Jameson Lucero

2022-06-29

Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk

In the middle of another problem, I came up with the following inequality which needed to be solve for $(a,b)$ :

$2x+3y+1\le a(x+2)+b(y+3)$

for all $(x,y)\in {\mathbb{R}}^{2}$ with ${x}^{2}+{y}^{2}\le 1.$.

Here the solution for $(a,b)$ must be a subset of the unit circle, and I believe that it is a singleton.

In the middle of another problem, I came up with the following inequality which needed to be solve for $(a,b)$ :

$2x+3y+1\le a(x+2)+b(y+3)$

for all $(x,y)\in {\mathbb{R}}^{2}$ with ${x}^{2}+{y}^{2}\le 1.$.

Here the solution for $(a,b)$ must be a subset of the unit circle, and I believe that it is a singleton.

Yair Boyle

Beginner2022-06-30Added 10 answers

Rewrite $2x+3y+1\le a(x+2)+b(y+3)$ as $(2-a)x+(3-b)y\le 2a+3b-1$.

Solve for the intersection of the line $(2-a)x+(3-b)y=2a+3b-1$ with ${x}^{2}+{y}^{2}=1$ to get that the $x$-coordinates of the intersection are

$\frac{-2{a}^{2}-3ab+5a+6b-2\pm \sqrt{-(b-3{)}^{2}(3{a}^{2}+12ab+8{b}^{2}-12)}}{{a}^{2}-4a+{b}^{2}-6b+13}.$

If this has two real solutions, then the inequality cannot be satisfied for all $(x,y)$ in the unit circle, so we need to find the intersection of $-(b-3{)}^{2}(3{a}^{2}+12ab+8{b}^{2}-12)\le 0$ and ${a}^{2}+{b}^{2}\le 1$. Simultaneously solving these equations, we find that the real solutions for $(a,b)$ are $(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}})$ and $(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}})$. It's easy to see that the first solution is valid and the second is invalid - plug in $(x,y)=(0,0)$, for instance.

Solve for the intersection of the line $(2-a)x+(3-b)y=2a+3b-1$ with ${x}^{2}+{y}^{2}=1$ to get that the $x$-coordinates of the intersection are

$\frac{-2{a}^{2}-3ab+5a+6b-2\pm \sqrt{-(b-3{)}^{2}(3{a}^{2}+12ab+8{b}^{2}-12)}}{{a}^{2}-4a+{b}^{2}-6b+13}.$

If this has two real solutions, then the inequality cannot be satisfied for all $(x,y)$ in the unit circle, so we need to find the intersection of $-(b-3{)}^{2}(3{a}^{2}+12ab+8{b}^{2}-12)\le 0$ and ${a}^{2}+{b}^{2}\le 1$. Simultaneously solving these equations, we find that the real solutions for $(a,b)$ are $(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}})$ and $(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}})$. It's easy to see that the first solution is valid and the second is invalid - plug in $(x,y)=(0,0)$, for instance.

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