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Difference between revisions of "2014 AMC 10A Problems/Problem 19"

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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>.
 
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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-.
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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}}</math>.
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SOMEONE HELP WITH THE CODE PLEASE
 
  
 
==Solution 2 (3D Coordinate Geometry)==
 
==Solution 2 (3D Coordinate Geometry)==
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The diagonal of the base of the cube with side length <math>4</math> is <math>4 \sqrt{2}</math>. Hence by similarity:
 
The diagonal of the base of the cube with side length <math>4</math> is <math>4 \sqrt{2}</math>. Hence by similarity:
  
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<math>XY = \sqrt{3^2 + \left(\frac{3}{10} \cdot 4 \sqrt{2} \right)^2} = \sqrt{\frac{225}{25} + \frac{6^2 \cdot 2}{25}} = \frac{\sqrt{99 \cdot 3}}{25} = \boxed{ \frac{3 \sqrt{33}}{25}}</math>.
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<math>XY = \sqrt{3^2 + \left(\frac{3}{10} \cdot 4 \sqrt{2} \right)^2} = \sqrt{\frac{225}{25} + \frac{6^2 \cdot 2}{25}} = \frac{\sqrt{99 \cdot 3}}{5} = \boxed{ \frac{3 \sqrt{33}}{5}}</math>.
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==Solution 4==
 
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We look at the graph from one direction, namely the side facing us. The diagram should be a 1*1 square stacked on a 2*2 square stacked on a 3*3 square stacked on a 4*4 square. Then, when we connect XY, we can see that since the sides of a square is parallel, we can use similar triangles on the overall big triangle and both the triangle with base the 3*3 square and the triangle with base the 2*2 square. PLEASE DO NOT TOUCH IT IS A WORK IN PROGRESS
 
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 11:33, 10 November 2024

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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}}$.

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 $XY$.

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

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

Solution 3

The diagonal of the base of the cube with side length $4$ is $4 \sqrt{2}$. Hence by similarity:

$XY = \sqrt{3^2 + \left(\frac{3}{10} \cdot 4 \sqrt{2} \right)^2} = \sqrt{\frac{225}{25} + \frac{6^2 \cdot 2}{25}} = \frac{\sqrt{99 \cdot 3}}{5} = \boxed{ \frac{3 \sqrt{33}}{5}}$.

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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