Determine the smallest radius a circle passing through EXACTLY three lattice points may have.
Problem
Source: Romania 2017 IMO TST 1, problem 4
Tags: geometry
gausskarl
18.03.2018 22:41
Sorry, I did not read the OP carefully
The answer is $\frac{5\sqrt{2}}{6}$. This is attained by taking, for example, $(0,0)$, $(1,2)$ and $(2,1)$. Now we prove that it is the minimum. First, notice that if we take two points $(x_i,y_i)$ and $(x_j, y_j)$ such that $$S=|x_i-x_j|+|y_i-y_j|\geq 4$$then the distance between them is larger than equal $2\sqrt{2}$, thus making the radius is at least $\sqrt{2}$.
On the other hand, if $S=1$ then there is clearly exist a fourth point belongs to the circle, so $S\in\{2,3\}$. Note that if we make two sides equals to $2$, a right triangle always appears, thus there exists a fourth point. In addition, no lattice triangle is equilateral (this is a classic result), so we have swept through all possible cases (Q.E.D).
. It is completely bash.
Kaskade
18.03.2018 22:44
Surely a circle through the unit square would pass through 4 points
PROF65
19.03.2018 02:03
Remark that if two point are on the same line or column then the circle will pass through the fourth point forming isoceles trapezoid so necessary the points $P,Q,R$ will occupy at least three columns and three lines,thus the least possibility is as below: $$\begin{array}{lcr}\cdot & P & \cdot \\ Q & \cdot & \cdot \\ \cdot & \cdot & R \end{array}$$and its radius $\frac{PR}{2\sin \angle QPR}$ coud be calculated easily using basic trigonometry but i will use Pick's theorem use $\frac{PR}{2\sin \angle QPR}=\frac{PR^2\cdot PQ }{2\sin \angle QPR \cdot PR\cdot PQ}=\frac{5\sqrt{2}}{4S} =\frac{5\sqrt{2}}{4(1+\frac{3}{2}-1)}=\frac{5\sqrt{2}}{6}=1,17..$ then the diameter is less than $\sqrt{10}$ , any point out of the mentioned square is distant from one of the point at least $\sqrt{10}$ finaly the point on the corner couldn't be on the circle because the triangle angle are different from $90^\circ$ RH HAS