Short and sweet...sour?

Geometry Level 3

Given that $\triangle ABC, AB=3 \text{ and } AC=5$ Let $O$ is circumcenter of $\triangle ABC$ . Find $\overrightarrow{AO} \cdot \overrightarrow{BC}$

If you think, it lacks sufficient details; give your answer as 999

The answer is 8.

1 solution

A Former Brilliant Member
Nov 30, 2018

Plot this triangle on Cartesian plane with $A = (0,3) , B=(0,0)$ and $C= (4,0)$

We know that Circumcenter of a Right Triangle lies at midpoint of the Hypotenuse i.e, $O = (\dfrac {(4-0)}{2} , \dfrac {(3-0)}{2}) = ( 2,1.5)$

$OA = \dfrac 12 \times$ Hypotenuse = $\dfrac 52$

$BC = 4$

$OA^{\displaystyle \rightarrow} • BC^{\rightarrow} = |OA| |BC| \cos \theta = \dfrac 52 \times 4 \times \dfrac 45 = \boxed{8}$

Sorry, I dont know yet how to use vectors in Latex. If anyone knows, please do let me know.

@Niraj Sawant You have only proved this for the case when ABC is a right angled triangle..........!! What about the general case??? What if the sided length of BC was 3 or pi or e........????

Aaghaz Mahajan - 2 years, 6 months ago

Short and sweet...sour?

The answer is 8.

1 solution

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