Solution to another S L Loney problem for IITJEE Advanced Mathematics

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are \theta and \phi. Prove that

4(1-\cos{\theta})(1-\cos{\phi})=\cos{\theta}+\cos{\phi}

Proof:

Let a, b, c be in AP. Hence, a<b<c, which in turn implies c is greatest and a is the least.

Hence, \theta=\angle{C} and \angle{A}.

Want:

4(1-\cos{A})(1-\cos{C})=\cos{A}+\cos{C}

Given:

b-a=c-b

(b-a)^{2}=(b-c)^{2}. Hence,

b^{2}-2ab+a^{2}=b^{2}+c^{2}-2bc

b^{2}+a^{2}-c^{2}=b^{2}-2bc+2ab

2b=a+c

4b^{2}=a^{2}+c^{2}+2ac

a^{2}+c^{2}-b^{2}=3b^{2}-2ac

a^{2}+b^{2}-c^{2}=b^{2}-2bc+2ab

b^{2}+c^{2}-a^{2}=b^{2}-2ab+2bc

Now, we have 1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc} and this is equal to the following:

\frac{2bc-(b^{2}+c^{2}-a^{2})}{2bc}=\frac{2bc-(b^{2}-2ab+2bc)}{2bc}

= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}.

Also, similarly, we have the following:

1-\cos{C}=1-\frac{a^{2}+b^{2}-c^{2}}{2ab}=1-\frac{b^{2}-2bc+2ab}{2ab}=\frac{2bc-b^{2}}{2ab}

\cos{A}+\cos{C}=\frac{b^{2}+c^{2}-a^{2}}{}+\frac{a^{2}+b^{2}-c^{2}}{2ab}

= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}

=\frac{a(b^{2}-2ab+2bc)+c(b^{2}-2bc+2ab)}{2abc}

=\frac{ab^{2}-2a^{2}b+2abc+b^{2}c-2bc^{2}+2abc}{2abc}

=\frac{ab^{2}+b^{2}c-2a^{2}b-2c^{2}b+4abc}{2abc}

= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}

=\frac{b(a+c)-2(a-c)^{2}}{2ac}

=\frac{2b^{2}-2(a-c)^{2}}{2ac}

=\frac{b^{2}-(a-c)^{2}}{2ac}

\frac{(b+a-c)}{(b-a+c)}{ac}

but it is given that b-a=c-b, hence, a+c=2b, a+c-b=b. So the above expression changes to

= \frac{(b+a-c)(c-b+c)}{ac}

= \frac{(2c-b)(a+b-c)}{ac}

= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}

= \frac{(2c-b)(a-b+c)}{ac}

= \frac{(2c-b)(2a-b)}{ac}

= \frac{4ac-2bc-2ab+b^{2}}{ac}

= 4(1-\cos{A})(1-\cos{C}).

QED.

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