Track / Stage 6 / 173 of 400 #1173 of 2000

Problem 1173

National olympiad, first round

Geometry Difficulty 6.2 Prove it

Frankin B.R.

Three circles touch each other externally and touch a fourth circle internally. Their centers were marked, and the circles themselves were erased. It turned out that it is impossible to determine which of the marked points is the center of the enclosing circle. Prove that the marked points form a rectangle.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $O$ and $R$ be the center and radius of the circumscribing circle, and $O_{1}, O_{2}, O_{3}$ and $R_{1}, R_{2}, R_{3}$ be the centers and radii of the other circles. Then

$OO_{i}=R-R_{i}(i=1,2,3), O_{i} O_{j}=R_{i}+R_{j}(i, j=1,2,3, i \neq j)$ . From this, $O O_{1}-O_{2} O_{3}=O O_{2}-O_{3} O_{1}=O O_{3}-O_{1} O_{2}=R$ $-R_{1}-R_{2}-R_{3}=d$ .

Let $d \neq 0$ , for example, $d>0$ . Then the distance from $O$ to any of the points $O_{1}, O_{2}, O_{3}$ is greater than the distance between the other two points. This uniquely determines $O$ , contrary to the condition. Indeed, if in each of the pairs $\left(\mathrm{OO}_{1}, \mathrm{O}_{2} \mathrm{O}_{3}\right),\left(\mathrm{OO}_{2}, \mathrm{O}_{1} \mathrm{O}_{3}\right)$ and $\left(\mathrm{OO}_{3}, \mathrm{O}_{1} \mathrm{O}_{2}\right)$ the longer segment is painted red and the shorter one blue, then $O$ is the only point where three segments of the same color meet. The same is true for $d < 0$ .

Therefore, $d=0$ , and in the non-self-intersecting quadrilateral formed by these points, opposite sides are equal and the diagonals are equal. Thus, it is a rectangle.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.