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Algebra Difficulty 4.8 AIME Prove it North Macedonia

Find all real numbers a,b,c,da, b, c, d such that
a+b+c+d=20a + b + c + d = 20
and
ab+ac+ad+bc+bd+cd=150. ab + ac + ad + bc + bd + cd = 150.

Solution

400=(a+b+c+d)2=a2+b2+c2+d2+2⋅150 so a2+b2+c2+d2=100. 400 = (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2 \cdot 150 \text{ so } a^2 + b^2 + c^2 + d^2 = 100.
Now
(a−b)2+(a−c)2+(a−d)2+(b−c)2+(b−d)2+(c−d)2==3(a2+b2+c2+d2)−2(ab+ac+ad+bc+bd+cd)==300−300=0 \begin{align*} (a - b)^2 + (a - c)^2 + (a - d)^2 + (b - c)^2 + (b - d)^2 + (c - d)^2 &= \\ &= 3(a^2 + b^2 + c^2 + d^2) - 2(ab + ac + ad + bc + bd + cd) = \\ &= 300 - 300 = 0 \end{align*}
Thus a=b=c=d=5a = b = c = d = 5.

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