Baltic Way 2011 · Shortlist problem

Geometry

Consider a right angled triangle $ABC$ with sides of length $3$ , $4$ , and $5$ . Determine the greatest possible radius of a circle that is tangent to two among the lines $BC$ , $CA$ , and $AB$ and that in addition passes through at least one of the points $A$ , $B$ , and $C$ .

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Topics

Circles and tangency · Geometric inequalities · Triangles and centers

Solutions

Solution

Consider a general triangle $ABC$ . Suppose we have a circle that touches the lines $AB$ and $AC$ . Since it cannot also pass through the point $A$ , we may suppose it passes through the point $C$ . The centre of the circle will then lie either on the internal, or the external, bisector of the angle at $A$ .

Assume the centre of the circle lies on the internal bisector. Then its radius is

r = b \tan \frac{A}{2} = 2R \sin B \tan \frac{A}{2},

where $R$ denotes the circumradius. The maximal radius is obtained when $A \ge B \ge C$ (the expression $\frac{\sin x}{\tan \frac{x}{2}} = 2 \cos^2 \frac{x}{2}$ is strictly decreasing for $0 \le x \le 180^\circ$ ).

Assume now the centre of the circle lies on the external bisector. Then its radius is

s = b \tan \frac{B+C}{2} = \frac{2R \sin B}{\tan \frac{A}{2}}.

The maximal radius is obtained when $B \ge C \ge A$ .

For the triangle at hand, $r$ is maximized by $b = 4$ and $A = 90^\circ$ , which gives $r = 4$ , and $s$ by $b = 5$ and $A$ the angle opposite the side of length $3$ . Then $\tan A = \frac{3}{4}$ , $\tan \frac{A}{2} = \frac{1}{3}$ , which produces the greatest radius $s = 15$ , which is thus the answer.