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 {(x2y)(x+my)=m22m3(y2x)(y+mx)=m22m3\begin{cases} \\(x-2y)(x+my)=m^2-2m-3 \\(y-2x)(y+mx)=m^2-2m-3 \end{cases} (x,yx,y are variables, mm is a parameter).

  • Solve the system when m=3m=-3.

  • Find mm so that the system has at least 1 solution (x0;y0)(x_0;y_0) satisfy x0>0x_0>0 and y0>0y_0>0.

2. Ugly Vieta's in Quadratic Equation:

Find a1a \geq 1 so that the equation ax2+(12a)x+1a=0ax^2+(1-2a)x+1-a=0 has 2 different solutions x1,x2x_1,x_2 satisfy x22ax1=a2a1x_2^{2}-ax_1=a^2-a-1.

3. Vieta Jumping, Where Are You?

Let x,yx,y be positive integers satisfy xyx2+y2+10xy|x^2+y^2+10.

  • Prove that x,yx,y are coprime odds.

  • Let k=x2+y2+10xyk=\dfrac{x^2+y^2+10}{xy}. Prove that 4k4|k and k12k \geq 12.

  • Bonus (not in the test): How many different values of kk are there?

4. Inequality:

Let xyzx \geq y \geq z be reals satisfy x+y+z=0x+y+z=0 and x2+y2+z2=6x^2+y^2+z^2=6.

  • Find S=(xy)2+(xy)(yz)+(yz)2S=(x-y)^2+(x-y)(y-z)+(y-z)^2.

  • Find the maximum value of P=(xy)(yz)(zx)P=|(x-y)(y-z)(z-x)|.

5. Geometry - Welcome to Russia:

Acute triangle ABCABC with BAC>45\angle BAC > 45^\circ. Draw squares ABMN,ACPQABMN, ACPQ outside the triangle ABCABC. AQAQ intersects BMBM at EE. ANAN intersects CPCP at FF.

  • Prove that triangle ABEABE and triangle ACFACF are similar. Prove that EFQNEFQN is concylic.

  • Let II be midpoint of segment EFEF. Prove that II is the circumcenter of triangle ABCABC.

  • MNMN intersects PQPQ at DD. (DMQ)(DMQ) and (DNP)(DNP) intersect at KDK \neq D. Tangents of (ABC)(ABC) at BB and CC intersect at JJ. Prove that D,A,K,JD, A, K, J are colinear.

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

6. Integral Sequence About Factorization:

For each positive integer xx greater than 11, let f(x)f(x) be the second greatest factor of xx.

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

Consider the following sequence {un}:{u0 is givenun+1=unf(un)\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 u0u_0, there exists an integer kk such that uk=1u_k=1.

  • (Is this too large?) Find kk if u0=22016×142017u_0=2^{2016} \times 14^{2017}.

Note by Tran Quoc Dat
3 years, 2 months ago

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