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Source: 2022 AIME II Problem 12

February 17, 2022
AMCAIMEAIME II

Problem Statement

Let a,b,x,a, b, x, and yy be real numbers with a>4a>4 and b>1b>1 such that x2a2+y2a216=(x20)2b21+(y11)2b2=1.\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1. Find the least possible value of a+b.a+b.