**A. 547.** A tetrahedron *OA*_{1}*A*_{2}*A*_{3} is given. For *i*=1,2,3 let *B*_{i} be a point in the interior of the edge *OA*_{i} and let *C*_{i} be a point on the ray *OA*_{i}, beyond *A*_{i}. Suppose that the polyhedron bounded by the six planes *OA*_{i+1}*A*_{i+2} and *B*_{i}*A*_{i+1}*A*_{i+2} (*i*=1,2,3) circumscribes a sphere, and the polyhedron bounded by the planes *B*_{i}*A*_{i+1}*A*_{i+2} and *C*_{i}*A*_{i+1}*A*_{i+2} also circumscribes a sphere. Prove that the polyhedron bounded by the planes *OA*_{i+1}*A*_{i+2} and *C*_{i}*A*_{i+1}*A*_{i+2} also circumscribes a sphere.

(5 points)

**B. 4393.** Consider the vertices of a regular triangle, the centre of the triangle and all points that divide its sides in a 1:2 ratio. What is the largest number of points that can be selected out the points listed above, such that no three of them form a regular triangle? (Suggested by

*A. Demeter,* Matlap, Kolozsvár

(4 points)

**B. 4398.** *A*_{1}, *B*_{1} and *C*_{1} are given points on sides *BC*, *CA* and *AB*, respectively, of the acute-angled triangle *ABC*. Mark a point *A*_{2} on the circle *AC*_{1}*B*_{1}. Let *B*_{2} be the intersection (different from *C*_{1}) of the line *C*_{1}*A*_{2} with the circle *BA*_{1}*C*_{1}. Furthermore, let *C*_{2} be the intersection (different from *A*_{1}) of the line *A*_{1}*B*_{2} and the circle *CB*_{1}*A*_{1}. Prove that the points *A*_{2}, *B*_{1} and *C*_{2} are collinear.

(4 points)

**C. 1099.** In a quadrilateral *ABCD*, *M* is the midpoint of line segment *AB*, and *N* is the midpoint of line segment *CD*. The length of diagonals *AC* and *BD* is , and they enclose an angle of 60^{o}. Find the distance *MN*.

Suggested by *G. Dávid,* Matlap, Kolozsvár

(5 points)

**K. 309.** The plane is divided into four parts by two intersecting lines. The measures of the four angles in degrees are 5*x*-9, 4*x*+27, *x*+*y*-30, *y*-*x*+3*z*, not necessarily in this order. What may be the four angles?

(6 points)

This problem is for grade 9 students only.

**K. 310.** We found a method for multiplying two numbers between 90 and 100: Add the two numbers, subtract 100 from the sum, and then write after the result the product of the differences of the original numbers from 100. E.g. in the case of 97 and 94, 97+94-100=91 and (100-97)(100-94)=18, so the product is 9118 and indeed, 97×94=9118. Why is this a correct method? How can it be extended so that it can be used for multiplying any two two-digit numbers?

(6 points)

This problem is for grade 9 students only.