## Solutions for problems "B" in February, 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.

**B.3522.**Solve the following equation on the set
of integers:
2*x*^{4}+*x*^{2}*y*^{2}+5*y*^{2}=*y*^{4}+10*x*^{2}. (*3* *points*)

**Solution.** If *A*=*x*^{2} and *B*=*y*^{2}, then by rearranging the equation and factoring we get (2*A*-*B*)(*A*+*B*-5)=0. Thus either 2*x*^{2}=*y*^{2}, and hence *x*=*y*=0, or *x*^{2}+*y*^{2}=5. Therefore, the equation has the following 9 solutions: *x*=*y*=0, *x*=\(\displaystyle pm\)2,*y*=\(\displaystyle pm\)1, valamint *x*=\(\displaystyle pm\)1,*y*=\(\displaystyle pm\)2.

**B.3523.**Given a semicircle, consider the
triangles in which the diameter of the semicircle is contained in one
side, and the other two sides are tangent to the semicircle. Which of
these triangles has minimum area? (*4* *points*)

**Solution.** If the triangle is reflected in the line containing the diameter, the sides of the resulting kite will all touch the given circle. We shall prove that out of all such kites, the square has minimum area. Then it follows that the triangle of minimum area is the isosceles right triangle whose hypotenuse contains the diameter of the semicircle. Let us prove a more general statement: Out of all quadrilaterals circumscribed around a given circle, the square has the minimum area. Let the radius of the circle be unity. Consider a quadrilateral whose sides touch the circle and whose angles are \(\displaystyle pi\)-2\(\displaystyle alpha\),\(\displaystyle pi\)-2\(\displaystyle beta\),\(\displaystyle pi\)-2\(\displaystyle gamma\),\(\displaystyle pi\)-2\(\displaystyle delta\), where the sum of the acute angles \(\displaystyle alpha\),\(\displaystyle beta\),\(\displaystyle gamma\),\(\displaystyle delta\) is \(\displaystyle pi\). Then the area of the quadrilateral is

\(\displaystyle \tan\alpha+\tan\beta+\tan\gamma+\tan\delta\le4\tan\Bigl({\alpha+\beta+\gamma+\delta\over4}\Bigr)=4\tan(\pi/4)=4\ ,\)

as the tangent function is strictly convex down on the interval (0,\(\displaystyle pi\)/2). For the same reason, equality only occurs in the case \(\displaystyle alpha\)=\(\displaystyle beta\)=\(\displaystyle gamma\)=\(\displaystyle delta\)=\(\displaystyle pi\)/4, that is, when all angles of the quadrilateral are right angles. That completes the proof.

**B.3524.**The sum of the real numbers *x*
and *y* is 1. Find the maximum value of
*xy*^{4}+*x*^{4}*y*. (*3* *points*)

**Solution.** As

*xy*^{4}+*x*^{4}*y*=*xy*(*x*^{3}+*y*^{3})=*xy*((*x*+*y*)^{3}-3*xy*(*x*+*y*))=*xy*(1-3*xy*) ,

the task is to find the maximum of the expression *A*(1-3*A*), given that the number *A* can be written in the form *x*(1-*x*). As

\(\displaystyle A(1-3A)={1\over12}-3\Bigl(A-{1\over6}\Bigr)^2\ ,\)

the the expression *A*(1-3*A*) assumes its maximum at the point *A*=1/6, and the value of the maximum is 1/12. It follows from the solution of the quadratic equation *x*(1-*x*)=1/6 that if \(\displaystyle x=(1\pm\sqrt{1/3})/2\), then *A*=1/6=*x*(1-*x*). Thus the maximum of the expression *A*(1-3*A*) on the given condition will also be 1/12.

**B.3525.**Prove that there are infinitely many
composite numbers in the sequence
1,31,331,3331,... . (*4* *points*)

**Solution.** 31 is a prime number. It follows from Fermat's little theorem that 31 is a factor of the number 10^{30}-1, and therefore it is also a factor of any number of the form 10^{30k}-1, where *k* is a positive integer. Thus it is also a factor of one third of such a number, that is, any number that consists of 30*k* digits of 3. By multiplying such a number by 100 and adding 31, the result is also divisible by 31, and all numbers obtained in this way belong to te sequence. Therefore, from 31 onwards, every 30th member of the sequence is a composite number, as it is divisible by 31.

