2023 AIME I Problems/Problem 13

Problem

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.

{insert diagram here}

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)

Solution 2 (no trig)

Let one of the vertices be at the origin and the three adjacent vertices be $u$ , $v$ , and $w$ . For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$ . Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21$ and $(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31$ . Since all of $u$ , $v$ , and $w$ have equal length, $u\cdot u=13$ , $v\cdot v=13$ , and $u\cdot v=-2.5$ . Symmetrically, $w\cdot w=13$ , $u\cdot w=-2.5$ , and $v\cdot w=-2.5$ . Hence the volume of the parallelepiped is given by $\sqrt{\operatorname{det}\begin{pmatrix}13&-2.5&-2.5\\-2.5&13&-2.5\\-2.5&-2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}15.5&-15.5&0\\-2.5&13&-2.5\\0&-15.5&15.5\end{pmatrix}}=\sqrt{15.5^2\operatorname\det\begin{pmatrix}1&-1&0\\-2.5&13&-2.5\\0&-1&1\end{pmatrix}}=\sqrt{15.5^2\cdot 8}$ .

For the other parallelepiped, the three diagonals involving the origin are of length $\sqrt{31}$ and the volume is $\sqrt{\operatorname{det}\begin{pmatrix}13&2.5&2.5\\2.5&13&2.5\\2.5&2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}10.5&-10.5&0\\2.5&13&2.5\\0&-10.5&10.5\end{pmatrix}}=\sqrt{10.5^2\operatorname\det\begin{pmatrix}1&-1&0\\2.5&13&2.5\\0&-1&1\end{pmatrix}}=\sqrt{10.5^2\cdot 18}$ .

Consequently, the answer is $\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}$ , giving $\boxed{125}$ .

~EVIN-

Solution 3 (No trig, no linear algebra)

Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting $P$ , and the one with three obtuse angles meeting $P'$ .

The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.

Denote the point with three acute angles meeting at it in $P$ as $A$ , and its neighbors $B$ , $C$ , and $D$ . Similarly, denote the point with three obtuse angles meeting at it in $P'$ as $A'$ , and its neighbors $B'$ , $C'$ , and $D'$ .

We have the following equations:

$\textrm{Height of }P\textrm{ from }ACD = \frac{\textrm{Vol}(ABCD) \cdot 3}{[ACD]},$ $\textrm{Height of }P'\textrm{ from }A'C'D' = \frac{\textrm{Vol}(A'B'C'D') \cdot 3}{[A'C'D']}.$

However, $ACD$ and $A'C'D'$ are both half the area of a rhombus with diagonals $\sqrt{31}$ and $\sqrt{21}$ , so our ratio is really

$\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')}.$

Because the diagonals of all of the faces are $\sqrt{31}$ and $\sqrt{21}$ , each edge of the parallelepipeds is $\sqrt{13}$ by the Pythagorean theorem.

We have $AB = AC = AD = \sqrt{13}$ , and $BC = CD = BD = \sqrt{21}$ . When we drop a perpendicular to the centroid of $BCD$ from $A$ (let's call this point $O$ ), we have $BO = \frac{\sqrt{21}}{\sqrt{3}} = \sqrt{7}$ . Thus,

$AB^2 - BO^2 = AO^2$ $13 - 7 = AO^2 = 6$ $AO = \sqrt{6}.$

The area of base $BCD$ is $\frac{21\sqrt{3}}{4}$ . Hence,

$\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}$ $= \frac{63\sqrt{2}}{12}.$

We can apply a similar approach to $A'B'C'D'$ .

$A'B' = A'C' = A'D' = \sqrt{13}$ , and $B'C' = C'D' = B'D' = \sqrt{31}$ . When we drop a perpendicular to the centroid of $B'C'D'$ from $A'$ (let's call this point $O'$ ), we have $B'O' = \frac{\sqrt{31}}{\sqrt{3}} = \sqrt{\frac{31}{3}}$ . Thus,

$A'B'^2 - B'O'^2 = A'O'^2$ $13 - \frac{31}{3} = A'O'^2$ $A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.$

The area of base $B'C'D'$ is $\frac{31\sqrt{3}}{4}$ . Hence,

$\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}$ $= \frac{186\sqrt{2}}{36}$ $= \frac{62\sqrt{2}}{12}.$

Finally,

\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12} = \frac{63}{62}.\] (Error compiling LaTeX. Unknown error_msg)

Our answer is $63 + 62 = \boxed{125}$ .

~mathboy100

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

Contents

Problem

Solution 1 (3-D Vector Analysis)

Solution 2 (no trig)

Solution 3 (No trig, no linear algebra)