## Solutions for advanced problems "A" in December, 2001 |

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. 278. ***P* is a point on the extension of
the diagonal *AC* of rectangle *ABCD* beyond *C*, such
that *BPD\(\displaystyle angle\)*=*CBP\(\displaystyle angle\)*. Find the ratio *PB*:*PC*.

**Solution 1.** Let the perpendicular projections of *P*
onto the lines *CD* and *AB* be *Q* and *R*,
respectively, and let *PAR\(\displaystyle angle\)*=*PCQ\(\displaystyle angle\)*=\(\displaystyle alpha\) and *BPD\(\displaystyle angle\)*=*CBP\(\displaystyle angle\)*=*BPR\(\displaystyle angle\)*=\(\displaystyle beta\). With the length of segment *PC*
chosen to be 1, let the length of segment *PB* be *x*. The
task is to determine *x*.

From the triangles *PCQ* and *PBR*, for example, and the
rectangle *CBRQ*, *x*sin\(\displaystyle beta\)=*BR*=*CQ*=cos, that is, .

From the triangles *PDQ*, *PCQ*, *PAR*, *PBR*,
and the rectangle *ARQD*,

\(\displaystyle \tan2\beta={DQ\over PQ}={AR\over\sin\alpha}={PR\cdot\cot\alpha\over\sin\alpha}={x\cdot\cos\beta\cdot\cot\alpha\over\sin\alpha}=x\cdot\cos\beta\cdot{\cos\alpha\over\sin^2\alpha}.\)

By rearranging and substituting \(\displaystyle \sin\beta={\cos\alpha\over x}\), we have

\(\displaystyle {x\cdot\cos\alpha\over\sin^2\alpha}={\tan2\beta\over\cos\beta}={2\sin\beta\over1-2\sin^2\beta}={2{\cos\alpha\over x}\over1-2{\cos^2\alpha\over x^2}},\)

*x*^{2}=2sin^{2}+2cos^{2}\(\displaystyle alpha\)=2, hence .

Thus the ratio *PB*:*PC* is always ,
independently of the angle \(\displaystyle alpha\).

**Solution 2.** This is an elementary proof for . Let *S* be the intersection of the lines *BC* and
*PD*. The triangle *BPS* is isosceles, according to the
given information. Let *R* be the point that is on the same side
of line *PB* as *S*, and for which the triangle *PBR*
is isosceles and right-angled, that is, *BR*=*PR* and
*PRB\(\displaystyle angle\)*=90^{o}.

Let, furthermore, *T* be the intersection of the circumscribed circle of rectangle *ABCD* with the line *PD*. *BD* is a diameter in the circle, therefore the segment *BT* is perpendicular to the line *PD*.

As *ACTD* is a cyclic quadrilateral, *PTC\(\displaystyle angle\)*=*PAD\(\displaystyle angle\)*=*PCS\(\displaystyle angle\)*. The angles of the triangles *PCS* and *PTC* are pairwise equal, thus the two triangles are similar, and *PC*:*PS*=*PT*:*PC*, that is, *PC*^{2}=*PS*^{.}*PT*.

As *PRB\(\displaystyle angle\)*=*PTB\(\displaystyle angle\)*=90^{o}, the quadrilateral *PRTB* is also cyclic, and *RTP\(\displaystyle angle\)*=*RBP\(\displaystyle angle\)*=*PRS\(\displaystyle angle\)*. The triangles *PRS* and *PTR* are also similar, thus with the above reasoning we have *PR*^{2}=*PS*^{.}*PT*.

We have obtained that *PC*^{2}=*PR*^{2}=*PS*^{.}*PT*, that is, *PC*=*PR*, and hence \(\displaystyle PB:PC=PB:PR=\sqrt2:1\).

**A. 279. **Are there such rational functions *f*
and *g* that (*f*(*x*))^{3}+(*g*(*x*))^{3}=*x*?

**Solution.** We shall prove that no rational functions of either real or complex coefficients satisfy this functional equation.

Assume, to the contrary, that there is a solution. Let the product of the denominators of *f*(*x*) and *g*(*x*) be the polynomial *c*(*x*). Then *f*(*x*)=*a*(*x*)/*c*(*x*) and *g*(*x*)=*b*(*x*)/*c*(*x*), with the appropriate polynomials *a* and *b*. By substituting these into (1), we have

(2) | a^{3}(x)+b^{3}(x)=x^{.}c^{3}(x). |

Consider a solution of the equation (2), such that *c*(*x*) is not 0, and its degree is a minimum. Let \(\displaystyle \varepsilon=\cos{2\pi\over3}+i\sin{2\pi\over3}\) be the first third root of unity. The left-hand side of (2) can be factorized:

(3) | (a(x)+b(x))^{.}(a(x)+\(\displaystyle varepsilon\)b(x))^{.}(a(x)+\(\displaystyle varepsilon\)^{2}b(x))=x^{.}c^{3}(x). |

We state that the three factors on the left-hand side are pairwise relative primes. We shall prove it for the first two factors, the proof is similar for the other pairs. Assume that the polynomials *a*(*x*)+*b*(*x*) and *a*(*x*)+\(\displaystyle varepsilon\)*b*(*x*) have a common factor of at least degree one: *d*(*x*). Then *d*(*x*) is also a factor of the polynomial \(\displaystyle {1\over1-\varepsilon}\big((a(x)+b(x))-(a(x)+\varepsilon b(x))\big)=b(x)\), and similarly, also a factor of *a*(*x*). *d*^{3}(*x*) is a factor of the left-hand side of (2), and therefore it also divides *x*^{.}*c*^{3}(*x*), that is, *d*(*x*) is a factor of *c*(*x*). But then by dividing the polynomials *a*(*x*), *b*(*x*), *c*(*x*) by *d*(*x*), we get a solution in which the degree of *c* is smaller. Therefore, the polynomials *a*(*x*)+*b*(*x*) and *a*(*x*)+\(\displaystyle varepsilon\)*b*(*x*) cannot have a common factor of at least degree one, that is, they are relative primes.