**B.3526.** *F* is the midpoint of side
*BC* of a rectangle *ABCD*, and *H* divides side
*CD* in the ratio 1:2, *CH* being the shorter part. How
large may the angle *HAF* be? (*4* *points*)

**Solution.**

Let the points *D* and *C* be fixed, and let *D*' and *H*' denote the reflections of *D* and *H* about the point *C*, respectively. Suppose that the angle *HAF* has a given measure \(\displaystyle gamma\). Where can the point *A* be then? The point *A* pont lies on the perpendicular drawn to *DC* at *D*, and it also lies on a circular arc connecting *H* and *D*' and centred on the perpendicular line drawn to *CD* at*H*'. The arc is the locus of the points where *HD*' subtends an angle \(\displaystyle gamma\). Thus the radius of the circle is at least *DH*'=*HD*', and in that case the central angle subtended by the segment *HD*' is \(\displaystyle pi\)/3, \(\displaystyle gamma\)=\(\displaystyle pi\)/6. If the radius of the circle is increased, the value of \(\displaystyle gamma\) will decrease monotonically to 0, and the point *A* can be constructed in each case. Therefore, the measure of the angle *HAF* can be any positive number not greater than \(\displaystyle pi\)/6.

**B.3527.** Given a trapezium in which the other
two sides are not parallel, construct, using a straight edge only, a
line whose segment between the non-parallel sides is divided into
three equal parts by the diagonals of the
trapezium. (*5* *points*)

**Solution.**

Label the vertices of the trapezium so that *A*,*B* should be the longer base and *C*,*D* should be the shorter base. Construct the intersection *E* of the sides *AD* and *BC* and the intersection *F* of the diagonals *AC* and *BD*. Consider the line through *F* parallel to the bases. Let *A*' and *B*' denote its intersections with the sides *AD* and *BC*, respectively. It follows easily from the equal areas of the triangles *AFD* and *BFC* that *A*'*F*=*FB*', as the area of triangle *AFD* is *m*^{.}*A*'*F*/2, and that of triangle *BFC* is *m*^{.}*FB*'/2, where *m* denotes the height of the trapezium. By central similitude, the line *EF* is also the median of triangle *ABE*, that is, it intersects *AB* at its midpoint *G*. The point *G* is easy to construct.

Construct the points *P* and *Q* where the lines *DG* and *CG* intersect the diagonals *AC* and *BD*, respectively. Now *AP*/*PC*=*AG*/*DC*=*GB*/*DC*=*BQ*/*QD*, thus *P* and *Q* divide the distance between the parallel lines *AB* and *CD* in the same ratio, that is *PQ* is parallel to the bases of the trapezium. Let *S* and *T*, respectively, denote the intersections of the line *PQ* with the sides *AD* and *BC*. Then *SP*/*PQ*=*AG*/*GB*=*PQ*/*QT*=1, that is the segment *ST* parallel to *AB* is divided by *P* and *Q* into three equal parts. Thus *PQ* is a line of the required property. Another line can be constructed by connecting the midpoint of *DC* (where line *EF* intersects the side *DC*) to the points *A* and *B*. The connecting lines intersect the diagonals at *P*' and *Q*'. With the above reasoning, it can be shown that the line *P*'*Q*' is also a solution of the problem.

It is clear that there is at most one such line on each side of the point *F*. As a line parallel to *AB* is gradually moving away from the point *E*, its segment between the two diagonals is growing while the other two segments are shrinking. Thus we have found all the solutions.

**B.3528.** Let \(\displaystyle alpha\) and \(\displaystyle beta\) be acute angles such that
sin^{2}\(\displaystyle alpha\)+sin^{2}\(\displaystyle beta\)<1. Prove that sin^{2}+sin^{2}\(\displaystyle beta\)<sin^{2}(+). (*4* *points*)

**Solution.**

The trigonometric Pythagorean formula says that sin^{2}\(\displaystyle alpha\)+sin^{2}(\(\displaystyle pi\)/2-\(\displaystyle alpha\))=1, thus it follows from the condition that \(\displaystyle beta\)<\(\displaystyle pi\)/2-\(\displaystyle alpha\), 0<\(\displaystyle alpha\)+\(\displaystyle beta\)<\(\displaystyle pi\)/2. As

sin^{2}(\(\displaystyle alpha\)+\(\displaystyle beta\))=sin^{2}\(\displaystyle alpha\)cos^{2}\(\displaystyle beta\)+sin^{2}\(\displaystyle beta\)cos^{2}\(\displaystyle alpha\)+2sin\(\displaystyle alpha\)sin\(\displaystyle beta\)cos\(\displaystyle alpha\)cos\(\displaystyle beta\) ,

the substitution of cos^{2}*x*=1-sin^{2}*x* and rearrangement shows that the statement is equivalent to

