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Difference between revisions of "2021 USAJMO Problems/Problem 4"

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(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.)
  
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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?
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==Solution 1==
  
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(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.)
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The answer is <math>128</math>, achievable by <math>A=(10,0), B=(0,-63), C=(-54,1)</math>. We now show the bound.
  
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==Solution 1 (Lcz's Solution)==
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We first do the following optimizations:
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We get that the answer is <math>128</math>.
 
  
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We want to make an optimization to get down to so we do WLOG, <math>A=(a,d)</math>, <math>B=(b,-e)</math>, <math>C=(-c,f)</math>, where one of <math>a,b</math> is <math>0</math> and one of <math>(d,f)</math> is <math>0</math>, and <math>a,b,c,d,e,f \geq 0</math>,and then we do casework shoelace, which there's two cases.
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-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>.
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Case 1: where <math>a=d=0</math>,  <math>wx-yz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
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Case 2 else or <math>(w+x)(y+z)-wz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
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-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.
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We can see that it's clear <math>63*64=4032<4042</math> so the sum is <math>127</math> or (a+d)(b+c)/leq <math>4042</math> so if the sum's less than <math>128</math> it is impossible to get an area of a triangle greater than <math>2016</math>. Thus done.
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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:
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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>.
 +
 
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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>.
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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.
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~Lcz
  
 
==See Also==
 
==See Also==
 
{{USAJMO newbox|year=2021|num-b=3|num-a=5}}
 
{{USAJMO newbox|year=2021|num-b=3|num-a=5}}
  
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[[Category:Olympiad Number Theory Problems]]
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[[Category:Olympiad Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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