**Source:** Grade 10 Selection Test for Gifted Math Students, Academic Year 2016 - 2017, High School for the Talented, HCMC, Vietnam.

Here I post the entire test. Among of them are some problems which I couldn't solve when doing the test (so hard!). Try to solve if you are interested.

**1. Type-2 Symmetric System of Equations:**

Given this system \(\begin{cases} \\(x-2y)(x+my)=m^2-2m-3 \\(y-2x)(y+mx)=m^2-2m-3 \end{cases}\) (\(x,y\) are variables, \(m\) is a parameter).

Solve the system when \(m=-3\).

Find \(m\) so that the system has at least 1 solution \((x_0;y_0)\) satisfy \(x_0>0\) and \(y_0>0\).

**2. Ugly Vieta's in Quadratic Equation:**

Find \(a \geq 1\) so that the equation \(ax^2+(1-2a)x+1-a=0\) has 2 different solutions \(x_1,x_2\) satisfy \(x_2^{2}-ax_1=a^2-a-1\).

**3. Vieta Jumping, Where Are You?**

Let \(x,y\) be positive integers satisfy \(xy|x^2+y^2+10\).

Prove that \(x,y\) are coprime odds.

Let \(k=\dfrac{x^2+y^2+10}{xy}\). Prove that \(4|k\) and \(k \geq 12\).

**Bonus (not in the test):**How many different values of \(k\) are there?

**4. Inequality:**

Let \(x \geq y \geq z\) be reals satisfy \(x+y+z=0\) and \(x^2+y^2+z^2=6\).

Find \(S=(x-y)^2+(x-y)(y-z)+(y-z)^2\).

Find the maximum value of \(P=|(x-y)(y-z)(z-x)|\).

**5. Geometry - Welcome to Russia:**

Acute triangle \(ABC\) with \(\angle BAC > 45^\circ\). Draw squares \(ABMN, ACPQ\) outside the triangle \(ABC\). \(AQ\) intersects \(BM\) at \(E\). \(AN\) intersects \(CP\) at \(F\).

Prove that triangle \(ABE\) and triangle \(ACF\) are similar. Prove that \(EFQN\) is concylic.

Let \(I\) be midpoint of segment \(EF\). Prove that \(I\) is the circumcenter of triangle \(ABC\).

\(MN\) intersects \(PQ\) at \(D\). \((DMQ)\) and \((DNP)\) intersect at \(K \neq D\). Tangents of \((ABC)\) at \(B\) and \(C\) intersect at \(J\). Prove that \(D, A, K, J\) are colinear.

**Clarification:** \((\cdot)\) denotes the circumcircle of the figure.

**6. Integral Sequence About Factorization:**

For each positive integer \(x\) greater than \(1\), let \(f(x)\) be the **second greatest factor** of \(x\).

For example, \(f(10)=5, f(45)=15, f(p)=1\) for any prime \(p\).

Consider the following sequence \(\left\{ u_n \right\}: \begin{cases} \\u_0 \text{ is given} \\u_{n+1}=u_n-f(u_n) \end{cases}\).

Prove that for any given value of \(u_0\), there exists an integer \(k\) such that \(u_k=1\).

(

**Is this too large?**) Find \(k\) if \(u_0=2^{2016} \times 14^{2017}\).

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