Maths Olympiad Prep

Track / Stage 4 / 246 of 340 #506 of 1964

Problem 506

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer

Given the sets
M={(x,y)∣x(x−1)⩽y(1−y)},N={(x,y)∣x2+y2⩽k}. \begin{array}{l} M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\ N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} . \end{array}

If M⊂NM \subset N, then the minimum value of kk is \qquad .
(2007, Shanghai Jiao Tong University Independent Admission Examination)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Notice that,
M={(x,y)∣ (x−12)2+(y−12)2⩽12} M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}

represents a disk with center (12,12)\left(\frac{1}{2}, \frac{1}{2}\right) and radius 22\frac{\sqrt{2}}{2}.
By M⊂N⇒k⩾2×12+22=2M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{1}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}
⇒kmin⁡=2.  \Rightarrow k_{\min }=2 \text {. }

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.