3. (3 points) From cards with letters, you can form the word WATERPIPE. How many words (not necessarily meaningful) can be formed from these cards, in which the letters R and P are adjacent?
Answer: $\frac{9!}{4!\cdot 2}=\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{2}=7560$

## Solution:

Let's replace the adjacent letters P and R with one letter, for example, the letter Sh, which is possible since they stand next to each other. Then we have 9 letters, and among them, 4 letters O, 2 letters B, and 2 letters D.

Thus, the number of ways to rearrange the letters in this word is $\frac{9!}{4!\cdot 2!\cdot 2!}$. Additionally, the letter Sh can be decoded in two ways: as PR and as RP, which doubles the number of ways. Therefore, we get $\frac{9!}{4!\cdot 2}$.

4. (3 points) Given a triangle $ABC$ with an area of $12 \sqrt{3}$ and an angle $C=120^{\circ}$. Point $D$ is the foot of the perpendicular dropped from point $C$ to side $AB$; points $E$ and $F$ are the feet of the perpendiculars dropped from point $D$ to sides $AC$ and $BC$, respectively. Find the perimeter of triangle $EFC$, given that it is isosceles.

Answer: $2 \sqrt{3}+3$

## Solution:

Triangle $EFC$ is isosceles. Since $\angle C$ in it is obtuse, point $C$ is its vertex, so $EC = FC$. Therefore, right triangles $ECD$ and $FCD$ are congruent by the leg and hypotenuse, which means $ED = FD$, i.e., point $D$ lies on the bisector of $\angle C$, hence triangle $ABC$ is also isosceles.

Thus, triangles $ABC$ and $EFC$ are similar. Let's find the similarity ratio. First, $CE: CD = \cos \angle ACD = \frac{1}{2}$. Second, $CD: AC = \cos \angle ACD = \frac{1}{2}$. Therefore, $CE: AC = \frac{1}{4}$.

Next, $12 \sqrt{3} = S_{ABC} = \frac{AC \cdot BC \cdot \sin \angle ACD}{2} = \frac{AC^2 \sqrt{3}}{4}$, from which $AC = 4 \sqrt{3}$. Hence, $AB = 2 AC \sin 60^{\circ} = 4 \sqrt{3}$. Therefore, $P_{ABC} = 8 \sqrt{3} + 12 \sqrt{3}$, and $P_{EFC} = \frac{1}{4} P_{ABC} = 2 \sqrt{3} + 3$.