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 USAJMO Problems/Problem 4"

(→‎Solution 1 (Lcz's Solution))
m (→‎Problem)
 
(3 intermediate revisions by one other user not shown)
Line 4: Line 4:
 
(A lattice point is a point <math>(x, y)</math> in the coordinate plane where <math>x</math> and <math>y</math> are both integers, not necessarily positive.)
 
(A lattice point is a point <math>(x, y)</math> in the coordinate plane where <math>x</math> and <math>y</math> are both integers, not necessarily positive.)
  
−
Carina has three pins, labeled <math>A, B</math>, and <math>C</math>, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance <math>1</math> away. What is the least number of moves that Carina can make in order for triangle <math>ABC</math> to have area 2021?
+
==Solution 1==
 +
 
 +
The answer is <math>128</math>, achievable by <math>A=(10,0), B=(0,-63), C=(-54,1)</math>. We now show the bound.
 +
 
 +
We first do the following optimizations:
  
−
(A lattice point is a point <math>(x, y)</math> in the coordinate plane where <math>x</math> and <math>y</math> are both integers, not necessarily positive.)
+
-if you have a point goes both left and right, we may obviously delete both of these moves and decrease the number of moves by <math>2</math>.
  
−
==Solution 1==
+
-if all of <math>A,B,C</math> lie on one side of the plane, for example <math>y>0</math>, we shift them all down, decreasing the number of moves by <math>3</math>, until one of the points is on <math>y=0</math> for the first time.
 +
 
 +
Now we may assume that <math>A=(a,d)</math>, <math>B=(b,-e)</math>, <math>C=(-c,f)</math> where <math>a,b,c,d,e,f \geq 0</math>. Note we may still shift all <math>A,B,C</math> down by <math>1</math> if <math>d,f>0</math>, decreasing the number of moves by <math>1</math>, until one of <math>d,f</math> is on <math>y=0</math> for the first time. So we may assume one of <math>(a,b)</math> and <math>(d,f)</math> is <math>0</math>, by symmetry. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:
  
−
The answer is <math>128</math>, achievable by <math>A=(10,0)</math>, B=(0,-63), C=(-54,1)<math>. We now show the bound.
 
  
−
We first do the following optimizations:
+
Case 1 (where <math>a=d=0</math>) if <math>wx-yz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
  
−
-if you have a point goes both left and right, we may obviously delete both of these moves and decrease the number of moves by </math>2<math>.
+
Case 2 (else) <math>wy+xy+xz=(w+x)(y+z)-wz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
  
−
-if all of </math>A,B,C<math> lie on one side of the plane, for example </math>y>0<math>, we shift them all down, decreasing the number of moves by </math>3<math>, until one of the points is on </math>y=0<math> for the first time.
 
  
−
Now we may assume that </math>A=(a,d)<math>, </math>B=(b,-e)<math>, </math>C=(-c,f)<math> where </math>a,b,c,d,e,f \geq 0<math>. Note we may still shift all </math>A,B,C<math> down by </math>1<math> if </math>d,f>0<math>, decreasing the number of moves by </math>1<math>, until one of </math>d,f<math> is on </math>y=0<math> for the first time. So we may assume one of </math>(a,b)<math> and </math>(d,f)<math> is </math>0<math>. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:
+
Note that <math>(m+n)^2=4mn+(m-n)^2</math> so if <math>m+n</math> is fixed then <math>mn</math> is maximized exactly when <math>|m-n|</math> is minimized. In particular, if <math>m+n \leq 127</math> then <math>mn-op \leq mn \leq 63*64 = 4032 <4042</math> as desired.
  
−
Case 1 (where </math>a=d=0<math>) if </math>wx-yz=4042<math>, find the minimum possible value of </math>w+x+y+z<math>.
 
−
Case 2 (else) </math>wy+xy+xz=(w+x)(y+z)-wz=4042<math>, find the minimum possible value of </math>w+x+y+z<math>.
 
  
−
Note that </math>(m+n)^2=4mn+(m-n)^2<math> so if </math>m+n<math> is fixed then </math>mn<math> is maximized exactly when </math>|m-n|<math> is minimized. In particular, if </math>m+n \leq 127<math> then </math>mn-op \leq mn \leq 63*64 = 4032 <4042$ as desired.
+
~Lcz
  
 
==See Also==
 
==See Also==

Latest revision as of 12:44, 21 April 2021

Problem

Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area 2021?

(A lattice point is a point $(x, y)$ in the coordinate plane where $x$ and $y$ are both integers, not necessarily positive.)

Solution 1

The answer is $128$, achievable by $A=(10,0), B=(0,-63), C=(-54,1)$. We now show the bound.

We first do the following optimizations:

-if you have a point goes both left and right, we may obviously delete both of these moves and decrease the number of moves by $2$.

-if all of $A,B,C$ lie on one side of the plane, for example $y>0$, we shift them all down, decreasing the number of moves by $3$, until one of the points is on $y=0$ for the first time.

Now we may assume that $A=(a,d)$, $B=(b,-e)$, $C=(-c,f)$ where $a,b,c,d,e,f \geq 0$. Note we may still shift all $A,B,C$ down by $1$ if $d,f>0$, decreasing the number of moves by $1$, until one of $d,f$ is on $y=0$ for the first time. So we may assume one of $(a,b)$ and $(d,f)$ is $0$, by symmetry. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:


Case 1 (where $a=d=0$) if $wx-yz=4042$, find the minimum possible value of $w+x+y+z$.

Case 2 (else) $wy+xy+xz=(w+x)(y+z)-wz=4042$, find the minimum possible value of $w+x+y+z$.


Note that $(m+n)^2=4mn+(m-n)^2$ so if $m+n$ is fixed then $mn$ is maximized exactly when $|m-n|$ is minimized. In particular, if $m+n \leq 127$ then $mn-op \leq mn \leq 63*64 = 4032 <4042$ as desired.


~Lcz

See Also

2021 USAJMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_USAJMO_Problems/Problem_4&oldid=152233"