Difference between revisions of "2007 AIME I Problems/Problem 13"
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− | == Solution == | + | == Solution 1== |
Note first that the intersection is a [[pentagon]]. | Note first that the intersection is a [[pentagon]]. | ||
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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>(\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>. | 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>. | ||
+ | |||
+ | == 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>. | ||
+ | |||
+ | 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. | ||
== See also == | == See also == |
Revision as of 15:19, 24 March 2007
Problem
A square pyramid with base and vertex
has eight edges of length 4. A plane passes through the midpoints of
,
, and
. The plane's intersection with the pyramid has an area that can be expressed as
. Find
.
Contents |
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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.
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 |