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

m (→‎See Also)
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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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==Solution 1 (Lcz's Solution)==
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==Solution 1==
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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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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.
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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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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.
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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 </math>2<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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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:
 +
 
 +
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.
  
 
==See Also==
 
==See Also==

Revision as of 16:12, 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

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$ (Error compiling LaTeX. Unknown error_msg)2$.

-if all of$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$0$. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:

Case 1 (where$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)(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.

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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