trigo

**prasum** · December 12th 2010, 06:59 AM

a triangle is formed by drawing tangents at A,B,C to the circumcircle of triangle ABC prove that the perimeter of this triangle is 2RtanAtanBtanC where R is the radius of circumcircle.

**snowtea** · December 12th 2010, 07:53 AM

See the diagram.

This shows 1/3 of the proof.

Consider the part of the external triangle opposite angle A. Call angle A theta.
The diagram shows that the length for the part of the triangle opposite angle A is
2*R*tan(theta) = 2*R*tan(A)

Note: The 2*theta is measured from the center of the circle.

Repeating the procedure for the parts opposite B and C and summing, we get:

Perimeter = 2*R*tan(A) + 2*R*tan(B) + 2*R*tan(C) = 2R(tan(A) + tan(B) + tan(C))

Is this the formula you want to prove?

**Soroban** · December 12th 2010, 09:21 AM

Hello, snowtea!

Lovely work!

You found that: . $\text{Perimeter}\:=\:2R\left(\tan A + \tan B + \tan C)$
. . which involves the sum of the tangents.

The original equation has the product of the tangents.

But not to worry . . . Here's a surprising theorem:

. . In $\Delta ABC\!:\;\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C$

Proof

$A + B + C \:=\:180^o \quad\Rightarrow\quad A + B \;=\;180^o - C$

Take tangents: . $\tan(A + B) \;=\;\tan(180^o - C)$

. . . . . . . . . $\dfrac{\tan A + \tan B}{1 - \tan A\tan B} \;=\;-\tan C$

. . . . . . . . . . $\tan A + \tan B \;=\;-\tan C + \tan A\tan B\tan C$

. . . . $\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C$

**prasum** · December 13th 2010, 05:48 AM

Originally Posted by Soroban

Hello, snowtea!

Lovely work!

You found that: . $\text{Perimeter}\:=\:2R\left(\tan A + \tan B + \tan C)$
. . which involves the sum of the tangents.

The original equation has the product of the tangents.

But not to worry . . . Here's a surprising theorem:

. . In $\Delta ABC\!:\;\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C$

Proof

$A + B + C \:=\:180^o \quad\Rightarrow\quad A + B \;=\;180^o - C$

Take tangents: . $\tan(A + B) \;=\;\tan(180^o - C)$

. . . . . . . . . $\dfrac{\tan A + \tan B}{1 - \tan A\tan B} \;=\;-\tan C$

. . . . . . . . . . $\tan A + \tan B \;=\;-\tan C + \tan A\tan B\tan C$

. . . . $\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C$

thanks

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