A Test for Grade 10 Gifted Math Students

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 $u_0=2^{2016} \times 14^{2017}$.