Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024)

We will show that there exists only one such triangle and it has sides of length $21$, $28$, $35$ and the two touching circles have radius equal to $5$.

In any right-angled triangle $ABC$ with hypotenuse $AB$ we denote $a=BC$, $b=AC$, $c=AB$. Clearly, two congruent circles with all tangent conditions from the statement exist. Denote them by $k_1(S_1,r)$ and $k_2(S_2,r)$ such that $k_1$ is tangent to $AC$ (and $k_2$ is tangent to $BC$). Moreover, denote the tangent points of these two circles to $AB$ by $T_1$ and $T_2$, respectively, as in the picture. Consider the incircle $k(S,\varrho )$ of the triangle $ABC$ and denote $D$ the tangent point of $k$ with $AB$.



In the first part of the solution we show that the radius $r$ of circles $k_1$ and $k_2$ is determined by the equation $$r=\frac{c(a+b-c)}{2(a+b)}(1)$$ In homothety with center $A$ and coefficient $r/\varrho$ which maps $k$ to $k_1$, the point $D$ is mapped to the point $T_1$. Therefore, $|A{T}_1|=|AD|\cdot r/\varrho$. Analogously, we get the equation $|B{T}_2|=|BD|\cdot r/\varrho$. Since $T_1{T}_2{S}_2{S}_1$ is a rectangle, we have $|T_1{T}_2|=|S_1{S}_2|=2r$. After plugging everything in the equation $c=|A{T}_1|+|T_1{T}_2|+|B{T}_2|$, by using the relation $|AD|+|BD|=c$ we obtain $$c=|AD|\cdot \frac{r}{\varrho }+2r+|BD|\cdot \frac{r}{\varrho }=2r+(|AD|+|BD|)\cdot \frac{r}{\varrho }=2r+\frac{cr}{\varrho }.$$

This yields $$r=\frac{c\varrho }{c+2\varrho }.$$ After using the well-known formula $\varrho =(a+b-c)/2$, we obtain the desired formula 1. Suppose that the lengths $a,b,c,r$ are integers and put $k=\gcd (a,b,c)$. Then $a=k{a}_1$, $b=k{b}_1$ and $c=k{c}_1$. Moreover, the Pythagorean theorem $a_1^2+b_1^2=c_1^2$ implies that the numbers $a_1,b_1,c_1$ are pairwise coprime and $a_1$ and $b_1$ have different parity. The number $c_1$ is odd and the number $a_1+b_1-c_1$ is even. After plugging in $a=k{a}_1$, $b=k{b}_1$ and $c=k{c}_1$ to 1 we obtain: $$r=\frac{k{c}_1(a_1+b_1-c_1)}{2(a_1+b_1)}=\frac{k}{a_1+b_1}\cdot c_1\cdot \frac{a_1+b_1-c_1}{2}.(2)$$ We will show that not only the second fraction, but also the first fraction of the right-hand side of 2 is an integer. Suppose that there is a prime $p$ that divides both $a_1+b_1$ and $c_1$. Then from the equation $(a_1+b_1{)}^2=c_1^2+2a_1{b}_1$ we have $p\mid a_1{b}_1$, which is a contradiction since the numbers $a_1,b_1$ and $c_1$ are pairwise coprime. Thus, the number $a_1+b_1$ is coprime to $c_1$ and also $a_1+b_1-c_1$. Since $r$ is an integer, from the equation 2 it follows, that the first fraction is an integer.

By the problem statement, $r$ is a prime. By 2 it is a product of three positive integers. Since $c_1>a_1\ge 1$, we necessarily have $c_1=r$ and the other two fractions in the right-hand side of 2 are equal to 1. Therefore, we have $k=a_1+b_1$, $a_1+b_1-c_1=2$ and also $k=a_1+b_1=c_1+2=r+2$. It follows that $$2a_1{b}_1=(a_1+b_1{)}^2-c_1^2=(r+2{)}^2-r^2=4r+4.$$ After dividing by two we obtain $$a_1{b}_1=2r+2=2(a_1+b_1-2)+2=2a_1+2b_1-2.$$ The last equation can be rewritten in the form $(a_1-2)(b_1-2)=2$. It easily follows that $\{a_1,b_1\}=\{3,4\}$, $c_1=r=a_1+b_1-2=5$, and $k=r+2=7$. For right-angled triangle with sides $7\cdot 3,7\cdot 4$ and $7\cdot 5$ we, indeed, have $r=5$. This completes the proof.