Example 20. The radius of the circumscribed circle around a right-angled triangle relates to the radius of the circle inscribed in this triangle as $5: 2$. Find the acute angles of the triangle.

Given (Fig. 22): $\triangle A B C, \widehat{A C B}=90^{\circ}, \omega_{1}$ - the circumscribed circle, $\omega_{2}$ - the inscribed circle, the ratio of the radius of circle $\omega_{1}$ to the radius of circle $\omega^{2}$ is $5: 2$.

To find: $\widehat{A B C}$ and $\widehat{B A C}$.

Solution. Since $\triangle A B C$ is a right-angled triangle, the point $O_{1}$, the center of the circumscribed circle $\omega_{1}$, is the midpoint of the hypotenuse $A B$. Since the circle $\omega_{2}$ is inscribed in triangle $A B C$, its center - point $O_{2}$ - is equidistant from the sides of triangle $A B C$, i.e., dropping perpendiculars $O_{2} D, O_{2} E$, and $O_{2} K$ from point $O_{2}$ to the sides of triangle $A B C$, we get: $\left|O_{2} D\right|=\left|O_{2} E\right|=\left|O_{2} K\right|$. Introducing an auxiliary parameter, let $|A B|=2 R$, then $\left|O_{2} E\right|=\frac{2}{5} R$. Since by construction $\left[O_{2} E\right] \|[D C]$ and $\left[O_{2} D\right] \|[C E]$, the quadrilateral $O_{2} D C E$ is a parallelogram and even, as it is easy to show, a square, i.e., $|D C|=|C E|=\left|O_{2} D\right|$. Introducing an auxiliary unknown, let $|A E|=x$. Then $|A K|=x$ as well, since $[A E]$ and $[A K]$ are segments of tangents drawn from point $A$ to the circle $\omega_{2}$. Now we will form an equation. In the right-angled triangle $A B C$ we have:

$$
|A C|^{2}+|B C|^{2}=|A B|^{2}
$$

Transitioning to the auxiliary parameter and the auxiliary unknown, we get the equation:

$$
\left(x+\frac{2}{5} R\right)^{2}+\left((2 R-x)+\frac{2}{5} R\right)^{2}=4 R^{2}
$$

i.e., $25 x^{2}-50 R x+24 R^{2}=0$, from which $x_{1}=\frac{4}{5} R ; x_{2}=\frac{6}{5} R$.

Thus, $|A E|=\frac{6}{5} R$ or $|A E|=\frac{4}{5} R$.

Since $|C E|=\left|O_{2} E\right|=\frac{2}{5}$, we get in the first case: $|A C|=\frac{6}{5} R$, and in the second case $|A C|=\frac{8}{5} R$.

In the first case, $|B C|=\sqrt{4 R^{2}-\left(\frac{6}{5} R\right)^{2}}=\frac{8}{5} R$, and in the second case $|B C|=\sqrt{R^{2}-\left(\frac{8}{5} R\right)^{2}}=\frac{6}{5} R$.

Thus, the triangles obtained in the first and second cases are congruent. Let's find the acute angles of one of them, for example, the one obtained in the first case. We get:

$$
\sin \widehat{A B C}=\frac{|A C|}{|A B|}=\frac{3}{5}, \sin \widehat{B A C}=\frac{|B C|}{|A B|}=\frac{4}{5}
$$

Therefore, the acute angles of the triangle are: $\arcsin \frac{3}{5}$ and $\arcsin \frac{4}{5}$.

#### Abstract

Note. If we take as the auxiliary unknown not the length of segment $A E$, but the length of the leg $A C$, then the resulting equation will be irrational.

Example 21. Prove that if the height and median, drawn from one vertex of an isosceles triangle, lie inside the triangle and form congruent angles with the lateral sides of the triangle, then this triangle is a right-angled triangle.

Given (Fig. 23): $\triangle A B C$, $[C K]$ - height, $[C M]-$ median, $\widehat{A C K}=\widehat{B C M}$

To prove: $\quad \triangle A B C-$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-038.jpg?height=351&width=672&top_left_y=247&top_left_x=1043)

Fig. 23 is a right-angled triangle.
Proof. We will prove that $\widehat{A C B}=90^{\circ}$. It is clear that the assumption that $\widehat{C A B}=90^{\circ}$ is inconsistent, because then the segment $C K$ would coincide with the side $A C$ and therefore must coincide with the side $B C$ to form the required congruent angles. However, the median cannot coincide with a side of the triangle.

