3. In a right-angled triangle $ABC$ with hypotenuse $AB$, let $I$ and $U$ be the center of the inscribed circle and the point of tangency of this circle with the leg $BC$, respectively. Determine the ratio $|AC|:|BC|$, if the angles $\angle CAU$ and $\angle CBI$ are equal.

(Jaroslav Zhouf)
SOLUTION. Let $V, W$ be the points of tangency of the incircle of triangle $ABC$ with sides $AC, AB$ respectively, and $\rho$ its radius. It is easy to see that the quadrilateral $CVIU$ is a square with side length $\rho$. From the axial symmetries with respect to the lines $AI$ and $BI$, and using the usual notation for the lengths of the legs of triangle $ABC$, we get

$$
|AW|=|AV|=b-\rho \quad \text{and} \quad |BW|=|BU|=a-\rho.
$$

From this, we obtain the expression for the length of the hypotenuse $AB$ in the form

$$
|AB|=|AW|+|BW|=a+b-2\rho,
$$

so by the Pythagorean theorem, the equality $a^{2}+b^{2}=(a+b-2\rho)^{2}$ holds.

![](https://cdn.mathpix.com/cropped/2024_04_17_3a6c0ba22278789571f3g-3.jpg?height=391&width=502&top_left_y=1044&top_left_x=777)

Only now do we use the condition of the problem, according to which the right triangles $ACU$ and $BU I$ are congruent in the acute angles at vertices $A$ and $B$ (marked in the figure). Since they are (generally) also congruent in the legs $CU$ and $UI$, in our situation, they are (by the SAS theorem) two congruent triangles. This leads to the equality $|AC|=|BU|$, i.e., $b=a-\rho$, so $\rho=a-b$. Substituting such a $\rho$ into the above modified equality from the Pythagorean theorem, we get

$$
a^{2}+b^{2}=(3b-a)^{2}.
$$

After expanding and simple rearrangement, we get $b(4b-3a)=0$, from which $3a=4b$.

Conclusion. For the desired ratio, we have $|AC|:|BC|=b:a=3:4$.

GUIDING AND ADDITIONAL PROBLEMS:

N1. Let $D, E, F$ be the points of tangency of the incircle of triangle $ABC$ with sides $BC, CA, AB$ respectively. Express the lengths of the segments into which points $D, E, F$ divide the individual sides using their lengths $a, b, c$. [It holds that $|AE|=|AF|=s-a, |BF|=|BD|=s-b$, $|CD|=|CE|=s-c$, where $2s=a+b+c$.]

N2. The lengths of the sides of a triangle are expressed by integers. Determine them if this triangle has a perimeter of 72 and its longest side is divided by the point of tangency of the inscribed circle into two segments in the ratio of lengths $3:4$. [61-C-I-2]

N3. Let $S$ be the midpoint of the base $AB$ of isosceles triangle $ABC$. Assume that the points of tangency of the incircles of triangles $ACS, BCS$ divide the base $AB$ into three equal segments. Determine the ratio $|AB|:|CS|$. [61-C-S-2]

N4. Find all right triangles with integer side lengths whose incircle has a radius of $2$. [69-C-S-3]

D1. Given a right triangle $ABC$ with hypotenuse $AB$. The orthogonal projection of the incircle of the given triangle onto the hypotenuse $AB$ is the segment $MN$. Prove that the center of the incircle of triangle $ABC$ is also the center of the circumcircle of triangle $MNC$. [Let $I$ be the center of the incircle and $U, V, W$ its orthogonal projections on sides $BC, CA$, $AB$ respectively. Triangles $IMW, INW$, and $IVC$ are clearly congruent isosceles right triangles (with legs of length $\rho$, the radius of the incircle, since $|MN|=2\rho$). Therefore, $|IM|=|IN|=|IC|$.]