Three Nice Polynomials of Degree III // Tre pæne polynomier af grad III

 We will solve Problem 7 (page 38, in the magazine from yesterday's Post) from Team Selection Test(TSTs) 1 for IMO.

" Problem 7.  Consider three polynomials of degree 3 with real 

coefficients $P,\;Q,\;R$  such that  $P(x)\leqslant Q(x)\leqslant R(x)$  for any real  $x$

  and, in addition, there exists a real number  $a$  such that  $P(a)=R(a).$

  Show that there exists a constant  $\lambda \in [0,\;1]$  with the property

  $Q=\lambda P+(1-\lambda)R.$  Does the statement remain true in the case 

                         when  $P,\;Q,\;R$  have degree 4?

{authors : } I. Cuculescu and L. Panaitopol, Bucharest"

ANSWER CiP

It is not entirely true for polynomials of degree 4. For example, if 

$P(x)=x^4\;,\;\;Q(x)=x^4+x^2\;,\;\;R(x)=x^4+2x^2$  then  $P(x)\leqslant Q(x)\leqslant R(x)$

  and we have $Q(x)=\lambda \cdot P(x)+(1-\lambda)\cdot R(x)$  but with  $\lambda =-1 \not \in [0,\;1]$

Solution CiP

                   Obviously  $P(a)=Q(a)=R(a).$  Let's define polynomials

$S(x)=R(x)-Q(x)\;\geqslant 0\;,\;\forall x\;\;;\;\;T(x)=R(x)-P(x)\;\geqslant 0\;,\;\forall x \tag{1}$

          We have  $S(a)=0$  therefore  $S(x)=(x-a)S_2(x)$ , for a certain polynomial  $S_2$  of degree 2.

Since when  $x$  passes through the value  $x=a$ , the polynomial  $S$  does not change sign, we must have 

$S(x)=(x-a)^2\cdot S_1(x) \tag{2}$

with  $S_1$-a first degree polynomial. But the polynomial  $S_1$  should have a root, through which if   $x$  passes, the expression in  (2)  changes sign again. Contradiction, so  $S_1$  is a constant.

          In exactly the same way results

$T(x)=(x-a)^2\cdot T_1(x) \tag{3}$

with  $T_1$  a constant.

          Let  $\lambda :=\frac{S(x)}{T(x)}\overset{(2)}{\underset{(3)}{=}}\frac{S_1}{T_1}=constant\;$.  Since  $\lambda=\frac{R(x)-Q(x)}{R(x)-P(x)}$  and  $R(x)-Q(x)\leqslant R(x)-P(x)$  we have  $0\leqslant \lambda \leqslant 1$  and 

 $Q(x)=\lambda \cdot P(x)+(1-\lambda )\cdot R(x).$

$\blacksquare$