Can't See The Forest From The Trees

Here are two methods to solve this.

Method 1

Let's call the radius of the trees $r$ . Through the center $O$ draw any line which intersects the boundary of the orchard in points $M$ and $N$ . Then draw the tangents to the circle in these two points. Now construct two lines $AD$ and $BC$ parallel to $MN$ at a distance $r$ . The rectangle $ABCD$ has the area $A=2r2R=4rR$ . By Minkowski's theorem if $A>4d(L)$ then the area $A$ contains at least two lattice points $P$ and $Q$ which are symmetric about $O$ , therefore the line $MN$ intersects these trees. The determinant of the lattice $d(L)$ is the n-dimensional volume of the fundamental parallelepiped, which in our case is $d(L)=d^2=25m^2$ . So $4r25m>4\cdot25m^2$ which gives $r>1m$ .

Method 2

The same result is obtained by drawing a radius tangent to the trees, then calculating the radius of the trees. Because of the symmetry of the orchard there are two cases to be analysed. In the first case the two trees that will come in discusion have the coordinates $M_1(5,5)$ and $M_2(15,10)$ . Using the formula for the distance between a point and a line $r=\frac{\left| {{y}_{0}}-m{{x}_{0}}-n \right|}{\sqrt{1+{{m}^{2}}}}$ and considering that the line goes through the origin: $\frac{\left| 5-5m-0 \right|}{\sqrt{1+{{m}^{2}}}}=\frac{\left| 10-15m-0 \right|}{\sqrt{1+{{m}^{2}}}}\Rightarrow {{m}_{1}}=\frac{1}{2}\,,\,{{m}_{2}}=\frac{3}{4}$ The first value has the trees on the same side and it's not good. The second value is the good one. Going back to the distance formula with this slope gives: $r=\frac{\left| 5-5\frac{3}{4} \right|}{\sqrt{1+{{\left( \frac{3}{4} \right)}^{2}}}}=\frac{5\frac{1}{4}}{\sqrt{\frac{25}{16}}}=1$ If $r=1$ then the line is tangent to the trees. So $r>1$ .

For the second case when $M_3(5,0)$ and $M_4(20,5)$ the same calculations give $m_3=\frac{1}{3}$ and $m_4=\frac{1}{5}$ . Going back with $m_4$ gives the radius $r=\sqrt{\frac{25}{26}}\approx 0.98$ . So it is not worthy looking to the fence like in case 2. So the final value is $r>1m$ .

I think the given answer is no correct. My answer is correct. My solution:-

The square lattice has 8 fold symmetry, so we shall look only at 45 degrees of the lattice.
Fig. 1 shows assumes trees as points and zones between critical trees are shown.
It is obvious that Top blue zone between (5,5) and (20,15), and Bottom pink zone between (5,0) and (20,5) needs widest blocking.
But Top zone is at a bigger slope, trees near (0,0) will block more than trees near (0,0) from the Bottom zone. So they need a shorter
radius than the the Bottom zone. Hence only bottom zone is critical.

In order to block full sight, there should be no gap between tangents to (5,0) and (20,5). Implies they have a common tangent.
Fig. 2 Shows this common tangent OE between lines OD and OF.
OD, OE, OF is intersected by up vertical from C at A, B, C.
OD, OE, OF is intersected by down vertical from D at D, E, F.
ABC and DEF form various similar right triangles with common angles at O. Hence the following ratios.
BC/EF=OC/OF=1/4, but OC=5, OF=20, and DF=5, also BC=DE= say h.
Therefore EF=4h , but DE+EF=h+4h=5h, also DE+EF=DF=5, implies h=1.
$Tan\theta$ =EF/OF=4/20, implies $Cos\theta=\sqrt {\dfrac{25}{26}}$ .
In Fig. 3, right triangles OBC and CBG have $\theta$ same.
$So\ r=h*Cos\theta=\sqrt{\dfrac {25}{26}}=\Large\ \ \color{#D61F06}{0.98}.$