## Solutions for advanced problems "A" in November, 2002 |

In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.

**A. 302.** Given the unit square
*ABCD* and the point *P* on the plane, prove that

\(\displaystyle 3AP+5CP+\sqrt5(BP+DP)\ge6\sqrt2.\)

**Solution 1** Let \(\displaystyle f(P)=3AP+5CP+\sqrt5(BP+DP)\) for any point *P* of the plane . Since each term is a convex (below) function of *P*, the function *f* is also convex. Thus, if a point is found where the function is differentiable and the gradient is the zero vector, the function has an absolute minimum there.

The gradient of the function is

\(\displaystyle \mathop{\rm grad}f=3{\overrightarrow{AP}\over|AP|}+ 5{\overrightarrow{CP}\over|CP|}+ \sqrt5\left({\overrightarrow{BP}\over|BP|}+ {\overrightarrow{DP}\over|DP|}\right).\)

Since the positions of the points *B* and *D* symmetrical, we should look for the minimum point on the diagonal *AC*.

By a little calculation, we can show that the minimum occurs at the point lying closer to *C* and dividing the diagonal *AC* in a 3:1 ratio. At that point, therefore, the gradient is 0, and it is easily obtained that the value of the function is \(\displaystyle 6\sqrt2\). This is the smallest value of the function.

**Solution 2 (This is the origin of the problem).** Let *A*=(0,0), *B*=(1,0), *C*=(1,1) and *D*=(0,1). Let *E*=(1,5/6), *F*=(0,1/2), *Q*=(3/4,3/4), *U*=(-3/10,2/5) and *V*=(3/2,1). It is easy to check that the points *U*,*F*,*Q*,*E*,*V* all lie on a line in this order. The points *A*,*Q*,*D*,*U* lie on a circle, and so do the points *B*,*V*,*C*,*Q*.

A Ptolemy's theorem states that *AU*^{.}*DP*+*DU*^{.}*AP*\(\displaystyle \ge\)*AD*^{.}*UP*=*UP*, and equality occurs if and only if the points *A*,*P*,*D*,*U* in this order are concyclic. Similarly, *BV*^{.}*CP*+*CV*^{.}*BP*\(\displaystyle \ge\)*BC*^{.}*VP*=*VP*; and equality occurs if and only if *B*,*V*,*C*,*P* are concyclic. Finally, *UP*+*VP*\(\displaystyle \ge\)*UV* from the triangle inequality.

The lengths of the line segments in the above inequalities are \(\displaystyle AU=CV={1\over2}\), \(\displaystyle DU={3\sqrt5\over10}\), \(\displaystyle BV={\sqrt5\over2}\) and \(\displaystyle UV={3\sqrt{10}\over5}\). Thus

\(\displaystyle {1\over2}\cdot DP+{3\sqrt5\over10}\cdot AP+ {\sqrt5\over2}\cdot CP+{1\over2}\cdot BP\ge UP+VP\ge{3\sqrt{10}\over5}.\)

Multiplication by \(\displaystyle 2\sqrt5\) yields the statement of the problem.

If *P*=*Q*, equality holds.

**A. 303.** *x*, *y* are
non-negative numbers, and
*x*^{3}+*y*^{4}\(\displaystyle \le\)*x*^{2}+*y*^{3}. Prove that
*x*^{3}+*y*^{3}\(\displaystyle \le\)2.

**Solution.**

2-*x*^{3}-*y*^{3}=3(*x*^{2}+*y*^{3}-*x*^{3}-*y*^{4})+(*x*-1)^{2}(2*x*+1)+(*y*-1)^{2}(3*y*^{2}+2*y*+1)\(\displaystyle \ge\)0.

**A. 304.** Find all functions
*R*^{+}\(\displaystyle \mapsto\)*R*^{+}, such that

*f*(*x*+*y*)+*f*(*x*)^{.}*f*(*y*)=*f*(*xy*)+*f*(*x*)+*f*(*y*)?

**Solution.** Substitute *x*=*y*=2: *f*(4)+*f*(2)^{2}=*f*(4)+2*f*(2). Hence *f*(2)=0.

Substitute *x*=*y*=1: *f*(2)+*f*^{2}(1)=3*f*(1), that is *f*^{2}(1)-3*f*(1)+2=0. This is a quadratic equation in *f*(1) that has two roots 1 and 2.

*Case 1:* *f*(1)=1. We shall prove that in this case the function *f* is additive as well as multiplicative, that is, *f*(*u*+*v*)=*f*(*u*)+*f*(*v*) and *f*(*uv*)=*f*(*u*)*f*(*v*) for all *u*,*v*. It follows from the equation (1) that it is enough to prove one of the two properties. Let us prove the additive property.

Substitute *y*=1 in the equation: *f*(*x*+1)+*f*(*x*)^{.}*f*(1)=2*f*(*x*)+*f*(1), that is, *f*(*x*+1)=*f*(*x*)+1 for all positive *x*.

Now let *u* and *v* be two arbitrary positive numbers. Apply the equation (1) to the number pairs *x*=*u*, *y*=*v*/*u* and *x*=*u*, *y*=*v*/*u*+1, too:

*f*(*u*+*v*/*u*)+*f*(*u*)^{.}*f*(*v*/*u*)=*f*(*v*)+*f*(*u*)+*f*(*v*/*u*),

and

*f*(*u*+*v*/*u*+1)+*f*(*u*)^{.}*f*(*v*/*u*+1)=*f*(*u*^{.}(*v*/*u*+1))+*f*(*u*)+*f*(*v*/*u*+1),

*f*(*u*+*v*/*u*)+1+*f*(*u*)(*f*(*v*/*u*)+1)=*f*(*u*+*v*)+*f*(*u*)+*f*(*v*/*u*)+1.

*f*(*u*+*v*)=*f*(*u*)+*f*(*v*) is obtained by subtracting the two equations.

It follows from the additive property that *f*(*n*)=*n* for all positive integers *n*, and it follows from the multiplicative property that *f*(*k*/*n*)=*f*(*k*)/*f*(*n*)=*k*/*n* for all integers *k*,*n*. Thus *f*(*q*)=*q* for all positive rational numbers *q*. Additivity also implies that the function is monotonic: *f*(*x*)<*f*(*x*)+*f*(*y*-*x*)=*f*(*y*) for *x*<*y*.

The only such function is the identity: *f*(*x*)=*x*.

*Case 2:* *f*(1)=2. We shall show that the function is constant: *f*(*u*)=2 for all positive *u*.

Substitute *y*=1 into the equation again: *f*(*x*+1)+2*f*(*x*)=2*f*(*x*)+2, thus *f*(*x*+1)=2. That proves the statement for *u*>1.

Now let *u* be an arbitrary real number. If *v* is a number such that *v*, *uv* and *u*+*v* are all greater than 1. The equation states that *f*(*u*+*v*)+*f*(*u*)^{.}*f*(*v*)=*f*(*uv*)+*f*(*u*)+*f*(*v*), that is, 2+*f*(*u*)^{.}2=2+*f*(*u*)+2. Hence *f*(*u*)=2.

Thus the equation is satisfied by two functions: *f*(*x*)=*x* and *f*(*x*)=2.