Difference between revisions of "1998 AIME Problems/Problem 11"
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<span style="font-size:100%">For non-asymptote version of image, see [[:Image:1998_AIME-11.png]]</span> | <span style="font-size:100%">For non-asymptote version of image, see [[:Image:1998_AIME-11.png]]</span> | ||
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The height of the triangles at the top/bottom is <math>\frac{20\sqrt{2} - 15\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}</math>. The [[Pythagorean Theorem]] gives that half of the base of the triangles is <math>\frac{15}{\sqrt{2}}</math>. We find that the middle [[rectangle]] is actually a [[square]], so the total area is <math>(15\sqrt{2})^2 + 4\left(\frac 12\right)\left(\frac 52\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right) = 525</math>. | The height of the triangles at the top/bottom is <math>\frac{20\sqrt{2} - 15\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}</math>. The [[Pythagorean Theorem]] gives that half of the base of the triangles is <math>\frac{15}{\sqrt{2}}</math>. We find that the middle [[rectangle]] is actually a [[square]], so the total area is <math>(15\sqrt{2})^2 + 4\left(\frac 12\right)\left(\frac 52\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right) = 525</math>. | ||
− | + | == Solution 2 == | |
First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or [[skew]]; they are both part of plane <math>PQR</math>, so they cannot be skew. Therefore, they are [[parallel]]. | First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or [[skew]]; they are both part of plane <math>PQR</math>, so they cannot be skew. Therefore, they are [[parallel]]. | ||
Latest revision as of 00:19, 6 May 2021
Problem
Three of the edges of a cube are and and is an interior diagonal. Points and are on and respectively, so that and What is the area of the polygon that is the intersection of plane and the cube?
Contents
Solution 1
For non-asymptote version of image, see Image:1998_AIME-11.png
This approach uses analytical geometry. Let be at the origin, at , at , and at . Thus, is at , is at , and is at .
Let the plane have the equation . Using point , we get that . Using point , we get . Using point , we get . Thus plane ’s equation reduces to .
We know need to find the intersection of this plane with that of , , , and . After doing a little bit of algebra, the intersections are the lines , , , and . Thus, there are three more vertices on the polygon, which are at .
We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are . The other two are of s with legs of length 15, so their hypotenuses are . So we have a hexagon with sides By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it .
The height of the triangles at the top/bottom is . The Pythagorean Theorem gives that half of the base of the triangles is . We find that the middle rectangle is actually a square, so the total area is .
Solution 2
First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane , so they cannot be skew. Therefore, they are parallel.
Let the cube's vertices be , , , , , , , and , with , , and on the bottom face as before, being the other bottom vertex, directly above , above , above , and above .
Clearly, the next vertex of the intersection (starting with , , ) will be somewhere on . Let it be , and have a distance of from D, and a distance of from .
Then, the next vertex will be somewhere on . It must be parallel to , so this implies that it has a distance of from , and thus a distance of from .
Now, the next vertex (call it ) will be somewhere on . The segment must be parallel to , so must have length , and must be .
Since , , and , we must have ; therefore,
We can now find that the hexagon has side lengths , , , , , and . Moreover, opposite angles of this must be equal (by symmetry), so segment divides the hexagon into two isosceles trapezoids. It is easy to find the length of (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or ), so it is now easy to finish the problem. From here, we can continue as in the first solution.
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.