1983 USAMO Problems/Problem 4
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 |
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