Let $\widehat{A C B}=x$. This is the unknown we are looking for. Introduce an auxiliary unknown: $\widehat{A C K}=\widehat{B C M}=\alpha$ - and an auxiliary parameter: $|C K|=h$. Then $\widehat{B C K}=x-\alpha, \widehat{M C K}=x-2 \alpha$. Since [CM] is a median, $|A M|=|B M|$, or $|A M|=|B K|-|M K|$. On the other hand, $|A M|=|A K|+|M K|$ and thus,

$$
|B K|-|M K|=|A K|+|M K|
$$

or

$$
2|M K|=|B K|-|A K|
$$

From the right triangles $A C K, M C K$ and $B C K$ we have respectively:

$|A K|=h \operatorname{tg} \alpha,|M K|=h \operatorname{tg}(x-2 \alpha),|B K|=h \operatorname{tg}(x-\alpha)$.

Thus, equation (21.1) becomes:

or

$$
2 h \operatorname{tg}(x-2 \alpha)=h \operatorname{tg}(x-\alpha)-h \operatorname{tg} \alpha
$$

$$
2 \operatorname{tg}(x-2 \alpha)=\operatorname{tg}(x-\alpha)-\operatorname{tg} \alpha
$$

We have obtained an equation that relates the unknown $x$ and the auxiliary unknown $\alpha$. Solving this equation, we get:

$$
\frac{2 \sin (x-2 \alpha)}{\cos (x-2 \alpha)}=\frac{\sin (x-2 \alpha)}{\cos (x-\alpha) \cos \alpha}
$$

Since $\triangle A B C$ is not isosceles, $x \neq 2 \alpha$, i.e., $\sin (x-2 \alpha) \neq 0$, and then $2 \cos (x-\alpha) \cos (x-2 \alpha)$, or $\cos x+\cos (x-2 \alpha)=$ $\cos (x-2 \alpha)$, i.e., $\cos x=0$, and therefore $x=90^{\circ}$. Thus, $\triangle A B C$ is a right triangle.[^1]

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-039.jpg?height=391&width=437&top_left_y=244&top_left_x=361)

Fig. 24

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-039.jpg?height=485&width=449&top_left_y=774&top_left_x=358)

Fig. 25

We have considered examples where, in solving problems of type $M$, an algebraic method was used, which required the introduction of an auxiliary parameter. In such cases, it is sometimes possible to find a purely geometric solution. In such cases, the introduction of an auxiliary parameter is naturally not required.

Let's return to the example 21 we solved using the algebraic method, introducing an auxiliary parameter (the length of the height). Let's solve this example purely geometrically.

Describe a circle $\omega (0,|O A|)$ around triangle $A B C$ (Fig. 24). If $O \in[A B]$, then $[A B]$ is the diameter of the circle $\omega$, and then $A \widehat{C B}=90^{\circ}$.

Assume that $O \notin[A B]$. Let $(C K) \cap \omega=D,(C M) \cap \omega=P$.

Construct the segment $D P$. (Points $D$ and $P$ are different because triangle $A B C$ is not isosceles, so its height $C K$ does not coincide with the median $CM$.) Since $\widehat{A C D}=\widehat{B C P}$, then $-A m D \cong$ $\cong \smile B n P$, i.e., $(D P) \|(A B)$, and therefore $\widehat{C D} P=\widehat{K} B=90^{\circ}$. Thus, $[C P]$ is the diameter of the circle $\boldsymbol{~ , ~ and ~} M \in[C P]$ and $|A M|=|B M|$.

Thus, the diameter $C P$ bisects the chord $A B$, and therefore $[C P] \perp[A B]$.

But $[C D] \perp[A B]$. Thus, two perpendiculars are dropped from point $C$ to the line $A B$: $C K$ and $C M$, i.e., the assumption that $O \notin[A B]$ leads to a contradiction. Thus, we have proved that $\widehat{A C B}=90^{\circ}$, i.e., triangle $A B C$ is a right triangle.

Consider another example, which is a problem of type $M$, but can also be solved without introducing an auxiliary parameter.