Difference between revisions of "1998 AIME Problems/Problem 11"
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This approach uses [[analytical geometry]]. Let <math>A</math> be at the origin, <math>B</math> at <math>(20,0,0)</math>, <math>C</math> at <math>(20,0,20)</math>, and <math>D</math> at <math>(20,20,20)</math>. Thus, <math>P</math> is at <math>\displaystyle (5,0,0)</math>, <math>Q</math> is at <math>(20,0,15)</math>, and <math>R</math> is at <math>\displaystyle (20,10,20)</math>. | This approach uses [[analytical geometry]]. Let <math>A</math> be at the origin, <math>B</math> at <math>(20,0,0)</math>, <math>C</math> at <math>(20,0,20)</math>, and <math>D</math> at <math>(20,20,20)</math>. Thus, <math>P</math> is at <math>\displaystyle (5,0,0)</math>, <math>Q</math> is at <math>(20,0,15)</math>, and <math>R</math> is at <math>\displaystyle (20,10,20)</math>. | ||
− | Let the plane <math>PQR</math> have the equation <math>ax + by + cz = d</math>. Using point <math>P</math>, we get that <math>5a = d</math>. Using point <math>Q</math>, we get <math>20a + 15c = d \Longrightarrow 4d + 15c = d \Longrightarrow d = -5c</math>. Using point <math>R</math>, we get <math>\displaystyle 20a + 10b + 20c = d \Longrightarrow 4d + 10b - 4d = d \Longrightarrow d = 10b</math>. Thus plane <math>PQR</math>’s equation reduces to <math>\frac{d}{5}x + \frac{d}{10}y - \frac{d}{5}z = d \Longrightarrow 2x + y - 2z = 10</math>. | + | Let the plane <math>PQR</math> have the equation <math>ax + by + cz = d</math>. Using point <math>P</math>, we get that <math>5a = d</math>. Using point <math>Q</math>, we get <math>20a + 15c = d \Longrightarrow 4d + 15c = d \Longrightarrow d = -5c</math>. Using point <math>R</math>, we get <math>\displaystyle 20a + 10b + 20c = d \Longrightarrow 4d + 10b - 4d = d \Longrightarrow d = 10b</math>. Thus plane <math>PQR</math>’s [[equation]] reduces to <math>\frac{d}{5}x + \frac{d}{10}y - \frac{d}{5}z = d \Longrightarrow 2x + y - 2z = 10</math>. |
We know need to find the intersection of this plane with that of <math>z = 0</math>, <math>z = 20</math>, <math>x = 0</math>, and <math>\displaystyle y = 20</math>. After doing a little bit of algebra, the intersections are the lines <math>y = -2x + 10</math>, <math>y = -2x + 50</math>, <math>y = 2z + 10</math>, and <math>z = x + 5</math>. Thus, there are three more vertices on the polygon, which are at <math>(0,10,0)(0,20,5)(15,20,20)</math>. | We know need to find the intersection of this plane with that of <math>z = 0</math>, <math>z = 20</math>, <math>x = 0</math>, and <math>\displaystyle y = 20</math>. After doing a little bit of algebra, the intersections are the lines <math>y = -2x + 10</math>, <math>y = -2x + 50</math>, <math>y = 2z + 10</math>, and <math>z = x + 5</math>. Thus, there are three more vertices on the polygon, which are at <math>(0,10,0)(0,20,5)(15,20,20)</math>. | ||
− | We can find the lengths of the sides of the polygons now. There are many (4) [[right triangle]]s with legs 5, 10, so their [[hypotenuse]]s are <math>5\sqrt{5}</math>. The other two are of <math>45-45-90 \triangle</math>s with legs of length 15, so their hypotenuses are <math>\displaystyle 15\sqrt{2} \displaystyle</math>. 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 <math>20\sqrt{2}</math>. | + | We can find the lengths of the sides of the polygons now. There are many (4) [[right triangle]]s with legs 5, 10, so their [[hypotenuse]]s are <math>5\sqrt{5}</math>. The other two are of <math>45-45-90 \triangle</math>s with legs of length 15, so their hypotenuses are <math>\displaystyle \displaystyle 15\sqrt{2} \displaystyle \displaystyle</math>. So we have a [[hexagon]] with sides <math>15\sqrt{2}, \displaystyle 5\sqrt{5}, 5\sqrt{5}, \displaystyle 15\sqrt{2}, 5\sqrt{5}, \displaystyle 5\sqrt{5}</math> 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 <math>20\sqrt{2}</math>. |