2sin^{2}\(\displaystyle alpha\)sin^{2}\(\displaystyle beta\)<2sin\(\displaystyle alpha\)sin\(\displaystyle beta\)cos\(\displaystyle alpha\)cos\(\displaystyle beta\)

Division by the positive quantity 2sin\(\displaystyle alpha\)sin\(\displaystyle beta\) leads to the inequalities sin\(\displaystyle alpha\)sin\(\displaystyle beta\)<cos\(\displaystyle alpha\)cos\(\displaystyle beta\), 0<cos(\(\displaystyle alpha\)+\(\displaystyle beta\)) which are obviously true, as we have shown above that 0<\(\displaystyle alpha\)+\(\displaystyle beta\)<\(\displaystyle pi\)/2.

**B.3529.** The sum of the first few terms in a
geometric progression is 11, the sum of their squares is 341, and the
sum of their cubes is 3641. Find the terms of the
sequence. (*5* *points*)

(Suggested by *Á.* *Besenyei,*
Budapest)

**Solution.**

Let *a* denote the first term of the sequence, let *q* be the common ratio, and let *n*be the number of terms. Let *Q*=*q*^{n}. *q*=1 is impossible, as in that case *an*=11, *a*^{2}*n*=341, and hence *a*=31, *n*=11/31, which contradicts to the condition *a*^{3}*n*=3641 Therefore, the following relationships are true: \(\displaystyle a{Q-1\over q-1}=11\)

, \(\displaystyle a^2{Q^2-1\over q^2-1}=341\)

, \(\displaystyle a^3{Q^3-1\over q^3-1}=3641.\)

Divide the second and third equalities by the first one. Then \(\displaystyle a{Q+1\over q+1}=31\)

, \(\displaystyle a^2{Q^2+Q+1\over q^2+q+-1}=331.\)

Hence *a*(*Q*-1)=11(*q*-1) és *a*(*Q*+1)=31(*q*+1), thus *a*=10*q*+21 and (from *a\(\displaystyle ne\)*0) \(\displaystyle Q={21q+10\over10q+21}\) can also be expressed in terms of *q*. Substitute the result into the equality *a*^{2}(*Q*^{2}+*Q*+1)=331(*q*^{2}+*q*+1) above. The possible values of *q* are the roots of the equation 2*q*^{2}+5*q*+2=0: *q*=-2 and *q*=-1/2. Hence either *a*=1, *Q*=-32, *n*=5, or *a*=16, *Q*=-1/32, *n*=5. Thus the solutions of the problem are the following two sequences: 1,-2,4,-8,16 and 16,-8,4,-2,1. It is easy to check that these sequences both satisfy the requirements.

**B.3530.** Evaluate .
(*5* *points*)

**Solution.** Let

\(\displaystyle A_n=\sum_{i=0}^{[n/2]}(-1)^i{n-i\choose i}\ .\)

for all positive integers *n*. The task is to determine the value of *A*_{2002}. By using the identity \(\displaystyle {n\choose k}={n-1\choose k-1}+{n-1\choose k}\), it is easy to check that *A*_{n}=*A*_{n-1}+*A*_{n-2} for all integers *n\(\displaystyle ge\)*3: If *k\(\displaystyle ge\)*2 then

\(\displaystyle A_{2k}=1+\sum_{i=1}^{k-1}(-1)^i\Bigl({2k-i-1\choose i}+{2k-i-1\choose i-1}\Bigr)+(-1)^k{k\choose k}=\)

\(\displaystyle =A_{2k-1}-\sum_{j=0}^{k-2}(-1)^j{2k-2-j\choose j}-(-1)^{k-1}{k-1\choose k-1}=A_{2k-1}-A_{2k-2}\ .\)

If *k\(\displaystyle ge\)*1, it is even simpler, as

\(\displaystyle A_{2k+1}=1+\sum_{i=1}^{k}(-1)^i\Bigl({2k-i\choose i}+{2k-i\choose i-1}\Bigr)=\)

\(\displaystyle =A_{2k}-\sum_{j=0}^{k-1}(-1)^j{2k-1-j\choose j}=A_{2k}-A_{2k-1}\ .\)

It is easy to calculate that *A*_{1}=*A*_{6}=1, *A*_{2}=*A*_{5}=0, and *A*_{3}=*A*_{4}=-1. Hence, by induction, it can be shown that *A*_{6k}=*A*_{6k+1}=1, *A*_{6k+2}=*A*_{6k+5}=0, and *A*_{6k+3}=*A*_{6k+4}=-1 for every positive integer *k*. As 2002=6^{.}333+4, the answer is *A*_{2002}=-1.

