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) | + | 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>. |
+ | |||
+ | 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 12:14, 16 April 2021
Problem
Carina has three pins, labeled , and
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance
away. What is the least number of moves that Carina can make in order for triangle
to have area 2021?
(A lattice point is a point in the coordinate plane where
and
are both integers, not necessarily positive.)
Carina has three pins, labeled , and
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance
away. What is the least number of moves that Carina can make in order for triangle
to have area 2021?
(A lattice point is a point in the coordinate plane where
and
are both integers, not necessarily positive.)
Solution 1 (Lcz's Solution)
We get that the answer is .
We want to make an optimization to get down to so we do WLOG, ,
,
, where one of
is
and one of
is
, and
,and then we do casework shoelace, which there's two cases.
Case 1: where
,
, find the minimum possible value of
.
Case 2 else or
, find the minimum possible value of
.
We can see that it's clear
so the sum is
or
so if the sum's less than
it is impossible to get an area of a triangle greater than
. Hence the answer must be at least
.
We now show that is achievable. Indeed, taking
is a valid solution, so we are done.
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.