A triangle has one side $a=6 \mathrm{~cm}$, the radius of the circumscribed circle is $r=5 \mathrm{~cm}$, and the radius of the inscribed circle is $\varrho=2 \mathrm{~cm}$. Determine the lengths of sides $b$ and $c$, as well as the area of the triangle, without using a function table.
I. solution: Let's find a relationship between the two unknown sides and the area. Using the formulas $\frac{a b c}{4 t}=r$ and $t=\varrho s$,

$$
b c=\frac{10 t}{3}
$$

and

$$
t=2 s=6+b+c
$$

from which

$$
b+c=t-6
$$

According to Heron's formula, since $s=\frac{t}{2}$,

$$
t=\sqrt{\frac{t}{2}\left(\frac{t}{2}-6\right)\left(\frac{t}{2}-b\right)\left(\frac{t}{2}-c\right)}
$$

that is,

$$
t^{2}=\frac{t}{2}\left(\frac{t}{2}-6\right)\left[\frac{t^{2}}{4}-\frac{t}{2}(b+c)+b c\right]
$$

Using (1) and (2), and dividing by $t^{2}$,

$$
1=\frac{1}{t} \frac{1}{2}\left(\frac{t}{2}-6\right)\left[\frac{t^{2}}{4}-\frac{t}{2}(t-6)+\frac{10 t}{3}\right]=\frac{1}{2}\left(\frac{t}{2}-6\right)\left(\frac{t}{4}-\frac{t-6}{2}+\frac{10}{3}\right)
$$

After removing the fractions and expanding the parentheses, we get the following quadratic equation for $t$:

$$
3 t^{2}-112 t+960=0
$$

The roots of this equation are $t_{1}=24, \quad t_{2}=\frac{40}{3}$.

If the value of $t$ is 24, then according to (1) and (2),

$$
b c=80, \quad b+c=18
$$

In this case, $b$ and $c$ can be determined as the two roots of the following quadratic equation:

$$
x^{2}-18 x+80=0
$$

From this, the lengths of the two sides are 8 and $10 \mathrm{~cm}$.

If $t=\frac{40}{3}$, then according to (1) and (2):

$$
b c=\frac{400}{9} \text { and } b+c=\frac{22}{3}=\frac{66}{9}
$$

$b$ and $c$ satisfy the following quadratic equation:

$$
9 y^{2}-66 y+400=0
$$

In this case, the discriminant of the equation is $66^{2}-4 \cdot 9 \cdot 400=4356-14400<0$, so there are no real roots.

Thus, the only solution to our problem is $t=24 \mathrm{~cm}^{2}$, and the two unknown sides are 8 and $10 \mathrm{~cm}$.

Remark: It can be observed that the lengths of the three sides of the triangle: $6,8,10$ form a Pythagorean triplet, so the triangle is right-angled by the converse of the Pythagorean theorem.

Klára Madarász (Szeged, Tömörkény Ig. II. o. t.)

II. solution: We will prove that the triangle is right-angled and then calculate its sides and area.

Let's calculate the radius of the inscribed circle of a right-angled triangle with a 6 cm leg and a 5 cm radius circumscribed circle (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_fcc81336250b5bcacdb3g-1.jpg?height=466&width=528&top_left_y=2185&top_left_x=777)

Figure 1

The hypotenuse of the triangle is $c=2 r=10 \mathrm{~cm}$, so by the Pythagorean theorem, the other leg is $8 \mathrm{~cm}$. In a right-angled triangle, the radius of the inscribed circle is equal to the length of the segment from the right-angle vertex to the point of tangency of the inscribed circle, which can be expressed using the semiperimeter $s$ and the differences of the sides. In this case, the radius of the inscribed circle is $s-c=12-10=2 \mathrm{~cm}$.

We will prove that the radius $r$ of the circumscribed circle, the radius $\varrho$ of the inscribed circle, and the side $a$ uniquely determine the triangle. The side $a$ already determines the angle $\alpha$ opposite to it in the circumscribed circle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_fcc81336250b5bcacdb3g-2.jpg?height=469&width=564&top_left_y=443&top_left_x=759)

Figure 2

The angle at the center $O$ of the inscribed circle, which is the intersection of the angle bisectors of $\beta$ and $\gamma$, is $180^{\circ}-\frac{\beta}{2}-\frac{\gamma}{2}=180^{\circ}-\frac{\beta}{2}-\frac{\gamma}{2}-\frac{\alpha}{2}+\frac{\alpha}{2}=180^{\circ}-90^{\circ}+\frac{\alpha}{2}=90^{\circ}+\frac{\alpha}{2}$. Knowing $\alpha$, we can construct the circle with a $90^{\circ}+\frac{\alpha}{2}$ visual angle over the side $a$, on which the point $O$ lies. The distance of the point $O$ from the side is the given $\varrho$, so the parallel line at a distance $\varrho$ from the side intersects the visual angle circle at two points. The angle bisectors can be drawn knowing the point $O$, and the sides of the triangle can be drawn with twice the angles.

Since the parallel line can intersect the visual angle circle at two points, we can get two triangles. However, these two triangles are symmetric with respect to the perpendicular bisector of the side $a$, so they are congruent. Thus, we have indeed proven that a side, the radius of the inscribed circle, and the radius of the circumscribed circle uniquely determine a triangle.

We have seen that when $r=5 \mathrm{~cm}$, the value of $\varrho$ is 2 for the triangle with sides 6, 8, and 10 cm, thus proving that the triangle in question is right-angled.

We have already calculated the sides. The area is half the product of the legs: $24 \mathrm{~cm}^{2}$.

László Börzsöny (Bp. V., Eötvös g. I. o. t.)

Remarks: 1. It can be seen that the problem can be solved using the thought process followed in the first solution even if different numerical values are given. 2. For a solution using trigonometric functions (but without a function table) with different numerical values, see the "Collection of Mathematical Circle Problems" booklet: problem 587 on page 218.