Download 2009-2010 Meet 1

SOML MEET 1
EVENT 4
Team Problem
1.
[15 Points]
NAME: __________________
NAME: __________________
NAME: __________________
TEAM: __________________
SCHOOL: __________________
A corner of a cube with volume 216 cubic inches is chopped off in such a
way that the cut runs through the three vertices adjacent to the vertex of
the corner chosen. If the corner chopped off was placed on a table with the
cut-face as the bottom, the geometric figure is a pyramid. Find the exact
value of the height (in inches) of the pyramid.
ANS: ________________ inches
SOML MEET 1
EVENT 4
Team Problem
1.
[15 Points]
KEY
A corner of a cube with volume 216 cubic inches is chopped off in such a way that
the cut runs through the three vertices adjacent to the vertex of the corner chosen. If
the corner chopped off was placed on a table with the cut-face as the bottom, the
geometric figure is a pyramid. Find the exact value of the height (in inches) of the
pyramid.
Solution:
Volume of the cube is 216 in3 and 63=216.
Lateral face of the pyramid is a right triangle
whose legs are 6” edges of the cube.
Therefore the base of the pyramid is an
equilateral triangle with sides 6 2 ”.
Due to the symmetry of a cube, the height
of this pyramid drops to the center of the
base and all we need to do is find the
dimensions of the shaded right triangle.
Let’s look more closely at the base:
The hypotenuse is an edge
of our cube, which is 6”.
The “leg” of the shaded isosceles triangle above is formed by
intersecting the bisectors of each angle of the base.
Since the base is an equilateral triangle, this creates a
300-1200-300 triangle or two 300-600-900 triangles:
The side opposite the 60° angle
of the darkest shaded triangle is
half of 6 2 ” = 3 2 ”
The sides of a 300-600-900 triangle satisfy the following ratio: 1: 3 :2.
Therefore
3 2 leg
=
which gives us:
2
3
leg = 2 6 ”
Finally our desired height, h, can be calculated from the
h2 + (2 6)2 = 62
Pythagorean Theorem:
h2 = 12
h=2 3
So the height of the pyramid is:
2 3”
ANS: 2 3 ”