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2023 AIME II Problems/Problem 14

Revision as of 16:21, 18 February 2023 by Facejo (talk | contribs) (→‎Solution (3-d vector analysis, analytic geometry + Calculus))

Problem

A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive intgers. Find $m+n.$

Solution (3-d vector analysis, analytic geometry + Calculus)

We introduce a Cartesian coordinate system to the diagram. We put the origin at $A$. We let the $z$-components of $B$, $C$, $D$ be positive. We set the $x$-axis in a direction such that $B$ is on the $x-O-z$ plane.

The coordinates of $A$, $B$, $C$ are $A = \left( 0, 0, 0 \right)$, $B = \left( x_B, 0 , 2 \right)$, $C = \left( x_C, y_C, 8 \right)$.

Because $AB \perp AC$, $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$. Thus, \[ x_B x_C + 16 = 0 . \hspace{1cm} (1) \]

Because $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$. Thus, \[ x_C^2 + y_C^2 + 8^2 = 2 \left( x_B^2 + 2^2 \right) . \hspace{1cm} (2) \]

Because plane $ABCD$ is perpendicular to plan $P$, $\hat z \cdot \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) = 0$. Thus, \[ \begin{vmatrix} 0 & 0 & 1 \\ x_B & 0 & 2 \\ x_C & y_C & 8 \end{vmatrix} = 0 . \hspace{1cm} (3) \]

Jointly solving (1), (2), (3), we get one solution $x_B = 4 \sqrt{2}$, $x_C = - 2 \sqrt{2}$, $y_C = 0$. Thus, the side length of the cube is $|AB| = \sqrt{x_B^2 + 2^2} = 6$.

Denote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\widehat{AB} \times \widehat{AP} = \widehat{AQ}$. Now, we compute the coordinates of $P$ and $Q$.

Because $|AB| = 6$, we have $\overrightarrow{AP} \times \overrightarrow{AQ} = 6 \overrightarrow{AB}$, $\overrightarrow{AQ} \times \overrightarrow{AB} = 6 \overrightarrow{AP}$, $\overrightarrow{AB} \times \overrightarrow{AP} = 6 \overrightarrow{AQ}$.

Hence, \begin{align*} \begin{bmatrix} \hat i & \hat j & \hat k \\ x_P & y_P & z_P \\ x_Q & y_Q & z_Q \end{bmatrix} & = 6 \left( 4 \sqrt{2} \hat i + 2 \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ x_Q & y_Q & z_Q \\ 4 \sqrt{2} & 0 & 2 \end{vmatrix} & = 6 \left( x_P \hat i + y_P \hat j + z_P \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 \sqrt{2} & 0 & 2 \\ x_P & y_P & z_P \end{vmatrix} & = 6 \left( x_Q \hat i + y_Q \hat j + z_Q \hat k \right) . \end{align*}

By solving these equations, we get \[ y_P^2 + y_Q^2 = 36 . ]\

In addition, we have $\overrightarrow{AC} = \overrightarrow{AP} + \overrightarrow{AQ}$. Thus, $P = \left( - \sqrt{2} , 3 \sqrt{2} , 4 \right)$, $Q = \left( - \sqrt{2} , - 3 \sqrt{2} , 4 \right)$.

Therefore, the volume of the water is \begin{align*} V = & 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ z_B u + z_P v + z_Q w \leq 7 \right\} dw dv du \\ & = 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w \leq 7 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w > 7 \right\} dw dv du . \end{align*}

Define $u' = 1 - u$, $v' = 1 - v$, $w' = 1 - w$. Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u' + 4 v' + 4 w' < 3 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u'=0}^1 \left( \int_{v'=0}^1 \int_{w'=0}^1 \mathbf 1 \left\{ v' + w' < \frac{3}{4} - \frac{u'}{2} \right\} dw' dv' \right) du' \\ & = 6^3 - 6^3 \int_{u'=0}^1 \frac{1}{2} \left( \frac{3}{4} - \frac{u'}{2} \right)^2 du' . \end{align*}

Define $u'' = \frac{3}{4} - \frac{u'}{2}$. Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u'' = 1/4}^{3/4} \left( u'' \right)^2 du'' \\ & = 6^3 - 6^3 \frac{1}{3} \left( \left(\frac{3}{4}\right)^3 - \left(\frac{1}{4}\right)^3 \right) \\ & = 216 - \frac{117}{4} \\ & = \frac{747}{4} . \end{align*}

Therefore, the answer is $747 + 4 = \boxed{\textbf{(751) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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