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Source: RMO Maharashtra and Goa 2019 P6

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272 posts

#1 h Nov 10, 2019, 12:37 PM • 3 Y

Y by Red-Apple, A4r43r457yhyt, Adventure10

Let $k$ be a positive real number. In the $X-Y$ coordinate plane, let $S$ be the set of all points of the form $(x,x^2+k)$ where $x\in\mathbb{R}$ . Let $C$ be the set of all circles whose center lies in $S$ , and which are tangent to $X$ -axis. Find the minimum value of $k$ such that any two circles in $C$ have at least one point of intersection.

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272 posts

#2 h Nov 10, 2019, 1:11 PM • 5 Y

Y by Red-Apple, ashrith9sagar_1, A4r43r457yhyt, Adventure10, Mango247

Answer
Solution
@below ,Yes there was a miscalculation. Is your solution same, if not then post it.

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#3 h Nov 10, 2019, 1:26 PM • 2 Y

Y by Adventure10, Mango247

I feel that the ans to this is $\frac{1}{4}$ .

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#5 h Nov 10, 2019, 2:26 PM • 4 Y

Y by ashrith9sagar_1, fjm30, Adventure10, Mango247

Solution: Note that the circle with coordinates $(x,x^2+k)$ has radius $x^2+k$ . So, the circles with centers $(x,x^2+k)$ and with center $(y,y^2+k)$ meet at some point for any $x,y$ . Triangle Inequality gives that this is equivalent to $(x^2+y^2+2k)^2 \geq (x^2-y^2)^2+(x-y)^2$ . Since squares are always positive, this implies $x^2+y^2+2k \geq |x-y|$ . The case is 'strongest' when $x,y$ have opposite signs. So we consider the case $|x|^2+|y|^2+2k \geq |x|+|y|$ which gives $2k \geq |x|-|x|^2+|y|-|y|^2$ . Maxima is achieved at $x=y=\frac{1}{2}$ which gives $k \geq \frac{1}{4}$ .
Now we check whether the inequality $(x^2+y^2+\frac{1}{2})^2 \geq (x^2-y^2)^2+(x-y)^2$ is true or not. Expanding and cancelling gives that this is equivalent to $4x^2y^2+\frac{1}{4} \geq 2xy$ which is obvious by AM-GM. Thus $k=\frac{1}{4}$ is the smallest value.

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#6 h Nov 11, 2019, 1:36 PM • 1 Y

Y by Adventure10

Mathotsav wrote:

Solution: Note that the circle with coordinates $(x,x^2+k)$ has radius $x^2+k$ . So, the circles with centers $(x,x^2+k)$ and with center $(y,y^2+k)$ meet at some point for any $x,y$ . Triangle Inequality gives that this is equivalent to $(x^2+y^2+2k)^2 \geq (x^2-y^2)^2+(x-y)^2$ . Since squares are always positive, this implies $x^2+y^2+2k \geq |x-y|$ . The case is 'strongest' when $x,y$ have opposite signs. So we consider the case $|x|^2+|y|^2+2k \geq |x|+|y|$ which gives $2k \geq |x|-|x|^2+|y|-|y|^2$ . Maxima is achieved at $x=y=\frac{1}{2}$ which gives $k \geq \frac{1}{4}$ .
Now we check whether the inequality $(x^2+y^2+\frac{1}{2})^2 \geq (x^2-y^2)^2+(x-y)^2$ is true or not. Expanding and cancelling gives that this is equivalent to $4x^2y^2+\frac{1}{4} \geq 2xy$ which is obvious by AM-GM. Thus $k=\frac{1}{4}$ is the smallest value.

Why is the triangle inequality $a+b\ge c$ ?

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#7 h Nov 11, 2019, 3:28 PM • 1 Y

Y by Adventure10

Maybe because degenerate triangle is possible?

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#9 h Nov 11, 2019, 10:59 PM • 1 Y

Y by Adventure10

Oh, I got it. Thanks

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#10 h Nov 12, 2019, 5:24 PM • 2 Y

Y by Adventure10, Mango247

that paper was too hard then previous rmo paper

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#12 h Jun 9, 2020, 10:43 AM

Y by

Red-Apple wrote:

that paper was too hard then previous rmo paper

[/quo

No I think it was easier than previous

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lifeismathematics

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lifeismathematics

#13 h Oct 5, 2023, 10:06 AM

Y by

Suppose $\mathcal{C}_{1}$ : a circle with center $(x,x^2+k)$ and radius $x^2+k$ (as it's tangent to $X$ axis)
$\mathcal{C}_{2}$ : a circle with center $(y, y^2+k)$ and radius $y^2+k$ (as it's tangent to $X$ axis)

Now since they intersect at at least one point, from triangle inequality we have $(x^2+k+y^2+k)^2 \geqslant (x-y)^2+(x^2-y^2)^2 \implies x^2(4k-1+4y^2)+2xy+4ky^2-y^2+4k^2 \geqslant 0$ $\forall$ $x \in \mathbb{R}$ (as choice of $\mathcal{C}_{1}$ , $\mathcal{C}_{2}$ is arbitrary). , So we have $\triangle_{x}=-16(-k^2+4k^3-2ky^2+8k^2y^2-y^4+4ky^4) \leqslant 0$ , so we get $-16(4k-1)(k+y^2)^2 \leqslant 0 \implies 4k-1 \geqslant 0 \implies k \geqslant \frac{1}{4}$ , which gives minimum such real $k$ is $\frac{1}{4}$ .

Let's verify the case we found, we need to check if the triangle inequality holds for $k=\frac{1}{4}$ , we get $\left(x^2+y^2+\frac{1}{2}\right)^2 \geqslant (x^2-y^2)^2+(x-y)^2 \implies 4x^2y^2 +\frac{1}{4} \geqslant 2xy$ which is true by $\mathrm{AM-GM}$ inequality, hence $k=\boxed{\frac{1}{4}}$ is the required minimum value. $\blacksquare$

This post has been edited 3 times. Last edited by lifeismathematics, Oct 5, 2023, 12:05 PM

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#14 h Oct 10, 2023, 10:32 AM

Y by

As no one decided to use Calc, here I go:
We are basically looking for a $k$ such that the sum of radius of any two circle is greater than or equal to the distence between the centers, making the circles intersecting or tangent. This can be written as:
$a^2+k+b^2+k \geq \sqrt{(a-b)^2+(a^2-b^2)^2}$
$\iff 2k \geq \sqrt{(a-b)^2+(a^2-b^2)^2}-a^2-b^2$
We are basically looking to minimize the RHS for positive reals $a,b$ . We can do so by letting $a+b=C$ where $C$ is a constant. Then the RHS will be:
$\sqrt{(C^2+1)(2a-C)^2}-2a^2+2Ca-C^2$
$= 2\sqrt{C^2+1}a+2Ca-2a^2-C\sqrt{C^2+1}-C^2$
We will minimize this using differentiation,:
$2\sqrt{C^2+1}+2C-4a=0\\ \iff a= \frac{\sqrt{C^2+1}+C}{2}$
This means the RHS at it's minima is equal to:
$C^2+1+C\sqrt{C^2+1}+2Ca-2a^2-C\sqrt{C^2+1}-C^2$
$= 1+C\sqrt{C^2+1}+C^2-\frac{1}{2}(2C^2+1+2C\sqrt{C^2+1})$
$= \frac{1}{2}$
Thus, $2k=\frac{1}{2}$ , hence, $k= \frac{1}{4}$

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