Given vectors $\mathbf{a}$ and $\mathbf{b},$ let $\mathbf{p}$ be a vector such that
$$\|\mathbf{p} - \mathbf{b}\| = 2 \|\mathbf{p} - \mathbf{a}\|.$$Among all such vectors $\mathbf{p},$ there exists constants $t$ and $u$ such that $\mathbf{p}$ is at a fixed distance from $t \mathbf{a} + u \mathbf{b}.$ Enter the ordered pair $(t,u).$
From $\|\mathbf{p} - \mathbf{b}\| = 2 \|\mathbf{p} - \mathbf{a}\|,$
$$\|\mathbf{p} - \mathbf{b}\|^2 = 4 \|\mathbf{p} - \mathbf{a}\|^2.$$This expands as
$$\|\mathbf{p}\|^2 - 2 \mathbf{b} \cdot \mathbf{p} + \|\mathbf{b}\|^2 = 4 \|\mathbf{p}\|^2 - 8 \mathbf{a} \cdot \mathbf{p} + 4 \|\mathbf{a}\|^2,$$which simplifies to $3 \|\mathbf{p}\|^2 = 8 \mathbf{a} \cdot \mathbf{p} - 2 \mathbf{b} \cdot \mathbf{p} - 4 \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2.$ Hence,
$$\|\mathbf{p}\|^2 = \frac{8}{3} \mathbf{a} \cdot \mathbf{p} - \frac{2}{3} \mathbf{b} \cdot \mathbf{p} - \frac{4}{3} \|\mathbf{a}\|^2 + \frac{1}{3} \|\mathbf{b}\|^2.$$We want $\|\mathbf{p} - (t \mathbf{a} + u \mathbf{b})\|$ to be constant, which means $\|\mathbf{p} - t \mathbf{a} - u \mathbf{b}\|^2$ is constant. This expands as
$$
\begin{align*}
\|\mathbf{p} - t \mathbf{a} - u \mathbf{b}\|^2 &= \|\mathbf{p}\|^2 + t^2 \|\mathbf{a}\|^2 + u^2 \|\mathbf{b}\|^2 - 2t \mathbf{a} \cdot \mathbf{p} - 2u \mathbf{b} \cdot \mathbf{p} + 2tu \mathbf{a} \cdot \mathbf{b} \\
&= \frac{8}{3} \mathbf{a} \cdot \mathbf{p} - \frac{2}{3} \mathbf{b} \cdot \mathbf{p} - \frac{4}{3} \|\mathbf{a}\|^2 + \frac{1}{3} \|\mathbf{b}\|^2 \\
&\quad + t^2 \|\mathbf{a}\|^2 + u^2 \|\mathbf{b}\|^2 - 2t \mathbf{a} \cdot \mathbf{p} - 2u \mathbf{b} \cdot \mathbf{p} + 2tu \mathbf{a} \cdot \mathbf{b} \\
&= \left( \frac{8}{3} - 2t \right) \mathbf{a} \cdot \mathbf{p} - \left( \frac{2}{3} + 2u \right) \mathbf{b} \cdot \mathbf{p} \\
&\quad + \left( t^2 - \frac{4}{3} \right) \|\mathbf{a}\|^2 + \left( u^2 + \frac{1}{3} \right) \|\mathbf{b}\|^2 + 2tu \mathbf{a} \cdot \mathbf{b}.
