Points $A, B, C, D$ lie on a straight line (in this order). $A B=123.4 \mathrm{~m}, C D=96.4 \mathrm{~m}$. The length of segment $B C$ cannot be measured. From a point $P$ lying on one side, the angles subtended by $A B, B C, C D$ are $25.7^{\circ}, 48.3^{\circ}, 32.9^{\circ}$, respectively. Calculate the length of segment $B C$.
I. Solution. First, we calculate the angle $P A D=\varphi$. Introduce the following notations: $A B=a, B C=x$, $C D=c$; the angles of view from $P$ are $\alpha, \beta, \gamma, \alpha+\beta+\gamma=\delta$, and $P B=p$ and $P C=q$. Using the sine theorem in the triangles $P A B, P B C$, and $P C D$, we get:

$$
\frac{a}{p}=\frac{\sin \alpha}{\sin \varphi}, \quad \frac{p}{q}=\frac{\sin (\varphi+\alpha+\beta)}{\sin (\varphi+\alpha)}, \quad \frac{q}{c}=\frac{\sin (\varphi+\delta)}{\sin \gamma}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_835eef190ea7d642fce4g-1.jpg?height=488&width=1198&top_left_y=396&top_left_x=453)

By multiplying the left and right sides of these three equalities, $p$ and $q$ cancel out, and among the remaining data, only $\varphi$ is unknown. In other words, we get a single-variable equation for $\varphi$:

$$
\frac{a}{c}=\frac{\sin \alpha \sin (\varphi+\alpha+\beta) \sin (\varphi+\delta)}{\sin \gamma \sin \varphi \sin (\varphi+\alpha)}
$$

which simplifies to:

$$
\frac{\sin (\varphi+\alpha+\beta) \sin (\varphi+\delta)}{\sin \varphi \sin (\varphi+\alpha)}=\frac{a \sin \gamma}{c \sin \alpha}=k
$$

where $k$ is a known number. By rearranging and applying the addition theorem, each term contains one of $\sin ^{2} \varphi, \sin \varphi \cos \varphi$, and $\cos ^{2} \varphi$:

$$
\begin{aligned}
& \sin ^{2} \varphi \cos (\alpha+\beta) \cos \delta+\sin \varphi \cos \varphi[\cos (\alpha+\beta) \sin \delta+\sin (\alpha+\beta) \cos \delta]+ \\
& \quad+\cos ^{2} \varphi \sin (\alpha+\beta) \sin \delta=k \sin ^{2} \varphi \cos \alpha+k \sin \varphi \cos \varphi \sin \alpha
\end{aligned}
$$

In the square brackets, $\sin (\delta+\alpha+\beta)=\sin (2 \delta-\gamma)$, so we have:

$$
\begin{gathered}
\sin (\alpha+\beta) \sin \delta \cos ^{2} \varphi+[\sin (2 \delta-\gamma)-k \sin \alpha] \cos \varphi \sin \varphi+ \\
+[\cos (\alpha+\beta) \cos \delta-k \cos \alpha] \sin ^{2} \varphi=0
\end{gathered}
$$

Dividing by $\sin ^{2} \varphi$ (since $\sin \varphi \neq 0$, as $P$ is not on the line $A D$), we get a quadratic equation in $\operatorname{ctg} \varphi$:

$$
\operatorname{ctg}^{2} \varphi+m \operatorname{ctg} \varphi+n=0
$$

where

$$
m=\frac{\sin (2 \delta-\gamma)-k \sin \alpha}{\sin (\alpha+\beta) \sin \delta}, \quad n=\frac{\cos (\alpha+\beta) \cos \delta-k \cos \alpha}{\sin (\alpha+\beta) \sin \delta}
$$

From the triangle $P B C$:

$$
\frac{x}{p}=\frac{\sin \beta}{\sin (\varphi+\alpha+\beta)}
$$

Combining this with the first equation of (1) to eliminate $p$:

$$
x=\frac{a \sin \beta \sin \varphi}{\sin \alpha \sin (\varphi+\alpha+\beta)}
$$

Numerically, $k=1.603, b=106.9^{\circ}, 2 \delta-\gamma=180.9^{\circ}, m=-0.7730, n=-1.658$, so the two values of $\operatorname{ctg} \varphi$ are $-0.958$ and 1.731. From these, $\varphi=133.77^{\circ}$ or 30.02°. The former would make the sum of the angles in the triangle $P B C$ greater than $180^{\circ}$, which is not possible. With the latter value of $\varphi$, we get $x=B C=109.5 \mathrm{~m}$.

