Page 5 - Curiosity July 2020

MATH CONCEPTS: FIND THE SHORTEST DISTANCE Rintu Nath
Four towns – A, B, C and D – they all lie at the corners of a square of side 100 km. Roads need to be constructed to connect them all using the shortest possible network of roads. Can you suggest a possible solution where the length of the road would be minimum?’
AB
CD
There are a number of ways that roads can be constructed connecting four towns A, B. C, and D. Try to connect all the four points with different lines and calculate total length of constructed road.
We can construct straight roads connecting A to B,
B to D and D to C. That will be 300 km of road.
It is possible to construct road network connecting all the four points A, B, C and D. The idea is to construct minimum road network. A single road PQ may be constructed that will reach nearest of all four towns A, B, C and D. From points P and Q, two roads each need to be constructed to reach each destination. Solution is given below:
Considering APB = 120° From triangle AMP
50 Km AM
B
AP= AM = 50 =57.73km sin 60° 0.866
60°
P Q
CD
However, we can do better than that. If we construct the two diagonals connecting four points the length
of the constructed road will be less than 300 km.
Using Pythagoras theorems,
AD2 = AC2 + CD2
AD = √AC2+CD2 = √1002+1002 = 100√2 km = BC Therefore AD + BC = 200√2 km, which is about 282 km. However, a still better solution exists!
AB
CD
CND
AM MP
MP =
= tan 60°
AM = 50 = 28.86 km = QN
Therefore, PQ = 100 – ( MP + QN) = 100 – ( 28.86 + 28.86) = 42.26 km
Length of the road = AP + BP + QC + QD + PQ = 4*57.73 + 42.26 = 273.18 km
Is it possible to increase length of PQ portion and get a better solution? Try it out.
Dr Rintu Nath is Scientist ‘F’ at Vigyan Prasar. Email: Rnath@vigyanprasar.gov.in
tan 60° √3
AB 3Km And, MP + PQ + QN = 100 km
July 2020
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