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. |
<asy> | <asy> | ||
Line 63: | Line 63: | ||
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>. | 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+( | + | <math>\sqrt{(0-4)^2+(10-0)^2+(0-4)^2}</math> |
We get the answer as <math>\sqrt{132} = 2\sqrt{33}</math>. | We get the answer as <math>\sqrt{132} = 2\sqrt{33}</math>. | ||
− | Continuing with solution 1, using similar | + | Continuing with solution 1, using similar triangles, we get the answer as <math>\textbf{(A)}\ \dfrac{3\sqrt{33}}5</math> |
~ghfhgvghj10 | ~ghfhgvghj10 |
Latest revision as of 21:23, 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
?
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 triangles, we get the answer as
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Video Solution
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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 |
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