Koundinya-Mathematics Teachers Hub

Wednesday, 14 March 2012

Here follows a problem which troubles (possibly)..

FIND THE SHORTEST LINE SEGMENT WITH ITS ENDS ON TWO SIDES OF A GIVEN TRIANGLE WHICH BISECTS THE TRIANGLE

i have a solution to this problem i will post the same however i wish someone comes out with a solution

... it's been days and no reply has come...so, here is the solution

consider the figure below

Let PQ be the shortest line segment such that Area of ΔCPQ is equal to Area of the Quadrilateral PQBA.

Let CP = x , CQ = y , PQ = u.

QM and BE are perpendiculars from Q and B on to CA, we can see that ΔCQM ||| ΔCBE, Then
${\color{black} \frac{QM}{CQ}=\frac{BE}{BC}=sin C}$

Hence we will have QM = ysin C and BE = a sin C

${\color{Yellow}\therefore 2\frac{QM X CP}{2}=\frac{2\mathit{xy}sin C}{2}=\frac{CA X BE}{2} = \frac{absin C}{2}}$

PQ bisects ΔABC implies

${\color{black} xy\ {} =\ {} \frac{ab}{2}\ {}or \ {} 2xy \ {}=\ {}ab\ {}\Rightarrow \ {} y\ {}=\ {}\frac{ab}{2x} }$

Using the Cosine formula for triangle CPQ

${\color{Yellow} \mathit{u}^{2}=PQ^{2}=CP^{2}+CQ^{2}-2.CP.CQ.cosC }$

${\color{black} =\ {}\mathit{x}^{2}+\mathit{y}^{2}-2\mathit{xy}cosC\ {}= \ {}\mathit{x}^{2}+\frac{a^{2}b^{2}}{4\mathit{x}^{2}}-abcosC}$

Let us assume this is equal to r. then

${\color{black} r\mathit{x}^{2}\ {}=\ {}\mathit{x}^{4}-(abcosC)\mathit{x}^{2}+\frac{a^{2}b^{2}}{4}}$

Upon simplification , we get
${\color{Yellow} (\mathit{x}^{2}-\frac{(abcosC)+r}{2})^{2}+\frac{a^{2}b^{2}}{4}\ {} =\ {}\frac{(abcosC+r)^{2}}{4} }$

Observation : Minimum value of R.H.S is reached when the first etrm of L.H.S is 0. ie.,
${\color{black} (\mathit{x}^{2}=\frac{(abcosC)+r}{2})^{2} }$

This implies that r = ab(1-cosC) and consequently we get

${\color{Yellow} \mathit{x} = \sqrt{\frac{ab}{2}}\ {}\Rightarrow \ {}\mathit{y}=\sqrt{\frac{ab}{2}} }$

Therefore length of PQ can be found as
${\color{Yellow} =\ {} \sqrt{\frac{(c-a+b)(c+a-b)}{2}} }$

Generalisation:

Find a+b-c, b+c-a, c+a-b, choose the least and next higher of them; for example if a+b-c and b+c-a are these two in that order, then P and Q must be chosen on BC and BA such that
${\color{black} BP,BQ\ {}=\ {} \sqrt{\frac{ca}{2}} }$
and the shortest length PQ that bisects triangle abc will then be
${\color{black} PQ=\ {} \sqrt{\frac{(a+b-c)(b+c-a)}{2}} }$

2 comments:

Vidyamanohar23 March 2012 at 23:51
A simple presentation for the same solution is
here .
1c's blog (Vamsee)24 March 2012 at 02:48
Both the methods are amazing(2nd methos is in the 1st comment)...lengthy or not lengthy, usage of two different methods (Algebra & Calculus) were put to reach to two different forms of solution...The Algebraic method though gives us extra information regarding choosing the points P and Q on the lines AB or BC or CA....KUDOS to both anyway!!!