Difference between revisions of "2007 AIME I Problems/Problem 13"
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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> | 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{ | + | Because <math>Q</math> is the midpoint of <math>\overline{BC}</math>, and <math>\overline{AB}\parallel\overline{DC}</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> | 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> | ||
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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> | <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> | + | Using this, we can get the area of <math>PYRQX = \sqrt{80}</math> so the answer is <math>\fbox{080}</math>. |
== See also == | == See also == |
Revision as of 21:55, 28 June 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 so the answer is
.
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.