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

This entry was written by Nalin Pithwa, posted on December 11, 2014 at 12:55 pm, filed under IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) and tagged CMI entrance, IITJEE maths, ISI entrance, plane trigonometry problem, RMO Maths, solutions of triangles. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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