Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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Four cubes with edge lengths <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math> are stacked as shown. What is the length of the portion of <math>\overline{XY}</math> contained in the cube with edge length <math>3</math>? | Four cubes with edge lengths <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math> are stacked as shown. What is the length of the portion of <math>\overline{XY}</math> contained in the cube with edge length <math>3</math>? | ||
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<asy> | <asy> | ||
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label("$4$", (4,2), W,fontsize(8pt)); | label("$4$", (4,2), W,fontsize(8pt)); | ||
</asy> | </asy> | ||
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+ | <math> \textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2 </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
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==Solution== | ==Solution== | ||
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+ | By Pythagorean Theorem in three dimensions, the distance <math>XY</math> is <math>\sqrt{4^2+4^2+10^2}=2\sqrt{33}</math>. | ||
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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>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/4FInNJ9wM6s | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 02:01, 16 February 2021
Contents
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 .
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.