Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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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: | ||
+ | <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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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. | ||
<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>. | <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>. | ||
Revision as of 20:04, 29 October 2024
Contents
[hide]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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-.
SOMEONE HELP WITH THE CODE PLEASE
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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Solution 3
The diagonal of the base of the cube with side length is . Hence by similarity:
.
Solution 4
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. .
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 |
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