The jealous engineer problem

 

One day Tony woke up and didn't find his wife in the house.

— Where were you?

— I wanted to work out and just walked around the block. Look at my pedometer.

— What distance does the pedometer show?

—  Really Tony? again with your jealousy?

Angry, Tony snatches the pedometer from her and he read a distance traveled of $805$ meters.

— How many laps did you do?

— One lap!

— East or North?

 — Are you serious, Tony?

— East or North?

— East!

Tony had been the engineer in charge of paving the block years ago, and he knew that the four streets that made up the block were the same distance from the church where he married his wife. In addition, he remembered that the corners formed the following sequence of angles until he got back to his house: $60°$, $135°$, $85°$, $80°$ and that if his wife went east she must have traveled $200$ meters on the first street before to reach the $60°$- corner, a distance that he also remembered perfectly.

Was the wife lying? Justify your answer.

El problema del ingeniero celoso

Un día, Tony se despertó y no encontró a su esposa en la casa.

— ¿Dónde estabas?

— Quería ejercitarme y solo recorrí la cuadra. Mira mi podómetro.

— ¿Qué distancia marca el podómetro?

— ¿En serio, Tony? ¿otra vez con tus celos?

Airado, Tony le arrebata el podómetro y lee una distancia recorrida de $805$ metros.

— ¿Cuántas vueltas diste?

— ¡Una! ¡una vuelta!

— ¿Este o norte?

— ¿Hablas en serio, Tony?

— ¿Este o norte?

— ¡Este!

Tony había sido el ingeniero a cargo del asfaltado de la cuadra años atrás y sabía que las cuatro calles que formaban la cuadra estaban a la misma distancia de la iglesia donde se casó con su esposa. Además, recordaba que las esquinas formaban la siguiente secuencia de ángulos hasta llegar de vuelta a su casa: $60°$, $135°$, $85°$, $80°$ y que si su esposa partió hacia el este debió haber recorrido $200$ metros en la primera calle antes de llegar a la esquina de $60°$, distancia que también recordaba a la perfección. 

¿Mentía la esposa? Justifica tu respuesta.

A generalization of Lami's theorem for 4 forces

 In physics, Lami's theorem is an equation relating the magnitudes of three coplanar, concurrent and non-collinear vectors, which keeps an object in static equilibrium, with the angles directly opposite to the corresponding vectors. According to the theorem

$$\frac{F_1}{\sin{\alpha}}=\frac{F_2}{\sin{\beta}}=\frac{F_3}{\sin{\gamma}}.\tag{1}$$

where $F_1$, $F_2$ and $F_3$ are the magnitudes of the three coplanar, concurrent and non-collinear vectors which keep the object in static equilibrium, and $\alpha$, $\beta$ and $\gamma$ are the angles directly opposite to the vectors (see Figure 1).

Figure 1. $\alpha$, $\beta$ and $\gamma$ are the angles directly opposite to the vectors $F_1$, $F_2$ and $F_3$

Lami's theorem is applied in static analysis of mechanical and structural systems. The theorem is named after Bernard Lamy. The proof of Lami's theorem is essentially based on the law of sines.

On the Internet there are hundreds of static equilibrium problems where they apply Lami's theorem to a three-force system, see for instance Dubey - Engineering Mechanics: Statics and Dynamics, section 3.10. Although Dubey's book is recent (2013) there is not a single equilibrium problem based on a four-force system. Coincidentally, the author of this note has come across questions on the Internet questioning the possibility of applying Lami's theorem for more than three forces. In this note we give a generalization of Lami's theorem for four forces.

Theorem 1 (Generalization). If four coplanar, concurrent and non-collinear forces act upon an object, and the object remains in static equilibrium, then

Figure 2. An example of the situation described in the theorem 1. 

$$AD\sin{\alpha'}+BC\sin{\gamma'}=AB\sin{\beta'}+CD\sin{\delta'}.\tag{2}$$

where $A$, $B$, $C$ and $D$ are the magnitudes of the four vectors and $\alpha'$, $\beta'$, $\gamma'$ and $\delta'$ are the angles between them (see Figure 2).

Proof. Consider the quadrilateral formed by the four vectors in such a manner that the head of one touches the tail of another (see Figure 3) and denote $\Delta$ its area. If $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles of the quadrilateral, then its area can be written as

Figure 3. Notice that if $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles of the quadrilateral formed by the four vectors, then $\alpha$ and $\alpha'$ are supplementary and similarly for $\beta'$, $\gamma'$ and $\delta'$

$$\Delta=\frac12AD\sin{\alpha}+\frac12BC\sin{\gamma}=\frac12AB\sin{\beta}+\frac12CD\sin{\delta}$$  

and as $\sin{\alpha'}=\sin{(\pi-\alpha)}=\sin{\alpha}$, and similarly for $\beta'$, $\gamma'$ and $\delta'$, the relation in $(2)$ follows.

$\square$

Theorem 1 is a generalization in the sense that if one of the vectors vanishes, the relation we obtain is that of Lami's theorem. Indeed, for instance suppose $C=0$, then the relation $(2)$ reduces to

$$D\sin{\alpha'}=B\sin{\beta'},$$

which is Lami's theorem.

Remark. A generalization of Lami's theorem is given by H. Shekhar. However, this generalization is different since it only considers cyclic polygons with an odd number of sides.