Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 2"

Latest revision as of 11:45, 12 July 2021

Problem

Two quadrilaterals are drawn on the plane such that they share no sides. What is the maximum possible number of intersections of the boundaries of the two quadrilaterals?

Solution

We find that it is possible to construct the maximal $\boxed{16}$ points, where each side of one quadrilateral intersects all four sides of the other quadrilateral.

~samrocksnature

Solution 2

Take two concave quadrilaterals. Call two lines "somewhat parallel" if the different in their slopes is less than $\frac{1}{2}$ . An arrow has approximately $4$ lines which are "somewhat" parallel, which means $2$ arrows that are $90^o$ to each other form $4 \cdot 4 = \boxed{16}$ intersections. $\linebreak$ ~Geometry285

Solution 3

A line can intersect $4$ other non-parallel lines $4$ times. If we draw $2$ quadrilaterals with non-parallel sides, it would then be possible to get $4 \cdot 4 = \boxed{16}$ intersections.

~Mathdreams

@@ Line 8: / Line 8: @@
 </center>
 ~samrocksnature
+==Solution 2==
+Take two concave quadrilaterals. Call two lines "somewhat parallel" if the different in their slopes is less than <math>\frac{1}{2}</math>. An arrow has approximately <math>4</math> lines which are "somewhat" parallel, which means <math>2</math> arrows that are <math>90^o</math> to each other form <math>4 \cdot 4 = \boxed{16}</math> intersections.
+<math>\linebreak</math>
+~Geometry285
+== Solution 3 ==
+A line can intersect <math>4</math> other non-parallel lines <math>4</math> times. If we draw <math>2</math> quadrilaterals with non-parallel sides, it would then be possible to get <math>4 \cdot 4 = \boxed{16}</math> intersections.
+~Mathdreams
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