Track / Stage 3 / 205 of 260 #845 of 2604

Problem 845

AMC 10/12, early questions

Geometry Difficulty 3.8 Find the answer China Mathematical Competition · China

Let $\triangle ABC$ be a given triangle. If $|\vec{BA} - t \vec{BC}| \ge |\vec{AC}|$ for any $t \in \mathbb{R}$ , then $\triangle ABC$ is ( ).

This was a multiple-choice question, but the options didn't survive into the source we have, so there is nothing here to pick from. Work it on paper and mark yourself against the solution below.

Official solution

Suppose $\angle ABC = \alpha$ . Since $|\vec{BA} - t \vec{BC}| \ge |\vec{AC}|$ , we have
$|\vec{BA}|^2 - 2t \vec{BA} \cdot \vec{BC} + t^2 |\vec{BC}|^2 \ge |\vec{AC}|^2.$
Let
$t = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BC}|^2},$
we get
$|\vec{BA}|^2 - 2|\vec{BA}|^2 \cos^2 \alpha + \cos^2 \alpha |\vec{BA}|^2 \ge |\vec{AC}|^2.$
That means $|\vec{BA}|^2 \sin^2 \alpha \ge |\vec{AC}|^2$ , i.e. $|\vec{BA}| \sin \alpha \ge |\vec{AC}|$ .

On the other hand, let point $D$ lie on line $BC$ such that $AD \perp BC$ . Then we have $|\vec{BA}| \sin \alpha = |\vec{AD}| \le |\vec{AC}|$ . Hence $|\vec{AD}| = |\vec{AC}|$ , and that means $\angle ACB = \frac{\pi}{2}$ .

Answer: C.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.