Difference between revisions of "1996 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | <center><asy> | + | $<center><asy> |
import three; | import three; | ||
size(250);defaultpen(0.7+fontsize(9)); | size(250);defaultpen(0.7+fontsize(9)); | ||
| Line 20: | Line 20: | ||
label("$6$",(unit*(r+1)/2,0,0),N); | label("$6$",(unit*(r+1)/2,0,0),N); | ||
label("$7$",(unit*r,unit*r/2,0),SW); | label("$7$",(unit*r,unit*r/2,0),SW); | ||
| − | </asy></center> | + | </asy></center><math> |
| − | (Figure not to scale) The area of the square shadow base is <math>48 + 1 = 49< | + | (Figure not to scale) The area of the square shadow base is </math>48 + 1 = 49<math>, and so the sides of the shadow are </math>7<math>. Using the similar triangles in blue, |
| − | <math>\frac {x}{1} = \frac {1}{6}< | + | </math>\frac {x}{1} = \frac {1}{6}<math>, and </math>\left\lfloor 1000x \right\rfloor = \boxed{166}$. |
== See also == | == See also == | ||
Revision as of 12:51, 13 April 2015
Problem
A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is 
centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed 
.
Solution
$
![[asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); draw(rightanglemark(P,(0,0,unit),(unit,0,unit),1.7));draw(rightanglemark((unit,0,unit),(unit,0,0),(r*unit,0,0),1.7)); label("$x$",(0,0,unit+unit/(r-1)/2),WSW); label("$1$",(unit/2,0,unit),N); label("$1$",(unit,0,unit/2),W); label("$1$",(unit/2,0,0),N); label("$6$",(unit*(r+1)/2,0,0),N); label("$7$",(unit*r,unit*r/2,0),SW); [/asy]](http://latex.artofproblemsolving.com/9/5/c/95ce6725473f91467d41a0e35cee50288617958d.png)
48 + 1 = 49
7
\frac {x}{1} = \frac {1}{6}
\left\lfloor 1000x \right\rfloor = \boxed{166}$.
See also
| 1996 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
