MathDB

1995 IberoAmerican

Part of IberoAmerican

Subcontests

(3)
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10th ibmo - chile 1995/q3.

Let r r and s s two orthogonal lines that does not lay on the same plane. Let AB AB be their common perpendicular, where A∈r A\in{}r and B∈s B\in{}s(*).Consider the sphere of diameter AB AB. The points M∈r M\in{r} and N∈s N\in{s} varies with the condition that MN MN is tangent to the sphere on the point T T. Find the locus of T T. Note: The plane that contains B B and r r is perpendicular to s s.

10th ibmo - chile 1995/q6.

A function f:N→Nf: \N\rightarrow\N is circular if for every p∈Np\in\N there exists n∈N, n≤pn\in\N,\ n\leq{p} such that fn(p)=pf^n(p)=p (ff composed with itself nn times) The function ff has repulsion degree k>0k>0 if for every p∈Np\in\N fi(p)≠pf^i(p)\neq{p} for every i=1,2,…,⌊kp⌋i=1,2,\dots,\lfloor{kp}\rfloor. Determine the maximum repulsion degree can have a circular function.
Note: Here ⌊x⌋\lfloor{x}\rfloor is the integer part of xx.
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10th ibmo - chile 1995/q1.

Find all the possible values of the sum of the digits of all the perfect squares.
[Commented by djimenez] Comment: I would rewrite it as follows: Let f:N→Nf: \mathbb{N}\rightarrow \mathbb{N} such that f(n)f(n) is the sum of all the digits of the number n2n^2. Find the image of ff (where, by image it is understood the set of all xx such that exists an nn with f(n)=xf(n)=x).

10th ibmo - chile 1995/q4.

In a m×nm\times{n} grid are there are token. Every token dominates every square on its same row (↔\leftrightarrow), its same column (↕\updownarrow), and diagonal (↘↖\searrow\hspace{-4.45mm}\nwarrow)(Note that the token does not \emph{dominate} the diagonal (↗↙\nearrow\hspace{-4.45mm}\swarrow), determine the lowest number of tokens that must be on the board to dominate all the squares on the board.
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Interesting problem of a triangle (equal chords)

The incircle of a triangle ABCABC touches the sides BCBC, CACA, ABAB at the points DD, EE, FF respectively. Let the line ADAD intersect this incircle of triangle ABCABC at a point XX (apart from DD). Assume that this point XX is the midpoint of the segment ADAD, this means, AX=XDAX = XD. Let the line BXBX meet the incircle of triangle ABCABC at a point YY (apart from XX), and let the line CXCX meet the incircle of triangle ABCABC at a point ZZ (apart from XX). Show that EY=FZEY = FZ.

10th ibmo - chile 1995/q2.

Let nn be a positive integer greater than 1. Determine all the collections of real numbers x_1,\ x_2,\dots,\ x_n\geq1\mbox{ and }x_{n+1}\leq0 such that the next two conditions hold:
(i) x112+x232+⋯+xnn−12=nxn+112x_1^{\frac12}+x_2^{\frac32}+\cdots+x_n^{n-\frac12}= nx_{n+1}^\frac12
(ii) x1+x2+⋯+xnn=xn+1\frac{x_1+x_2+\cdots+x_n}{n}=x_{n+1}