RMO 2016 Delhi Region

Problem 1

We can see that as we move the point $P$ on the circumference of the circle$[$excluding $X$ and $Y],$the $\angle XPY=\angle XP_1Y$ remains constant.So this shows that $AB=A_1B_1.$Now we use extended sin rule to complete the problem.
Let the circum-radius of $\triangle PAB$ be $R$ and $\triangle P_1A_1B_1$ be $R_1.$
In $\triangle PAB, \frac{AB}{sin\angle P}=2R$ and in $\triangle P_1A_1B_1,\frac{A_1B_1}{sin\angle P_1}=\frac{AB}{sin\angle P}=2R_1.$ Therefore $2R=2R_1\Rightarrow\boxed {R=R_1}.$Hence Proved.

Problem 5 1)its the easiest- use the identity ${ (x+y+z) }^{ 2 }+{ (x+y-z) }^{ 2 }+{ (x-y+z) }^{ 2 }+{ (-x+y+z) }^{ 2 }={ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }$ which gives infinite solutions in integers x,y,z.u can check it satisfis all other cond.s 2)solve this one other way round, assume 2016 = ${ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }$ then it implies 504 = ${ (x) }^{ 2 }+{ (y) }^{ 2 }+{ (z) }^{ 2 }$ by inspection x=22 y=4 z=6, then use the identity Problem 2 -we obtain nth term in terms of n,a,b then use strong induction.

Can anyone post complete solution to 2nd and last problem?

Problem 4

By the Cauchy-Schwarz inequality, $\left(\dfrac{1}{a}+1+c\right)(a^{3}+b^{2}+c) \geq (a+b+c)^{2}=9$

$\implies (ac+a+1)(a^{3}+b^{2}+c) \geq 9a$

Thus, we get

$f(a,b,c)=\quad \sum {\dfrac{(ac+a+1)a}{(ac+a+1)({a}^{3}+{b}^{2}+{c})}} \leq \quad \sum \dfrac{a(ac+a+1)}{9a}$

$\implies f(a,b,c) \leq \dfrac{1}{9}[(ac+a+1)+(ba+b+1)+(cb+c+1)]$

$=\dfrac{1}{9}[(a+b+c)+(1+1+1)+(ab+bc+ca)] \leq \dfrac{1}{9}[3+3+3]=\boxed{1}$

Here we used the identity $(a+b+c)^{2} \geq 3(ab+bc+ca)$ to get $ab+bc+ca \leq 3$

Solution to Problem 1. Let $P_{1}$ and $P_{2}$ be the two points on $\odot\omega_{1}$ as $P$ varies across the circumference of $\odot \omega_{1}.$ Similarly, let $P_{1}X$ and $P_{1}Y$ meet $\odot\omega_{2}$ at $A_{1}$ , $B_{1}$ and $P_{2}X$ and $P_{2}Y$ meet $\odot\omega2$ at $A_{2}$ , $B_{2}$ respectively.

Clearly , $\angle XP_{1}Y = \angle XP_{2}Y$ and $\angle XB_{1}Y = \angle XB_{2}Y$

Adding these two $eq^{n}$ we get:: $\angle A_{1}XB_{1} = \angle A_{2}XB_{2}$ => $A_{1}B_{1} = A_{2}B_{2}.$

Also, $\Delta XP_{1}Y \sim \Delta A_{1}P_{1}B_{1}$ => $\dfrac{Radius XP_{1}Y}{Radius A_{1}P_{1}B_{1}}$ = $\dfrac{XY}{A_{1}B_{1}}$ ........[ $1$ ]

Similarly, $\Delta XP_{2}Y \sim \Delta A_{2}P_{2}B_{2}$

=> $\dfrac{Radius XP_{2}Y}{Radius A_{2}P_{2}B_{2}}$ = $\dfrac{XY}{A_{2}B_{2}}$

=> $\dfrac{Radius XP_{1}Y}{Radius A_{2}P_{2}B_{2}}$ = $\dfrac{XY}{A_{1}B_{1}}$ ........[ $2$ ]

[Because $\Delta XP_{1}Y$ and $\triangle XP_{2}Y$ have the same circumcircle. & $A_{1}B_{1} = A_{2}B_{2}$]

From $eq^{n}$ $[1]$ and $[2]$ we get that :: Radius $A_{1}P_{1}B_{1}$ = Radius $A_{2}P_{2}B_{2}.$

$K.I.P.K.I.G$

Diagram to problem 1.........

Diagram to problem $3$

Need a solution for q 2 and 6

Please explain the statement above the concluding statement