MathDB
1994 Chile Classification / Qualifying NMO VI

Source:

October 7, 2021
algebrageometrycombinatoricsnumber theorychilean NMO

Problem Statement

p1. The Fernández family, made up of father, mother and their three children: Pablo, Claudio and Carolina, subscribed for the year 1994 to the city newspaper. Subscription started on Saturday January 11st. The family agreed that each member would get up early to pick up the newspaper, alternately starting with paper and in the order indicated. How many days on Sunday it will be time for each family member?
p2. Given three lines in the plane, concurrent at a point OO, consider three consecutive angles that are formed between them (obviously they add up to 180o180^o). Take a point PP in the plane, out of those lines, and let AA, BB, CC be their orthogonal projections on the three lines. Prove that the △ABC\vartriangle ABC has the same angles as those that the lines form to each other.
p3. Consider ten positive, not necessarily distinct, integers that add up to 9595. Find the smallest possible value of the sum of its squares.
p4. In △ABC\vartriangle ABC, with right angle at CC, the altitude hc=CDh_c = CD is drawn. Let r,r1,r2r,r_1,r_2 be the inradii, of the triangles △ABC\vartriangle ABC, △ADC\vartriangle ADC, △BCD\vartriangle BCD, respectively. Prove that r+r1+r2=hcr + r_1 + r_2 = h_c
p5. Consider 8n−48n- 4 points on a grid, arranged in the shape of a cross, as shown in the figure. Determine the maximum number of squares with vertices at the 8n−48n-4 points. In the figure, they have outlined some cases. https://cdn.artofproblemsolving.com/attachments/c/d/e7090c54eaf2bd3749cdee2ff77172ada7866a.jpg
p6. Let AA and DD be two opposite vertices of a regular hexagon. A frog begins to jump to starting at AA. From any vertex in the hexagon, other than DD, the frog can jump to any vertex adjacent, but as soon as it reaches DD, it stops jumping (stops there). Let P(n)P (n) be the number of different paths, with exactly nn jumps, from AA to DD. Prove that: ∙\bullet P(n)=0P (n) = 0 if nn is even ∙\bullet P(2n+1)=2⋅3n−1P (2n + 1) = 2\cdot 3^{n- 1}
p7. Consider the product of all positive integers multiples of 66, which are less than thousand. Find the number of zeros this product ends with.
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