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China Mathematical Competition

Geometry Difficulty 4.8 AIME Prove it China

Given the line $L: x + y - 9 = 0$ and the circle $M: 2x^2 + 2y^2 - 8x - 8y - 1 = 0$ , point $A$ is on $L$ and points $B, C$ are on $M$ ; $\angle BAC = 45^\circ$ and the line $AB$ is through the center of $M$ . Then the range of the $x$ coordinate of point $A$ is ______.

Solution

Suppose that $A(a, 9-a)$ . Then the distance from the center of $M$ to the line $AC$ is
$d = |AM| \times \sin \angle BAC \\ = \sqrt{(a-2)^2 + (9-a-2)^2} \times \sin 45^\circ \\ = \sqrt{2a^2 - 18a + 53} \times \frac{\sqrt{2}}{2}.$
On the other hand, since the line $AC$ intercepts $M$ , it follows that $d \le$ the radius of $M = \sqrt{\frac{17}{2}}$ , i.e.
$\sqrt{2a^2 - 18a + 53} \times \frac{\sqrt{2}}{2} \le \sqrt{\frac{17}{2}}$
The solution is $3 \le a \le 6$ .

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