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>. |
− | + | [[Image:AIME I 2007-13.png]] | |
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− | == Solution 1== | + | == Solution == |
+ | === Solution 1 === | ||
Note first that the intersection is a [[pentagon]]. | 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. <math>A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E( | + | 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(0,2\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>. |
− | + | <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); | ||
+ | draw(A--B--C--D--A--E--B--E--C--E--D); | ||
+ | label("A",A, SE); | ||
+ | label("B",B,(1,0,0)); | ||
+ | label("C",C, SE); | ||
+ | label("D",D, W); | ||
+ | label("E",E,N); | ||
+ | label("P",P, NW); | ||
+ | label("Q",Q,(1,0,0)); | ||
+ | label("R",R, S); | ||
+ | label("Y",Y,NW); | ||
+ | label("X",X,NE); | ||
+ | draw(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)+blue); D(P--(0,-P.y),linetype("6 6") + linewidth(0.7) + red); | ||
+ | MP("\color{blue}{3\sqrt{2}}",(X+Y)/2); | ||
+ | MP("2\sqrt{2}",(Q+R)/2); | ||
+ | MP("\color{red}{\sqrt{\frac{5}{2}}}",(0,-P.y/2),E); | ||
+ | MP("\color{red}{\sqrt{\frac{5}{2}}}",(0,2*P.y/5),E); | ||
+ | </asy> | ||
+ | </center> | ||
− | == Solution 2== | + | 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=== | ||
Use the same coordinate system as above, and let the plane determined by <math>\triangle PQR</math> intersect <math>\overline{BE}</math> at <math>X</math> and <math>\overline{DE}</math> at <math>Y</math>. Then the line <math>\overline{XY}</math> is the intersection of the planes determined by <math>\triangle PQR</math> and <math>\triangle BDE</math>. | Use the same coordinate system as above, and let the plane determined by <math>\triangle PQR</math> intersect <math>\overline{BE}</math> at <math>X</math> and <math>\overline{DE}</math> at <math>Y</math>. Then the line <math>\overline{XY}</math> is the intersection of the planes determined by <math>\triangle PQR</math> and <math>\triangle BDE</math>. | ||
Note that the plane determined by <math>\triangle BDE</math> has the equation <math>x=y</math>, and <math>\overline{PQ}</math> can be described by <math>x=2(1-t)-t,\ y=t,\ z=t\sqrt{2}</math>. It intersects the plane when <math>2(1-t)-t=t</math>, or <math>t=\frac{1}{2}</math>. This intersection point has <math>z=\frac{\sqrt{2}}{2}</math>. Similarly, the intersection between <math>\overline{PR}</math> and <math>\triangle BDE</math> has <math>z=\frac{\sqrt{2}}{2}</math>. So <math>\overline{XY}</math> lies on the plane <math>z=\frac{\sqrt{2}}{2}</math>, from which we obtain <math>X=\left( \frac{3}{2},\frac{3}{2},\frac{\sqrt{2}}{2}\right)</math> and <math>Y=\left( -\frac{3}{2},-\frac{3}{2},\frac{\sqrt{2}}{2}\right)</math>. The area of the pentagon <math>EXQRY</math> can be computed in the same way as above. | Note that the plane determined by <math>\triangle BDE</math> has the equation <math>x=y</math>, and <math>\overline{PQ}</math> can be described by <math>x=2(1-t)-t,\ y=t,\ z=t\sqrt{2}</math>. It intersects the plane when <math>2(1-t)-t=t</math>, or <math>t=\frac{1}{2}</math>. This intersection point has <math>z=\frac{\sqrt{2}}{2}</math>. Similarly, the intersection between <math>\overline{PR}</math> and <math>\triangle BDE</math> has <math>z=\frac{\sqrt{2}}{2}</math>. So <math>\overline{XY}</math> lies on the plane <math>z=\frac{\sqrt{2}}{2}</math>, from which we obtain <math>X=\left( \frac{3}{2},\frac{3}{2},\frac{\sqrt{2}}{2}\right)</math> and <math>Y=\left( -\frac{3}{2},-\frac{3}{2},\frac{\sqrt{2}}{2}\right)</math>. The area of the pentagon <math>EXQRY</math> can be computed in the same way as above. | ||
