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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
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
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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