Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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=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.( | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Now we can use the distance formula in 3D, which is <math>\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}</math> and plug it in for the distance of <math>XY</math>. | ||
+ | |||
+ | <math>\sqrt{(0-4)^2+(y_{1}-y_{2})+(z_{1}-z_{2})}</math> | ||
+ | |||
+ | We get the answer as <math>\sqrt{132} = 2\sqrt{33}</math>. | ||
+ | |||
+ | Continuing with solution 1, using similar right triangles, we get the answer as <math>\textbf{(A)}\ \dfrac{3\sqrt{33}}5</math> | ||
+ | |||
+ | ~ghfhgvghj10 | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:19, 25 November 2022
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
?
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.(
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 right 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.