When you say that the problems are easy, and if you are not the USAMO winner, then you are not aware of how wrong you are… ]]>

Problem #1 is the type of trick problem I encountered during my undergraduate calculus examine. Clever minds can easily see the solution.

Problem #2 required the appropriate replacement and rearrangement to see an obvious solution. Not tricky like Problem #1 but far less intuitive.

I ignored Problems #3 and #4 because I am weak in geometry. Although I may be able to solve each problem, I do not remember enough of my geometry to solve without searching a text.

Problem #5 may be provable through brute force. But still less labour intensive than doing a 4-variable Truth Table.

]]>This is a standard MSU exam problem (they try to fail the students and always arrange solutions that do not belong to the domain of definition). You must, first, specify the domain of definition, which in this case is determined by the inequalities 0<sin(-x)0, and only then solve the equation sin(-x)=sin(x/2)+sin(3x/2)=2sin((x/2+3x/2)/2)cos((x/2-3x/2)/2)=2sin(x)cos(x/2). Thus, cos(x/2)=-1/2, hence x=4pi/3+4kpi or 8pi/3+4kpi. However, the first series of solutions doesn’t belong to the domain of definition. Hence x=8pi/3+4kpi.

The second problem is also easy, but one must be careful about the signs. The third problem is very easy.

The difficult problems are the last two. In the last problem you have an ellipse symmetric with respect to x=y (i.e. x=y is one of its axis). Thus, if (x,y) is a solution then so is (y,x). Hence, there is a line x=a that intersects the ellipse at points (a,b) and (a,c), and {a,b,c}={r,s,t}. If you draw the ellipse, you see that there are two points of intersection if and only if -1<x<7. Furthermore, a<min{b,c} if x<1, if x=1 then a=b (contradiction), if x=7 then b=c (contradiction), and if 1<x<7 then either b<1 or c<1. In any (admissible) event -1<min{a,b,c}<1. I have no idea how kids recently graduated from a high school are supposed to solve this. I tried also to look at problem 4. There is a lot of symmetry in this problem (in particular the center of the base of the cone lies on the line passing through the point of tangency of the smaller sphere with P and perpendicular to the interval joining the two points of tangency of the equiradial spheres. I tried to compute the angle (taking the symmetries into account it is not difficult), but got no solutions (most likely there is a mistake in my computations).

As I grew up in Moscow, I suppose I know the (not very hidden) agenda, and will not comment on this.

]]>