 ≡ 2(๐๐๐ 10) 20122 ≡ 4(๐๐๐ 10) 20123 ≡ 8(๐๐๐ 10) 20124 ≡ 6(๐๐๐ 10) 20125 ≡ 2(๐๐๐ 10) 20126 ≡ 4(๐๐๐ 10) 20127 ≡ 8(๐๐๐ 10) 20128 ≡ 6(๐๐๐ 10) โฎ ≡ โฎ (โฎ) We can notice that the unit digit repeats its sequence which 2, 4, 8, and 6 and since 2012 is divisible by 4 then, 20122012 ≡ 6 (๐๐๐ 10) Answer: 6 14th Annual FTTS MATH CHALLENGE~2012 HS 3&4 P a g e | 23