Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

3. Let points $A$ and $B$ lie on the parabola $y^{2}=6 x$ and the circle $\odot C:(x-4)^{2}+y^{2}=1$ , respectively. Then the range of $|A B|$ is $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

3. $[\sqrt{15}-1,+\infty)$ .

Since $|A B|_{\text {min }}=|A C|_{\text {min }}-1$ , we only need to consider the range of $|A C|$ .
Notice,
$\begin{array}{l} |A C|^{2}=(x-4)^{2}+y^{2} \\ =(x-4)^{2}+6 x \\ =x^{2}-2 x+16 \\ =(x-1)^{2}+15 . \end{array}$

Also, $x \geqslant 0$ , so $|A C|_{\min }=\sqrt{15}$ .
Therefore, the range of $|A B|$ is $[\sqrt{15}-1,+\infty)$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.