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

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