4. Circles with radii $\mathrm{a}>\mathrm{b}>0$ touch each other externally. Three common tangents are drawn to them. Find the perimeter of the triangle formed by them.
Answer: $P=\frac{4 a \sqrt{a b}}{a-b}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a6fe9b83749c6c6f2c11g-3.jpg?height=754&width=1521&top_left_y=554&top_left_x=336)

According to the theorem of tangent segments, the given triangle ABC is isosceles, and all common tangent segments are equal, i.e., $\mathrm{CE}=\mathrm{CF}=\mathrm{CD}=\mathrm{DB}=\ldots=\mathrm{h} / 2$, where $\mathrm{h}=\mathrm{FE}$ is the height of the rectangular trapezoid O1FEO2. Since the bases of the trapezoid are equal to a and b, and the lateral side O1O2=a+b, then by the Pythagorean theorem $h=2 \sqrt{a b}$.

By drawing the height in the trapezoid O1FEO2 from vertex O2, we get $\cos \angle O 2 O 1 F=\frac{a-b}{a+b}=\sin \angle E A O 2$. From this, in the right triangle ADC, we conclude that $A C=\frac{C D}{\sin \angle E A O 2}=\frac{\sqrt{a b}(a+b)}{a-b}$.

Therefore, $P=2 A C+2 C D=\frac{2 \sqrt{a b}(a+b)}{a-b}+2 \sqrt{a b}=\frac{4 a \sqrt{a b}}{a-b}$.

P.S. There are various methods to solve this problem, but in any case, all formulas for segment lengths must be derived!

## Criteria:

| 7 | The answer is correct, and the solution is fully justified. |
| :--- | :--- |
| $6-5$ | The answer is correct, but the solution is not fully justified, or the answer is incorrect due to one arithmetic error. |
| 5 | The answer is incorrect, and there are several errors in the formulas or transformations, but the overall idea of the solution is correct. |

## $4-3$

3 The answer is incorrect or missing, but there are initial steps in the solution.

2-1 There are minor advances in the solution. Special cases have been considered.