Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Find the answer

G3 (1-6, Czechoslovakia) Two planes $P$ and $Q$ intersect at line $p$ . It is known that point $A$ is in plane $P$ but not in plane $Q$ , and point $C$ is in plane $Q$ but not in plane $P$ , and neither is on $p$ . Construct a quadrilateral $A B C D$ such that $A B / / C D, A D = B C$ , and point $B$ is in plane $P$ , point $D$ is in plane $Q$ , and the quadrilateral has an inscribed circle.

Solution

First, prove that $A B$ and $C D$ are both parallel to $p$ . This is because if we set a point $M$ on $p$ , there is only one line $l$ through $M$ that is parallel to $A B$ and $C D$ . Since $l$ is both in $P$ and in $Q$ , it follows that $l = p$ .

As shown in Figure 4-4, draw lines $A B$ and $C D$ parallel to $p$ through points $A$ and $C$ , respectively.

In the plane formed by $A B$ and $C D$ , draw $C E \perp A B$ , with the foot of the perpendicular at $E$ . It is easy to see that
$\begin{aligned} A E & =\frac{1}{2}(A B+C D) \\ & =\frac{1}{2}(A D+B C) \\ & =A D . \end{aligned}$

Therefore, a circle can be drawn with $A$ as the center and $A E$ as the radius, intersecting $C D$ at $D$ , and then further determining $B$ .

When $A E > C E$ , the problem has two solutions; when $A E = C E$ , the problem has one solution (in this case, the trapezoid $A B C D$ becomes a square); when $A E < C E$ , there is no solution.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.