Do you know this property?

Geometry Level 4

Consider triangle A B C ABC . The equation of side B C BC is y = x y=x . The centroid of triangle is ( 8 , 3 ) (8,3) and circumcentre is ( 9 , 3 ) (9,3) . If circumradius of triangle is R R , then find the value of R 5 \dfrac{ R }{ \sqrt{ 5 } } .


The answer is 3.

Rishi Sharma by Rishi Sharma , 32, India

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2 solutions

Rishi Sharma
May 11, 2017

Property-1 :- Centroid of a traingle divides the line joining orthocentre and circumcentre in the ratio 2 : 1 2:1 .

Property-2 :-The image of orthocentre through any side of the triangle lie on it's circumcircle.

Let's O ( a , b ) O(a , b) be the orthocentre. Using property-1. We get a = 6 , b = 3 a=6 , b=3 . Image of O O through side BC will be ( 3 , 6 ) (3,6) . According to property 2 this point lies on circumcircle of triangle. Hence R = ( 9 − 3 ) 2 + ( 3 − 6 ) 2 = 3 5 R=\sqrt{ { (9-3) }^{ 2 }+{ (3-6) }^{ 2 } } =3\sqrt{ 5 }

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well i did in the same manner .. nice soln,

Samarth Singh - 4 years, 1 month ago

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Thanks. Try this one

Rishi Sharma - 4 years, 1 month ago

Or, you could find the foot of perpendicular from O on BC (say M). Then using the property that centroid divides median in the ratio 2:1, you could find A by section formula.Hence, compute OA=R.

Indraneel Mukhopadhyaya - 4 years, 1 month ago

Nice problem! Its always good to see "geometry" in coordinate geometry

Harsh Shrivastava - 4 years, 1 month ago

L e t G b e t h e C e n t r o i d , O t h e c i r c u m c e n t r e , M ( m , n ) t h e m i d p o i n t o f B C . O M i s ⊥ t o B C , ⟹ s l o p e o f O M i s − 1. ∴ Y O M = − ( X − 9 ) + 3. S o M i s a t i n t e r s e c t i o n o f B C a n d O M ⟹ n = m a n d n = − ( m − 9 ) + 3. ∴ M ( 6 , 6 ) . S i n c e m i d i a n A M = 3 ∗ O M . A ( 6 + 3 ∗ ( 8 − 6 ) , 6 + 3 ∗ ( 3 − 6 ) ) = A ( 12 , − 3 ) . S o R = A O = ( 12 − 9 ) 2 + ( − 3 − 3 ) 2 = 3 5 . 3 5 5 = 3 Let~G~be~the~Centroid ,~~~O~the~circumcentre ,~~~M(m,n)~the~midpoint~of~BC. \\ OM~is~\bot~to~BC,~\implies~slope~of~OM~is~ -1.\\ \therefore~Y_{OM}=-(X-9)+3.\\ So~M~is~at~intersection~of~BC~and~OM~\implies~n=m~~and~n=-(m-9)+3.\\ \therefore~M(6,6).\\ Since~midian~AM=3*OM.~~~~~~A \big(6+3*(8-6),6+3*(3-6) \big )=A(12,-3).\\ So~R=AO=\sqrt{(12-9)^2+(-3-3)^2}=3\sqrt5.\\ \dfrac{3\sqrt5}{\sqrt5}=\Large~~~\color{#D61F06}{3}\\

I saw Mr. Indraneel Mukhopadhyaya's comment after posting this. I am retaining my posting since he has not given details.

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