Difference between revisions of "2008 AIME I Problems/Problem 15"
(→Solution 1) |
|||
Line 42: | Line 42: | ||
The length of the perpendicular from <math>P</math> to <math>MN</math> in <math>\triangle{MNP}</math> is <math>\frac{\sqrt{17}}{\sqrt{2}}</math>, and the length of the perpendicular from <math>R</math> to <math>MN</math> in <math>\traignel{MNR}</math> is <math>\frac{\sqrt{51}}{\sqrt{2}}</math>. Adding those two lengths, <math>PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}</math>. (Alternatively, we could have used that <math>\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}</math>.) | The length of the perpendicular from <math>P</math> to <math>MN</math> in <math>\triangle{MNP}</math> is <math>\frac{\sqrt{17}}{\sqrt{2}}</math>, and the length of the perpendicular from <math>R</math> to <math>MN</math> in <math>\traignel{MNR}</math> is <math>\frac{\sqrt{51}}{\sqrt{2}}</math>. Adding those two lengths, <math>PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}</math>. (Alternatively, we could have used that <math>\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}</math>.) | ||
− | Drop a [[perpendicular]] from <math>R</math> to | + | Drop a [[perpendicular]] from <math>R</math> to the side of the square containing <math>M</math> and let the intersection be <math>G</math>. |
<cmath> | <cmath> | ||
Line 52: | Line 52: | ||
pathpen=black; | pathpen=black; | ||
real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2)); | real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2)); | ||
− | pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45); | + | pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0); |
D(2*N--P--2*M); D(N--R--M); D(P--R); | D(2*N--P--2*M); D(N--R--M); D(P--R); | ||
D((R.x,2*N.y)--R--(2*M.x,R.y)); | D((R.x,2*N.y)--R--(2*M.x,R.y)); | ||
Line 60: | Line 60: | ||
MP("\sqrt{17}",M/2,S); | MP("\sqrt{17}",M/2,S); | ||
D(N--M,dashed); | D(N--M,dashed); | ||
+ | D(G--R,dashed); | ||
MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE); | MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE); | ||
− | MP("G", | + | MP("G",G,SW); |
</asy><asy>import three; import math; | </asy><asy>import three; import math; | ||
size(300); | size(300); |
Revision as of 20:33, 5 January 2014
Problem
A square piece of paper has sides of length . 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
from the corner, and they meet on the diagonal at an angle of
(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
, where
and
are positive integers,
, and
is not divisible by the
th power of any prime. Find
.
![[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)--(s,0)); D((0,0)--r*dir(45)); D((r*dir(45).x,2*s)--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); MP("\mathrm{cut}",((0,s)+r*dir(45))/2,N); MP("\mathrm{cut}",((s,0)+r*dir(45))/2,E); MP("\mathrm{fold}",(r*dir(45).x,s+r/2*dir(45).y),E); MP("\mathrm{fold}",(s+r/2*dir(45).x,r*dir(45).y));[/asy]](http://latex.artofproblemsolving.com/2/1/3/21378bf620713fc38bd4747453f6451f9faf16bd.png)
Solution
import three; import math; size(500); pathpen=black; real r = (51^0.5-17^0.5)/200, h=867^0.25/100; triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0); triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h); D(B--F--H--cycle); D(B--F--G--cycle); D(G--I--H); D(B--I); D(A--B--C--D--cycle); triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h); triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h); D(Ia--I--Ic); D(Fa--F--Fc--Fd--cycle); D(A--Fa); D(C--Fc); D(D--Fd); (Error making remote request. Unknown error_msg)
Solution 1
In the original picture, let be the corner, and
and
be the two points whose distance is
from
. Also, let
be the point where the two cuts intersect.
Using (a 45-45-90 triangle),
.
is equilateral, so
. (Alternatively, we could find this by the Law of Sines.)
The length of the perpendicular from to
in
is
, and the length of the perpendicular from
to
in $\traignel{MNR}$ (Error compiling LaTeX. Unknown error_msg) is
. Adding those two lengths,
. (Alternatively, we could have used that
.)
Drop a perpendicular from to the side of the square containing
and let the intersection be
.
![[asy]import cse5; size(200); pathpen=black; real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2)); pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0); D(2*N--P--2*M); D(N--R--M); D(P--R); D((R.x,2*N.y)--R--(2*M.x,R.y)); MP("30^\circ",R-(0.25,1),SW); MP("30^\circ",R-(1,0.5),SW); MP("\sqrt{17}",N/2,W); MP("\sqrt{17}",M/2,S); D(N--M,dashed); D(G--R,dashed); MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE); MP("G",G,SW); [/asy]](http://latex.artofproblemsolving.com/f/3/d/f3d1876820ecbfcf6cfb1b201e8d429431680538.png)
import three; import math; size(300); pathpen=black; real r=(51^0.5-17^0.5)/200, h=867^0.25/100; triple A=(1,0,0), Ap=A+(r,-r,h), F=A+(0,0,h), N=(1,-r,h), M=(1+r,0,h); D(A--Ap--M--cycle); D(A--Ap--N--cycle); D(N--F--M); D(A--F); D(F--Ap,dashed); MP("\sqrt{34}",(A+Ap)/2,W); MP("\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}",(F+N)/2,N); MP("\frac{\sqrt{51}+\sqrt{17}}{2}",(A+7*M)/8,ENE); MP("\sqrt[4]{867}",(F+A)/2,E); MP("A",A,E); MP("A'",Ap,W); MP("F",F,E); D(rightanglemark(A,F,M,0.05)); D(rightanglemark(Ap,N,A,0.05)); D(rightanglemark(Ap,M,F,0.05)); (Error making remote request. Unknown error_msg)
Let be the smaller square base of the tray and let
be the larger square, such that
, etc, are edges. Let
be the foot of the perpendicular from
to plane
.
We know and
. Now, use the Pythagorean Theorem on triangle
to find
:
The answer is .
Solution 2
In the final pyramid, let be the smaller square and let
be the larger square such that
, etc. are edges.
It is obvious from the diagram that .
Let and
be the positive
and
axes in a 3-d coordinate system such that
has a positive
coordinate. Let
be the angle made with the positive
axis. Define
and
analogously.
It is easy to see that if , then
. Furthermore, this means that
.
We have that , so
.
It is easy to see from the Law of Sines that .
Now, .
It follows that the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.