**B.3531.** The inscribed circle of a triangle
*ABC* is centred at *O*, and touches the sides at
*A*_{1}, *B*_{1}, *C*_{1} (in
customary notation). The lines *A*_{1}*O*,
*B*_{1}*O*, *C*_{1}*O* intersect
the segments *B*_{1}*C*_{1},
*C*_{1}*A*_{1},
*A*_{1}*B*_{1} at the points
*A*_{2}, *B*_{2}, *C*_{2},
respectively. Prove that the lines *AA*_{2},
*BB*_{2} és *CC*_{2} are
concurrent. (*5* *points*)

(Suggested by *Á.* *Besenyei,* Budapest)

**Solution.** The following statement will be used several times in the proof:

If *X*' is a point on the side *YZ* of the triangle *XYZ*,and \(\displaystyle eta\)_{Y}, \(\displaystyle eta\)_{Z}, \(\displaystyle delta\)_{Y}, \(\displaystyle delta\)_{Z} denote the angles *XYZ*, *XZY*, *YXX*', *ZXX*', respectively then

\(\displaystyle {YX'\over ZX'}={\sin\delta_Y\over\sin\delta_Z}\cdot{\sin\eta_Z\over\sin\eta_Y}\ ,\)

as it follows from the sine rule that *YX*'/*XX*'=sin\(\displaystyle delta\)_{Y}/sin\(\displaystyle eta\)_{Y}, and *ZX*'/*XX*'=sin\(\displaystyle delta\)_{Z}/sin\(\displaystyle eta\)_{Z}. In the special case of *XY*=*XZ*, \(\displaystyle eta\)_{Y}=\(\displaystyle eta\)_{Z}, and thus *YX*'/*ZX*'=sin\(\displaystyle delta\)_{Y}/sin\(\displaystyle delta\)_{Z}.

As the measures of the angles *C*_{1}*A*_{1}*B*_{1}, *C*_{1}*B*_{1}*A*_{1}, *A*_{1}*C*_{1}*C*_{2} and *B*_{1}*C*_{1}*C*_{2} are (with the conventional notation) (\(\displaystyle beta\)+\(\displaystyle gamma\))/2, (\(\displaystyle alpha\)+\(\displaystyle gamma\))/2, \(\displaystyle beta\)/2 and \(\displaystyle alpha\)/2, respectively. With the above statement, it follows that

\(\displaystyle {A_1C_2\over B_1C_2}={\sin(\beta/2)\over\sin(\alpha/2)}\cdot{{\sin{\alpha+\gamma\over2}}\over{\sin{\beta+\gamma\over2}}}\ .\)

Let\(\displaystyle gamma\)_{B} and \(\displaystyle gamma\)_{A} denote the angles *A*_{1}*CC*_{2} and *B*_{1}*CC*_{2}, respectively, and let *C*_{3} be the intersection of the line *CC*_{2} with the segment *AB*. By using *CA*_{1}=*CB*_{1} and using the above statement again, we have

\(\displaystyle {AC_3\over C_3B}={\sin\gamma_A\over\sin\gamma_B}\cdot{\sin\beta\over\sin\alpha}={B_1C_2\over A_1C_2}\cdot{\sin\beta\over\sin\alpha}={\sin(\alpha/2)\over\sin(\beta/2)}\cdot{\sin{\beta+\gamma\over2}\over{\sin{\alpha+\gamma\over2}}}\cdot{\sin\beta\over\sin\alpha}\ .\)

Similarly,

\(\displaystyle {BA_3\over A_3C}={\sin(\beta/2)\over\sin(\gamma/2)}\cdot{\sin{\gamma+\alpha\over2}\over{\sin{\beta+\alpha\over2}}}\cdot{\sin\gamma\over\sin\beta}\)

and

\(\displaystyle {CB_3\over B_3A}={\sin(\gamma/2)\over\sin(\alpha/2)}\cdot{\sin{\alpha+\beta\over2}\over{\sin{\gamma+\beta\over2}}}\cdot{\sin\alpha\over\sin\gamma}\ ,\)

where *A*_{3} and *B*_{3} denote the intersections of the lines *AA*_{2} and *BB*_{2} with the sides *BC* and *AC*, respectively. Therefore, by putting all together we get

\(\displaystyle {AC_3\over C_3B}\cdot{BA_3\over A_3C}\cdot{CB_3\over B_3A}=1\ ,\)

and thus the statement of the problem follows immediately from Ceva's theorem.