2018 AMC 8 Problems/Problem 24
Contents
Problem
In the cube with opposite vertices and and are the midpoints of segments 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 1
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
Solution 2
We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of the square are of length . Let be the origin and let be the positive direction, be the positive direction, and be the positive direction. We find that .
Notice that so is a rhombus. Furthermore, by the distance formula, .
By the Law of Cosines on we have . By the Law of Cosines on we have .
Bretschneider's formula states given a quadrilateral with sides then where . Using this formula, we find that
Using Bretschneider's formula again, we can find that .
The answer is thus so we circle answer choice .
~franzliszt
Solution 3
We will use the following
Theorem: Suppose we have a quadrilateral with edges of length (in that order) and diagonals of length . Bretschneider's formula states that the area .
Proof. Here is one of my favorite proofs of this with vector geometry.
Suppose a quadrilateral has sides such that and that the diagonals of the quadrilateral are and . The area of any such quadrilateral is .
Lagrange's Identity states that . Therefore:
Then if represent (and are thus the side lengths) while represent (and are thus the diagonal lengths), the area of a quadrilateral is:
Back to the problem. We use vectors. WLOG suppose the cube has sides of length . Let be the origin and let be the direction, be the direction, and be the direction.
Then and . Let . Dot product gives
Hence, .
Now, notice that so .
Using Bretschneider's formula, we obtain .
Using Bretschneider's formula again, we can find that .
Hence, the answer is so we circle answer choice .
~franzliszt (again)
Note
Problem 21 of the 2008 AMC 10A 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.
Video Solution
https://www.youtube.com/watch?v=04pV_rZw8bg ~ Happytwin
Video Solution
https://www.youtube.com/watch?v=ji9_6XNxyIc ~ MathEx
Video Solution
https://youtu.be/FDgcLW4frg8?t=2823
~ pi_is_3.14
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.