\end{align*}
$$
The only non-constant terms in this expression are $\left( \frac{8}{3} - 2t \right) \mathbf{a} \cdot \mathbf{p}$ and $\left( \frac{2}{3} + 2u \right) \mathbf{b} \cdot \mathbf{p}.$ We can them make them equal 0 by setting $2t = \frac{8}{3}$ and $2u = -\frac{2}{3}.$ These lead to $t = \frac{4}{3}$ and $u = -\frac{1}{3},$ so $(t,u) = \boxed{\left( \frac{4}{3}, -\frac{1}{3} \right)}.$
{
"problem": "Given vectors $\\\\\\\\mathbf{a}$ and $\\\\\\\\mathbf{b},$ let $\\\\\\\\mathbf{p}$ be a vector such that\\\\n\\\\\\\\[\\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{b}\\\\\\\\| = 2 \\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{a}\\\\\\\\|.\\\\\\\\]Among all such vectors $\\\\\\\\mathbf{p},$ there exists constants $t$ and $u$ such that $\\\\\\\\mathbf{p}$ is at a fixed distance from $t \\\\\\\\mathbf{a} + u \\\\\\\\mathbf{b}.$ Enter the ordered pair $(t,u).$",
"level": "Level 5",
"type": "Precalculus",
"solution": "From $\\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{b}\\\\\\\\| = 2 \\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{a}\\\\\\\\|,$\\\\n\\\\\\\\[\\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{b}\\\\\\\\|^2 = 4 \\\\\\\\|\\\\\\\\mathbf{p} - \\\\\\\\mathbf{a}\\\\\\\\|^2.\\\\\\\\]This expands as\\\\n\\\\\\\\[\\\\\\\\|\\\\\\\\mathbf{p}\\\\\\\\|^2 - 2 \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} + \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2 = 4 \\\\\\\\|\\\\\\\\mathbf{p}\\\\\\\\|^2 - 8 \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} + 4 \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2,\\\\\\\\]which simplifies to $3 \\\\\\\\|\\\\\\\\mathbf{p}\\\\\\\\|^2 = 8 \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - 2 \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} - 4 \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2.$ Hence,\\\\n\\\\\\\\[\\\\\\\\|\\\\\\\\mathbf{p}\\\\\\\\|^2 = \\\\\\\\frac{8}{3} \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - \\\\\\\\frac{2}{3} \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} - \\\\\\\\frac{4}{3} \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + \\\\\\\\frac{1}{3} \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2.\\\\\\\\]We want $\\\\\\\\|\\\\\\\\mathbf{p} - (t \\\\\\\\mathbf{a} + u \\\\\\\\mathbf{b})\\\\\\\\|$ to be constant, which means $\\\\\\\\|\\\\\\\\mathbf{p} - t \\\\\\\\mathbf{a} - u \\\\\\\\mathbf{b}\\\\\\\\|^2$ is constant. This expands as\\\\n\\\\\\\\begin{align*}\\\\n\\\\\\\\|\\\\\\\\mathbf{p} - t \\\\\\\\mathbf{a} - u \\\\\\\\mathbf{b}\\\\\\\\|^2 &= \\\\\\\\|\\\\\\\\mathbf{p}\\\\\\\\|^2 + t^2 \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + u^2 \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2 - 2t \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - 2u \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} + 2tu \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{b} \\\\\\\\\\\\\\\\\\\\n&= \\\\\\\\frac{8}{3} \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - \\\\\\\\frac{2}{3} \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} - \\\\\\\\frac{4}{3} \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + \\\\\\\\frac{1}{3} \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2 \\\\\\\\\\\\\\\\\\\\n&\\\\\\\\quad + t^2 \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + u^2 \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2 - 2t \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - 2u \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} + 2tu \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{b} \\\\\\\\\\\\\\\\\\\\n&= \\\\\\\\left( \\\\\\\\frac{8}{3} - 2t \\\\\\\\right) \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p} - \\\\\\\\left( \\\\\\\\frac{2}{3} + 2u \\\\\\\\right) \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p} \\\\\\\\\\\\\\\\\\\\n&\\\\\\\\quad + \\\\\\\\left( t^2 - \\\\\\\\frac{4}{3} \\\\\\\\right) \\\\\\\\|\\\\\\\\mathbf{a}\\\\\\\\|^2 + \\\\\\\\left( u^2 + \\\\\\\\frac{1}{3} \\\\\\\\right) \\\\\\\\|\\\\\\\\mathbf{b}\\\\\\\\|^2 + 2tu \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{b}.\\\\n\\\\\\\\end{align*}The only non-constant terms in this expression are $\\\\\\\\left( \\\\\\\\frac{8}{3} - 2t \\\\\\\\right) \\\\\\\\mathbf{a} \\\\\\\\cdot \\\\\\\\mathbf{p}$ and $\\\\\\\\left( \\\\\\\\frac{2}{3} + 2u \\\\\\\\right) \\\\\\\\mathbf{b} \\\\\\\\cdot \\\\\\\\mathbf{p}.$ We can them make them equal 0 by setting $2t = \\\\\\\\frac{8}{3}$ and $2u = -\\\\\\\\frac{2}{3}.$ These lead to $t = \\\\\\\\frac{4}{3}$ and $u = -\\\\\\\\frac{1}{3},$ so $(t,u) = \\\\\\\\boxed{\\\\\\\\left( \\\\\\\\frac{4}{3}, -\\\\\\\\frac{1}{3} \\\\\\\\right)}.$"
}