An investor has two rectangular pieces of land of size $120\times 100$. a. On the first land, she want to build a house with a rectangular base of size $25\times 35$ and nines circular flower pots with diameter $5$ outside the house. Prove that even the flower pots positions are chosen arbitrary on the land, the remaining land is still sufficient to build the desired house. b. On the second land, she want to construct a polygonal fish pond such that the distance from an arbitrary point on the land, outside the pond, to the nearest pond edge is not over $5$. Prove that the perimeter of the pond is not smaller than $440-20\sqrt{2}$.
Problem
Source: Vietnam MO 2018 1st day 3rd problem
Tags: geometry, inequalities, combinatorial geometry
dungxibo123
11.01.2018 11:09
a. Replace a nines circular flower pots with diameter $5$ by a square rectangular $5$ Will be easier
gausskarl
11.01.2018 16:59
Part a. is quite easy. Just partition the land into $10$ smaller rectangular cells of size $30\times 40$. Then by Pigeon Hole, there exists a cell which interior not contain any of the flower pots centers.
dungxibo123
11.01.2018 18:11
I have one solution is quite difficult than your Gausskarl. My solution: Narrow the square $120 \times 120$ into square $(120-17.5\times \sqrt{2})\times (120-17.5\times \sqrt{2})$ We'll prove that in the smaller circle will exist at least 1 point out of the range of 9 circles with each's radius is $(4.5+17.5\times \sqrt{2})\times (4.5+17.5\times \sqrt{2})$ Because the range of square $(120-17.5\times \sqrt{2})\times (120-17.5\times \sqrt{2})$ is larger than total range of 9 circles with each's radius is $(4.5+17.5\times \sqrt{2})\times (4.5+17.5\times \sqrt{2})$ so we have the thing need to prove
Idea_lover
16.01.2018 09:04
What about part b. people ?
ThE-dArK-lOrD
16.01.2018 16:10
Let the coordinate of the vertices of the rectangular piece of land are $(0,0),(100,0),(0,120)$ and $(100,120)$.
From each vertex, there must exists a point whose distance to that vertex is at most $5$.
Let those $4$ points for those $4$ vertices are $(x_1,y_1),(100-x_2,y_2),(x_3,120-y_3)$ and $(100-x_4,120-y_4)$.
Note that $x_i^2+y_i^2\leq 5^2\implies x_i+y_i\leq 5\sqrt{2}$ for all $i\in \{ 1,2,3,4\}$.
The pond must connect $2$ points correspond to two consecutive vertices of the rectangle.
Note that these $4$ sides of the pond are disjoint.
So, the perimeter of the pond is at least
\begin{align*}
& \sqrt{(100-x_2-x_1)^2+(y_1-y_2)^2} +\sqrt{(x_1-x_3)^2+(120-y_1-y_3)^2}+\sqrt{(x_2-x_4)^2+(120-y_2-y_4)^2} +\sqrt{(100-x_3-x_4)^2+(y_3-y_4)^2}\\
& \geq (100-x_1-x_2)+(120-y_1-y_3)+(120-y_2-y_4)+(100-x_3-x_4)\\
& = 440-\sum_{i=1}^{4}{(x_i+y_i)}\\
& \geq 440-20\sqrt{2}.
\end{align*}
Idea_lover
16.01.2018 16:17
ThE-dArK-lOrD wrote:
Let the coordinate of the vertices of the rectangular piece of land are $(0,0),(100,0),(0,120)$ and $(100,120)$.
From each vertex, there must exists a point whose distance to that vertex is at most $5$.
Let those $4$ points for those $4$ vertices are $(x_1,y_1),(100-x_2,y_2),(x_3,120-y_3)$ and $(100-x_4,120-y_4)$.
Note that $x_i^2+y_i^2\leq 5^2\implies x_i+y_i\leq 5\sqrt{2}$ for all $i\in \{ 1,2,3,4\}$.
The pond must connect $2$ points correspond to two consecutive vertices of the rectangle.
Note that these $4$ sides of the pond are disjoint.
So, the perimeter of the pond is at least
\begin{align*}
& \sqrt{(100-x_2-x_1)^2+(y_1-y_2)^2} +\sqrt{(x_1-x_3)^2+(120-y_1-y_3)^2}+\sqrt{(x_2-x_4)^2+(120-y_2-y_4)^2} +\sqrt{(100-x_3-x_4)^2+(y_3-y_4)^2}\\
& \geq (100-x_1-x_2)+(120-y_1-y_3)+(120-y_2-y_4)+(100-x_3-x_4)\\
& = 440-\sum_{i=1}^{4}{(x_i+y_i)}\\
& \geq 440-20\sqrt{2}.
\end{align*}
Beautiful solution ThE-dArK-lOrD, thanks !