Product of the Radii

Geometry Level 4

For A B C \triangle ABC , the perimeter is 29 6 \dfrac{29}{6} and the products of the three sides is 4.

The circumradius and inradius of A B C \triangle ABC are R R and r r respectively.

If the quantity R × r R \times r can be written in the form m n \dfrac{m}{n} , where m m and n n are coprime positive integers, find m + n m+n .


The answer is 41.

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4 solutions

Sam Bealing
Apr 17, 2016

We use the two well known formulae for the area of a triangle:

A = a b c 4 R , A = a + b + c 2 r A=\frac{abc}{4R},A=\frac{a+b+c}{2} r

a b c 4 R = a + b + c 2 r R r = a b c 2 ( a + b + c ) \frac{abc}{4R}=\frac{a+b+c}{2} r \Rightarrow R r=\frac{abc}{2(a+b+c)}

The question gives us:

a + b + c = 29 6 , a b c = 4 a+b+c=\frac{29}{6}, abc=4

R r = 4 2 29 6 = 12 29 m = 12 , n = 29 R r=\frac{4}{2 \frac{29}{6}}= \frac{12}{29} \Rightarrow m=12,n=29

m + n = 12 + 29 = 41 m+n=12+29=\boxed{41}

Note: An example traingle where the statement is true is 4 3 , 3 2 , 2 \frac{4}{3},\frac{3}{2},2 . (Just to show the configuration is possible)

Moderator note:

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.

Niranjan Khanderia - 5 years, 1 month ago

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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 R=3,r=2 then we could follow through the same working and get the product of the radii to be 6 6 but actually no such triangle exists because R 2 r R \geq 2r . I realise my naming of a , b a,b at the end may have been slightly ambiguous so I've edited that.

Sam Bealing - 5 years, 1 month ago

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Thank you and a good explanation.

Niranjan Khanderia - 5 years, 1 month ago
Akshay Sharma
Apr 28, 2016

Let ( a , b , c ) (a,b,c) , \triangle , S S , R R , r r be the sides,area of triangle,semi-perimeter,circumradius and inradius respectively.

We know that , R = a b c 4 ( i ) R=\frac{a b c}{4\triangle} \cdots (i)

a n d and

r = S ( i i ) r=\frac{\triangle}{S} \cdots (ii)

Now, ( i ) × ( i i ) (i)\times (ii)

R × r = ( a b c 4 ) × ( S ) R\times r=\left (\frac{a b c}{4\triangle}\right ) \times \left( \frac{\triangle}{S}\right )

Substituting the values of given parameters, we get

R × r = 12 29 \boxed{R\times r=\frac{12}{29}}

m + n = 41 \implies \boxed {m+n=41}

I think you mean m + n = 41 m+n=41 .

Sam Bealing - 5 years, 1 month ago

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Thanks for pointing it out, I have edited it.

Akshay Sharma - 5 years, 1 month ago
Dipan Laha
Apr 21, 2016

Let A be the area of the triangle and s be the semi-perimeter( a + b + c 2 \frac{a+b+c}{2} ) We know that, r = A s \frac{A}{s} * * R = a b c 4 A \frac{abc}{4A} From here, R * r = a b c 4 A \frac{abc}{4A} * A s \frac{A}{s} = 4 4 \frac{4}{4} * 12 29 \frac{12}{29} = 12 29 \frac{12}{29} From here, m = 12, n = 29 Thus, m + n = 12 + 29 = 41 Note : - Since abc = 4 and ( a + b + c 2 \frac{a+b+c}{2} ) = 29 12 \frac{29}{12}

Aakash Khandelwal
Apr 19, 2016

We have

= s r \triangle = sr

= a b c / 4 R \triangle= abc/4R .

From both equations R r = a b c / 4 s Rr= abc/4s .

Hence R r = 1 / s = 12 / 29 Rr=1/s= 12/29 .

Therefore answer= 41 \boxed{41}

P l e a s e N o t e Please-Note :: All are in usual notations

Same method used.

Niranjan Khanderia - 5 years, 1 month ago

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