Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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==Note== | ==Note== | ||
− | In the 2008 AMC 10A, Question 21 (https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_21) was nearly identical to this question, except that in this question | + | In the 2008 AMC 10A, Question 21 (https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_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== | ==See Also== |
Revision as of 18:21, 22 December 2018
Contents
[hide]Problem 24
In the cube with opposite vertices and and are the midpoints of edges 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 (https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_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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