Revision as of 14:15, 25 November 2022

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)

Video Solution

https://youtu.be/4FInNJ9wM6s

~IceMatrix

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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Difference between revisions of "2014 AMC 10A Problems/Problem 19"

Revision as of 14:15, 25 November 2022

Contents

Problem

Solution

Solution 2 (3D Coordinate Geometry)

Video Solution

See Also

@@ Line 35: / Line 35: @@
 =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>.
-However, the question is asking us about the ratio of <math>XY</math> to 3. So we get <math>\textbf{(C)}\ \dfrac{2\sqrt{33}}3</math>.
-~ghfhgvghj10
 ==Video Solution==