II OM - III - Task 1

A beam of length $ a $ has been suspended horizontally by its ends on two parallel ropes of equal length $ b $. We rotate the beam by an angle $ \varphi $ around a vertical axis passing through the center of the beam. By how much will the beam be raised?
When solving geometric problems, a properly executed drawing is an important aid to our imagination. We represent spatial figures through mappings onto the drawing plane. There are various ways of such mapping. In elementary geometry, we most often draw an oblique parallel projection of the figure; in many cases, it is convenient to use the method of orthogonal projections onto two perpendicular planes, i.e., the so-called Monge's method (compare, for example, problem no. 23 on page 113).
We will present the solution to our problem in two variants, using once one and once the other of the mentioned mapping methods.
When using the method of oblique parallel projection, we will take as the projection plane - the plane passing through the beam $ AB $ and through the suspension points $ M $ and $ N $. We therefore draw the quadrilateral $ ABNM $ "in natural size" (fig. 59 a and b). Let $ S $ denote the midpoint of the beam. After twisting the beam, it will take the position $ CD $. The midpoint of the segment $ CD $ lies on the projection plane; let us assume that this is the point $ T $.
The position of the projection of point $ C $ depends on the direction of projection; for the oblique projection of point $ C $, we can take any point $ C $, for example, as in figure 59 a or 59 b. The projection of point $ D $ will be the point symmetric to point $ C $ with respect to point $ T $.
Calculating the sought length $ ST = x $ is simple. We draw a segment $ TK $ parallel and equal to segment $ SA $; then






So $ AM = b $, and segment $ KM $ is the leg of the right triangle $ KMC $ with the hypotenuse $ MC = b $ and the leg $ KC $. Segment $ KC $ is the base of the isosceles triangle $ KTC $, in which $ TK = TC = \frac{1}{2} AB = \frac{1}{2} a $, $ \measuredangle KTC = \varphi $.
Therefore,






Finally, we get






If $ b < a $, the angle of twist $ \varphi $ cannot be greater than the angle $ \varphi_0 $ given by the formula






For the value $ \varphi = \varphi_0 $, we have $ x = b $. Further increase of the angle is not possible without stretching the lines.
If $ b \geq a $, the maximum value of $ \varphi $ is $ 180^\circ $. For the value $ \varphi = 180^\circ $, the lines cross if $ b > a $, and overlap each other if $ b = a $.
In the above solution, the goal was to calculate the elevation of the beam after its twisting. The figure in parallel projection was only a possibly simple illustration necessary for this calculation. If we want the drawing to be a graphical solution to the problem, i.e., to give the correct length of segment $ ST $ given the lengths $ a $, $ b $, and the angle $ \varphi $, it should be done a bit differently. Specifically, the point $ T $, which we chose arbitrarily in figures 59 a and b, must be constructed from the given quantities $ a $, $ b $, $ \varphi $.
For this purpose, note that in the right triangle $ KMC $, we know the hypotenuse $ MC = MA = b $ and the leg $ KC $, which is the base of the isosceles triangle $ KCT $, where $ TK = TC = \frac{1}{2} a $, and $ \measuredangle KTC = \varphi $. From these data, we can construct the triangle to find the length of $ KM $ and the length of $ ST = AM - KM $.
The construction is shown in figure 60.
We construct triangle $ ASP $, where $ AS = SP = \frac{1}{2}a $, $ \measuredangle ASP = \varphi $. We draw a semicircle with diameter $ AM $ and a chord $ AL = AP $ in it. We measure on the line $ MA $ a segment $ MK $ equal to segment $ ML $. Point $ K $ determines the level of point $ T $; the sought elevation will be determined by segment $ TS = KA $.
The projection of the twisted beam $ C $ will be drawn as before, choosing point $ C $ arbitrarily.