Difference between revisions of "2011 AIME I Problems/Problem 13"
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Thus the distance from <math>A</math> to the plane is <math>11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}</math>, and <math>33+294+3=\boxed{330}</math>. | Thus the distance from <math>A</math> to the plane is <math>11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}</math>, and <math>33+294+3=\boxed{330}</math>. | ||
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+ | ==Solution 3== | ||
+ | [[File:2011 AIME I 13.png|500px]] | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 11:37, 8 September 2022
Problem
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled . The three vertices adjacent to vertex
are at heights 10, 11, and 12 above the plane. The distance from vertex
to the plane can be expressed as
, where
,
, and
are positive integers. Find
.
Solution 1
Set the cube at the origin with the three vertices along the axes and the plane equal to , where
. The distance from a point
to a plane with equation
is
so the (directed) distance from any point
to the plane is
. So, by looking at the three vertices, we have
, and by rearranging and summing,
Solving the equation is easier if we substitute , to get
, or
. The distance from the origin to the plane is simply
, which is equal to
, so
.
Solution 2
Let the vertices with distance be
, respectively. An equilateral triangle
is formed with side length
. We care only about the
coordinate:
. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so
. Designate the midpoint of
as
. Notice that median
is parallel to the plane because the
and vertex
have the same
coordinate,
, and the median contains
and the
. We seek the angle
of the line:
through the centroid
perpendicular to the plane formed by
,
with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular
to
. Since
makes a
right triangle, the orthogonal line makes the same right triangle rotated
. Therefore,
.
It is also known that the centroid of is a third of the way between vertex
and
, the vertex farthest from the plane. Since
is a diagonal of the cube,
. So the distance from the
to
is
. So, the
from
to the centroid is
.
Thus the distance from to the plane is
, and
.
Solution 3
Video Solution
https://youtube.com/watch?v=Wi-aqv8Ron0
See also
2011 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.