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Final Round of National Olympiad

Estonia geometry

Problem

The midpoints of sides , and of a triangle are , and , respectively. The centers of circles , and are , and , respectively, and the centers of circles , , are , , , respectively. No two of the given six circles intersect in two points nor are they inside each other. Circles , and touch each other externally.

i) Prove that the sum of the radii of circles , and does not exceed one quarter of the perimeter of the triangle .

ii) Prove that if the sum of the radii of circles , and equals one quarter of the perimeter of the triangle then the triangle is equilateral.

problem
Solution
Let the radii of the circles , , be , , , and the radii of the circles , , be , , , respectively (Fig. 28). By assumptions, which sum up to . The assumptions also imply inequalities which sum up to .

i) Altogether, we obtain the inequality , which implies as desired.

Figure 28

ii) Suppose that . Then all inequalities above must hold as equalities. The equalities imply , analogously . Denoting , we get where summing side-by-side gives , i.e., . Analogously, . Thus all sides of the triangle have length .

Techniques

TrianglesTangentsDistance chasingOptimization in geometry