Subcontests
(8)Circumcircle is tangent to another incircle
Given a triangle ABC. Its incircle is tangent to BC,CA,AB at D,E,F respectively. Let K,L be points on CA,AB respectively such that K=A=L,∠EDK=∠ADE,∠FDL=∠ADF. Prove that the circumcircle of AKL is tangent to the incircle of ABC. 2011 divides a_{k-1}a_k - k for all k = 1,2,...,2010
Let a sequence of integers a0,a1,a2,⋯,a2010 such that a0=1 and 2011 divides ak−1ak−k for all k=1,2,⋯,2010. Prove that 2011 divides a2010+1. 1x2xn blocks showing a total of larger than 2011
[asy]
draw((0,1)--(4,1)--(4,2)--(0,2)--cycle);
draw((2,0)--(3,0)--(3,3)--(2,3)--cycle);
draw((1,1)--(1,2));
label("1",(0.5,1.5));
label("2",(1.5,1.5));
label("32",(2.5,1.5));
label("16",(3.5,1.5));
label("8",(2.5,0.5));
label("6",(2.5,2.5));
[/asy]
The image above is a net of a unit cube. Let n be a positive integer, and let 2n such cubes are placed to build a 1×2×n cuboid which is placed on a floor. Let S be the sum of all numbers on the block visible (not facing the floor). Find the minimum value of n such that there exists such cuboid and its placement on the floor so S>2011. Permutations of (1,...,n) such that sum of a_i/i is integer
For each positive integer n, let sn be the number of permutations (a1,a2,⋯,an) of (1,2,⋯,n) such that 1a1+2a2+⋯+nan is a positive integer. Prove that s2n≥n for all positive integer n. Sum of all numbers possible by removing digits of n
For a number n in base 10, let f(n) be the sum of all numbers possible by removing some digits of n (including none and all). For example, if n=1234, f(n)=1234+123+124+134+234+12+13+14+23+24+34+1+2+3+4=1979; this is formed by taking the sums of all numbers obtained when removing no digit from n (1234), removing one digit from n (123, 124, 134, 234), removing two digits from n (12, 13, 14, 23, 24, 34), removing three digits from n (1, 2, 3, 4), and removing all digits from n (0). If p is a 2011-digit integer, prove that f(p)−p is divisible by 9.Remark: If a number appears twice or more, it is counted as many times as it appears. For example, with the number 101, 1 appears three times (by removing the first digit, giving 01 which is equal to 1, removing the first two digits, or removing the last two digits), so it is counted three times.