p1. On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$. p2. Consider a triangle $\triangle ABC$ and a point $D$ in segment $BC$. The triangles $\triangle ABD$ and $\triangle ADC$ are similar in ratio $\frac{1}{\sqrt3}$. Determine the angles of the triangle $\triangle ABC$. p3. Consider a horizontal line $L$ with $20$ different points $P_1, P_2,..., P_{20}$ on her. For each pair of points $P_i$,$P_j$ a circle is drawn such that the segment $P_iP_j$ is a diameter. Determine the maximum number of intersections between circles that can occur, considering only those that occur strictly above $L$. p4. The bottle in the figure has a circular base, and the bottom of it is a perfect cylinder. The upper part is not very well defined. With the aid of a graded ruler (with which you can measure distances), and a water tap, propose a method that allows you to estimate very precisely the total volume of the bottle. p5. A quadrilateral $ABCD$ is inscribed in a circle. Suppose that $|DA| =|BC|= 2$ and$ |AB| = 4$. Let $E $be the point of intersection of lines$ BC$ and $DA$. Suppose that $\angle AEB = 60^o$ and that $|CD| <|AB|$. Calculate the radius of the circle. p6. Determine all triples of positive integers $(p, n, m)$, with $p$ a prime number, which satisfy the equation $p^m- n^3 = 27$
Problem
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Tags: algebra, geometry, combinatorics, number theory, chilean NMO
BackToSchool
26.10.2021 16:58
Rewrite the equation and factorize as:
\begin{align*}
p^m- n^3 = 27 \Rightarrow p^m &= n^3+27 \\
\Rightarrow p^m &= (n+3)(n^2-3n+9)
\end{align*}
If $p=2$, then $n+3$ and $n^2-3n+9$ should be exponents of $2$.
Let $n+3=2^k$ for some integer $k$, we have $n=2^k-3$ .
\begin{align*}
p^m &= (n+3)(n^2-3n+9) \\
\Rightarrow 2^m &= 2^k[(2^k-3)(2^k-6)+9)]
\end{align*}However, $2\nmid (2^k-3)(2^k-6)+9$. Contradiction!
If $p=3$, then $n+3$ and $n^2-3n+9$ should be exponents of $3$.
Let $n+3=3^k \rightarrow n=3^k-3$ for some integer $k \ge 2$.
\begin{align*}
p^m &= (n+3)(n^2-3n+9) \\
\Rightarrow 3^m &=3^k[(3^k-3)^2-3(3^k-3)+9] \\
&=3^k(3^{2k}-9\cdot 3^k+27) \\
&=3^{k+3}\cdot (3^{2k-3}-3^{k-1}+1)
\end{align*}We have $3^{2k-3}-3^{k-1}=0$ or $-3^{k-1}+1=0$.
$$3^{2k-3}-3^{k-1} =0 \Rightarrow 2k-3 = k-1 \Rightarrow k = 2, n = 6, m = 5 \Rightarrow (p,n,m) = (3, 6, 5) $$$$-3^{k-1}+1 = 0 \Rightarrow k-1 = 0 \Rightarrow k = 1, n=0, m=3 \Rightarrow (p,n,m) = (3, 0, 3) $$However, the latter is not a positive integer solution.
BackToSchool
27.10.2021 08:45
For any four points $A_1,A_2, A_3, A_4$ in order (say $L\to R$), two circles with line segments $\overline {A_1A_3}$ and $\overline {A_2A_4}$ as diameters shall intersect each other at two points, one above line $L$ and the other one below the line $L$. Here is looking for the number of the intersections above the line $L$.
It turns to find the number of ways to choose $4$ points out of 20, which equals $\frac {20!}{4!16!}=\boxed {4845}$