Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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Note that <math>EJCI</math> is a rhombus by symmetry. | Note that <math>EJCI</math> is a rhombus by symmetry. | ||
− | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt | + | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= s\sqrt 3</math> and <math>JI= s\sqrt 2</math>. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is <math>\frac{s^2\sqrt 6}{2}</math>. This gives <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> |
==Note== | ==Note== |
Revision as of 22:45, 9 September 2019
Contents
Problem 24
In the cube with opposite vertices and and are the midpoints of vertices and respectively. Let be the ratio of the area of the cross-section to the area of one of the faces of the cube. What is
Solution
Note that is a rhombus by symmetry. Let the side length of the cube be . By the Pythagorean theorem, and . Since the area of a rhombus is half the product of its diagonals, the area of the cross section is . This gives . Thus
Note
In the 2008 AMC 10A, Question 21 was nearly identical to this question, except that in this question you have to look for the square of the area, not the actual area.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
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