In a △ A B C sides of triangle a , b , c are in an arithmetic progression and m = tan 2 A tan 2 C . Find 9 m .
Note: a is the opposite side to ∠ A .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
thanks sir , nice and short:)
Since the sides are in arithmetic progression, then
b = a + d , c = a + 2 d
where d is the common difference.
Let s 1 = sin ∠ A , s 2 = sin ∠ B , s 3 = sin ∠ C
c 1 = cos ∠ A , c 2 = cos ∠ B , c 3 = cos ∠ C
Applying the Sines Law,
s 1 a = s 2 a + d = s 3 a + 2 d
It follows that
s 2 = s 1 ( 1 + e ) and s 3 = s 1 ( 1 + 2 e )
where e = a d
We now have,
tan 2 θ = 1 + cos θ sin θ
Therefore,
m = ( 1 + c 1 ) ( 1 + c 3 ) s 1 s 3
We'll now develop expression for c 1 , and c 3 , using the Law of Cosines:
c 1 = 2 ( a + d ) ( a + 2 d ) ( ( a + d ) 2 + ( a + 2 d ) 2 − a 2 )
Dividing top and bottom by a 2 , we get:
c 1 = 2 ( 1 + e ) ( 1 + 2 e ) ( ( 1 + e ) 2 + ( 1 + 2 e ) 2 − 1 )
= 2 ( 1 + e ) ( 1 + 2 e ) ( 1 + 6 e + 5 e 2 ) = 2 ( 1 + 2 e ) ( 1 + 5 e )
And, for c 3 we have,
c 3 = 2 ( a ) ( a + d ) ( a 2 + ( a + d ) 2 − ( a + 2 d ) 2 )
= 2 ( 1 + e ) ( 1 + ( 1 + e ) 2 − ( 1 + 2 e ) 2 )
= 2 ( 1 + e ) ( 1 − 2 e − 3 e 2 ) = 2 ( 1 − 3 e )
Now,
m = ( 1 + c 1 ) ( 1 + c 3 ) s 1 s 3 = ( 1 + c 1 ) ( 1 + c 3 ) s 1 2 ( 1 + 2 e )
= ( 1 + c 1 ) ( 1 + c 3 ) ( 1 − c 1 2 ) ( 1 + 2 e )
= ( 1 + c 3 ) ( 1 − c 1 ) ( 1 + 2 e )
Using the above expressions of c 1 and c 3 , we have
1 − c 1 = 2 ( 1 + 2 e ) ( 1 − e )
1 + c 3 = 2 3 ( 1 − e )
Hence,
m = 3 ( 1 − e ) ( 1 + 2 e ) ( 1 − e ) ( 1 + 2 e ) = 3 1
Therefore,
9 m = 9 ( 3 1 ) = 3
oh yes nice solution!! did the same:)
Problem Loading...
Note Loading...
Set Loading...
Recall that the incircle is the intersection of the angle bisectors. Let r be the inradius.
Then, tan 2 A = s − a r and tan 2 C = s − c r .
Recall that the formula for the inradius, given by heron's formula is r = s ( s − a ) ( s − b ) ( s − c ) .
Thus tan 2 A × tan 2 C = ( s − a ) ( s − c ) r 2 = s ( s − b ) .
With a = a , b = a + d , c = a + 2 d , we get that s = 2 a + b + c = 2 3 ( a + d ) .
Hence s s − b = 2 3 ( a + d ) 2 3 ( a + d ) − ( a + d ) = 3 1 .