0
$\begingroup$

Let $A_1,A_2,.....,A_{18}$ be the vertices of a regular polygon with $18$ sides. How many of the triangles $\Delta A_iA_jA_k,1\le i<j<k\le 18$, are isosceles but not equilateral?

A. $63$

B. $70$

C. $126$

D. $144$

I attempted to use the following result:

$$ \#\text{isosceles} = \begin{cases} n^2/2 - (5/3) n & \text{for } n\equiv 0 \pmod 6 \\ n^2/2 - (1/2) n & \text{for } n\equiv 1, 5 \pmod 6 \\ n^2/2 - n & \text{for } n\equiv 2, 4 \pmod 6 \\ n^2/2 - (7/6) n & \text{for } n\equiv 3 \pmod 6 \end{cases}$$

But using this I end up with:

$\dfrac{18^2}{2}-\dfrac{5}{3}\cdot 18=132$

Hence none of the options are matching.Please help with a plausible solution.

Thus it is clear that I am lagging in some knowledge about this somewhere.Also I request you to give me proper theoretical basis to sole any problem of this genre.

$\endgroup$
  • $\begingroup$ Users one request please no need to upvote or downvote the attempts. See with constant downvotes, be it intentional or otherwise bans the user to ask questions. I believe I haven't posted anything that is worth no effort and hence all the downvotes $\endgroup$ – Saradamani Apr 23 '18 at 9:51
4
$\begingroup$

There are $18$ ways to choose the apex of the triangle. If the apex is at vertex $0$ the two other vertices can be at one of the pairs $\pm1$, $\ldots$, $\pm5$, $\pm7$, $\pm8$. As a consequence there are $18\cdot7=126$ possibilities in all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.