Difference between revisions of "2007 AIME I Problems/Problem 13"
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== Problem == | == Problem == | ||
− | A [[square pyramid]] with [[base]] <math>ABCD</math> and [[vertex]] <math>E</math> has eight [[ | + | A [[square]] [[pyramid]] with [[base]] <math>ABCD</math> and [[vertex]] <math>E</math> has eight [[edge]]s of length <math>4</math>. A [[plane]] passes through the [[midpoint]]s of <math>AE</math>, <math>BC</math>, and <math>CD</math>. The plane's intersection with the pyramid has an area that can be expressed as <math>\sqrt{p}</math>. Find <math>p</math>. |
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|__TOC__ | |__TOC__ | ||
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− | | | + | |<asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); |
− | + | triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2; | |
+ | D(A--B--C--D--A--E--B--E--C--E--D); MP("A",A); MP("B",B,(1,0,0)); MP("C",C); MP("D",D); MP("E",E,N); D(MP("P",P)); D(MP("Q",Q,(1,0,0))); D(MP("R",R)); | ||
+ | </asy> <!-- Asymptote replacement for Image:AIME_I_2007-13.png --> | ||
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Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. <math>A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,0,2\sqrt{2})</math>. Using the coordinates of the three points of intersection (<math>(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)</math>), it is possible to determine the equation of the plane. The equation of a plane resembles <math>ax + by + cz = d</math>, and using the points we find that <math>2a = d \Longrightarrow d = \frac{a}{2}</math>, <math>-2b = d \Longrightarrow d = \frac{-b}{2}</math>, and <math>-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}</math>. It is then <math>x - y + 2\sqrt{2}z = 2</math>. | Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. <math>A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,0,2\sqrt{2})</math>. Using the coordinates of the three points of intersection (<math>(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)</math>), it is possible to determine the equation of the plane. The equation of a plane resembles <math>ax + by + cz = d</math>, and using the points we find that <math>2a = d \Longrightarrow d = \frac{a}{2}</math>, <math>-2b = d \Longrightarrow d = \frac{-b}{2}</math>, and <math>-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}</math>. It is then <math>x - y + 2\sqrt{2}z = 2</math>. | ||
− | Write the equation of the lines and substitute to find that the other two points of intersection on <math>\overline{BE}</math>, <math>\overline{DE}</math> are <math>(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2})</math>. To find the area of the pentagon, break it up into pieces (an [[isosceles triangle]] on the top, an [[isosceles trapezoid]] on the bottom). Using the [[distance formula]] (<math>\sqrt{a^2 + b^2 + c^2}</math>), it is possible to find that the area of the triangle is <math>\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}</math>. The trapezoid has area <math>\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}(2\sqrt{2} + 3\sqrt{2}) = \frac{5\sqrt{5}}{2}</math>. In total, the area is <math>4\sqrt{5} = \sqrt{80}</math>, and the solution is <math>080</math>. | + | <center> <asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); |
+ | triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2); | ||
+ | D(A--B--C--D--A--E--B--E--C--E--D); MP("A",A); MP("B",B,(1,0,0)); MP("C",C); MP("D",D); MP("E",E,N); D(MP("P",P)); D(MP("Q",Q,(1,0,0))); D(MP("R",R)); D(MP("Y",Y,NW)); D(MP("X",X,NE)); D(P--X--Q--R--Y--cycle,linetype("6 6")+linewidth(0.7)); | ||
+ | </asy> <asy>pointpen = black; pathpen = black+linewidth(0.7); | ||
+ | pair P = (0, 2.5^.5), X = (3/2^.5,0), Y = (-3/2^.5,0), Q = (2^.5,-2.5^.5), R = (-2^.5,-2.5^.5); | ||
+ | D(MP("P",P,N)--MP("X",X,NE)--MP("Q",Q)--MP("R",R)--MP("Y",Y,NW)--cycle); D(X--Y,linetype("6 6") + linewidth(0.7)); D(P--(0,-P.y),linetype("6 6") + linewidth(0.7)); | ||
+ | MP("3\sqrt{2}",(X+Y)/2); MP("2\sqrt{2}",(Q+R)/2); MP("\sqrt{\frac{5}{2}}",(0,-P.y/2),E); MP("\sqrt{\frac{5}{2}}",(0,2*P.y/5),E); | ||
+ | </asy> | ||
+ | |||
+ | </center> | ||
+ | |||
+ | Write the equation of the lines and substitute to find that the other two points of intersection on <math>\overline{BE}</math>, <math>\overline{DE}</math> are <math>\left(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2}\right)</math>. To find the area of the pentagon, break it up into pieces (an [[isosceles triangle]] on the top, an [[isosceles trapezoid]] on the bottom). Using the [[distance formula]] (<math>\sqrt{a^2 + b^2 + c^2}</math>), it is possible to find that the area of the triangle is <math>\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}</math>. The trapezoid has area <math>\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}\left(2\sqrt{2} + 3\sqrt{2}\right) = \frac{5\sqrt{5}}{2}</math>. In total, the area is <math>4\sqrt{5} = \sqrt{80}</math>, and the solution is <math>\boxed{080}</math>. | ||
=== Solution 2=== | === Solution 2=== |
Revision as of 21:19, 11 June 2008
Problem
A square pyramid with base and vertex has eight edges of length . A plane passes through the midpoints of , , and . The plane's intersection with the pyramid has an area that can be expressed as . Find .
import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2; D(A--B--C--D--A--E--B--E--C--E--D); MP("A",A); MP("B",B,(1,0,0)); MP("C",C); MP("D",D); MP("E",E,N); D(MP("P",P)); D(MP("Q",Q,(1,0,0))); D(MP("R",R)); (Error compiling LaTeX. 79ebb2a541577ae09ecbe3d167a6828243c28f28.asy: 7.2: no matching function 'D(void(flatguide3))') |
Solution
Solution 1
Note first that the intersection is a pentagon.
Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. . Using the coordinates of the three points of intersection (), it is possible to determine the equation of the plane. The equation of a plane resembles , and using the points we find that , , and . It is then .
import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2); D(A--B--C--D--A--E--B--E--C--E--D); MP("A",A); MP("B",B,(1,0,0)); MP("C",C); MP("D",D); MP("E",E,N); D(MP("P",P)); D(MP("Q",Q,(1,0,0))); D(MP("R",R)); D(MP("Y",Y,NW)); D(MP("X",X,NE)); D(P--X--Q--R--Y--cycle,linetype("6 6")+linewidth(0.7)); (Error compiling LaTeX. 7d026cc334f620864c6fd4def338ba54880874a0.asy: 7.2: no matching function 'D(void(flatguide3))')
Write the equation of the lines and substitute to find that the other two points of intersection on , are . To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula (), it is possible to find that the area of the triangle is . The trapezoid has area . In total, the area is , and the solution is .
Solution 2
Use the same coordinate system as above, and let the plane determined by intersect at and at . Then the line is the intersection of the planes determined by and .
Note that the plane determined by has the equation , and can be described by . It intersects the plane when , or . This intersection point has . Similarly, the intersection between and has . So lies on the plane , from which we obtain and . The area of the pentagon can be computed in the same way as above.
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |