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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 22:20, 17 July 2016
Problem
Six segments 
and 
are given in a plane. These are congruent to the edges 
and 
, respectively, of a tetrahedron 
. Show how to construct a segment congruent to the altitude of the tetrahedron from vertex 
with straight-edge and compasses.
Solution
Throughout this solution, we denote the length of a segment 
by 
.
In this solution, we employ several lemmas. Two we shall take for granted: given any point and a line 
not passing through , we can construct a line 
through parallel to ; and given any point on a line , we can construct a line through perpendicular to .
Lemma 1: If we have two segments and 
on the plane with non-zero length, we may construct a circle at either endpoint of whose radius is 
.
Proof: We can construct arbitrarily many copies of by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of like this until we create a circle of radius and center 
that intersects segment . We can then take this intersection point 
and draw a line through it perpendicular to , and draw a circle with center passing through , and consider its intersection 
with . Note that 
and 
. Take an endpoint 
of : then draw a line through parallel to , and a line through parallel to 
. Let these two lines intersect at 
. Then 
is a rectangle, so 
. Our desired circle is then a circle centered at through .
Lemma 2: Given three collinear points , 
, 
in this order, if 
and 
with 
, then we can construct a segment of length 
.
Proof: From Lemma 1, we can construct a circle through with radius 
, and then construct a perpendicular through to 
: these two objects intersect at 
and 
. Both 
and 
have length , from the Pythagorean Theorem.
Proof of the original statement: Note that we can construct a triangle 
congruent to triangle 
by applying Lemma 1 to segments 
, 
, and . Similarly, we can construct 
and 
outside triangle such that 
and 
.
Let 
be a point outside of the plane containing 
through such that 
. Then the altitudes of triangles 
and 
to segment 
are congruent, as are the altitudes of triangles 
and 
to segment 
. However, if we project the altitudes of and from onto the plane, their intersection is the base of the altitude of tetrahedron 
from . In addition, these altitude projections are collinear with the altitudes of triangles and . Therefore, the altitudes of and from and intersect at the base 
of the altitude of from . In summary, we can construct by constructing the perpendiculars from and to and respectively, and taking their intersection.
Let 
be the intersection of 
with . Then the altitude length we seek to construct is, from the Pythagorean Theorem, 
. We can directly apply Lemma 2 to segment 
to obtain this segment. This shows how to construct a segment of length 
.
See Also
| 1983 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
