Eva inscribed a circle in a given triangle. Then she drew three segments that touched the inscribed circle and formed three smaller triangles within the original triangle, see the figure. The perimeters of these three smaller triangles were $12 \mathrm{~cm}, 14 \mathrm{~cm}$, and $16 \mathrm{~cm}$.

Determine the perimeter of the original triangle.

(E. Novotná)

![](https://cdn.mathpix.com/cropped/2024_04_17_25f31dcc0efac6090fc3g-3.jpg?height=440&width=440&top_left_y=1208&top_left_x=814)
Let's denote the vertices of the original triangle as $A, B, C$, the points of tangency of the inscribed circle as $D, E, F$, and its center as $S$. The endpoints of the added segments are $G, H, I, J, K, L$, and their points of tangency with the circle are $M, N, O$.

![](https://cdn.mathpix.com/cropped/2024_04_17_25f31dcc0efac6090fc3g-3.jpg?height=531&width=528&top_left_y=1905&top_left_x=770)

Segments $H D$ and $H M$ are tangents from point $H$ to the circle, so angles $H D S$ and $H M S$ are right angles. The corresponding triangles $H D S$ and $H M S$ share the side $H S$ and have congruent legs $S D$ and $S M$, which are radii of the inscribed circle. By the Pythagorean theorem, the legs $H D$ and $H M$ are also congruent, i.e., $|H D| = |H M|$.

For the same reason, $|G F| = |G M|$, so the perimeter of triangle $A H G$ is

$$
|A H| + |H M| + |M G| + |G A| = |A H| + |H D| + |F G| + |G A| = |A D| + |A F|.
$$

This means that the perimeter of the corner triangle $A H G$ is equal to the sum of the distances from point $A$ to the points of tangency on sides $A B$ and $A C$.

The same property holds for the perimeters of the other two corner triangles. Therefore, the perimeter of triangle $A B C$ is

$$
12 + 14 + 16 = 42 \text{ (cm)}.
$$

Scoring suggestion. 2 points for recognizing $|H D| = |H M|$ and its application; 1 point for its justification (an appeal to symmetry about the line $H S$ can also be accepted); 2 points for expressing the perimeter of the corner triangle $(*)$; 1 point for calculating the perimeter of triangle $A B C$.