AIME 2008 I · 第 15 题

Solution 1

In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\sqrt{17}$ from $P$. Also, let $R$ be the point where the two cuts intersect.

Using $\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}$. $\triangle{MNR}$ is equilateral, so $MR = NR = \sqrt{34}$. (Alternatively, we could find this by the Law of Sines.)

The length of the perpendicular from $P$ to $MN$ in $\triangle{MNP}$ is $\frac{\sqrt{17}}{\sqrt{2}}$, and the length of the perpendicular from $R$ to $MN$ in $\triangle{MNR}$ is $\frac{\sqrt{51}}{\sqrt{2}}$. Adding those two lengths, $PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}$. (Alternatively, we could have used that $\sin 75^{\circ} = \sin (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}$.)

Drop a perpendicular from $R$ to the side of the square containing $M$ and let the intersection be $G$.

$$\begin{aligned}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\ MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{aligned}$$

Let $A'B'C'D'$ be the smaller square base of the tray and let $ABCD$ be the larger square, such that $AA'$, etc, are edges. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$.

We know $AA'=MR=\sqrt{34}$ and $A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}$. Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$:

$$\begin{aligned}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{aligned}$$ The answer is $867 + 4 = \boxed{871}$.

Solution 2

In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$, etc. are edges.

It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$.

Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coordinate. Let $\alpha$ be the angle made with the positive $x$ axis. Define $\beta$ and $\gamma$ analogously.

It is easy to see that if $P: = (x,y,z)$, then $x = AA'\cdot \cos\alpha$. Furthermore, this means that $\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

We have that $\alpha = \beta = 105^\circ$, so $\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}$.

It is easy to see from the Law of Sines that $\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}$.

Now, $z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}$.

It follows that the answer is $867 + 4 = \boxed{871}$.~Shen Kislay Kai

Solution 3 (Vectors)

Let $A$ be a corner of the square. Let $B$ and $C$ be the points where the cuts start from the edges, and $O$ be the point where the cuts meet. Let $X$ and $Y$ be points such that $OX=OY=1,$ $\overline{OX} \parallel \overline{AB}$ and $\overline{OY} \parallel \overline{AC},$ respectively. Let $P$ be the point such that the cuts $OB$ and $OC$ meet at $P$ when folded.

By the AAS congruence theorem, $\triangle{AOC} \cong \triangle{AOB}.$ Hence, it follows that $\triangle{OBC}$ is equilateral and $OB=OC=BC=OP.$ Since $\triangle{ABC}$ is an isosceles right triangle, we have $OP=OB=OC=BC=AB\cdot\sqrt{2}=\sqrt{17}\cdot\sqrt{2}=\sqrt{34}.$ Thus, $OP=\sqrt{34}.$

Also, note that $\angle{POX}=\angle{BOX}=\angle{AOX}-\angle{AOB}=105^\circ.$

Furthermore, let $O=(0,0,0)$ be the origin, $X=(1,0,0),$ and $Y=(0,1,0).$ Let $\overrightarrow{P}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}$ denote the vector at $P,$ $\overrightarrow{X}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ denote the vector at $X,$ and $\overrightarrow{Y}=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ denote the vector at $Y.$ It suffices to solve for $p_3.$

By the properties of the dot product for vectors,

$$\overrightarrow{P}\cdot\overrightarrow{X}=\overrightarrow{P}\cdot\overrightarrow{Y}=\|\overrightarrow{P}\|\|\overrightarrow{X}\|\cos\angle{POX}=\|\overrightarrow{P}\|\|\overrightarrow{Y}\|\cos\angle{POY}.$$ Since $\angle{POX}=\angle{POY}=105^\circ,$ $\|\overrightarrow{P}\|=OP=\sqrt{34},$ and $\|\overrightarrow{X}\|=\|\overrightarrow{Y}\|=1,$ the above equations simplify to

$$\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}\cdot\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\sqrt{34}\cos 105^\circ.$$ This leads to $p_1=p_2=\sqrt{34}\cos 105^\circ.$ Also, since $\|\overrightarrow{P}\|=\sqrt{34},$ we have \begin{align*} p_3^2&=34-(p_1^2+p_2^2)\\ &=34-2\cdot(\sqrt{34}\cos 105^\circ)^2\\ &=34(1-2\cos^2 105^\circ)\\ &=-34\cos 210^\circ\\ &=34\cos 30^\circ\\ &=34\frac{\sqrt{3}}{2}\\ &=17\sqrt{3}\\ &=\sqrt{867}.\\ \end{align*}

Thus, $p_3=\sqrt[4]{867}$ and $m+n=\boxed{871}.$

~ReticulatedPython