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Thread: Find number of different triangles possible?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    Find number of different triangles possible?

    The sides of a triangle are a, b and c where a, b and c are integers and $\displaystyle a\leq b\leq c$. If c is given, show that the number of different triangles is $\displaystyle \frac{1}{4}c(c + 2)$ or $\displaystyle \frac{1}{4}(c + 1)^2$, according as c is even or odd. Also, show that the number of isoceles or equilateral triangle is $\displaystyle \frac{1}{2}(3c - 2)$ or $\displaystyle \frac{1}{2}(3c - 1)$, according as $\displaystyle c$ is even or odd.
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  2. #2
    Junior Member
    May 2009
    The important fact here is that in a triangle with sides $\displaystyle a,b,c$, it must be true that $\displaystyle a+b>c, a+c>b,$ and $\displaystyle b+c>a.$

    The last two inequalities are obvious given that $\displaystyle a\leq b\leq c$, so we just need to see how many ways we can choose $\displaystyle a$ and $\displaystyle b$ so that $\displaystyle a+b>c.$

    Suppose $\displaystyle c$ is even.

    Then if $\displaystyle a=1$, then $\displaystyle b$ must equal $\displaystyle c$.
    if $\displaystyle a=2$, then $\displaystyle b$ can equal $\displaystyle c$ or $\displaystyle c-1$.
    if $\displaystyle a=c/2$, then b can equal $\displaystyle c/2+1, ... , c$.

    (So far we have $\displaystyle 1+2+\cdots+c/2=\frac{1}{8}c(c+2)$ ways to choose $\displaystyle a$ and $\displaystyle b$.

    Doing the same thing when $\displaystyle a>c/2$, we find that there are $\displaystyle \frac{1}{8}n(n+2)$ more ways to choose, giving a total of $\displaystyle \frac{1}{4}c(c+2)$ triangles.

    The other cases of this problem are not much different.
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