## Solutions of Triangles — a tricky IITJEE problem

Question (IITJEE 1978). Suppose $p_{1},p_{2},p_{3}$ are the altitudes through vertices A, B, C of a triangle ABC with area $\Delta$.

Prove that:

$\frac {1}{p_{1}} + \frac {1}{p_{2}} - \frac {1}{p_{3}}=\frac {2ab}{(a+b+c) \Delta} \cos^{2} (C/2)$

Proof:

The RHS looks daunting. But, if we bring the factor $\Delta$ in its denominator to the LHS, then the problem unfolds itself. Since, $\Delta =(1/2)ap_{1}=(1/2)bp_{2}=(1/2)cp_{3}$, the problem is equivalent to the following:

prove: $a+b-c=\frac {4ab}{a+b+c}\cos^{2}(C/2)$

If we write $2\cos^{2}(C/2)$ as $1+\cos C$, and multiply both the sides of by $a+b+c$, then the problem is reduced to  proving that

$(a+b)^{2}-c^{2}=2ab+2ab \cos C$.

which is the same as the cosine formula 🙂 🙂 🙂

More later…

Nalin Pithwa