Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Prove it Croatia

Let aa, bb, cc and dd be four pairwise distinct real numbers. If aa and bb are solutions of the equation x2−10cx−11d=0x^2 - 10c x - 11d = 0, and cc and dd are solutions of the equation x2−10ax−11b=0x^2 - 10a x - 11b = 0, determine the sum a+b+c+da + b + c + d.

Solution

From Vieta's formulae we have a+b=10ca + b = 10c and c+d=10ac + d = 10a. By adding these equations we get
a+b+c+d=10(a+c). a + b + c + d = 10(a + c).
Since aa is a solution of the equation x2−10cx−11d=0x^2 - 10c x - 11d = 0, and d=10a−cd = 10a - c, it follows that
0=a2−10ac−11d=a2−10ac−11(10a−c)=a2−110a+11c−10ac. 0 = a^2 - 10a c - 11d = a^2 - 10a c - 11(10a - c) = a^2 - 110a + 11c - 10a c.
Analogously, we get
c2−110c+11a−10ac=0. c^2 - 110c + 11a - 10a c = 0.
By subtracting these equations it follows that
(a−c)(a+c−121)=0. (a - c)(a + c - 121) = 0.
Since a≠ca \neq c, we get a+c=121a + c = 121. Therefore, a+b+c+d=10⋅121=1210a + b + c + d = 10 \cdot 121 = 1210.

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