Question

Consider a convex polygon which has \[44\] diagonals, then calculate the number of triangles joining the vertices of a polygon in which exactly one side is common in the triangle and polygon.

Answer 1

Hint: Here, we will assume that a general polygon consists of \[n\]sides. We will use the combination formula which is \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] where \[n\] is the total number of items in the set and \[r\]is the number of items we have selected from that set.
Then using combinatorics, we will get the number of lines by selecting two vertices and the number of triangles by selecting three vertices.

Complete step-by-step answer:
Step 1: Let’s consider a convex polygon which is having \[n\](number of vertices or sides).
Now, the number of lines which we can draw from any two vertices will be the same as the number of possible ways of selecting (Combinations) of vertices i.e.:
\[{}^n{C_2}\]
Step2: Now, these total lines will consist of both the number of sides and diagonals.
Now as we have assumed that the total number of sides is \[n\]then,
\[ \Rightarrow {}^n{C_2} = n + 44{\text{ }}\](\[\because \]total number of lines equals the number of possible ways of selecting vertices) …… (1)
Solving the above equation (1) by using the combination formula which is \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] for\[n\]:
\[ \Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}{\text{ = }}n + 44\]…. (2)
We know that \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)!\] so, by putting the value of \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)!\] and \[2! = 2 \times 1 = 2\]in the above equation (2):
\[\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2 \times \left( {n - 2} \right)!}} = n + 44\]
Dividing the numerator and denominator of the LHS by \[\left( {n - 2} \right)!\]:
\[ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = n + 44\].
Now, by taking \[2\] on the RHS side and multiplying it with the term \[n + 44\]and on the LHS side multiplying \[n\]with the term \[n - 1\]:
\[n\left( {n - 1} \right) = 2\left( {n + 44} \right)\]
\[ \Rightarrow {n^2} - n = 2n + 88\]
Taking all terms on the LHS side we form a quadratic equation \[a{x^2} + bx + c = 0\]:
\[{n^2} - n - 2n - 88 = 0\]
By adding the coefficient of \[n\] we get:
\[ \Rightarrow {n^2} - 3n - 88 = 0\]…. (3)
Step 3: Finally, solving the quadratic equation (3) by comparing the equation with \[a{x^2} + bx + c = 0\] we find possible values of \[n\].
Now we find two numbers that multiply to give us the product of \[a\] and \[c\] also by adding, it gives us \[b\]
where \[a = 1\], \[b = - 3\]and \[c = - 88\].
Through intuition, these numbers are \[ - 11\] and \[8\]. So, we have that the quadratic equation \[{n^2} - 3n - 88 = 0\] becomes:
\[ \Rightarrow {n^2} - 11n{\text{ + 8}}n - 88 = 0\]
By taking \[n\] common from first and \[8\] common from the last two terms:
\[
  {n^2} - 11n{\text{ + 8}}n - 88 = 0 \\
   \Rightarrow n\left( {n - 11} \right) + 8\left( {n - 11} \right) = 0 \\
 \]
By taking \[\left( {n - 11} \right)\] common from the above expression \[n\left( {n - 11} \right) + 8\left( {n - 11} \right) = 0\] we get:
\[\left( {n + 8} \right)\left( {n - 11} \right) = 0\]
Solving \[\left( {n + 8} \right)\left( {n - 11} \right) = 0\] we get \[n = - 8,11\] .
From here we get the value of \[n\] i.e.:
\[ \Rightarrow n = 11\] \[\left( {\because n{\text{ can't be negative}}} \right)\]
Step 4: Similarly, by using step 1, the number of triangles formed by vertices is equaled to the number of possible ways of selecting (combination) of \[3\]vertices, which is:
\[ \Rightarrow {}^n{C_3}\]
Step 5: By subtracting the above numbers of ways to form a triangle from the triangle which are having \[2\]sides in common, we will get those triangles having exactly one common side.
Now, the number of triangles that have two common sides is the number of vertices of the polygon as each vertex is there in adjacent common sides.
Therefore, triangles having exactly one common side are \[{}^n{C_3} - n\] …. (4)
By substituting the value of \[n{\text{ = 11}}\]as calculated in step 3 in the above term (4) we will get our answer:
\[{}^n{C_3} - n{\text{ = }}{}^{11}{C_3} - 11\]
By using the combination formula to solve the above term \[{}^{11}{C_3} - 11\]
\[
  {}^{11}{C_3} - 11 = \dfrac{{11!}}{{\left( {11 - 3} \right)! \times 3!}} - 11 \\
   = \dfrac{{11 \times 10 \times 9 \times 8!}}{{8! \times 3!}} - 11 \\
 \]
Solving it by using normal multiplication and division:
\[
  \dfrac{{11 \times 10 \times 9 \times 8!}}{{8! \times 3!}} - 11 = 165 - 11 \\
   = 154 \\
 \]

So, the total triangles having exactly one side common are \[154\].

Note: Students get confused to solve these types of problems because here we don’t know the exact number of sides and vertices. You should find the value of \[n\]first which is the total number of vertices or sides.
Also, you should remember that in step no 3 the quadratic equation will factorize to \[\left( {n + 8} \right)\left( {n - 11} \right) = 0\]. Now when we are finding the value of \[n\] which is the total number of vertices or sides that its value can’t be negative so we can’t take \[n = - 8\].