11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits of $n$ and its last three digits coincide with the last three digits of $n$. For example, if $\mathrm{n}=1234567890$, then the answer must be 12890.

| param1 | Answer |
| :---: | :---: |
| 2016 | 59989 |
| 664 | 49989 |
| 1580 | 39989 |
| 4000 | 79989 |

## Condition

Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer such that $S(n)+S(n+1)=2016$. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits of $n$ and its last three digits coincide with the last three digits of $n$. For example, if $\mathrm{n}=1234567890$, then the answer must be 12890.

## Answer.

59989

## Solution

If the number $n$ does not end in 9, then $S(n+1)=S(n)+1$, and the sum $S(n)+S(n+1)$ is odd, which means it cannot equal 2016. Therefore, the number $n$ must end in 9. If it ends in one 9, then $S(n+1)=S(n)-8$, and thus $S(n)=1012$. The smallest number with such a sum of digits is 599...989 (the total number of 9s is 111). If the number of 9s at the end of $n$ is not less than two, then by similar reasoning, the sum $S(n)$ will be larger, and the number $n$ will also be larger. Therefore, the answer is 59989.

12. Let $A C$ be the largest side of triangle $A B C$. Points $K$ and $M$ are chosen on side $A C$ such that $A M=A B$ and $C K=C B$. It is given that the radius of the circumcircle of triangle $K B M$ equals param1, and the radius of the incircle of triangle $A B C$ equals param2. Find the length of $B T$ squared if $T$ is the point where the incircle of $A B C$ touches its side $B C$.

| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 7 | 5 | 24 |
| 8 | 5 | 39 |
| 9 | 7 | 32 |
| 11 | 8 | 57 |
| 13 | 9 | 88 |

## Condition

Let $A C$ be the largest side of triangle $A B C$. Points $K$ and $M$ are chosen on side $A C$ such that $A M=A B$ and $C K=C B$. It is given that the radius of the circumcircle of triangle $K B M$ equals 7, and the radius of the incircle of triangle $A B C$ equals 5. Find the length of $B T$ squared if $T$ is the point where the incircle of $A B C$ touches its side $B C$.

## Answer.

24
# Solution

Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, which is also the bisector of the isosceles triangle $BAM$. This means that point $I$ lies on the perpendicular bisector of segment $BM$. Similarly, it lies on the perpendicular bisector of segment $BK$. Therefore, it is the center of the circumcircle of triangle $KBM$. It is known that $B T=\frac{B C+B A-A C}{2}=\frac{C K+A M-A C}{2}=\frac{K M}{2}$. Triangle $K I M$ is isosceles: $І \mathrm{~K}=\mathrm{I} M$ as radii of the circumcircle of triangle $\mathrm{K} B \mathrm{~L}$. Let the incircle touch side $AC$ at point $S$. Then $K M=2 \mathrm{KS}=$ $=2 \sqrt{I K^{2}-\mid S^{2}}=2 \sqrt{R^{2}-r^{2}}$, so $B T^{2}=R^{2}-r^{2}=24$.