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Logarithm Question Define [r] to be the greatest integer less than or equal to r Find the value of [ log3 (1+ 1/1) + log3(1+1/2) +log3(1+1/3) + …… log3(1+1/2017) ] The one inside the bracket is simply log32018 so our answer is 6 Trigonometry Calculus has proven that the function f(x) = 2x + sinx is a strictly increasing function, that is if x>y, then f(x)>f(y). Find all pairs (x,y) that satisfy sin x + x = y and sin y + y = x Clearly sin x = - sin y Let k be any integer, we have two cases, for an integer k. Case 1: y = -x + 2kpi , then the first equation yields 2x+sinx=2k*pi. Now, note that x = k*pi satisfies the first condition, with y=k*pi. Now, since 2x+sin x is increasing, then no other value of x satisfies our equation Case 2: y= (2k+1)pi +x, then |y-x|=|2k+1|pi >= pi But |y-x| = |sinx| from our first equality, a contradiction Thus the ordered pairs, (k * pi, k * pi) for any integer k satisfy the two equations. Functional Equations Let f(x) be a function with domain R/{0} that satisfies: xf(xy)+f(-y)=xf(x) for all x and y in the domain. Find the value of f(2017)/f(5), if f(5) > 0 Substitute x=1 we get f(y)+f(-y)=f(1) or f(x)+f(-x)=f(1) Substitute y=-1 we get xf(-x)+2=xf(x) or f(x)-f(-x)=f(1)/x Systems of two equations, we get that f(x) = ½(f(1)+f(1)/x) = c + c/x for some real c. It is easy to see that this satisfies the conditions in the equality. Since f(5) > 0, c cannot be 0 hence: f(2017)/f(5) = (1+1/2017)/(1+1/5) = 5045/6051 Trigonometry (difficult) Let P be an interior point of acute triangle ABC with AB=c, AC=b, and BC=a, that satisfies: angle APB = angle CBA + angle ACB AND angle BPC = angle ACB + angle BAC. Find BA/AP in terms of a,b, and c. Denote angles BAC, ABC, ACB, to be A, B, C respectively, A+B+C =180 Angle APC = 360 – (B+C) – (C+A) = A+B Let m = angle ABP, we get the following: Angle BAP=180 – (B+C) – m = A-m Angle PAC = A – (A-m) = m Angle ACP = 180 – m – (A+B) = C-m Angle PCB = C- (C-m) =m And Angle PBC = B-m (*) Sine law on ABP leads us to AP/sin m = AB/sin (B+C) =AB/sinA (**) Sine law on BCP gives us BP/sin m = BC /sin(C+A) = BC/(sinB) (***) Sine law on triangle ABC gives us BC/sinA = AC/sinB From these three equations, we get that AP/BP = (ABsin m/sinA)/(BC sin m/sinB) = (AB/BC)(sinB/sinA) = (AB/BC)(AC/BC) = bc/a2

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