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
18.03.2018 22:41
Sorry, I did not read the OP carefully
. It is completely bash.
18.03.2018 22:44
Surely a circle through the unit square would pass through 4 points
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