olymmath

KMO 2001

Korean Mathematical Olympiad 2001.

Other problems

Problem 2

Let PP be a given point inside a convex quadrilateral O1O2O3O4O_1O_2O_3O_4 . For each i=1,2,3,4i = 1,2,3,4 , consider the lines ll that pass through PP and meet the rays OiOi−1O_iO_{i-1} and OiOi+1O_iO_{i+1} (where O0=O4O_0 = O_4 and O5=O1O_5 = O_1 ) at distinct points Ai(l)A_i(l) and Bi(l)B_i(l) , respectively. Denote fi(l)=PAi(l)⋅PBi(l)f_i(l) = PA_i(l) \cdot PB_i(l) . Among all such lines ll , let lil_i be the one that minimizes fif_i . Show that if l1=l3l_1 = l_3 and l2=l4l_2 = l_4 , then the quadrilateral O1O2O3O4O_1O_2O_3O_4 is a parallelogram.

Problem 3

Let x1,x2,⋯ ,xnx_1,x_2, \cdots,x_n and y1,y2,⋯ ,yny_1,y_2, \cdots ,y_n be arbitrary real numbers satisfying x12+x22+⋯+xn2=y12+y22+⋯+yn2=1x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2=1 . Prove that (x1y2−x2y1)2≤2∣1−∑k=1nxkyk∣(x_1y_2-x_2y_1)^2 \le 2\left|1-\sum_{k=1}^n x_ky_k\right| and find all cases of equality.

Problem 4

For given positive integers nn and NN , let PnP_n be the set of all polynomials f(x)=a0+a1x+⋯+anxnf(x)=a_0+a_1x+\cdots+a_nx^n with integer coefficients such that:

[list] (a) ∣aj∣≤N|a_j| \le N for j=0,1,⋯ ,nj = 0,1, \cdots ,n ; (b) The set {j∣aj=N}\{ j \mid a_j = N\} has at most two elements. [/list]

Find the number of elements of the set {f(2N)∣f(x)∈Pn}\{f(2N) \mid f(x) \in P_n\} .

Problem 5

In a triangle ABCABC with ∠B<45∘\angle B < 45^{\circ} , DD is a point on BCBC such that the incenter of △ABD\triangle ABD coincides with the circumcenter OO of △ABC\triangle ABC . Let PP be the intersection point of the tangent lines to the circumcircle ω\omega of △AOC\triangle AOC at points AA and CC . The lines ADAD and COCO meet at QQ . The tangent to ω\omega at OO meets PQPQ at XX . Prove that XO=XDXO=XD .

Problem 6

For a positive integer n≥5n \ge 5 , let ai,bi(i=1,2,⋯ ,n)a_i,b_i (i = 1,2, \cdots ,n) be integers satisfying the following two conditions:

[list] (a) The pairs (ai,bi)(a_i,b_i) are distinct for i=1,2,⋯ ,ni = 1,2,\cdots,n ; (b) ∣a1b2−a2b1∣=∣a2b3−a3b2∣=⋯=∣anb1−a1bn∣=1|a_1b_2-a_2b_1| = |a_2b_3-a_3b_2| = \cdots = |a_nb_1-a_1b_n| = 1 . [/list]

Prove that there exist indices i,ji,j such that 1<∣i−j∣<n−11<|i-j|<n-1 and ∣aibj−ajbi∣=1|a_ib_j-a_jb_i|=1 .