MathDB

2012 Dutch Mathematical Olympiad

Part of Dutch Mathematical Olympiad

Subcontests

(5)
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numbers from 1 to n in each column and row of a nxn board, grey cells

We number the columns of an n×nn\times n-board from 11 to nn. In each cell, we place a number. This is done in such a way that each row precisely contains the numbers 11 to nn (in some order), and also each column contains the numbers 11 to nn (in some order). Next, each cell that contains a number greater than the cell's column number, is coloured grey. In the figure below you can see an example for the case n=3n = 3.
[asy] unitsize(0.6 cm);
int i;
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.8)); fill(shift((1,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), gray(0.8)); fill(shift((0,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), gray(0.8));
for (i = 0; i <= 3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); }
label("11", (0.5,3.5)); label("22", (1.5,3.5)); label("33", (2.5,3.5)); label("33", (0.5,2.5)); label("11", (1.5,2.5)); label("22", (2.5,2.5)); label("11", (0.5,1.5)); label("22", (1.5,1.5)); label("33", (2.5,1.5)); label("22", (0.5,0.5)); label("33", (1.5,0.5)); label("11", (2.5,0.5)); [/asy]
(a) Suppose that n=5n = 5. Can the numbers be placed in such a way that each row contains the same number of grey cells? (b) Suppose that n=10n = 10. Can the numbers be placed in such a way that each row contains the same number of grey cells?
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perpendiculars given, parallel wanted

We are given an acute triangle ABCABC and points DD on BCBC and EE on ACAC such that ADAD is perpendicular to BCBC and BEBE is perpendicular to ACAC. The intersection of ADAD and BEBE is called HH. A line through HH intersects line segment BCBC in PP, and intersects line segment ACAC in QQ. Furthermore, KK is a point on BEBE such that PKPK is perpendicular to BEBE, and LL is a point on ADAD such that QLQL is perpendicular to ADAD. Prove that DKDK and ELEL are parallel. [asy] unitsize(1 cm);
pair A, B, C, D, E, H, K, L, P, Q;
A = (0,0); B = (6,0); C = (3.5,4); D = (A + reflect(B,C)*(A))/2; E = (B + reflect(A,C)*(B))/2; H = extension(A, D, B, E); P = extension(H, H + dir(-10), B, C); Q = extension(H, H + dir(-10), A, C); K = (P + reflect(B,E)*(P))/2; L = (Q + reflect(A,D)*(Q))/2;
draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(K--P--Q--L); draw(rightanglemark(B,D,A,5)); draw(rightanglemark(B,E,A,5)); draw(rightanglemark(P,K,B,5)); draw(rightanglemark(A,L,Q,5));
dot("AA", A, SW); dot("BB", B, SE); dot("CC", C, N); dot("DD", D, NE); dot("EE", E, NW); dot("HH", H, N); dot("KK", K, SW); dot("LL", L, SE); dot("PP", P, NE); dot("QQ", Q, NW); [/asy]
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