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Algebra Difficulty 3.9 AMC 10/12 Find the answer China

Given vectors a⃗=(1+2m,1−2m)\vec{a} = (1 + 2^m, 1 - 2^m), b⃗=(4m−3,4m+5)\vec{b} = (4^m - 3, 4^m + 5), suppose mm is real. Then the minimum of the dot product of a⃗⋅b⃗\vec{a} \cdot \vec{b} is ______.

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

Solution

Let t=2mt = 2^m, and then a⃗=(1+t,1−t)\vec{a} = (1 + t, 1 - t), b⃗=(t2−3,t2+5)\vec{b} = (t^2 - 3, t^2 + 5). Thus,
a⃗⋅b⃗=(1+t)(t2−3)+(1−t)(t2+5)=2(t−2)2−6≥−6. \begin{aligned} \vec{a} \cdot \vec{b} &= (1 + t)(t^2 - 3) + (1 - t)(t^2 + 5) \\ &= 2(t - 2)^2 - 6 \geq -6. \end{aligned}
When t=2t = 2, namely, m=1m = 1, a⃗⋅b⃗\vec{a} \cdot \vec{b} takes the minimum −6-6.

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