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I.S.I. and C.M.I. Entrance Math Olympiad

Real Surds – Problem 2 Pre RMO 2017

Suppose \(a,b\) are positive real numbers such that \(a\sqrt{a}+b\sqrt{b}=183\). \(a\sqrt{b}+b\sqrt{a}=182\). Find \(\frac{9}{5}(a+b)\).
This problem will use the following elementary algebraic identity: $$ (x+y)^3 = x^3 + y^3 + 3x^2y + 3xy^2 $$ Can you identify what is x and what is y?
Set \( x = \sqrt a, y = \sqrt b \).  Then the given information translates to $$ x^3 + y^3 = 183 , x^2y + xy^2 = 182 $$ This implies \( (x + y)^3 \\ = ( \sqrt a + \sqrt b)^3 \\ = x^3 + y^3 + 3(x^2y + xy^2) \\ = 183 + 3 \times 182 \\=  729 \) Finally taking cube root on both sides, we have \( \sqrt a + \sqrt b = 9 \)
Note that \( \sqrt a b + a \sqrt b = 182 \\ \Rightarrow \sqrt a \sqrt b ( \sqrt a + \sqrt b ) = 182 \\ \Rightarrow \sqrt {ab} \times 9 = 182 \) So at this point we know \( ( \sqrt a + \sqrt b ) = 9, \sqrt{ab} = \frac{182}{9} \). It should be easy to find the value of \( \frac{9}(5) (a+b) \) from these relations.  
\( a+b \\ = (\sqrt a + \sqrt b )^2  – 2 \sqrt{ab} \\ = 9^2 – 2 \times \frac {182}{9} \\ = \frac{365}{9} \) Hence \( \frac {9}{5}(a+b) =  \frac {9}{5} \times \frac {365}{9} = 73 \)

By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

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