Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it Ireland

Show that the number
(25112525310363)+1251252+53+10363 \left( \frac{251}{\frac{1}{\sqrt{252-5\sqrt{3}}}} - \frac{10\sqrt{3}}{63} \right) + \frac{1}{\frac{251}{\sqrt{252+5\sqrt{3}}}} + \frac{10\sqrt{3}}{63}
is an integer and find its value.

Solution

Let a3=252a^3 = 252 and b3=250b^3 = 250, then ab=2523/2503=10633ab = \sqrt[3]{252^3/250} = 10\sqrt[3]{63}. Using that 251=a3+b32251 = \frac{a^3+b^3}{2} and 1=a3b321 = \frac{a^3-b^3}{2}, the given expression inside the bracket becomes

a3+b32a3b32ab+a3b32a3+b32+ab \frac{\frac{a^3+b^3}{2}}{\frac{a^3-b^3}{2} - ab} + \frac{\frac{a^3-b^3}{2}}{\frac{a^3+b^3}{2} + ab}

The second term is obtained from the first by replacing bb with b-b. Therefore, we consider only one of the two terms. The first term is equal to

a3+b3a3b3ab2ab=a3+b3a2+ab+b22ab=a3+b3a2ab+b2=a+b. \frac{a^3 + b^3}{\frac{a^3 - b^3}{a-b} - 2ab} = \frac{a^3 + b^3}{a^2 + ab + b^2 - 2ab} = \frac{a^3 + b^3}{a^2 - ab + b^2} = a + b.

Hence, the second term is equal to aba-b and the given number is equal to

((a+b)+(ab))3=(2a)3=8252=2016. ((a+b) + (a-b))^3 = (2a)^3 = 8 \cdot 252 = 2016.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.