Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2008 AIME I Problems/Problem 15

Revision as of 16:12, 23 March 2008 by I like pie (talk | contribs) (Added Picture (Asymptote))

Problem

A square piece of paper has sides of length 100. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form $\sqrt[n]{m}$, where $m$ and $n$ are positive integers, $m<1000$, and $m$ is not divisible by the $n$th power of any prime. Find $m+n$. [asy]import cse5; size(200); pathpen=black; real s=sqrt(17); real r=(sqrt(51)+s)/(sqrt(2)); D((0,2*s)--(0,0)--(2*s,0)); D((0,s)--(r*dir(45))); D((s,0)--(r*dir(45))); D((0,0)--(r*dir(45))); D((r*dir(45))--((r*dir(45)).x,2*s)); D((r*dir(45))--(2*s,(r*dir(45)).y)); MP("30^\circ",r*dir(45)-(0.25,1),SW); MP("30^\circ",r*dir(45)-(1,0.5),SW); MP("\sqrt{17}",(0,s/2),W); MP("\sqrt{17}",(s/2,0),S);[/asy]

Solution

Solution 1

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 $867 + 4 = \boxed{871}$.

Solution 2

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 $MR$ and $NR$ are also $\sqrt{34}$.

The length of the perpendicular from $P$ to $MN$ in $\triangle{MNP}$ is $\frac{\sqrt{17}}{\sqrt{2}}$.

The length of the perpendicular from $R$ to $MN$ in $\traignel{MNR}$ (Error compiling LaTeX. Unknown error_msg) is $\frac{\sqrt{51}}{\sqrt{2}}$.

Adding those two lengths, $PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}$.

Drop a perpendicular from $R$ to line $MN$ and let the intersection be $G$.

\[PG=PR\div\sqrt{2}=\frac{\sqrt{17}+\sqrt{51}}{2}\]

\[MG=PG-PM=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\]

Now, move to 3D:

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. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$.

We know $AA'=MR=\sqrt{34}$ and $AB=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}$.

Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$:

\begin{align*}\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{17\sqrt{3}}\\ AF&=\sqrt[4]{867}\end{align*}

The answer is $867 + 4 = \boxed{871}$.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&oldid=24281"