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Trignometry

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29 Sep 2008 20:27:33 IST

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880

Good Trig Inequality

None

Given to me by a good friend and a past [lets hope that changes :)] goiitian:

If a,b and c are the sides opposite to angles A, B and C in $\bigtriangleup ABC$ and h₁, h₂ and h₃ are the altitudes perpendicular to a,b and c respectively, prove that

$\frac{a^2}{h_1^2} + \frac{b^2}{h_2^2} + \frac{c^2}{h_3^2} \ge 4<br/>$

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Comments (7)

Decoder

Blazing goIITian

Joined: 1 Apr 2007

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29 Sep 2008 22:41:47 IST

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will this require ..inequalities..like a+b >c ...??

or we have to proceed like general inequalities..and then apply SOTS..

Decoder

Blazing goIITian

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30 Sep 2008 13:20:10 IST

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the most funda-vella method is equating all these..we get condition for an equilateral triangle..putting known relations ..we get 4..

other method is putting value of a/h in terms of sides and area and circumradius..
later circumradiur r get cancelled ..and we r just to calculate the minimum value of a known inequality..

but in this prove i neither used sots nor jensen inequality for trigonometric ratios...
but triangle inequalities always go with added conditions...

am i right sir ??

Dipanjan

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Joined: 30 Jul 2008

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30 Sep 2008 15:13:31 IST

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Hari Shankar

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30 Sep 2008 15:44:54 IST

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We have $h_1 = b \sin C = c \sin B; h_2 = a \sin C = c \sin A; h_3 = a \sin B = b \sin A$

Hence, $\frac{a^2}{h_1^2} + \frac{b^2}{h_2^2} = \frac{a^2}{b^2 \sin^2 C} + \frac{b^2}{a^2 \sin^2 C} \ge \frac{2}{\sin^2 C}$ (from AM-GM) ....................1

Likewise $\frac{b^2}{h_2^2} + \frac{c^2}{h_3^2} \ge \frac{2}{\sin^2 A}$ .................2

$\frac{a^2}{h_1^2} + \frac{c^2}{h_3^2} \ge \frac{2}{\sin^2 B}$ ...................3

Hence $\frac{a^2}{h_1^2} + \frac{b^2}{h_2^2} + \frac{c^2}{h_3^2} \ge \frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C}$

Now, we use the well known inequality that $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$

From AM-GM, we have $(\sin^2 A + \sin^2 B + \sin^2 C) \left(\frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C} \right) \ge 9$

and from the last two inequalities $\frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C} \ge 4$ and we are done.

Decoder

Blazing goIITian

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1 Oct 2008 16:09:53 IST

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nice..amazingly i tried by both methods..second one was too far..one is discouraged till he reach there...

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

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10 Oct 2008 20:40:16 IST

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http://downloads.shaastra.org/2008/onlinemath/omc_ps1.pdf

Is the friend Pardesi? :P

Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

11 Oct 2008 09:37:16 IST

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that you are a 'ganga'ster is now proved without doubt. thanx for leaking the source.

You should be courtmarshalled man!

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