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Difference between revisions of "2023 AIME II Problems/Problem 14"

(→‎Solution 2 (3D Geometry))
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~sgdzw
 
~sgdzw
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==Solution 3 (Simple)==
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[[File:2023 AIME II 14.png|400px|right]]
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Denote <math>h(X)</math> the distance from point <math>X</math> to <math>\mathcal{P},  h(A) = 0, h(B) = 2,</math>
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<math>h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7,  AB = a, AC = a \sqrt{2}.</math>
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Let slope <math>AB</math> to <math>\mathcal{P}</math> be <math>\alpha.</math> Notation is shown in the diagram.
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<cmath>\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB} \frac {AC}{h(C)} = \frac {\sqrt(2)}{4} \implies a = 6.</cmath>
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Let <math>S = GI \cap CD \implies h(S) = h(G) = 7.</math>
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<math>h(C) – h(G) = 8 – 7 = 1, h(D)- h(I) = 10 – 7 = 3.</math>
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<math>h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies \frac {DI}{DE} =</math>
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<math>\frac {h(D) – h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.</math>
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Similarly <math>CG = \frac {3}{2} \implies SD = 9.</math>
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Let the volume of water be <math>V,</math> volume of the pyramid <math>SCGJ</math> be <math>U.</math>
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It is clear that <math>U + V = 27U = \frac {SD}{6} DI^2 = \frac {243}{8} \implies V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {117}{4}</math>  from which <math>\boxed{751}.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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== See also ==
 
== See also ==

Revision as of 13:09, 4 March 2023

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 $m+n.$

Diagram

[asy] //Made by Djmathman size(250); defaultpen(linewidth(0.6)); pair A = origin, B = (6,3), X = rotate(40)*B, Y = rotate(70)*X, C = X+Y, Z = X+B, D = B+C, W = B+Y; pair P1 = 0.8*C+0.2*Y, P2 = 2/3*C+1/3*X, P3 = 0.2*D+0.8*Z, P4 = 0.63*D+0.37*W; pair E = (-20,6), F = (-6,-5), G = (18,-2), H = (9,8); filldraw(E--F--G--H--cycle,rgb(0.98,0.98,0.2)); fill(A--Y--P1--P4--P3--Z--B--cycle,rgb(0.35,0.7,0.9)); draw(A--B--Z--X--A--Y--C--X^^C--D--Z); draw(P1--P2--P3--P4--cycle^^D--P4); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); label("$\mathcal P$",(-13,4.5)); [/asy]

Solution (3D 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)

Solution 2 (3D Geometry)

[asy] defaultpen(linewidth(0.6)); pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0); pair P1 = (-5, 0), P2 = (10, 0); draw(A--B--D--C--A); filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9)); draw(C--H, dotted); draw(B--G, dotted); draw(P1--P2); dot("$A$",A,S); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,N); dot("$H$",H,S); dot("$G$",G,S); dot("$P$",P,SE); dot("$Q$",Q,E); label("$\mathcal P$",(11, 0)); [/asy]

Let's first view the cube from a direction perpendicular to $ABDC$, as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$, we have \[\frac{CA}{CH} = \frac{AB}{AG}.\] We know $AB = x$, $AG = \sqrt{x^2-2^2}$, $AC = \sqrt{2}x$, $CH = 8$. Plug them into the above equation, we get \[\frac{\sqrt{2}x}{8} = \frac{x}{\sqrt{x^2-2^2}}.\] Solving this we get the cube's side length $x = 6$, and $AC = 6\sqrt{2}.$

Let $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\mathcal P$. Notice that $C$ is $8$ meters from $\mathcal P$, this means \[CP = \frac{1}{8}CA = \frac{3\sqrt{2}}{4}.\] Similarly, \[DQ = \frac{3}{8}CA = \frac{9\sqrt{2}}{4}.\]

[asy] defaultpen(linewidth(0.6)); pair A = (0, 0), C = (0, 2*6), X = (6, 6), Y = (-6, 6), P = (0, 1.75*6), I = (-0.25*6, 1.75*6), J = (0.25*6, 1.75*6); pair P1 = (-8, 0), P2 = (8, 0); draw(A--X--C--Y--A); filldraw(A--X--J--I--Y--cycle,rgb(0.35,0.7,0.9)); draw(P1--P2, dotted); dot("$A$",A,S); dot("$C$",C,N); dot("$P$",P,S); label("$\mathcal P$",(10, 0)); [/asy]

Now, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$. To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$, as shown above. The area of the smaller base is simply \[S_1 = CP^2 = \Bigl(\frac{3\sqrt{2}}{4}\Bigr)^2 = \frac{9}{8}.\] Similarly, the area of the larger base is \[S_2 = DQ^2 = \Bigl(\frac{9\sqrt{2}}{4}\Bigr)^2 = \frac{81}{8}.\]

Finally, applying the formula for a frustum's volume,

\[V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot 6 \cdot \Bigl(\frac{9}{8} + \sqrt{\frac{9}{8}\cdot\frac{81}{8}} + \frac{81}{8}\Bigl) = \frac{117}{4}.\]

The water's volume is thus \[6^3 - \frac{117}{4} = \frac{747}{4},\] giving $\boxed{751}$.

~sgdzw

Solution 3 (Simple)

2023 AIME II 14.png

Denote $h(X)$ the distance from point $X$ to $\mathcal{P}, h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \sqrt{2}.$

Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB} \frac {AC}{h(C)} = \frac {\sqrt(2)}{4} \implies a = 6.\] Let $S = GI \cap CD \implies h(S) = h(G) = 7.$ $h(C) – h(G) = 8 – 7 = 1, h(D)- h(I) = 10 – 7 = 3.$ $h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies \frac {DI}{DE} =$ $\frac {h(D) – h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.$

Similarly $CG = \frac {3}{2} \implies SD = 9.$

Let the volume of water be $V,$ volume of the pyramid $SCGJ$ be $U.$ It is clear that $U + V = 27U = \frac {SD}{6} DI^2 = \frac {243}{8} \implies V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {117}{4}$ from which $\boxed{751}.$

vladimir.shelomovskii@gmail.com, vvsss


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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