p1. Determine which of the following numbers is greater : $99!$ or $50^{99}$. p2. Let $x , y$ be positive real numbers with $x > y$ that satisfy the relation $x^2 +y^2 = axy$ where $a$ is a real number greater than $2$. Find all possible values of $a$ that make $\frac{x + y}{x - y}$ an integer. p3. In the figure below $ABCD$ is a square. Segment $AE$ is $3$ units and segment $EF$ is $1$ unit. The angles $\angle AED$ and $\angle BFA$ are right. Calculate the length of the segments $FC$ p4. A design $X$ is an array of the digits $1,2,..., 9$ in the shape of an $X$, for example, We will say that a design $X$ is balanced if the sum of the numbers of each of the diagonals match. Determine the number of designs $X$ that are balanced.
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Tags: algebra, geometry, combinatorics, number theory, chilean NMO
BackToSchool
02.09.2022 02:25
parmenides51 wrote: p1. Determine which of the following numbers is greater : $99!$ or $50^{99}$.
\begin{align*}
99! &= 99 \times 98 \times 97 \times \cdots \times 3 \times 2 \times 1 \\
&= (99 \times 1) \times (98 \times 2) \times (97 \times 3) \times \cdots \times (51 \times 49) \times 50 \\
&= (50^2 - 49^2) \times (50^2 - 48^2) \times (50^2 - 47^2) \times \cdots \times (50^2 - 1^2) \times 50 \\
&< 50^ {99}
\end{align*}
BackToSchool
02.09.2022 05:39
parmenides51 wrote: p3. In the figure below $ABCD$ is a square. Segment $AE$ is $3$ units and segment $EF$ is $1$ unit. The angles $\angle AED$ and $\angle BFA$ are right. Calculate the length of the segments $FC$
Note that
$$\begin{cases} \angle AED &= \angle BFA = 90^{o} \\ \angle ADE &= \angle BAF = 90^{o} - \angle DAE \\ AD &= BA \end{cases} \implies \triangle AED \cong \triangle BAF \text { by AAS}$$$$\implies BF = AE = 3$$$$\implies AB = \sqrt {3^2 + 4^2} = 5$$Now we can use coordinates to calculate the length of $FC$
$$A(0,0), B(5,0), C(5,5), D(0,5), F(\frac{16}{5}, \frac{12}{5})$$$$FC = d = \sqrt {(5-\frac{16}{5})^2 + (5- \frac{12}{5})^2 } =\sqrt{10}$$
The.Math.Terminator
02.09.2022 12:54
p2. I think that $a$ can not be real, only integer, because in this case we have infinite many solutions.
${{\left( \frac{x+y}{x-y} \right)}^{2}}=\frac{{{x}^{2}}+{{y}^{2}}+2xy}{{{x}^{2}}+{{y}^{2}}-2xy}=\frac{a+2}{a-2}=1+\frac{4}{a-2}\in \mathbb{N}\Rightarrow a-2\in \left\{ 1,2,4 \right\}\Rightarrow a\in \left\{ 3,4,6 \right\}$