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 $(t,u) = \\\\boxed{\\\\left( \\\\frac{4}{3}, -\\\\frac{1}{3} \\\\right)}.$"
}