[[Image:1998_AIME-11b.png]] | [[Image:1998_AIME-11b.png]] | ||
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=== Solution 2 === | === 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]]. |
Let the cube's vertices be <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math>, with <math>A</math>, <math>B</math>, and <math>C</math> on the bottom face as before, <math>H</math> being the other bottom vertex, <math>D</math> directly above <math>C</math>, <math>E</math> above <math>B</math>, <math>F</math> above <math>A</math>, and <math>G</math> above <math>H</math>. | Let the cube's vertices be <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math>, with <math>A</math>, <math>B</math>, and <math>C</math> on the bottom face as before, <math>H</math> being the other bottom vertex, <math>D</math> directly above <math>C</math>, <math>E</math> above <math>B</math>, <math>F</math> above <math>A</math>, and <math>G</math> above <math>H</math>. | ||
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Now, the next vertex (call it <math>Y</math>) will be somewhere on <math>AF</math>. The segment must be parallel to <math>QR</math>, so <math>FY</math> must have length <math>2x</math>, and <math>AY</math> must be <math>20 - 2x</math>. | Now, the next vertex (call it <math>Y</math>) will be somewhere on <math>AF</math>. The segment must be parallel to <math>QR</math>, so <math>FY</math> must have length <math>2x</math>, and <math>AY</math> must be <math>20 - 2x</math>. | ||
− | Since <math>\displaystyle DX \parallel AP</math>, <math>DR \parallel AY</math>, and <math>RX \parallel PY</math>, we must have <math>\displaystyle \triangle APY \sim \triangle DXR</math>; therefore, | + | Since <math>\displaystyle DX \parallel AP</math>, <math>DR \parallel AY</math>, and <math>RX \parallel PY</math>, we must have <math>\displaystyle \triangle APY \displaystyle \sim \displaystyle \triangle \displaystyle DXR \displaystyle</math>; therefore, |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
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<math>x=5</math></div> | <math>x=5</math></div> | ||
− | We can now find that the hexagon has side lengths <math>\displaystyle 15\sqrt {2}</math>, <math>5\sqrt {5}</math>, <math>5\sqrt {5}</math>, <math>15\sqrt {2}</math>, <math>5\sqrt {5}</math>, and <math>5\sqrt {5}</math>. Moreover, opposite angles of this must be equal (by symmetry), so segment <math>RY</math> divides the hexagon into two isosceles | + | We can now find that the hexagon has side lengths <math>\displaystyle \displaystyle 15\sqrt {2} \displaystyle \displaystyle</math>, <math>5\sqrt {5}</math>, <math>5\sqrt {5}</math>, <math>\displaystyle \displaystyle 15\sqrt {2} \displaystyle \displaystyle </math>, <math>5\sqrt {5}</math>, and <math>5\sqrt {5}</math>. Moreover, opposite angles of this must be equal (by symmetry), so segment <math>RY</math> divides the [[hexagon]] into two [[isosceles trapezoid]]s. It is easy to find the length of <math>RY</math> (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or <math>20\sqrt {2}</math>), so it is now easy to finish the problem. From here, we can continue as in the first solution. |
== See also == | == See also == |
Revision as of 13:04, 8 September 2007
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
[hide]Solution
Solution 1
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 many (4) right triangles with legs 5, 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 |