# Pre RMO more algebra problems for practice

Question I:

I rode one third of a journey at 10kmph, one third more at 9, and the rest at 8 kmph; if I had ridden half the journey at 10kmph, and the other half at 8 kmph, I should have been half a minute longer on the way: what distance did I ride?

Question 2:

The express train leaves Bristol at 3pm and reaches London at 6pm; the ordinary train leaves London at 1:30pm and arrives at Bristol at 6pm. If both trains travel uniformly, find the time when they will meet.

Question 3:

Solve (a) $0.\dot{6}x + 0.75x-0.1\dot{6} = x - 0.58\dot{3}x+5$

Solve (b) $\frac{37}{x^{2}-5x+6} + \frac{4}{x-2} = \frac{7}{3-x}$

Question 4:

Simplify: $(1+x)^{2} \div \{ 1 + \frac{x}{1-x+ \frac{x}{1+x+x^{2}}}\}$

Question 5:

Find the square root of $\frac{4a^{2}-12ab-6bc+4ac+9b^{2}+c^{2}}{4a^{2}+9c^{2}-12ac}$

Question 6:

Find the square root of $4a^{4}+9(a^{2}+\frac{1}{a^{2}})+12a(a^{2}+1)+18$

Question 7:

Solve the following system of equations:

$\frac{1}{3}(x+y)+2z=21$

$3x - \frac{1}{2}(y+z) = 65$

$x + \frac{1}{2}(x+y-z) = 38$

Question 8:

A number consists of three digits, the right hand one being zero. If the left hand and middle digits be interchanged the number is diminished by 180; if the left hand digit be halved and the other two digits are interchanged, the number is diminished by 336; find the number.

Question 9:

$\frac{2}{x^{2}+xy+y^{2}}$, $\frac{-4x}{x^{3}-y^{3}}$, $\frac{x^{2}}{y^{2}(x-y)^{2}}$, and $\frac{-x^{2}}{x^{3}y-y^{4}}$

Question 10:

Simplify:

$\frac{a^{3}+b^{3}}{a^{4}-b^{4}} - \frac{a+b}{a^{2}-b^{2}} -\frac{1}{2} \{ \frac{a-b}{a^{2}+b^{2}} - \frac{1}{a-b} \}$

More later,
Nalin Pithwa

# Another Romanian Mathematical Olympiad problem

Ref: Romanian Mathematical Olympiad — Final Round, 1994

Ref: Titu Andreescu

Problem:

Let M, N, P, Q, R, S be the midpoints of the sides AB, BC, CD, DE, EF, FA of a hexagon. Prove that

$RN^{2}=MQ^{2}+PS^{2}$ if and only if MQ is perpendicular to PS.

Proof:

Let a, b, c, d, e, f be the coordinates of the vertices of the hexagon. The points M, N, P, Q, R, and S have the coordinates

$m=\frac{a+b}{2}$, $n=\frac{b+c}{2}$, $=\frac{c+d}{2}$,

$q=\frac{d+e}{2}$, $r=\frac{e+f}{2}$, $s=\frac{f+a}{2}$, respectively.

Using the properties of the real product of complex numbers, (please fill in the gaps here), we have

$RN^{2}=MQ^{2}+PS^{2}$

if and only if

$(e+f-b-c).(e+f-b-c) = (d+e-a-b).(d+e-a-b)+(f+a-c-d).(f+a-c-d)$

That is,

$(d+e-a-b).(f+a-c-d)=0$

hence, MQ is perpendicular to PS, as claimed. QED.

More later,

Nalin Pithwa