Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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Let the length of the segment <math>XY</math> that is inside the cube with side length <math>3</math> be <math>x</math>. By similar triangles, <math>\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}</math>, giving <math>x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-</math>. | Let the length of the segment <math>XY</math> that is inside the cube with side length <math>3</math> be <math>x</math>. By similar triangles, <math>\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}</math>, giving <math>x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-</math>. | ||
| − | =Solution 2 (3D Coordinate Geometry)= | + | ==Solution 2 (3D Coordinate Geometry)== |
Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin. | Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin. | ||
Revision as of 20:23, 28 October 2023
Problem
Four cubes with edge lengths 
, 
, 
, and 
are stacked as shown. What is the length of the portion of 
contained in the cube with edge length ?
![[asy] dotfactor = 3; size(10cm); dot((0, 10)); label("$X$", (0,10),W,fontsize(8pt)); dot((6,2)); label("$Y$", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10)); draw((1, 10)--(1.5,10.5)); draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7)); draw((2,9)--(3,10)); draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4)); draw((3,7)--(4.5,8.5)); draw((4.5,6)--(6,6)--(6,2)--(4,0)); draw((4,4)--(6,6)); label("$1$", (1,9.5), W,fontsize(8pt)); label("$2$", (2,8), W,fontsize(8pt)); label("$3$", (3,5.5), W,fontsize(8pt)); label("$4$", (4,2), W,fontsize(8pt)); [/asy]](http://latex.artofproblemsolving.com/e/4/1/e41120e351bda42fada2fe2803e19a131bc30b96.png)
Solution
By Pythagorean Theorem in three dimensions, the distance 
is 
.
Let the length of the segment that is inside the cube with side length be 
. By similar triangles, 
, giving 
.
Solution 2 (3D Coordinate Geometry)
Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin.
![[asy] dotfactor = 3; size(10cm); dot((0, 10)); label("$X(0,10,0)$", (0,10),W,fontsize(8pt)); dot((6,2)); label("$Y(4,0,4)$", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10)); draw((1, 10)--(1.5,10.5)); draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7)); draw((2,9)--(3,10)); draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4)); draw((3,7)--(4.5,8.5)); draw((4.5,6)--(6,6)--(6,2)--(4,0)); draw((4,4)--(6,6)); label("$1$", (1,9.5), W,fontsize(8pt)); label("$2$", (2,8), W,fontsize(8pt)); label("$3$", (3,5.5), W,fontsize(8pt)); label("$4$", (4,2), W,fontsize(8pt)); label("$D(0,0,0)$", (0,0), W,fontsize(8pt)); [/asy]](http://latex.artofproblemsolving.com/d/2/6/d26d45a4b5b55aa7a59dcd355353f8ce6476fd3c.png)
Now we can use the distance formula in 3D, which is 
and plug it in for the distance of .
We get the answer as 
.
Continuing with solution 1, using similar triangles, we get the answer as 
~ghfhgvghj10
Video Solution
~IceMatrix
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
