[Anish Ray]

Solution of the 5th problem

Given, $$\frac{(a+b)-(c+d)}{\sqrt{2}}>\sqrt{a^2+b^2}-\sqrt{c^2+d^2}$$

$$=> (a-d)+(b-c)>\sqrt{2}\{\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\}$$

Since a>b>c>d>0,both the LHS and RHS of the above in-equation is positive we can square both the sides of the in-equation,

So,Squaring both sides we get

$$(a-d)^2+(b-c)^2+2(a-d)(b-c)>2(a^2+b^2+c^2+d^2)-4\sqrt{(a^2+b^2)(c^2+d^2)}$$

$$=> 2(a-d)(b-c)+4\sqrt{(a^2+b^2)(c^2+d^2)}>a^2+b^2+c^2+d^2+2ad+2bc$$

\(\ =(a+d)^2+(b+c)^2=2(b+c)^2\), Since, a+d=b+c.

$$=> 2(a+d-2d)(b-c)+4\sqrt{(a^2+b^2)(c^2+d^2)}>2(b+c)^2$$

$$=> 2(b+c-2d)(b-c)+4\sqrt{(a^2+b^2)(c^2+d^2)}>2(b+c)^2$$

$$=> 2(b^2-c^2)-4d(b-c)+4\sqrt{(a^2+b^2)(c^2+d^2)}>2(b+c)^2$$

$$=> 4\sqrt{(a^2+b^2)(c^2+d^2)}>2(b+c)^2-2(b^2-c^2)+4d(b-c)$$

$$=> 4\sqrt{(a^2+b^2)(c^2+d^2)}>4c(b+c)+4d(b-c)=4c(a+d)+4d(b-c)=4(ac+bd)$$

$$=> \sqrt{(a^2+b^2)(c^2+d^2)}>ac+bd$$

, which is Cauchy-Schwarz's inequality.So, $$\frac{(a+b)-(c+d)}{\sqrt{2}}>\sqrt{a^2+b^2}-\sqrt{c^2+d^2}$$

is true.

[Anish Ray]

RMO 2017 Maharashtra and Goa Region Paper

\(\ 1\).(\(\ 16\) marks)Consider a chessboard of size \(\ 8\) units\(\ \times8\) units (i.e., each small square on the board has a side length of \(\ 1\) unit).Let \(\ S\) be the set of all the \(\ 81\) vertices of all the squares on the board.What is the number of line segments whose vertices are in \(\ S\),and whose length is a positive integer. (The segments need not be parallel to the sides of the board.)

\(\ 2\).(\(\ 16\) marks)For any positive integer \(\ n\), let \(\ d(n)\) denotes the number of positive divisors of \(\ n\); and let \(\ \phi(n)\) denotes the number of elements from the set \(\ \{1,2,...,n\}\) that are co-prime to \(\ n\).(For example \(\ d(12)=6 \) and \(\ \phi(12)=4\).)

Find the smallest positive integer \(\ n\) such that \(\ d(\phi(n))=2017\).

\(\ 3\).(\(\ 16\) marks)Let \(\ P(x)\) and \(\ Q(x)\) be polynomials of  degree \(\ 6\) and degree \(\ 3\) respectively,such that:
\(\ P(x)>Q(x)^2+Q(x)+x^2-6\), for all \(\ x\in\mathbb{R}\).

If all the roots of \(\ P(x)\) are real numbers, then prove that there exist two roots of \(\ P(x)\), say alpha,beta, such that \(\ |\alpha-\beta|<1\).

\(\ 4\).(\(\ 16\) marks)Let \(\ l_1,l_2,l_3,\dots,l_{40}\) be forty parallel lines.As shown in the diagram, let m be another line that intersects the line \(\ l_1\) to \(\ l_{40}\) in the points \(\ A_1,A_2.A_3,\dots,A_40\) respectively.Similarly let n be another line that intersects the lines \(\ l_1\) to \(\ l_40\) in the points \(\ B_1,B_2,B_3,\dots,B_{40}\) respectively.

Given that \(\ A_1B_1=1\), \(\ A_{40}B_{40}=14\), and the areas of the \(\ 39\) trapeziums \(\ A_1B_1B_2A_2\),\(\ A_2B_2B_3A_3,\dots\),\(\ A_{39}B_{39}B_{40}A_{40}\) are all equal; then count the number of segments \(\ A_iB_i\) whose length is a positive integer; where \(\ i\in\{1,2,\dots,40\}\).

\(\ 5\).(\(\ 18\) marks)If \(\ a,b,c,d\in\mathbb{R}\) such that \(\ a>b>c>d>0\)  and \(\ a+d=b+c\);

then prove that :

$$\frac{(a+b)-(c+d)}{\sqrt{2}}>\sqrt{a^2+b^2}-\sqrt{c^2+d^2}$$

\(\ 6\).(\(\ 18\) marks)Let \(\ \triangle{ABC}\) be acute-angled; and let \(\ \Gamma\) be its circumcircle.Let \(\ D\) be a point on minor arc \(\ BC\) of \(\ \Gamma\).Let \(\ E\) and \(\ F\) be points on line \(\ AD\) and \(\ AC\) respectively, such that \(\ BE\perp AD\) and \(\ DF\perp AC\).Prove that \(\ EF\parallel BC\) if and only if \(\ D\) is the midpoint of \(\ BC\).

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