Let r be the radius of a circle through three points of a parallelogram. Show that any point inside the parallelogram is a distance ≤ r from at least one of its vertices.
Solution
Take the parallelogram to be ABCD. Take r to be the circumradius of ABC. Any point inside the parallelogram lies inside ABC or ADC, which is congruent. Take it to be inside ABC. Then as the diagram shows it must lie inside or on the perimeter of one of the six right-angled triangles with vertex O, each of which has hypoteneuse r. So if is sufficient to show that a point inside or on a right-angled triangle with hypoteneuse r is a distance ≤ r from the vertices at either end of the hypoteneuse.
If P lies on RS, then the result is obvious. Otherwise ∠PRS ≤ ∠TRS and ∠PSR ≤ ∠TSR, so ∠PRS + ∠PSR ≤ ∠TRS + ∠TSR = 90o. Hence ∠RPS ≥ 90o, so ∠RPS must be the largest angle of the triangle RPS. Hence PS and PR ≤ RS, which establishes the result.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03