Math IOQM

Greatest Integer Function properties: 1. x-1 < [x] <= x < [x] + 1 2. for any integer 'n', [n+x] = n + [x] 3. [-x] 3a.  [-x] = -[x] - 1 if x is not integer 3b. [-x] = -[x] if x is integer 4. [x + y] >= [x] + [y] 4a. In general: [x + y + z ... -w] >= [x] + [y] + [z] -[w] 5. [xy] >= [x][y] and generalize like above Ex1: Solve for x. 2[x] = x + 2{x} Where [] => greatest integer function and {} = fractional part function. Solution: Let x = p + q where p = [x] and q = {x} where 0 <= q < 1. 2p = p + q + 2q = p + 3q => p = 3q Only 3 values of q can give an integer =>  q = 0 => p = 0 q = 1/3 => p =1 q = 2/3 => p = 2 So 3 solutions: 0+0, 1 + 1/3, 2 + 2/3 x = 0,4/3,8/3 Ex2: 36{x}^2 = 6[x]x Solution: Let x = p + q where p = [x] and q = {x} where 0 <= q < 1. => 36q^2 = 6p(p + q) => 6q^2 = p^2 + pq => p^2 + pq - 6q^2 = 0 Factorize to get: p^2 + 3pq - 2pq - 6q^2 = 0 p(p + 3q) -2q(p + 3q) = 0 => (p - 2q)(p + 3q) =0 p = 2q or p = -3q ...

  Statement of the identity cot ⁡  ⁣ ( θ 2 ) = 1 + cos ⁡ θ sin ⁡ θ \boxed{\displaystyle \cot\!\left(\tfrac{\theta}{2}\right)=\frac{1+\cos\theta}{\sin\theta}} This relates the cotangent of half an angle to the sine and cosine of the whole angle (with θ ≢ 2 k π \theta\not\equiv 2k\pi so sin ⁡ θ ≠ 0 \sin\theta\neq0 ). Proof using double-angle formulas Start from double-angle expressions sin ⁡ θ    =    2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 ) , 1 + cos ⁡ θ    =    2 cos ⁡ 2  ⁣ ( θ 2 ) . \sin\theta \;=\; 2\sin\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr), \quad 1+\cos\theta \;=\; 2\cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr). Form the ratio 1 + cos ⁡ θ sin ⁡ θ \dfrac{1+\cos\theta}{\sin\theta} 1 + cos ⁡ θ sin ⁡ θ    =    2 cos ⁡ 2  ⁣ ( θ 2 ) 2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 )    =    2 cos ⁡ 2  ⁣ ( θ 2 ) 2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 )    =    cos ⁡  ⁣ ( θ 2 ) sin ⁡  ⁣ ( θ 2 ) . \frac{1+\cos\theta}{\sin\theta} \;=\; \frac{2\cos^{2}\!\bigl(\tfrac{\theta}{2}...

 In a right triangle if inradius  = r, hypotenuse  = c and legs are a,b then:  r = (a + b - c)/2 Proof: From the center of the incircle, the perpendiculars are dropped on the 2 legs (a,b). It creates a square and splits the base in 2 parts r, a-r and similarly the other leg is split into r,b-r. Since 2 tangents from one point are equal => c = a - r + b - r => r = (a + b - c)/2 Circumradius of a right triangle = c/2 where c is hypotenuse. Proof: Easy by co-ordinate geometry. A = (0,0) B = (a,0) C = (0,b) Midpoint of BC = a/2,b/2 OA = OB = OC = sqrt(a^2/4 + b^2/4) => O is the circumcenter and OA = Hypotenuse/2  = Circumradius.

 Q1. The solution of 12 + x 7 x + 12 + x 7 12 = 64 3 ( x 7 ) \frac{\sqrt[7]{12 + x}}{x} + \frac{\sqrt[7]{12 + x}}{12} = \frac{64}{3} \left(\sqrt[7]{x}\right) is of the form a b \frac{a}{b} , where a , b a, b are natural numbers with gcd ⁡ ( a , b ) = 1 \gcd(a, b) = 1 ; then b − a b - a is equal to: A) 115  B) 114  C) 113  D) 125 Solution:  (x+12)^(1/7)[1/x + 1/12] = 64.x^(1/7)/3  (x+12)^(1/7)[(x + 12)/12x] = 64.x^(1/7)/3 (x+12)^(8/7) = 12.64.x^(8/3)/3 [(x+12)/x]^(8/7) = 256 = 2^8 => [(x+12)/x]^(1/7) = 2 =>  (x+12)/x = 2^7 = 128 => x + 12 = 128x => 127x = 12 => x = 12/127 => a = 12, b = 127 => b - a = 115 = Answer Q2. The value of ( 52 + 6 43 ) 3 / 2 − ( 52 − 6 43 ) 3 / 2 (52 + 6\sqrt{43})^{3/2} - (52 - 6\sqrt{43})^{3/2} is: A) 858  B) 918  C) 758  D) 828 Solution: Note that: 52 + 6.sqrt(43) = (sqrt(43)  + 3)^2 52 - 6.sqrt(43) = (sqrt(43)  - 3)^2 [52 + 6.sqrt(43)]^(1/2) = (sqrt(43)  + 3) [52 - 6.sqrt(43)]^(1/2) ...

Q1. (Combinatorics) "How many ordered quadruples (a, b, c, d) of positive odd integers are there that satisfy the equation a + b + c + 2d = 15?" Solution: Answer: 34 Why Rewrite every odd variable in terms of non-negative integers a = 2 x + 1 ,    b = 2 y + 1 ,    c = 2 z + 1 ,    d = 2 w + 1 , x , y , z , w ≥ 0 a = 2x+1,\; b = 2y+1,\; c = 2z+1,\; d = 2w+1,\qquad x,y,z,w \ge 0 Plug into the equation ( 2 x + 1 ) + ( 2 y + 1 ) + ( 2 z + 1 ) + 2 ( 2 w + 1 ) = 15 ⇒ 2 x + 2 y + 2 z + 4 w + 5 = 15 ⇒ 2 ( x + y + z + 2 w ) = 10 ⇒ x + y + z + 2 w = 5 (2x+1)+(2y+1)+(2z+1)+2(2w+1)=15 \Rightarrow 2x+2y+2z+4w+5 = 15 \Rightarrow 2(x+y+z+2w)=10 \Rightarrow x+y+z+2w = 5 Count solutions for x + y + z + 2 w = 5 x+y+z+2w=5 Because w w is an integer, list its possible values and use stars-and-bars for the remaining sum: w w equation for x + y + z x+y+z # solutions ( n + 2 2 ) \binom{n+2}{2} 0 x + y + z = 5 x+y+z = 5 ( 7 2 ) = 21 \binom{7}{2}=21 1 x + y ...