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
Hence we will have QM =
ysin C and BE = a sin C
PQ bisects ΔABC implies
Using the Cosine formula for triangle CPQ
Let us assume this is equal to r. then
\mathit{x}^{2}+\frac{a^{2}b^{2}}{4}} "{\color{black} r\mathit{x}^{2}\ {}=\ {}\mathit{x}^{4}-(abcosC)\mathit{x}^{2}+\frac{a^{2}b^{2}}{4}}")
Upon simplification , we get
+r}{2})^{2}+\frac{a^{2}b^{2}}{4}\ {} =\ {}\frac{(abcosC+r)^{2}}{4} } "{\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.,
+r}{2})^{2} } "{\color{black} (\mathit{x}^{2}=\frac{(abcosC)+r}{2})^{2} }")
This implies that
r = ab(1-cosC) and
consequently we get
Therefore length of PQ can be found as
(c+a-b)}{2}} } "{\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
and the shortest length PQ that bisects triangle abc will then be
(b+c-a)}{2}} } "{\color{black} PQ=\ {} \sqrt{\frac{(a+b-c)(b+c-a)}{2}} }")
