In a triangle, $b+c=5$ units, $\beta-\gamma=20^{\circ}$, and the difference between the projections of sides $b$ and $c$ on side $a$ is 1 unit. Calculate the sides of the triangle.
I. solution. Let's assume that using the usual notation, the foot of the altitude from vertex $A$ is $A_{0}$, which falls inside the side $BC$, so $\beta$ $\gamma = \angle A_{0}CA$. In the triangle $AB^{\prime}C$, the side $B^{\prime}C$ is the difference of the projections, thus 1 unit, the angle at vertex $A$ is $\beta - \gamma$, and the sum of the sides meeting at $A$ is $AC + AB^{\prime} = AC + AB = b + c$. Let's write the cosine theorem for the side $B^{\prime}C$ of this triangle. With the substitution $c = 5 - b$,

\[
\begin{gathered}
1 = b^2 + (5 - b)^2 - 2b(5 - b) \cos 20^{\circ}, \\
\left(1 + \cos 20^{\circ}\right) b^2 - 5\left(1 + \cos 20^{\circ}\right) b + 12 = 0, \\
b = \frac{5}{2} + \sqrt{\frac{25}{4} - \frac{12}{1 + \cos 20^{\circ}}} \approx 2.5 + \sqrt{6.25 - 6.187} \approx \\
\approx 2.5 + \sqrt{0.063} \approx 2.5 + 0.251 = 2.751 \text{ units, and thus } \\
c = \frac{5}{2} - \sqrt{\frac{25}{4} - \frac{12}{1 + \cos 20^{\circ}}} \approx 2.249 \text{ units }
\end{gathered}
\]

(we immediately discarded the other root, which is the value we obtained for $c$, because $b > \gamma$ is impossible).

The side $a$ is equal to the sum of the projections: $a = BA_{0} + A_{0}C$. Therefore, by equating the expressions for $AA_{0}$ from the right triangles $AB A_{0}$ and $AC A_{0}$, we get an equation for $a$:

\[
\begin{gathered}
AA_{0}^2 = AB^2 - BA_{0}^2 = AC^2 - A_{0}C^2 \\
A_{0}C^2 - BA_{0}^2 = AC^2 - AB^2 \\
(BA_{0} + A_{0}C)(A_{0}C - BA_{0}) = (AC + AB)(AC - AB)
\end{gathered}
\]

thus, considering (1) and (2),

\[
a \cdot 1 = a = 5 \cdot 2 \sqrt{\frac{25}{4} - \frac{12}{1 + \cos 20^{\circ}}} \approx 2.51 \text{ units. }
\]

According to all this, the sides are: $a \approx 2.51, b \approx 2.751, c \approx 2.249$ units.

![](https://cdn.mathpix.com/cropped/2024_05_02_3c3f903ea1510d8fc0fdg-1.jpg?height=580&width=640&top_left_y=2120&top_left_x=729)

If $A_{0}$ falls on the extension of side $BC$ (beyond $B$ due to $b > c$, see the second diagram), then the difference of the projections of $b$ and $c$ is the side $BC = a$, and the angle subtended by $A$ is $\alpha$ instead of $\beta - \gamma$, which differs from it if $\beta \neq 90^{\circ}$. In this case, our previous calculations are not applicable. However, we will show that this case is not possible.

In the case to be ruled out, $\beta > 90^{\circ}, \gamma > 70^{\circ}$ would hold, so $b > c > AA_{0} = A_{0}C \tan \gamma > BC \tan \gamma = \tan \gamma > \tan 70^{\circ} > 2.7$, and thus $b + c > 2 \cdot 2.7 = 5.4$, which is impossible because $b + c = 5$. Therefore, our assumption is correct.

Lídia Rejtő (Budapest, Berzsenyi D. High School III. grade)

Note. We calculated the fraction under the square root to 4 significant figures using logarithms, but in the difference, only 2 significant figures remained, so we could not write more than 3 significant figures for the square root in any case. However, with ordinary division, we can write 5 significant figures for the quotient because the divisor 1.9397 also has 5 significant figures, so we get one more digit under the square root:

\[
\sqrt{6.25 - 6.1865} = \sqrt{0.0635} \approx 0.252
\]

thus $b \approx 2.752, c \approx 2.248, a \approx 2.52$ units.

II. solution. We use from the second part of the first solution that $\angle B^{\prime}C B^{\prime}A = 180^{\circ} - \beta > 90^{\circ}$, so $\delta = 119.78^{\circ}$, and thus $\angle B^{\prime}CD = \gamma = 50.22^{\circ}, \beta = 70.22^{\circ}$. Now we can calculate $b$ and $c$ from the triangle $AB^{\prime}C$, and then $a$ from the original triangle. This way, we avoided using the cosine theorem.

Mária Treer (Budapest, Kaffka M. High School II. grade)

Several solutions used trigonometric identities to utilize the difference $\beta - \gamma$. One such solution is outlined in the

III. solution. Let the radius of the circumcircle of the triangle be $R$, so

\[
b + c = 2R \sin \beta + 2R \sin \gamma = 2R(\sin \beta + \sin \gamma) = 4R \sin \frac{\beta + \gamma}{2} \cos 10^{\circ} = 5
\]

On the other hand, the difference of the projections $b^{\prime}$ and $c^{\prime}$ (using $\beta < 90^{\circ}$):

\[
\begin{aligned}
b^{\prime} - c^{\prime} = b \cos \gamma - c \cos \beta & = 2R(\sin \beta \cos \gamma - \sin \gamma \cos \beta) = 2R \sin (\beta - \gamma) = \\
& = 4R \sin 10^{\circ} \cos 10^{\circ} = 1
\end{aligned}
\]

From these,

\[
\sin \frac{\beta + \gamma}{2} = 5 \sin 10^{\circ}, \quad \frac{\beta + \gamma}{2} = 60.22^{\circ}
\]

(because it is an acute angle). We can proceed as in the second solution.

János Kemenes (Budapest, Konyves Kálmán High School IV. grade)

Note. The relationship $(\beta + \gamma) / 2 = 180^{\circ} - \delta$ obtained by combining the second and third solutions can also be derived elementarily:

\[
180^{\circ} - \delta = \angle BB^{\prime}D = \angle BB^{\prime}A - \angle DB^{\prime}A = \beta - \frac{\beta - \gamma}{2} = \frac{\beta + \gamma}{2}.
\]