In triangle $ABC$, side $AB$ is 1 unit, and the angles on it are $15^{\circ}$ and $60^{\circ}$. Determine the lengths of the other two sides of the triangle without using trigonometric functions.
I. Solution. Let's choose the labels of the triangle's vertices such that the angle at $A$ is $60^{\circ}$, and the angle at $B$ is $15^{\circ}$. Then the angle at $C$ is $180^{\circ}-\left(60^{\circ}+15^{\circ}\right)=105^{\circ}$. Let's take point $D$ on side $AB$ such that $\angle ACD=60^{\circ}$, point $E$ on side $BC$ such that $\angle CDE=90^{\circ}$, and finally point $F$ on side $AB$ such that $\angle DEF=120^{\circ}$ (Figure 1).

Then, by simple calculation, we can determine the angles of triangles $ACD$, $CDE$, $DEF$, and $EFB$:

$\angle ADC=180^{\circ}-(\angle DAC+\angle ACD)=60^{\circ}$; $\angle DCE=\angle ACB-\angle ACD=45^{\circ}$, $\angle DEC=180^{\circ}-(\angle CDE+\angle DCE)=45^{\circ}$; $\angle EDF$

Thus, triangle $ACD$ is equilateral, and triangles $CDE$, $DEF$, and $EFB$ are isosceles. If we denote the length of $AC$ by $x$, then $AC=AD=CD=DE=EF=FB=x$. Let the midpoint of $AD$ be $M$, and the midpoint of $DF$ be $N$. Since triangle $ADC$ is equilateral, $CM$ is perpendicular to $AD$, $AM=MD=\frac{x}{2}$, and $CM=\frac{\sqrt{3}}{2} \cdot x$. Segment $EN$ is also perpendicular to $DF$, so the corresponding angles of triangles $CMD$ and $DNE$ are equal. Since $CD=DE$ also holds, the two triangles are congruent, so $DN=CM=\frac{\sqrt{3}}{2} \cdot x$, and since $N$ is the midpoint of $DF$, $FN=DN=\frac{\sqrt{3}}{2} \cdot x$. Using these, we get

$$
1=AB=AD+DN+NF+FB=x+\frac{\sqrt{3}}{2} \cdot x+\frac{\sqrt{3}}{2} \cdot x+x=(2+\sqrt{3}) x .
$$

Thus, $x=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$. The length of side $BC$ can be determined from the right triangle $BCM$ using the Pythagorean theorem:

$$
BC^{2}=CM^{2}+MB^{2}=\left(\frac{\sqrt{3}}{2} \cdot x\right)^{2}+\left(1-\frac{x}{2}\right)^{2}=x^{2}-x+1=6-3 \sqrt{3}
$$

Therefore, the lengths of the other two sides of triangle $ABC$ are $2-\sqrt{3}$ and $\sqrt{6-3 \sqrt{3}}$.

Julia Poto (Budapest, Veres P. Gymnasium, 9th grade)

II. Solution. We can assume that in triangle $ABC$, the angle at $A$ is $60^{\circ}$, and the angle at $B$ is $15^{\circ}$. The angle at $C$ of the triangle is obtuse, so the foot of the altitude from $B$, point $T$, is on the extension of side $AC$ beyond $C$ (Figure 2). Let $AC=x$ and $CT=y$.

Since $\angle BAT=60^{\circ}$ and $\angle ATB=90^{\circ}$, $\angle ABT=30^{\circ}$. Reflecting the triangle $BAT$ over line $BT$, the reflection and the original triangle together form an equilateral triangle with side length $AB=1$. Thus, $AT=\frac{1}{2}$ and $BT=\frac{\sqrt{3}}{2}$. Line $BC$ bisects angle $ABT$ because $\angle CBT=\angle ABT-\angle ABC=30^{\circ}-15^{\circ}=15^{\circ}=\angle ABC$. Using the angle bisector theorem, we get

$$
\frac{TC}{CA}=\frac{BT}{BA}, \quad \text { that is } \quad \frac{y}{x}=\frac{\sqrt{3}}{2}
$$

We know that $AT=AC+CT=\frac{1}{2}$, so

$$
x+y=\frac{1}{2}
$$

Solving the system of equations (1) and (2),

$$
x=2-\sqrt{3} \quad \text { and } \quad y=\frac{2 \sqrt{3}-3}{2}
$$

follows. The length of side $BC$ is then obtained from triangle $BCT$ using the Pythagorean theorem:

$$
BC^{2}=CT^{2}+TB^{2}=\left(\frac{2 \sqrt{3}-3}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}=6-3 \sqrt{3}
$$

Therefore, the lengths of the other two sides of triangle $ABC$ are $2-\sqrt{3}$ and $\sqrt{6-3 \sqrt{3}}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_1cb0fa89f99d9e3fd8e0g-2.jpg?height=704&width=1138&top_left_y=142&top_left_x=469)