p1. For how many natural numbers $n$, the numbers $n, 1 + 2n, 1 + 4n$ are all prime? p2. Find all the solutions in the real numbers of the following system: $x + y + z = 1 $ $x^2 + y^2 + z^2 = 1 $ $x^3 + y^3 + z^3 = 1$ p3. Consider a triangle $\vartriangle ABC$, right at $C$ and a point $P$ at a distance $4$ from vertex $A$, $7$ from vertex $B$, and $1$ from vertex $C$. What are the smallest and largest lengths that side $AC$ can have? p4. Let $ j, k$ be two positive integers. A pile of pencils, all of different colors, is distributed first in $j$ drawers and then the same pile in $k$ drawers. Determine the minimum number of pencils needed in the stack to guarantee that no matter how they are distributed, there will be two colors together in the same drawer both times. PS. Problem 1 was also used as Juniors p1.
Problem
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Tags: chilean NMO, geometry, algebra, number theory, combinatorics
The.Math.Terminator
04.09.2022 19:52
If $n\ne 2\Rightarrow n=3k\pm 1\Rightarrow $ one of the numbers $1\text{ }+\text{ }2n,\text{ }1\text{ }+\text{ }4n$ is divisible by 3
The.Math.Terminator
04.09.2022 19:56
Let ${{S}_{k}}={{x}^{k}}+{{y}^{k}}+{{z}^{k}}$ $S=xy+yz+zx,P=xyz$
Just use ${{(x+y+z)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xy+yz+zx)$ and ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)$ and determinate the ${{S}_{1}},S,P$ and write the equation ${{t}^{3}}-{{S}_{1}}{{t}^{2}}+St-P=0$ and solve this equation.
soryn
07.09.2022 17:16
P3 is interesting
soryn
02.10.2022 23:56
Plecase,any solution for p3!!
DouDragon
03.10.2022 00:07
Squaring the first then subtracting the second from that, we find $2(xy + yz + zx) = 0$ meaning $xy + yz + zx = 0.$ Multiplying the first two then subtracting the third from that, we find $x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2 = 0.$ Cubing the first then subtracting the third from that, we find $3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 6xyz = 0$ meaning $xyz = 0.$ Since $x + y + z = 1, xy + yz + zx = 0,$ and $xyz = 0,$ we can construct the polynomial $t^3 - t^2 = 0$ of which $x,y,$ and $z$ are the roots. It factors as $t \cdot t \cdot (t - 1)$ meaning that of $x,y,$ and $z,$ one is $1$ and the other two are $0.$ Plugging into the original system, we find that it works.
parmenides51
03.10.2022 00:16
an idea using analytic geometry for p3, incomplete
Let $A(0,0),C(O,a),B(b,a), D(b,0)$ and $P(x,y)$
We have the circles $C_1$: $x^2+y^2=4^2$
$C_2$: $x^2+(y-a)^2=1^2$
$C_2$: $(x-b)^2+(y-a)^2=7^2$
wanted is min and max of $a\ge 0$ under the conditions that those 3 circles are concurrent, that is their system of equations has a unique solution
I can't go any further than this
Shroud666
04.10.2022 09:45
Solution :
Consider P to be a Fixed Point Say (0,0)
Now draw 3 Concentric Circles of Radius {1,4,7}
Assume any Point Randomly on smallest circle as C (For the sake of simplicity let it be (1,0) ( Now drawing any 2 perpendicular lines passing through C is bound to intersect the remaining 2 circles ( Since C is inside of those) at A pair each for each circle ( Total 8 Points ) )
These points are all The possible Configurations of A and B . ( Sorry Can't add a diagram yet )
Now to Maximise and Minimise AC Not we only have two consider The X axis as one line and x = 1 as the other perpendicular Line )
So AC ( max ) is 4 + 1 = 5
Anand AC ( min ) is 4-1 = 3
soryn
04.10.2022 22:39
Thank you,Shroud666!
PM.MATHLEAGUE.VA.CHAMP
05.10.2022 04:35
WLOG, let the solutions be the roots of monic polynomial $p(n)=n^3+bn^2+cn+d$. $x+y+z=-b=1\implies b=-1$ by Vieta's. $(x+y+z)^2-2(xy+yz+xz)=x^2+y^2+z^2=1\implies xy+yz+xz=c=0$, by Vieta's. We now have simplified to $p(n)=n^3-n^2+d$; we can finish with Newton's sums ($x^0+y^0+z^0=3$ obviously for reference). Thus, $1(1)+1(-1)+1(0)+3(d)=0\implies d=0\implies p(n)=n^3-n^2=n^2(n-1)=0\implies n\in(0,1)\implies (x,y,z)$ is some permutation of $\boxed{(0,0,1)}$