Disk inequality

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1098 posts

#1 h Jul 20, 2021, 9:03 PM • 3 Y

Y by centslordm, Abidabi, iker_tz

In the plane, there are $n \geqslant 6$ pairwise disjoint disks $D_{1}, D_{2}, \ldots, D_{n}$ with radii $R_{1} \geqslant R_{2} \geqslant \ldots \geqslant R_{n}$ . For every $i=1,2, \ldots, n$ , a point $P_{i}$ is chosen in disk $D_{i}$ . Let $O$ be an arbitrary point in the plane. Prove that $O P_{1}+O P_{2}+\ldots+O P_{n} \geqslant R_{6}+R_{7}+\ldots+R_{n}.$ (A disk is assumed to contain its boundary.)

This post has been edited 1 time. Last edited by popcorn1, Jul 20, 2021, 9:13 PM

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#2 h Jul 20, 2021, 9:23 PM • 1 Y

Y by centslordm

Double posted https://artofproblemsolving.com/community/c6h2625916p22698473

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#3 h Jul 20, 2021, 9:28 PM • 2 Y

Y by centslordm, GorgonMathDota

Denote the center of disk $i$ as $O_i$ , for all $i = 1, 2, \dots, n$ .
Claim 01. (IGO 2017/2) Given any six non-intersecting disk, each having radius at least $r$ . Then the radius of the circle intersecting all of the six disks is at least $r$ .
Proof. Consider $OO_i$ for each $i$ . By Pigeon Hole Principle, there exists two circles with center $O_i, O_j$ such that $\angle O_i O O_j \le 60^{\circ}$ . Suppose that the circle intersecting all of the six disks has radius $r_0 < r$ . By Law of Cosine on $\triangle OO_i O_j$ , we have
$O_iO_j^2 = OO_i^2 + OO_j^2 - 2 \cdot OO_i \cdot OO_j \cdot \cos O_i OO_j \le OO_i^2 + OO_j^2 - OO_i \cdot OO_j$ WLOG $OO_i \ge OO_j$ , then we have
$OO_i \cdot OO_j \ge OO_j^2 \ \text{and} \ O_i O_j > R_i + R_j \ge R_i + r > R_i + r_0 \ge OO_i$ from which this forces $O_i O_j^2 \le OO_i^2 + (OO_j^2 - OO_i \cdot OO_j) \le OO_i^2 < O_i O_j^2$ , which is a contradiction.

Now, we just prove the inequality by induction on $n \ge 6$ , the base case is trivial. To do this, applying the lemma to the six largest disks, and we will have find index $i$ and $j$ such that $OP_i \ge R_j \ge R_6$ . Delete this disk and apply the induction hypothesis, and we get
$OP_1 + OP_2 + \dots + OP_n \ge R_6 + R_7 + \dots + R_n$

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#4 h Jul 20, 2021, 9:47 PM • 2 Y

Y by centslordm, Aryan-23

Proposed by Mohammad Ali Abam and Morteza Saghafian from Iran

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#5 h Jul 20, 2021, 9:51 PM • 2 Y

Y by centslordm, bladekill97

Dadgarnia wrote:

Proposed by Mohammad Ali Abam and Morteza Saghafian from Iran

:coolspeak:

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#6 h Jul 20, 2021, 11:51 PM

Y by

I contend $OP_i\ge R_6$ for some $i\le6$ . This suffices by induction.

Drop the condition $R_1\ge R_2\ge\cdots\ge R_6$ , and instead number the disks $\mathcal D_1$ , \ldots, $\mathcal D_6$ counterclockwise with respect to $O$ . Let the center of $\mathcal D_i$ be $O_i$ .

Since $\angle O_1OO_2+\cdots+\angle O_6OO_1=360^\circ$ , for some index $i$ we have $\angle O_iOO_{i+1}\le60^\circ$ . Without loss of generality $OO_i\ge OO_{i+1}$ . Evidently $\angle O_iOO_{i+1}$ is not the largest angle in $\triangle O_iOO_{i+1}$ , i.e.\ $\overline{O_iO_{i+1}}$ is not the longest side, so $OO_i\ge O_iO_{i+1}\ge R_i+R_{i+1}.$ This allows us to conclude $OP_i\ge OO_i-R_i\ge R_{i+1}\ge\min\{R_1,\ldots,R_6\}.$

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#7 h Jul 21, 2021, 12:26 AM • 3 Y

Y by ElaineHuang, Anthony-Cho, Want-to-study-in-NTU-MATH

Let $C_i$ denote the center of $D_i$ .
First, we can delete some disks $D_i$ with $i\geq 6$ and $\overline{OC_i}\geq 2R_i$ , if the inequality holds in this case, then it holds for all cases.
Let $A_i=\{j|\overline{OC_j}>R_i+R_j,\ j<i\}$ .
If there is 6 disks $D_{j_1}, D_{j_2}, \dots, D_{j_6},\ j_1, j_2, j_3, j_4, j_5, j_6<i$ such that $\{j_1, j_2, j_3, j_4, j_5, j_6\}\cap A_i=\emptyset$ , then (WLOG suppose that $D_{j_1}, D_{j_2}, \dots, D_{j_6}$ is clockwise order to $O$ ) there exists two adjacent disk $D_{j_k}, D_{j_{k+1\mod 6}}$ (WLOG suppose that is $D_{j_1}$ and $D_{j_2}$ ) such that $\angle C_{j_1}OC_{j_2}\leq 60^o$ .
But $\overline{C_{j_1}C_{j_2}}>R_{j_1}+R_{j_2}>\max(R_{j_1}+R_i, R_{j_2}+R_i)>\max(\overline{OC_{j_1}}, \overline{OC_{j_2}})$ , which is a contradiction.
$\therefore |A_i|\geq i-5$
From $i=6$ to $n$ , we can choose one element $j_i$ from $A_i$ such that it hasn't be chosen.
As we know $OP_{j_i}\geq R_i$ , therefore $\overline{OP_1}+\overline{OP_2}+\cdots+\overline{OP_n}\geq\overline{OP_{j_6}}+\overline{OP_{j_7}}+\cdots+\overline{OP_{j_n}}\geq R_6+R_7+\cdots+R_n$ .

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awesomeming327.

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awesomeming327.

#8 h May 19, 2023, 10:46 PM

Y by

We proceed with induction. $n\le 5$ is clearly true. Let $O_i$ be the center of $D_i$ . Then, there exist $1\le i<j\le 6$ such that $\angle O_iOO_j\le 60^\circ$ . Thus, $O_iO_j$ is not the longest side of that triangle. Without loss of generality, $OP_i+R_i\ge OO_i\ge O_iO_j\ge R_i+R_j$ so $OP_i\ge R_j\ge R_6$ . Remove $D_i$ and the result is still true. Therefore, we can add in $OP_i$ to the left side and $R_6$ to the right and it is still true.

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