Gábor Székely (Budapest, Madách I. g. II. o. t.)

Remark: From (2), we can also derive an equation for the trigonometric functions of $2 \varphi$ by eliminating the fractions and using the identity $2 \sin u \sin v=\cos (u-v)-\cos (u+v)$:

$$
a \sin \gamma[\cos \alpha-\cos (2 \varphi+\alpha)]=c \sin \alpha[\cos \gamma-\cos (2 \varphi+2 \delta-\gamma)]
$$

From this, using the addition theorem, we get an equation of the form:

$$
J \cos 2 \varphi+K \sin 2 \varphi=L
$$

We saw in problem 1085.1 that its solution is:

$$
\cos 2 \varphi=\frac{J L \mp K \sqrt{J^{2}+K^{2}-L^{2}}}{J^{2}+K^{2}}, \quad \sin 2 \varphi=\frac{K L \pm J \sqrt{J^{2}+K^{2}-L^{2}}}{J^{2}+K^{2}}
$$

The calculation is somewhat more difficult than the one above.

II. Solution. Express $p$ from the triangles $P A B$ and $P B D$, and $q$ from the triangles $P C D$ and $P A C$. From the equality of the expressions:

$$
\begin{aligned}
& (p=) \quad \frac{a \sin \varphi}{\sin \alpha}=\frac{(x+c) \sin (\varphi+\delta)}{\sin (\beta+\gamma)} \\
& (q=) \quad \frac{c \sin (\varphi+\delta)}{\sin \gamma}=\frac{(x+a) \sin \varphi}{\sin (\alpha+\beta)}
\end{aligned}
$$

By multiplying the left and right sides of the two equalities, the factors containing $\varphi$ cancel out. Among the remaining data, only $x$ is unknown, which can be calculated from the equation:

$$
(x+a)(x+c)=\frac{a c \sin (\alpha+\beta) \sin (\beta+\gamma)}{\sin \alpha \sin \gamma}=M
$$

Thus,

$$
\begin{gathered}
x^{2}+(a+c) x-(M-a c)=0 \\
x=B C=\frac{1}{2}\left[-(a+c) \pm \sqrt{(a+c)^{2}+4(M-a c)}\right]
\end{gathered}
$$

In our case, $M > a c$, because

$$
\begin{array}{ll}
0^{\circ}1 \\
0^{\circ}1
\end{array}
$$

and the product of numbers greater than 1 is greater than 1. Thus, the constant term in (4), or the product of the roots, is negative, so the roots are real, and one is negative, the other positive. Only the latter can be used, yielding $x=109.6 \mathrm{~m}$.

Jenő Lehel (Budapest, Apáczai Csere J. gyak. g. IV. o. t.)

Remarks: 1. We get a similar equation if we express the double area of the triangles $P A B, P B C, P C D$, and $P A D$ using the common height $P P^{\prime}$ from $P$ and the bases on the line $A D$, and the sides and the included angles from $P$, and multiply the corresponding sides of the resulting equalities. This yields the equation:

$$
\frac{x(x+a+c)}{a c}=\frac{\sin \beta \sin (\alpha+\beta+\gamma)}{\sin \alpha \sin \gamma}
$$

István Gálfi (Budapest, Kandó K. hír. ip. t. III. o. t.)

2. (3) and (5) follow directly from Pappus's theorem, which states that if the lines $t, u, v, w$ passing through a point $S$ are intersected by a line at points $T, U, V, W$, respectively, then

$$
\frac{T U}{U V}: \frac{T W}{W V}=\frac{\sin \angle T S U}{\sin \angle U S V}: \frac{\sin \angle T S W}{\sin \angle W S V} .
$$

István Somogyi (Ajka, Bródy I. g. II. o. t.)[^0]


[^0]:    ${ }^{1}$ K.M.L. $23(1961 / 11) 131$. o.