Triangle sides in A.P

Geometry Level 3

In a A B C \triangle ABC sides of triangle a , b , c a,b,c are in an arithmetic progression and m = tan A 2 tan C 2 m= \tan \frac A2 \tan \frac C2 . Find 9 m 9m .

Note: a a is the opposite side to A \angle A .


The answer is 3.

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2 solutions

Calvin Lin Staff
Nov 11, 2016

Recall that the incircle is the intersection of the angle bisectors. Let r r be the inradius.

Then, tan A 2 = r s a \tan \frac{ A}{2} = \frac{ r} { s-a } and tan C 2 = r s c \tan \frac{ C}{2} = \frac{ r } { s-c } .

Recall that the formula for the inradius, given by heron's formula is r = ( s a ) ( s b ) ( s c ) s r = \frac{ \sqrt{ (s-a)(s-b)(s-c) } } { \sqrt{s} } .

Thus tan A 2 × tan C 2 = r 2 ( s a ) ( s c ) = ( s b ) s \tan \frac{A}{2} \times \tan \frac{ C}{2} = \frac{ r^2 } { (s-a)(s-c) } = \frac{ (s-b) } { s } .

With a = a , b = a + d , c = a + 2 d a = a , b = a + d, c = a + 2d , we get that s = a + b + c 2 = 3 ( a + d ) 2 s = \frac{a+b+c} {2} = \frac{ 3 ( a+d) } { 2 } .

Hence s b s = 3 ( a + d ) 2 ( a + d ) 3 ( a + d ) 2 = 1 3 \frac{ s-b } { s } = \frac{ \frac{ 3 (a+d) } { 2 } - (a+d) } { \frac{ 3 (a+d ) } { 2} } = \frac{1}{3} .

thanks sir , nice and short:)

Rakshit Joshi - 4 years, 7 months ago
Hosam Hajjir
Nov 9, 2016

Since the sides are in arithmetic progression, then

b = a + d , c = a + 2 d b = a + d, c = a + 2d

where d is the common difference.

Let s 1 = sin A , s 2 = sin B , s 3 = sin C s_1 = \sin \angle A, s_2 = \sin \angle B, s_3 = \sin \angle C

c 1 = cos A , c 2 = cos B , c 3 = cos C c_1 = \cos \angle A, c_2 = \cos \angle B, c_3 = \cos \angle C

Applying the Sines Law,

a s 1 = a + d s 2 = a + 2 d s 3 \dfrac{a}{s_1} = \dfrac{a + d}{s_2} = \dfrac{a + 2 d}{s_3}

It follows that

s 2 = s 1 ( 1 + e ) s_2 = s_1 (1 + e) and s 3 = s 1 ( 1 + 2 e ) s_3 = s_1 (1 + 2 e)

where e = d a e = \dfrac{d}{ a}

We now have,

tan θ 2 = sin θ 1 + cos θ \tan \dfrac{\theta}{2} = \dfrac{\sin \theta}{1+\cos \theta}

Therefore,

m = s 1 s 3 ( 1 + c 1 ) ( 1 + c 3 ) m = \dfrac{s_1 s_3}{ (1 + c_1)(1 + c_3) }

We'll now develop expression for c 1 c_1 , and c 3 c_3 , using the Law of Cosines:

c 1 = ( ( a + d ) 2 + ( a + 2 d ) 2 a 2 ) 2 ( a + d ) ( a + 2 d ) c_1 = \dfrac{ ( (a+d)^2 + (a + 2d)^2 - a^2 )}{ 2 (a + d)(a + 2 d) }

Dividing top and bottom by a 2 a^2 , we get:

c 1 = ( ( 1 + e ) 2 + ( 1 + 2 e ) 2 1 ) 2 ( 1 + e ) ( 1 + 2 e ) c_1 = \dfrac{( (1 + e)^2 + (1 + 2 e)^2 - 1)}{ 2 (1 + e)(1 + 2 e)}

= ( 1 + 6 e + 5 e 2 ) 2 ( 1 + e ) ( 1 + 2 e ) = ( 1 + 5 e ) 2 ( 1 + 2 e ) = \dfrac{(1 + 6 e + 5 e^2) }{ 2 (1 + e)(1 + 2 e)} = \dfrac{(1 + 5e)}{2(1 + 2e)}

And, for c 3 c_3 we have,

c 3 = ( a 2 + ( a + d ) 2 ( a + 2 d ) 2 ) 2 ( a ) ( a + d ) c_3 = \dfrac{(a^2 + (a + d)^2 - (a + 2d)^2 )}{ 2 (a)(a+d)}

= ( 1 + ( 1 + e ) 2 ( 1 + 2 e ) 2 ) 2 ( 1 + e ) =\dfrac{ (1 + (1 + e)^2 - (1 + 2e)^2 )}{ 2 (1 + e)}

= ( 1 2 e 3 e 2 ) 2 ( 1 + e ) = ( 1 3 e ) 2 = \dfrac{(1 - 2 e - 3 e^2)}{ 2(1 + e)} = \dfrac{(1 - 3e)}{ 2}

Now,

m = s 1 s 3 ( 1 + c 1 ) ( 1 + c 3 ) = s 1 2 ( 1 + 2 e ) ( 1 + c 1 ) ( 1 + c 3 ) m = \dfrac{s_1 s_3}{ (1 + c_1)(1 + c_3) } = \dfrac{s_1^2(1 + 2e)}{ (1 + c_1)(1 + c_3)}

= ( 1 c 1 2 ) ( 1 + 2 e ) ( 1 + c 1 ) ( 1 + c 3 ) = \dfrac{(1 - c_1^2) (1 + 2e) }{ (1 + c_1)(1 + c_3) }

= ( 1 c 1 ) ( 1 + 2 e ) ( 1 + c 3 ) = \dfrac{(1 - c_1) (1 + 2e)}{ (1 + c_3) }

Using the above expressions of c 1 c_1 and c 3 c_3 , we have

1 c 1 = ( 1 e ) 2 ( 1 + 2 e ) 1 - c_1 = \dfrac{(1 - e)}{2(1 + 2e)}

1 + c 3 = 3 ( 1 e ) 2 1 + c_3 = \dfrac{3(1 - e)}{2}

Hence,

m = ( 1 e ) ( 1 + 2 e ) 3 ( 1 e ) ( 1 + 2 e ) = 1 3 m = \dfrac{(1 - e)(1 + 2e)} {3(1 - e)(1 + 2 e)} = \dfrac{1}{3}

Therefore,

9 m = 9 ( 1 3 ) = 3 9 m = 9 (\dfrac{1}{3} ) = 3

oh yes nice solution!! did the same:)

Rakshit Joshi - 4 years, 7 months ago

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