Difference between revisions of "2007 AIME I Problems/Problem 13"
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Revision as of 20:11, 4 July 2013
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 making remote request. Unknown error_msg) |
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 making remote request. Unknown error_msg)
![[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]](http://latex.artofproblemsolving.com/2/b/8/2b8b9e40ce0004d1e035b9700a3edf3a2deebbcc.png)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.