Solve the system of equation for (x,y)∈R {√x2+y2+√(x−4)2+(y−3)2=53x2+4xy=24 Explain your answer
Problem
Source: 2017 Azerbaijan Junior/Senior National Olympiad
Tags: system of equations, algebra
28.04.2022 16:46
x=2,y=1.5
28.04.2022 16:55
triangle inequality √x2+y2+√(x−4)2+(y−3)2≥√(4−x+x)2+(y+3−y)2=5 equality only for x4−x=y3−y 3x−xy=4y−xy 3x=4y 8xy=24 xy=3 y2=94 x,y=(−32,−2) doesnt satisfy x,y=(2,32)
03.05.2022 18:11
StarLex1 wrote: triangle inequality √x2+y2+√(x−4)2+(y−3)2≥√(4−x+x)2+(y+3−y)2=5 equality only for x4−x=y3−y 3x−xy=4y−xy 3x=4y 8xy=24 xy=3 y2=94 x,y=(−32,−2) doesnt satisfy x,y=(2,32) Yeah a solution with minkowki is very good,but there is a decent solution like taking the first expression to the right and squaring both sides,this will lead a very good position
18.07.2022 22:19
05.03.2023 22:06
Just take the first expression to the RHS and square both side!
31.07.2023 18:00
√x2+y2+√(x−4)2+(y−3)2=5, this represents that the sum of the distances from a point (x,y) to (0,0) and (3,4), the distance between 3,4 and origin is 5 so the point lies on y=3x4, subsituting into second equation, (x,y)=(2,1.5) (note that the x coord needs to be positive)
31.08.2024 02:30
√x2+y2=5−√(x−4)2+(y−32)⇒x2+y2=25+x2−8x+16+y2−6y+9−10√(x−4)2+(y−3)250−8x−6y=10√(x−4)2+(y−3)2Divide both side by 2 and square both sides 625+16x2+9y2−200x−150y+24xy=25x2−200x+25y2−150y+6259x2+16y2=24xy⇒(3x−4y)2=0⇒3x=4y3x2+4xy=3x2+3x2=6x2=24⇒x=2y=1,5
31.08.2024 03:36
It's just Minkowski's Inequality. The rest follows directly.
15.11.2024 17:02
Minkowski theorem kills this problem.
13.12.2024 10:40
Good solution with Minkovski