https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Floweytf&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T15:14:56ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_13&diff=1481372007 AIME I Problems/Problem 132021-03-02T00:08:50Z<p>Floweytf: Added a solution</p>
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<div>== Problem ==<br />
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>.<br />
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[[Image:AIME I 2007-13.png]]<br />
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== Solution ==<br />
=== Solution 1 ===<br />
Note first that the intersection is a [[pentagon]].<br />
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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(0,2\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>.<br />
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<center> <br />
<asy>import three; <br />
pointpen = black; <br />
pathpen = black+linewidth(0.7); <br />
currentprojection = perspective(2.5,-12,4);<br />
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);<br />
draw(A--B--C--D--A--E--B--E--C--E--D); <br />
label("A",A, SE); <br />
label("B",B,(1,0,0)); <br />
label("C",C, SE); <br />
label("D",D, W); <br />
label("E",E,N); <br />
label("P",P, NW); <br />
label("Q",Q,(1,0,0)); <br />
label("R",R, S); <br />
label("Y",Y,NW); <br />
label("X",X,NE); <br />
draw(P--X--Q--R--Y--cycle,linetype("6 6")+linewidth(0.7)); <br />
</asy><br />
<asy><br />
pointpen = black; <br />
pathpen = black+linewidth(0.7); <br />
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); <br />
D(MP("P",P,N)--MP("X",X,NE)--MP("Q",Q)--MP("R",R)--MP("Y",Y,NW)--cycle); <br />
D(X--Y,linetype("6 6") + linewidth(0.7)+blue); D(P--(0,-P.y),linetype("6 6") + linewidth(0.7) + red);<br />
MP("\color{blue}{3\sqrt{2}}",(X+Y)/2); <br />
MP("2\sqrt{2}",(Q+R)/2); <br />
MP("\color{red}{\sqrt{\frac{5}{2}}}",(0,-P.y/2),E); <br />
MP("\color{red}{\sqrt{\frac{5}{2}}}",(0,2*P.y/5),E); <br />
</asy><br />
</center><br />
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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>.<br />
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=== Solution 2===<br />
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>.<br />
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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.<br />
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=== Solution 3 ===<br />
<center> <br />
<asy>import three; <br />
import math;<br />
pointpen = black; <br />
pathpen = black+linewidth(0.7); <br />
currentprojection = perspective(2.5,-12,4);<br />
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);<br />
draw(A--B--C--D--A--E--B--E--C--E--D); <br />
draw(B--H--Q, linetype("6 6")+linewidth(0.7)+blue);<br />
draw(X--H, linetype("6 6")+linewidth(0.7)+blue);<br />
draw(D--I--R, linetype("6 6")+linewidth(0.7)+blue);<br />
draw(Y--I, linetype("6 6")+linewidth(0.7)+blue);<br />
label("A",A, SE); <br />
label("B",B,NE); <br />
label("C",C, SE); <br />
label("D",D, W); <br />
label("E",E,N); <br />
label("P",P, NW); <br />
label("Q",Q,(1,0,0)); <br />
label("R",R, S); <br />
label("Y",Y,NW); <br />
label("X",X,NE); <br />
label("H",H,NE);<br />
label("I",I,S);<br />
draw(P--X--Q--R--Y--cycle,linetype("6 6")+linewidth(0.7)); <br />
</asy><br />
</center> <br />
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><br />
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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>. <br />
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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> <br />
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Using the same principle, we can get that <math>|\triangle{IYR}|=\frac{1}{6}|\triangle{PHI}|</math><br />
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Therefore, the area of <math>PYRQX</math> is <math>\frac{2}{3}\cdot|\triangle{PHI}|</math><br />
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<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><br />
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Using this, we can get the area of <math>PYRQX</math><br />
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== See also ==<br />
{{AIME box|year=2007|n=I|num-b=12|num-a=14}}<br />
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[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Floweytf