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Geometry Difficulty 3.8 AMC 10/12 Find the answer China

Let △ABC\triangle ABC be a given triangle. If ∣BA⃗−tBC⃗∣≥∣AC⃗∣|\vec{BA} - t \vec{BC}| \ge |\vec{AC}| for any t∈Rt \in \mathbb{R}, then △ABC\triangle ABC is ( ).

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

Solution

Suppose ∠ABC=α\angle ABC = \alpha. Since ∣BA⃗−tBC⃗∣≥∣AC⃗∣|\vec{BA} - t \vec{BC}| \ge |\vec{AC}|, we have
∣BA⃗∣2−2tBA⃗⋅BC⃗+t2∣BC⃗∣2≥∣AC⃗∣2. |\vec{BA}|^2 - 2t \vec{BA} \cdot \vec{BC} + t^2 |\vec{BC}|^2 \ge |\vec{AC}|^2.
Let
t=BA⃗⋅BC⃗∣BC⃗∣2, t = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BC}|^2},
we get
∣BA⃗∣2−2∣BA⃗∣2cos⁡2α+cos⁡2α∣BA⃗∣2≥∣AC⃗∣2. |\vec{BA}|^2 - 2|\vec{BA}|^2 \cos^2 \alpha + \cos^2 \alpha |\vec{BA}|^2 \ge |\vec{AC}|^2.
That means ∣BA⃗∣2sin⁡2α≥∣AC⃗∣2|\vec{BA}|^2 \sin^2 \alpha \ge |\vec{AC}|^2, i.e. ∣BA⃗∣sin⁡α≥∣AC⃗∣|\vec{BA}| \sin \alpha \ge |\vec{AC}|.

On the other hand, let point DD lie on line BCBC such that AD⊥BCAD \perp BC. Then we have ∣BA⃗∣sin⁡α=∣AD⃗∣≤∣AC⃗∣|\vec{BA}| \sin \alpha = |\vec{AD}| \le |\vec{AC}|. Hence ∣AD⃗∣=∣AC⃗∣|\vec{AD}| = |\vec{AC}|, and that means ∠ACB=π2\angle ACB = \frac{\pi}{2}.

Answer: C.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.