olymmath

HMMT 2002 K6

  • Calculus
  • Answer

Determine the positive value of aa such that the parabola y=x2+1y=x^{2}+1 bisects the area of the rectangle with vertices (0,0),(a,0),(0,a2+1)(0,0),(a, 0),\left(0, a^{2}+1\right), and (a,a2+1)\left(a, a^{2}+1\right).

Dataset text NuminaMath olympiads. Quoted with source.

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\sqrt{3}

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