For △ A B C , the perimeter is 6 2 9 and the products of the three sides is 4.
The circumradius and inradius of △ A B C are R and r respectively.
If the quantity R × r can be written in the form n m , where m and n are coprime positive integers, find m + n .
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Thanks for verifying that such a triangle could exist. It is not immediately apparent why that must be the case, so we have to do that for completeness.
For example, there isn't a triangle whose perimeter is 1 and product of all 3 sides is 2.
Thanks for your Note. You could avoid confusion by not naming the sides with a, b. How did you get the sides ? I did it this way:
4=2 * 2, but we have 6 in denominator, so others would be 2/3 and 3/2 with a multiplier 2 to one of them. It can not be with 3/2, so it must be with 2/3. It works so stop trials.
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The reason I gave an example of sides was just to show that a triangle with the given radii actually existed. For example if R = 3 , r = 2 then we could follow through the same working and get the product of the radii to be 6 but actually no such triangle exists because R ≥ 2 r . I realise my naming of a , b at the end may have been slightly ambiguous so I've edited that.
Let ( a , b , c ) , △ , S , R , r be the sides,area of triangle,semi-perimeter,circumradius and inradius respectively.
We know that , R = 4 △ a b c ⋯ ( i )
a n d
r = S △ ⋯ ( i i )
Now, ( i ) × ( i i )
R × r = ( 4 △ a b c ) × ( S △ )
Substituting the values of given parameters, we get
R × r = 2 9 1 2
⟹ m + n = 4 1
I think you mean m + n = 4 1 .
Let A be the area of the triangle and s be the semi-perimeter( 2 a + b + c ) We know that, r = s A * * R = 4 A a b c From here, R * r = 4 A a b c * s A = 4 4 * 2 9 1 2 = 2 9 1 2 From here, m = 12, n = 29 Thus, m + n = 12 + 29 = 41 Note : - Since abc = 4 and ( 2 a + b + c ) = 1 2 2 9
We have
△ = s r
△ = a b c / 4 R .
From both equations R r = a b c / 4 s .
Hence R r = 1 / s = 1 2 / 2 9 .
Therefore answer= 4 1
P l e a s e − N o t e :: All are in usual notations
Same method used.
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We use the two well known formulae for the area of a triangle:
A = 4 R a b c , A = 2 a + b + c r
4 R a b c = 2 a + b + c r ⇒ R r = 2 ( a + b + c ) a b c
The question gives us:
a + b + c = 6 2 9 , a b c = 4
R r = 2 6 2 9 4 = 2 9 1 2 ⇒ m = 1 2 , n = 2 9
m + n = 1 2 + 2 9 = 4 1
Note: An example traingle where the statement is true is 3 4 , 2 3 , 2 . (Just to show the configuration is possible)