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2014 AMC 10A Problems/Problem 19

Revision as of 14:20, 25 November 2022 by Ghfhgvghj10 (talk | contribs) (→‎Solution 2 (3D Coordinate Geometry))

Problem

Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\overline{XY}$ contained in the cube with edge length $3$?

[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]

$\textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2$

Solution

By Pythagorean Theorem in three dimensions, the distance $XY$ is $\sqrt{4^2+4^2+10^2}=2\sqrt{33}$.

Let the length of the segment $XY$ that is inside the cube with side length $3$ be $x$. By similar triangles, $\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}$, giving $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-$.

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 $\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}$ and plug it in for the distance of $XY$.

$\sqrt{(0-4)^2+(10-0)^2+(0-4)}$

We get the answer as $\sqrt{132} = 2\sqrt{33}$.

Continuing with solution 1, using similar triangles, we get the answer as $\textbf{(A)}\ \dfrac{3\sqrt{33}}5$

~ghfhgvghj10

Video Solution

https://youtu.be/4FInNJ9wM6s

~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

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