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

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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>.
 
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>.
 
Case 2 else or <math>(w+x)(y+z)-wz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
 
Case 2 else or <math>(w+x)(y+z)-wz=4042</math>, find the minimum possible value of <math>w+x+y+z</math>.
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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We can see that it's clear <math>63*64=4032<4042</math> so the sum is <math>127</math> or <math>(a+d)(b+c)\leq 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>. Hence the answer must be at least <math>128</math>.
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We now show that <math>128</math> is achievable. Indeed, taking <math>A=(5, 1), B=(-52, 0), C=(0,-70)</math> is a valid solution, so we are done.
  
 
==See Also==
 
==See Also==

Revision as of 13:14, 16 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.)

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 (Lcz's Solution)

We get that the answer is $128$.

We want to make an optimization to get down to so we do WLOG, $A=(a,d)$, $B=(b,-e)$, $C=(-c,f)$, where one of $a,b$ is $0$ and one of $(d,f)$ is $0$, and $a,b,c,d,e,f \geq 0$,and then we do casework shoelace, which there's two cases. Case 1: where $a=d=0$, $wx-yz=4042$, find the minimum possible value of $w+x+y+z$. Case 2 else or $(w+x)(y+z)-wz=4042$, find the minimum possible value of $w+x+y+z$. We can see that it's clear $63*64=4032<4042$ so the sum is $127$ or $(a+d)(b+c)\leq 4042$ so if the sum's less than $128$ it is impossible to get an area of a triangle greater than $2016$. Hence the answer must be at least $128$.

We now show that $128$ is achievable. Indeed, taking $A=(5, 1), B=(-52, 0), C=(0,-70)$ is a valid solution, so we are done.

See Also

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

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