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Difference between revisions of "2023 AIME I Problems/Problem 13"

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such as the one shown below.
 
such as the one shown below.
  
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==Solution 1 (3-d Vector Analysis)==
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==Solution 1 (3-D Vector Analysis)==
  
 
Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>.
 
Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>.

Revision as of 14:27, 8 February 2023

Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$. The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. A parallelepiped is a solid with six parallelogram faces such as the one shown below.

Solution 1 (3-D Vector Analysis)

Denote $\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}$. Denote by $d$ the length of each side of a rhombus.

Now, we put the solid to the 3-d coordinate space. We put the bottom face on the $x-O-y$ plane. For this bottom face, we put a vertex with an acute angle $2 \alpha$ at the origin, denoted as $O$. For two edges that are on the bottom face and meet at $O$, we put one edge on the positive side of the $x$-axis. The endpoint is denoted as $A$. Hence, $A = \left( d , 0 , 0 \right)$. We put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$. Hence, $B = \left( d \cos 2 \alpha , d \sin 2 \alpha , 0 \right)$.

For the third edge that has one endpoint $O$, we denote by $C$ its second endpoint. We denote $C = \left( u , v , w \right)$. Without loss of generality, we set $w > 0$. Hence, \[ u^2 + v^2 + w^2 = d^2 . \hspace{1cm} (1) \]

We have \begin{align*} \cos \angle AOC & = \frac{\overrightarrow{OA} \cdot \overrightarrow{OC}}{|OA| \cdot |OC|} \\ & = \frac{u}{d} , \hspace{1cm} (2) \end{align*} and \begin{align*} \cos \angle BOC & = \frac{\overrightarrow{OB} \cdot \overrightarrow{OC}}{|OB| \cdot |OC|} \\ & = \frac{u \cos 2 \alpha + v \sin 2 \alpha}{d} . \hspace{1cm} (3) \end{align*}

Case 1: $\angle AOC = \angle BOC = 2 \alpha$ or $2 \left( 90^\circ - \alpha \right)$.

By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \\ & = \pm d \cos 2 \alpha \tan \alpha . \end{align*}

Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} . \hspace{1cm} (4) \end{align*}

Case 2: $\angle AOC = 2 \alpha$ and $\angle BOC = 2 \left( 90^\circ - \alpha \right)$, or $\angle BOC = 2 \alpha$ and $\angle AOC = 2 \left( 90^\circ - \alpha \right)$.

By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \mp d \cos 2 \alpha \frac{1 + \cos 2 \alpha}{\sin 2 \alpha} \\ & = \mp d \cos 2 \alpha \cot \alpha . \end{align*}

Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} . \hspace{1cm} (5) \end{align*}

We notice that $(4) > (5)$. Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.

Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is \begin{align*} \frac{(4)}{(5)} & = \frac{\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha}} {\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha}} \\ & = \sqrt{\frac{\tan^2 2 \alpha - \tan^2 \alpha}{\tan^2 2 \alpha - \cot^2 \alpha}} . \end{align*}

Recall that $\tan \alpha = \frac{\sqrt{21}}{\sqrt{31}}$. Thus, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{\sqrt{21 \cdot 31}}{5}$. Plugging this into the equation above, we get \begin{align*} \frac{(4)}{(5)} & = \frac{63}{62}. \end{align*}

Therefore, the answer is $63 + 62 = \boxed{\textbf{(125) }}$.

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

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