JEE Novice - (6)

We know that $c^2=a^2+b^2-2ab\times \cos\theta$ ,but $\cos\theta$ would be negative for obtuse angles $\theta$ hence, $c^2>a^2+b^2$ for obtuse angled triangles.

I forgot the second case ;(

This seems like a nice and simple solution, although I'm having difficulty understanding how you figured out both the upper and lower bounds in both cases because your method seems to give only a single bound in each case (lower bound in first case and upper bound in second case). You also didn't analyze the case of angle $A$ being obtuse, if I understand your solution correctly. Anyway, I didn't know that we can extend the Pythagorean Theorem for obtuse triangles like that.

I did it in quite a complicated way using the law of cosines and then solving a few quadratic inequalities and comparing bounds. Also used the fact that a linear polynomial is little-o of a quadratic polynomial.

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Here's my approach (if anyone cares):

Denote the triangle as $\Delta\textrm{ABC}$ where the sides $AB,BC,CA$ are (w.l.o.g) $k,11,15$ respectively. By the Law of Cosines, we have,

$\cos(A)=\frac{104+k^2}{30k}~,~\cos(B)=\frac{k^2-104}{22k}~,~\cos(C)=\frac{346-k^2}{330}$

We use the fact that the cosine of any obtuse angle $\in (-1,0)$ and check the three cases respectively.

If $A$ is obtuse, you get $0\lt (k+26)(k+4)\lt 30k$ . Even at $k=0$ (which cannot be a valid $k$ ). the MHS is greater than RHS and using the "little-o" fact, we can easily conclude that there are no positive real solutions for $k$ when $A$ is obtuse. (this covers positive integral solutions too).

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If $B$ is obtuse, you get $0\lt (k+26)(k-4)\lt 22k$ . Considering the LHS-MHS part, you get the solution set for $k$ as $(-\infty,-26)\cup (4,\infty)$ . Now, taking into account the RHS (upper bound) and noting that the upper bound doesn't hold for $k=11$ (but holds for $k=10$ ), we can again use the "little-o" argument to get the solution set of $k$ as $(4,11)$ .

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If $C$ is obtuse, you get $0\lt 676-k^2\lt 330$ . The LHS-MHS part gives us $k\lt 26$ and the MHS-RHS part gives us $k\gt 18$ . Combining the two, we have $k\in (18,26)$ .

Now, we simply collect the solution sets from the possible cases which is the following set (interval union): $(4,11)\cup (18,26)$ . We can easily count the number of positive integral solutions for $k$ to be $13$ , and we're done.

$\huge Correct!$

Best question best solution