Difference between revisions of "2008 AIME I Problems/Problem 15"
m (Changed the centering) |
(→Solution 1) |
||
(One intermediate revision by one other user not shown) | |||
Line 40: | Line 40: | ||
Using <math>\triangle{MNP}</math> (a 45-45-90 triangle), <math>MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}</math>. <math>\triangle{MNR}</math> is [[equilateral triangle|equilateral]], so <math>MR = NR = \sqrt{34}</math>. (Alternatively, we could find this by the [[Law of Sines]].) | Using <math>\triangle{MNP}</math> (a 45-45-90 triangle), <math>MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}</math>. <math>\triangle{MNR}</math> is [[equilateral triangle|equilateral]], so <math>MR = NR = \sqrt{34}</math>. (Alternatively, we could find this by the [[Law of Sines]].) | ||
− | 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>\triangle{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>\ | + | 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>\triangle{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>\sin 75^{\circ} = \sin (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}</math>.) |
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>. | 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>. | ||
Line 63: | Line 63: | ||
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",G,SW); | MP("G",G,SW); | ||
− | </asy><center> | + | </asy></center> |
Let <math>ABCD</math> be the smaller square base of the tray and let <math>A'B'C'D'</math> be the larger square, such that <math>AA'</math>, etc, are edges. Let <math>F</math> be the foot of the perpendicular from <math>A</math> to plane <math>A'B'C'D'</math>. | Let <math>ABCD</math> be the smaller square base of the tray and let <math>A'B'C'D'</math> be the larger square, such that <math>AA'</math>, etc, are edges. Let <math>F</math> be the foot of the perpendicular from <math>A</math> to plane <math>A'B'C'D'</math>. |
Latest revision as of 18:09, 20 October 2018
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 .
Solution
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 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 .
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.