Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m
 
(8 intermediate revisions by 3 users not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
We find that it is possible to construct the maximal <math>\boxed{16}</math> points, where each side of one quadrilteral intersects all four sides of the other quadrilateral.
+
We find that it is possible to construct the maximal <math>\boxed{16}</math> points, where each side of one quadrilateral intersects all four sides of the other quadrilateral.
 
<center>
 
<center>
−
[[File:Invites2.png|500px]]
+
[[File:Invites2D.png|500px]]
 
</center>
 
</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
 +
 +
==See also==
 +
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
 +
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

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.

Invites2D.png

~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

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_2&oldid=158245"