The product of three polynomials that are pairwise relative primes can only have the form *x*^{.}*c*^{3}(*x*) if two of them are perfect cubes, and one is *x* times a perfect cube. As the equation is symmetrical in the polynomials *b*(*x*), \(\displaystyle varepsilon\)*b*(*x*), \(\displaystyle varepsilon\)^{2}*b*(*x*), we can assume that the first factor is *x* times a perfect cube, that is, with appropriate polynomials *u*(*x*), *v*(*x*), *w*(*x*),

*a*(*x*)+*b*(*x*)=*x*^{.}*u*^{3}(*x*),

*a*(*x*)+\(\displaystyle varepsilon\)*b*(*x*)=*v*^{3}(*x*),

*a*(*x*)+\(\displaystyle varepsilon\)^{2}*b*(*x*)=*w*^{3}(*x*),

and

*c*(*x*)=*u*(*x*)*v*(*x*)*w*(*x*).

It is easy to show that

(4) | (1+\(\displaystyle varepsilon\)^{2})(a(x)+\(\displaystyle varepsilon\)b(x))+(1+\(\displaystyle varepsilon\))(a(x)+\(\displaystyle varepsilon\)^{2}b(x))=a(x)+b(x). |

On the left-hand side, each term is a perfect cube:

(1+\(\displaystyle varepsilon\)^{2})(*a*(*x*)+\(\displaystyle varepsilon\)*b*(*x*))=*d*^{3}(*x*)

and

(1+\(\displaystyle varepsilon\))(*a*(*x*)+\(\displaystyle varepsilon\)^{2}*b*(*x*))=*e*^{3}(*x*)

with appropriate polynomials *d*(*x*) and *e*(*x*). By substitution of these into (4),

*d*^{3}(*x*)+*e*^{3}(*x*)=*x*^{.}*u*^{3}(*x*).

Thus the polynomials *d*(*x*), *e*(*x*), *u*(*x*) represent another solution of the equation. If the degree of *u*(*x*) is smaller than the degree of *c*(*x*) it is contradiction, as we assumed above that the degree of *c*(*x*) was a minimum. On the other hand, the degree of *u*(*x*) can only be greater than or equal to the degree of *c*(*x*)=*u*(*x*)*v*(*x*)*w*(*x*) if *v*(*x*) and *w*(*x*) are constants. But then the polynomials *a*(*x*) and *b*(*x*) are also constant, and thus there is a constant on the left-hand side of (2), while the right-hand side is a polynomial of at least degree one. This is also contradiction.

**A. 280. **For each positive integer *n*, let
*f*_{n}(\(\displaystyle vartheta\))=sin\(\displaystyle vartheta\)^{.}sin(2)^{.}sin(4\(\displaystyle vartheta\))^{.}...^{.}sin(2^{n}). For
all real \(\displaystyle vartheta\) and all *n*, prove that

IMC 8, Prague, 2001

**Solution.** Let us allow *n*=0, too. Then obviously, *f*_{0}(\(\displaystyle vartheta\))=sin\(\displaystyle vartheta\).

It is easy to show that on the right-hand side of (5) we have \(\displaystyle \left({\sqrt3\over2}\right)^n\), and the statement to prove is:

(6) | \(\displaystyle |f_n(\vartheta)|<\left({\sqrt3\over2}\right)^n\) |

The proof is by mathematical induction. In the cases *n*=1 and *n*=2, the statement is trivial.

\(\displaystyle |\sin\vartheta|<{2\over\sqrt3},\)

and

|sin\(\displaystyle vartheta\)^{.}sin2\(\displaystyle vartheta\)|<1.

Now let *n*>3, and let us assume that the statement is true for all smaller values. If \(\displaystyle |\sin\vartheta|\le{\sqrt3\over2}\) then it follows from the assumption that

\(\displaystyle |f_n(\vartheta)|=|\sin\vartheta|\cdot|f_{n-1}(2\vartheta)|<{\sqrt3\over2}\cdot\left({\sqrt3\over2}\right)^{n-1}=\left({\sqrt3\over2}\right)^n.\)

If \(\displaystyle |\sin\vartheta\){\sqrt3\over2}|">, then \(\displaystyle |\cos\vartheta|<{1\over2}\),

\(\displaystyle |\sin\vartheta\cdot\sin2\vartheta|={1\over2}|\cos\vartheta-\cos3\vartheta|<{1\over2}\left({1\over2}+1\right)={3\over4},\)

and

\(\displaystyle |f_n(\vartheta)|=|\sin\vartheta\cdot\sin2\vartheta|\cdot|f_{n-2}(2\vartheta)|<{3\over4}\cdot\left({\sqrt3\over2}\right)^{n-2}=\left({\sqrt3\over2}\right)^n.\)