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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.

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

Note by Tran Quoc Dat
3 months, 4 weeks ago