1983 USAMO Problems/Problem 4


Six segments $S_1, S_2, S_3, S_4, S_5,$ and $S_6$ are given in a plane. These are congruent to the edges $AB, AC, AD, BC, BD,$ and $CD$, respectively, of a tetrahedron $ABCD$. Show how to construct a segment congruent to the altitude of the tetrahedron from vertex $A$ with straight-edge and compasses.


Throughout this solution, we denote the length of a segment $S$ by $|S|$.

In this solution, we employ several lemmas. Two we shall take for granted: given any point $A$ and a line $\ell$ not passing through $A$, we can construct a line $\ell'$ through $A$ parallel to $\ell$; and given any point $A$ on a line $\ell$, we can construct a line $\ell'$ through $A$ perpendicular to $\ell$.

Lemma 1: If we have two segments $S$ and $T$ on the plane with non-zero length, we may construct a circle at either endpoint of $S$ whose radius is $|T|$.

Proof: We can construct arbitrarily many copies of $T$ 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 $T$ like this until we create a circle of radius $|T|$ and center $P_1$ that intersects segment $S$. We can then take this intersection point $P_2$ and draw a line $\ell$ through it perpendicular to $S$, and draw a circle with center $P_2$ passing through $P_1$, and consider its intersection $P_3$ with $\ell$. Note that $P_2P_3\perp S$ and $|P_2P_3|=|T|$. Take an endpoint $P_4$ of $S$: then draw a line through $P_3$ parallel to $S$, and a line through $P_4$ parallel to $P_2P_3$. Let these two lines intersect at $P_5$. Then $P_2P_3P_5P_4$ is a rectangle, so $|P_4P_5|=|T|$. Our desired circle is then a circle centered at $P_4$ through $P_5$.

Lemma 2: Given three collinear points $A$, $B$, $C$ in this order, if $|AB|=a$ and $|BC|=b$ with $a>b$, then we can construct a segment of length $\sqrt{a^2-b^2}$.

Proof: From Lemma 1, we can construct a circle through $C$ with radius $a$, and then construct a perpendicular through $B$ to $AC$: these two objects intersect at $D$ and $E$. Both $BD$ and $DE$ have length $\sqrt{a^2-b^2}$, from the Pythagorean Theorem.

Proof of the original statement: Note that we can construct a triangle $B'C'D'$ congruent to triangle $BCD$ by applying Lemma 1 to segments $S_4$, $S_5$, and $S_6$. Similarly, we can construct $A_B$ and $A_C$ outside triangle $B'C'D'$ such that $A_BC'D'\cong ACD$ and $A_CB'D'\cong ABD$.

Let $A'$ be a point outside of the plane containing $S_1$ through $S_6$ such that $A'B'C'D'\cong ABCD$. Then the altitudes of triangles $A_BC'D'$ and $A'C'D'$ to segment $C'D'$ are congruent, as are the altitudes of triangles $A_CB'D'$ and $A'B'D'$ to segment $B'D'$. However, if we project the altitudes of $A'C'D'$ and $A'B'D'$ from $A'$ onto the plane, their intersection is the base of the altitude of tetrahedron $A'B'C'D'$ from $A'$. In addition, these altitude projections are collinear with the altitudes of triangles $A_BC'D'$ and $A_CB'D'$. Therefore, the altitudes of $A_BC'D'$ and $A_CB'D'$ from $A_B$ and $A_C$ intersect at the base $X'$ of the altitude of $A'B'C'D'$ from $A'$. In summary, we can construct $X'$ by constructing the perpendiculars from $A_B$ and $A_C$ to $C'D'$ and $B'D'$ respectively, and taking their intersection.

Let $Y'$ be the intersection of $A_BX'$ with $C'D'$. Then the altitude length we seek to construct is, from the Pythagorean Theorem, $\sqrt{|A_BY'|^2-|X'Y'|^2}$. We can directly apply Lemma 2 to segment $A_BY'X'$ to obtain this segment. This shows how to construct a segment of length $A'X'$.

See Also

1983 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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