View source for MIE 2016

Note: Anyone that solve any of the problems can post your solutions.

===Problem 1===
Choose the correct answer.

(a) <math>\sqrt{2016}-\sqrt{2015}<\sqrt{2017}-\sqrt{2016}<(2\sqrt{2016})^{-1}</math>

(b) <math>\sqrt{2017}-\sqrt{2016}<\sqrt{2016}-\sqrt{2015}<(2\sqrt{2016})^{-1}</math>

(c) <math>\sqrt{2017}-\sqrt{2016}<(2\sqrt{2016})^{-1}<\sqrt{2016}-\sqrt{2015}</math>

(d) <math>\sqrt{2016}-\sqrt{2015}<(2\sqrt{2016})^{-1}<\sqrt{2017}-\sqrt{2016}</math>

(e) <math>(2\sqrt{2016})^{-1}<\sqrt{2017}-\sqrt{2016}<\sqrt{2016}-\sqrt{2015}</math>

===Problem 2===
The following system has <math>k</math> integer solutions. We can say that:

<math>\begin{cases}\frac{x^2-2x-14}{x}>3\\\\x\leq12\end{cases}</math>


(a) <math>0\leq k\leq 2</math>

(b) <math>2\leq k\leq 4</math>

(c) <math>4\leq k\leq6</math>

(d) <math>6\leq k\leq8</math>

(e) <math>k\geq8</math>

===Problem 3===
Let <math>Z_1</math> and <math>Z_2</math> be complex numbers such that <math>Z_2</math> is a pure imaginary number and <math>|Z_1-Z_2|=|Z_2|</math>. For any values of <math>Z_1</math> and <math>Z_2</math> that satisfies these conditions we have:

(a) <math>\mbox{Im}(Z_2)>0</math>

(b) <math>\text{Im}(Z_2)\leq0</math>

(c) <math>|Z_1|\leq2|Z_2|</math>

(d) <math>\text{Re}(Z_1)\geq0</math>

(e) <math>\text{Re}(Z_1)\geq\text{Im}(Z_2)</math>

===Problem 4===
In the expansion of 

<math>\left(x\sin2\beta+\frac{1}{x}\cos2\beta\right)^{10}</math>

the independent term (in other words, the term without <math>x</math>) is equal to <math>63/256</math>. With <math>\beta</math> being a real number such that <math>0< \beta<\pi/8</math> and <math>x\neq0</math>, the value of <math>\beta</math> is:


(a) <math>\frac{\pi}{9}</math>

(b) <math>\frac{\pi}{12}</math>

(c) <math>\frac{\pi}{16}</math>

(d) <math>\frac{\pi}{18}</math>

(e) <math>\frac{\pi}{24}</math>

===Problem 5===
Compute <math>\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}</math>, knowing that <math>\sin\alpha\cos\alpha=\frac{1}{5}</math>.


(a) <math>\frac{22}{21}</math>

(b) <math>\frac{23}{22}</math>

(c) <math>\frac{25}{23}</math>

(d) <math>\frac{13}{12}</math>

(e) <math>\frac{26}{25}</math>

===Problem 6===
Let <math>A</math> be <math>A=\begin{vmatrix}1&a&-2\\a-2&1&1\\2&-3&1\end{vmatrix}</math> with <math>a\in\mathbb{R}</math>. We know that <math>\text{det}(A^2-2A+I)=16</math>. The sum of the values of <math>a</math> that satisfies this condition is:

(a) <math>0</math>

(b) <math>1</math>

(c) <math>2</math>

(d) <math>3</math>

(e) <math>4</math>

Note: <math>\text{det}(X)</math> is the determinant of the matrix <math>X</math>.

===Problem 7===
The product of the real roots of the following equation is equal to:

<math>y^{\log_3\sqrt{3y}}=y^{\log_33y}-6,~~y>0</math>


(a) <math>\frac{1}{3}</math>

(b) <math>\frac{1}{2}</math>

(c) <math>\frac{3}{4}</math>

(d) <math>2</math>

(e) <math>3</math>

===Problem 8===
Let <math>f(x)=\sqrt{|x-1|+|x-2|+|x-3|+...+|x-2017|}</math>. The minimum value of <math>f(x)</math> is in the interval:

(a) <math>(-\infty,1008]</math>

(b) <math>(1008,1009]</math>

(c) <math>(1009,1010]</math>

(d) <math>(1010,1011]</math>

(e) <math>(1011,+\infty)</math>

===Problem 9===
Let <math>x</math>, <math>y</math> and <math>z</math> be complex numbers that satisfies the following system:

<math>\begin{cases}x+y+z=7\\x^2+y^2+z^2=25\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\end{cases}</math>


Compute <math>x^3+y^3+z^3</math>.


(a) <math>210</math>

(b) <math>235</math>

(c) <math>250</math>

(d) <math>320</math>

(e) <math>325</math>

===Problem 10===
A hexagon is divided into 6 equilateral triangles. How many ways can we put the numbers from 1 to 6 in each triangle, without repetition, such that the sum of the numbers of three adjacent triangles is always a multiple of 3? Solutions obtained by rotation or reflection are differents, thus the following figures represent two distinct solutions.

[[File:MIE_2016_problem_10.png]]


(a) <math>12</math>

(b) <math>24</math>

(c) <math>36</math>

(d) <math>48</math>

(e) <math>96</math>

===Problem 11===
Let <math>(a_1,a_2,a_3,a_4,...)</math> be an arithmetic progression and <math>(b_1,b_2,b_3,b_4,...)</math>, an geometric progression of integer terms, of ratio <math>r</math> and <math>q</math>, respectively, where <math>r</math> and <math>q</math> are positive integers, with <math>q>2</math> and <math>b_1>0</math>. We also know that <math>a_1+b_2=3</math> and <math>a_4+b_3=26</math>. The value of <math>b_1</math> is:

(a) <math>1</math>

(b) <math>2</math>

(c) <math>3</math>

(d) <math>4</math>

(e) <math>5</math>



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