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Mathematics puzzle 1:
Date 06-12-2013 | Views  4714



Mathematics Puzzle 1

by Vivek Singh | Aug 28, 2012

Sohe towns of Akbar, Bablu, and Garima are equidistant from each other. If a car is three miles from Akbar and four miles from Bablu, what is the maximum possible distance of the car from Garima? Assume the land is flat. (As published on Facebook/BharatTutors.)

Solution:

Let the point P denote the car, A denote Akbar, B denote Bablu, and C denote Garima.
Its clear that, P must be on the opposite side of AB to C, for otherwise we could reflect P in AB, thereby increasing CP, while keeping AP and BP the same.
Also, P must be on the same side of AC as B, for otherwise we could reflect P in AC, and then extend AB so that BP = 4, thereby increasing CP.
Similarly, P must be on the same side of AB as C.
Equilateral triangle ABC, with point P such that AP = 3 and BP = 4.

Puzzle 1 at BharatTutors


Hence quadrilateral APBC is convex, and with diagonals AB and CP, so that we may apply Ptolemy's Inequality, which states that:
AB.CP less than or equal to AP.BC + BP.AC, with equality if, and only if, APBC is cyclic.

Since AB = BC = AC, we get CP less than or equal to AP + BP = 7, with equality if P lies on the arc AB of the (unique) circumcircle of triangle ABC.
It is clear that equality can occur, as, for any side length, AP/BP increases continuously from 0 without limit as P moves anticlockwise along the arc AB (omitting the end point B) Hence at some point AP/BP will reach the value 3/4.

Therefore, the maximum possible distance of the car from Garima is 7 miles.

Stay tuned for more puzzles!

(source: www.qbyte.org)

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