Geometrically invalid

Geometry Level 5

Let O O and G G denote the circumcenter and orthocenter of the triangle A B C ABC , respectively. If A G O = 9 0 \angle AGO = 90^\circ , find ( B C ) 2 (BC)^2 .

Note : Image drawn not necessarily up to scale.


The answer is 325.

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.

1 solution

Mark Hennings
Sep 18, 2016

If A , B , C A,B,C have position vectors a , b , c \mathbf{a},\mathbf{b},\mathbf{c} with respect to the circumcentre O O as origin (so that a , b , c \mathbf{a},\mathbf{b},\mathbf{c} all have the same modulus R R , the circumradius), then the position vectors of G G and M M (the midpoint of B C BC are g = 1 3 ( a + b + c ) m = 1 2 ( b + c ) \mathbf{g} \; = \; \tfrac13(\mathbf{a}+\mathbf{b}+\mathbf{c}) \hspace{1cm} \mathbf{m} = \tfrac12(\mathbf{b}+\mathbf{c}) and we require that g ( a m ) = 0 \mathbf{g}\cdot(\mathbf{a}-\mathbf{m}) = 0 . Thus 0 = ( a + b + c ) ( 2 a b c ) = 2 a 2 + a ( b + c ) b + c 2 = a b + a c 2 b c = 1 2 [ 2 R 2 A B 2 ] + 1 2 [ 2 R 2 A C 2 ] [ 2 R 2 B C 2 ] \begin{array}{rcl} 0 & = & (\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot (2\mathbf{a} - \mathbf{b} - \mathbf{c}) \; = \; 2|\mathbf{a}|^2 + \mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) - |\mathbf{b}+\mathbf{c}|^2 \\ & = & \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} - 2 \mathbf{b}\cdot\mathbf{c} \\ & = & \tfrac12\big[2R^2 - AB^2\big] + \tfrac12\big[2R^2 - AC^2\big] - \big[2R^2 - BC^2\big] \end{array} and hence B C 2 = 1 2 ( A B 2 + A C 2 ) BC^2 = \tfrac12(AB^2 + AC^2) . In our case, we deduce that B C 2 = 325 BC^2 = \boxed{325} .

I solved using simple Pythagoras theorm

Aditya Kumar - 4 years, 9 months ago

Log in to reply

May I know your approach please?

Mahadi Hasan - 4 years, 8 months ago

This question was there in pre rmo in north delhi region

Aditya Kumar - 4 years, 9 months ago

Nice use of vectors. I understood it properly. Thanks.

Priyanshu Mishra - 4 years, 8 months ago

Log in to reply

Even I solved it using vectors.But why do you think that this is geometrically invalid?

Indraneel Mukhopadhyaya - 4 years, 8 months ago

Log in to reply

Itsnothing. No title was coming to my mind, so i put that without any reason.

Do you have a good title for this problem.

Priyanshu Mishra - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...