Application of Ruler and Compass Construction in Transportation Optimization Problem
——A case of Optimum Transporting Route Design
In real life, we often face such a problem: Some cargo needs to be conveyed from location a to b. Transportation like water way, road , railway, airway, etc, is the option, whose cost is different from each other. Then we need to design a best plan to save as much cost as possible. This kind of problem usually can be changed in to the geometric model—we can find the most economical way to transport the cargo by drawing the auxiliary lines, which will be discussed in the following with the ruler and compass construction
1、Background. The distance between warehouse A and the straight line BC beside the straight river is 10km,AC is perpendicular to BC in point C, the distance from wharf B to C is 30 km. Some cargo needs to be transported from A to B. The unit cost for road is twice as much as it for waterway. Then what is the best route to achieve the lowest whole cost?
2、Build geometric model, to find the resolution with auxiliary lines by ruler and compass construction.
(1)、Build geometric model.(As following drawing)
To know the most saving route means to find a turning point H at straight line BC, which makes the cost of rout AH-HB lowest. Thus, the problem is turned to find a point H, which conforms to the above demand, at straight line BC.
(2)、Draw auxiliary lines by ruler and compass construction:
a) Casually get point P at line segment BC, get the mid-point F of line segment BC with ruler and compass construction.
b)Draw a circle: P is the center of circle, PF is radius.
c)Draw tangent line BG of circle P passing point B, the point of tangency is D, link PD, then PD⊥BG,and PD=PF=BP.
d)Draw AE⊥BG passing point A,the foot point is E,AE and BC meet at point H, then H is the needed point. Route AH-HB is the most saving route.
(3)、Certification of the most saving route AH-HB.
a) As P is a casual point at line segment BC, to certify AH-HB as the most saving route means to prove the cost of route AP-PB is higher than that of route AH-HB.
b)PD=PF=BP,the cost of water way (line segment BP) is half of the cost of road (line segment PD). Then the cost of BP should be the same of PD, the cost of route AP-PB is equal to that of route AP-PD.
c)PD⊥BG,HE⊥BG,so PD∥HE,△BDP∽△BEH,so HE=HB,as same to b) the cost of route AH-HB is equal to that of route AH-HE can be proved.
d)As both route AP-PD and AH-HE are by road, the cost can be tested by measure the length of the line segments.
e)Draw PI⊥AE crossing point P,foot point is I,as PD⊥BG,HE⊥BG,then the quadrangle PDEI is rectangle,so PD=IE,in RT△API, hypotenuse AP> right-angle side AI, then route AP-PD is longer than route AH-JE, that is the cost of route AP-PD is larger than that of route AH-HE. Thus, route AH-HB as the most saving route is proved.
(4)Calculate the length of line segment HC.
△BEH∽△ACH,so ==,set the length of HC as x, according to the Pythagorean theorem, in RT△ACH,102+x2=4x2,so x=,the distance from point H to C is km.
(5)General study of this kind of problem – how to draw auxiliary lines when the cost of road is m times as much as that of water way(m>1).
By observing the method of drawing auxiliary lines when m=2, we can get the general way of drawing auxiliary lines when m>1 and m is an integer:
a) Get point P at line segment BC, divide BC equally into m parts with ruler and compass construction.
b)Draw a circle: P is the center of circle, BC is radius.
c)Draw tangent line BG of circle P passing point B, the point of tangency is D, link PD, then PD⊥BG,and PD =BC。
d)Draw AE⊥BG passing point A,the foot point is E,AE and BC meet at point H, then H is the needed point. Route AH-HB is the most saving route.
In a similar way,when m>1 and m is a grade, the way to draw auxiliary lines is:
a) Get point P at line segment BC, divide BC equally into n parts with ruler and compass construction, n is equal to element of m.( e.g., draw a radiation BL not overlapping BC, then r is the radius, draw the arc on BL, finally we get BQ=QR=RS=ST, link PT, draw parallel lines of PT passing point QRS separately, crossing over BP with point UVW—the quartering of BP is finished.)
b)Draw a circle: P is the center of circle, BP is radius.
c)Draw tangent line BG of circle P passing point B, the point of tangency is D, link PD, then PD⊥BG,and PD =BC。
d)Draw AE⊥BG passing point A,the foot point is E,AE and BC meet at point H, then H is the needed point. Route AH-HB is the most saving route.
3、Summary
When facing different kind of transportation ,the starting point of designing the most saving route is to find one or more turning points between the two locations, then , integrating different transportation via the turning point, the whole most saving route can be found. The purpose to draw auxiliary lines in such kind of problem is to turn different routes that containing various transportations to the routes that only containing one kind of transportation. Finally , with planimetry principle, we can get the shortest route among the routes with one kind of transportation. By means of geometric model, drawing auxiliary lines with the ruler and compass construction, the solving process can be simplified and the turning point can be found immediately.