Question

If \[a + b + c = 1\] and \[a,b,c\] are all distinct positive real, then prove that \[(1 - a)(1 - b)(1 - c) > 8abc\].

Answer 1

Hint: To find the required number of sides for a convex polygon we use the formula for the sum of all the interior angles of a polygon of n sides: \[(n - 2) \times {180^0}.\] where n is the number of sides of a regular convex polygon. Then after substituting the given value in the formula for the sum of the interior angles of a convex polygon to get the required value for n.

Complete step by step answer:
Formula used: The sum of all interior angles of a polygon of n sides = \[(n - 2) \times {180^0}.\] …………………. (i)
Where n is the number of sides of the convex polygon.
Substituting the given value for the sum of the interior angles in eqn (i).we get
\[{1440^0} = (n - 2) \times {180^0}\]
\[ \Rightarrow (n - 2) = \dfrac{{{{1440}^0}}}{{{{180}^0}}}\]
\[ \Rightarrow (n - 2) = 8\]
\[ \Rightarrow n = 10\]
From the above solution, we have found that the required number of sides of the polygon will be 10.

Hence, option (B) is the correct answer.

Note:
In a convex polygon, all interior angles are less than \[{180^0}\] which is exactly opposite of the concave polygon. So, it is recommended that you do not use the same formula for concave polygons as well. We also note that the value for n involved in the above formula will always be a positive integer or natural number.
In order to tackle such kinds of questions, we have to remember all the necessary properties regarding a convex polygon.
Some of them are:-
(i) All the internal angles of a convex polygon is always less than \[{180^0}\].
(ii) All the diagonals of a regular convex polygon lie inside of the polygon.
(iii) The intersection of two convex polygons will form a convex polygon.
(iv) Sum of all the interior angles of a regular convex polygon of n sides = \[(n - 2) \times {180^0}.\]