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

Question

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 $x^2+y^2\le1.$

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

Since I have no clue to solve it in this form, I tried to substitute some points of the closed unit disk and make a system of inequalities. But it seems like there should be a general way of solving these type of problems.

Answer 1

Rewrite $2x+3y+1\le a(x+2)+b(y+3)$ as $(2-a)x+(3-b)y\leq 2a+3b-1$. 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 $$\frac{-2a^2-3ab+5a+6b-2\pm\sqrt{-(b-3)^2(3a^2+12ab+8b^2-12)}}{a^2-4a+b^2-6b+13}.$$ 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 $-(b-3)^2(3a^2+12ab+8b^2-12)\leq 0$ and $a^2+b^2\leq 1$. 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.

Answer 2

We want to find all $(a, b) = (\cos Z, \sin Z)$ where $Z\in[0,2\pi)$ such that the below inequality is true for all $|(x,y)|\leq 1$:

$$2x+3y+1\leq a(x+2)+b(y+3)$$ $$(2-a)x + (3-b)y \leq 2a + 3b - 1$$ Note that $3-b=3-\sin Z > 0$.

$$\Big(\frac{2-a}{3-b}\Big)x + y \leq \frac{2a}{3-b} + \frac{3 b - 1}{3-b}$$ $$y \leq \Big(\frac{a-2}{3-b}\Big)x+\frac{2a-4}{3-b} + \frac{3b +3}{3-b}$$ $$y \leq \Big(\frac{a-2}{3-b}\Big)\big(x+2\big) + \frac{3b+3}{3-b}$$

This inequality is linear in $(x, y)$. We want to make sure that the solution set to this inequality includes the entire unit disk. We can visualize this overlap by converting back to $(a, b) = (\cos Z, \sin Z)$ and varying the "$Z$" slider in this Desmos graph (click to view): https://www.desmos.com/calculator/fjsd7oglsu.

The point $(\cos Z, \sin Z)=(a, b)$ always appears to lie on the line (can check) as well as the unit circle (clearly). We wanted the unit disk to lie entirely within the solution set of the inequality, but we are unlucky for nearly all values of $Z$. If there is such a solution, we can see from the Desmos graph that it will satisfy $0<Z<\frac{\pi}{2}$.

The boundary line of the solution set varies continuously with $Z$ (trig functions are continuous). To find any exact solutions if they exist, we should start by finding $Z\in\big(0,\frac{\pi}{2}\big)$, i.e. $(a,b)$ in the first quadrant, for which the line is tangent to the unit circle. The boundary line is:

$$(2-a)x + (3-b)y = 2a + 3b - 1$$

$$(2-a, \, 3-b) \cdot (x, y) = 2 a + 3 b - 1$$

This is parallel to the line

$$(2-a, \, 3-b) \cdot (x, y) = 0$$

Therefore, $(2-a, \, 3-b)$ is orthogonal to our original line. But we also want the radius, $(a, b)$, to be orthogonal to the line (since this is what it means to be tangent to a circle). In 2 dimensions, if U and V are both orthogonal to a third vector W, then U and V must be parallel, i.e. their cross product must be zero.

$$(2-a, \, 3-b) \times (a, b) = 0 $$ $$(2-a)b - (3-b)a = 0$$ $$2b-ab-3a+ab = 0$$ $$2b - 3a = 0$$

Therefore, $Z = \arctan \big(\frac{3}{2}\big)$, and $(\cos Z, \sin Z) = (\cos \arctan 1.5, \sin \arctan 1.5)$.

With the variables $(a, b)$ as used in the question, this means $b=\frac{3a}{2}$. Therefore:

$$a^2+b^2=1$$ $$a^2+\big(\frac{3a}{2}\big)^2=1$$ $$4a^2+9a^2=4$$ $$a^2=\frac{4}{13}$$ $$a=\frac{2}{\sqrt{13}}$$

Therefore, $(a, b)=\big(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\big)$. Since this is the only tangent point in the first quadrant, and since any other points clearly do not satisfy the inequality, this must be the only point that satisfies the inequality.

Answer 3

Since $(a,b)$ lies on the unit circle and $(a,b)$ satisfies the linear inequality, it must be tangent to the circle. So first find the tangent line to the unit circle at $(\cos t, \sin t)$. This is:

$$y - \sin t = (-\cot t)(x - \cos t) \implies y = -(\cot t)x + \csc t.$$

Now the equality case of the condition $2x+3y+1 = a(x+2)+b(y+3)$ rearranges to $(3-b)y = (a-2)x$ $ + (2a + 3b - 1)$. Using the identity $\csc^2 t - \cot^2 t = 1$ and comparing coefficients:

$$\frac{1}{(3-b)^2} \left((2a + 3b - 1)^2 - (a-2)^2 \right) = 1$$ $$(4a^2 + 12ab + 9b^2 - 4a - 6b + 1) - (a^2 - 4a + 4) = 9 - 6b + b^2$$ $$3a^2 + 12ab + 8b^2 - 12 = 0$$

and now substitute $12 = 12(a^2+b^2)$:

$$-9a^2 + 12ab - 4b^2 = 0$$ $$-(3a - 2b)^2 = 0 \implies b = 3a/2$$

thus $(3b/2)^2 + b^2 = 1 \implies a = \frac{2}{\sqrt{13}}, b = \frac{3}{\sqrt{13}}$. We discard the negative solutions as substituting $(x,y) = (0,0)$ into the condition we need $1 ≤ 2a + 3b$.