Track / Stage 3 / 127 of 260 #127 of 1964

Problem 127

AMC 10/12, early questions

Geometry Difficulty 3.4 Find the answer

Given that point $A$ is a fixed point on the circle $O: x^2 + y^2 = 4$ , and point $B$ is a moving point on circle $O$ . If it satisfies $|\vec{AO} + \vec{BO}| = |\vec{AO} - \vec{BO}|$ , then $\vec{AO} \cdot \vec{AB} = \boxed{4}$ .

Your answer

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

Official solution

Analysis

This question examines the square of a vector being equal to the square of its magnitude and involves the operation of the dot product of vectors, making it a medium-level question.

From $|\vec{AO} + \vec{BO}| = |\vec{AO} - \vec{BO}|$ , we can deduce that $\vec{AO} \cdot \vec{BO} = 0$ , which means $\text{AO BO}$ . Therefore, $\triangle AOB$ is an isosceles right triangle with side length $2$ , from which we can derive the result.

Solution

Given $|\vec{AO} + \vec{BO}| = |\vec{AO} - \vec{BO}| \Rightarrow (\vec{AO} + \vec{BO})^2 = (\vec{AO} - \vec{BO})^2 \Rightarrow \vec{AO} \cdot \vec{BO} = 0$ ,

$\text{AO BO}$ ,
$\therefore \triangle AOB$ is an isosceles right triangle with side length $2$ ,

Thus, $\vec{AO} \cdot \vec{AB} = |\vec{AO}||\vec{AB}|\cos 45^{\circ} = 2 \times 2\sqrt{2} \times \frac{\sqrt{2}}{2} = 4$ .
Therefore, the answer is $\boxed{4}$ .

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