Solve the system of equation for $(x,y) \in \mathbb{R}$ $$\left\{\begin{matrix} \sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2}=5\\ 3x^2+4xy=24 \end{matrix}\right.$$ 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 $\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2} \geq \sqrt{(4-x+x)^2+(y+3-y)^2}= 5$ equality only for $\frac{x}{4-x} = \frac{y}{3-y}$ $3x-xy=4y-xy$ $3x=4y$ $8xy = 24$ $xy=3$ $y^2 = \frac{9}{4}$ $x,y = (\frac{-3}{2},-2)$ doesnt satisfy $x,y = (2,\frac{3}{2})$
03.05.2022 18:11
StarLex1 wrote: triangle inequality $\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2} \geq \sqrt{(4-x+x)^2+(y+3-y)^2}= 5$ equality only for $\frac{x}{4-x} = \frac{y}{3-y}$ $3x-xy=4y-xy$ $3x=4y$ $8xy = 24$ $xy=3$ $y^2 = \frac{9}{4}$ $x,y = (\frac{-3}{2},-2)$ doesnt satisfy $x,y = (2,\frac{3}{2})$ 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
$\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2}=5$, 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 $y=\frac{3x}{4}$, subsituting into second equation, $(x,y)=(2, 1.5)$ (note that the x coord needs to be positive)
31.08.2024 02:30
$$\sqrt{x^2+y^2}\hspace{1mm}=\hspace{1mm}5\hspace{1mm}-\hspace{1mm}\sqrt{(x-4)^2+(y-3^2)}\Rightarrow x^2\hspace{1mm}+\hspace{1mm}y^2\hspace{1mm}=\hspace{1mm}25\hspace{1mm}+\hspace{1mm}x^2\hspace{1mm}-\hspace{1mm}8x\hspace{1mm}+\hspace{1mm}16\hspace{1mm}+\hspace{1mm}y^2\hspace{1mm}-\hspace{1mm}6y\hspace{1mm}+\hspace{1mm}9\hspace{1mm}-\hspace{1mm}10\sqrt{(x-4)^2+(y-3)^2}$$$$50\hspace{1mm}-\hspace{1mm}8x\hspace{1mm}-\hspace{1mm}6y\hspace{1mm}=\hspace{1mm}10\sqrt{(x-4)^2+(y-3)^2}$$Divide both side by 2 and square both sides $$625\hspace{1mm}+\hspace{1mm}16x^2\hspace{1mm}+\hspace{1mm}9y^2\hspace{1mm}-\hspace{1mm}200x\hspace{1mm}-\hspace{1mm}150y\hspace{1mm}+\hspace{1mm}24xy\hspace{1mm}=\hspace{1mm}25x^2\hspace{1mm}-\hspace{1mm}200x\hspace{1mm}\hspace{1mm}+\hspace{1mm}25y^2\hspace{1mm}-\hspace{1mm}150y\hspace{1mm}+\hspace{1mm}625$$$$9x^2\hspace{1mm}+\hspace{1mm}16y^2\hspace{1mm}=\hspace{1mm}24xy\Rightarrow (3x-4y)^2=0\Rightarrow 3x=4y$$$$3x^2\hspace{1mm}+\hspace{1mm}4xy\hspace{1mm}=\hspace{1mm}3x^2\hspace{1mm}+\hspace{1mm}3x^2\hspace{1mm}=\hspace{1mm}6x^2\hspace{1mm}=\hspace{1mm}24\Rightarrow x=2\hspace{3mm}y=1,5$$
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