+ | |||
+ | === Solution 3 === | ||
+ | <center> | ||
+ | <asy>import three; | ||
+ | import math; | ||
+ | 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), H=(4,2,0), I=(-2,-4,0); | ||
+ | draw(A--B--C--D--A--E--B--E--C--E--D); | ||
+ | draw(B--H--Q, linetype("6 6")+linewidth(0.7)+blue); | ||
+ | draw(X--H, linetype("6 6")+linewidth(0.7)+blue); | ||
+ | draw(D--I--R, linetype("6 6")+linewidth(0.7)+blue); | ||
+ | draw(Y--I, linetype("6 6")+linewidth(0.7)+blue); | ||
+ | label("A",A, SE); | ||
+ | label("B",B,NE); | ||
+ | label("C",C, SE); | ||
+ | label("D",D, W); | ||
+ | label("E",E,N); | ||
+ | label("P",P, NW); | ||
+ | label("Q",Q,(1,0,0)); | ||
+ | label("R",R, S); | ||
+ | label("Y",Y,NW); | ||
+ | label("X",X,NE); | ||
+ | label("H",H,NE); | ||
+ | label("I",I,S); | ||
+ | draw(P--X--Q--R--Y--cycle,linetype("6 6")+linewidth(0.7)); | ||
+ | </asy> | ||
+ | </center> | ||
+ | Extend <math>\overline{RQ}</math> and <math>\overline{AB}</math>. The point of intersection is <math>H</math>. Connect <math>\overline{PH}</math>. <math>\overline{EB}</math> intersects <math>\overline{PH}</math> at <math>X</math>. Do the same for <math>\overline{QR}</math> and <math>\overline{AD}</math>, and let the intersections be <math>I</math> and <math>Y</math> | ||
+ | |||
+ | Because <math>Q</math> is the midpoint of <math>\overline{BC}</math>, and <math>\overline{AB}\parallel\overline{BC}</math>, so <math>\triangle{RQC}\cong\triangle{HQB}</math>. <math>\overline{BH}=2</math>. | ||
+ | |||
+ | Because <math>\overline{BH}=2</math>, we can use mass point geometry to get that <math>\overline{PX}=\overline{XH}</math>. <math>|\triangle{XHQ}|=\frac{\overline{XH}}{\overline{PH}}\cdot\frac{\overline{QH}}{\overline{HI}}\cdot|\triangle{PHI}|=\frac{1}{6}\cdot|\triangle{PHI}|</math> | ||
+ | |||
+ | Using the same principle, we can get that <math>|\triangle{IYR}|=\frac{1}{6}|\triangle{PHI}|</math> | ||
+ | |||
+ | Therefore, the area of <math>PYRQX</math> is <math>\frac{2}{3}\cdot|\triangle{PHI}|</math> | ||
+ | |||
+ | <math>\overline{RQ}=2\sqrt{2}</math>, so <math>\overline{IH}=6\sqrt{2}</math>. Using the law of cosines, <math>\overline{PH}=\sqrt{28}</math>. The area of <math>\triangle{PHI}=\frac{1}{2}\cdot\sqrt{28-18}\cdot6\sqrt{2}=6\sqrt{5}</math> | ||
+ | |||
+ | Using this, we can get the area of <math>PYRQX</math> | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:08, 1 March 2021
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 .
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 .
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.
Solution 3
Extend and . The point of intersection is . Connect . intersects at . Do the same for and , and let the intersections be and
Because is the midpoint of , and , so . .
Because , we can use mass point geometry to get that .
Using the same principle, we can get that
Therefore, the area of is
, so . Using the law of cosines, . The area of
Using this, we can get the area of
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.