Thread: Circle inscribed of triangle

The circle O is an inscribed circle of $\displaystyle \Delta $ABC and points P, Q and R are the points of tangency of sides BC, CA and AB respectively.
AB = AC = 13, BC = 10.

The scalar product of two vectors $\displaystyle \vec{AB}.\vec{AO}, \vec{AB}.\vec{BC}$?

Any tips?


Look at the picture.

Originally Posted by provasanteriores

The circle O is an inscribed circle of $\displaystyle \Delta $ABC and points P, Q and R are the points of tangency of sides BC, CA and AB respectively.
AB = AC = 13, BC = 10.
The scalar product of two vectors $\displaystyle \vec{AB}.\vec{AO}, \vec{AB}.\vec{BC}$?
Any tips? Look at the picture.

From the given, it follows that $|BP|=|PC|=5~$ or $\Delta BAC$ is isosceles.

$\vec{u} \cdot \vec{v} = |u| \, |v| \, \cos{\theta}$, where $\theta$ is the angle between $\vec{u} \text{ and } \vec{v}$

note triangles BPA and CPA are 5-12-13 right triangles ... circle radius is rather simple to determine.

Originally Posted by skeeter

$\vec{u} \cdot \vec{v} = |u| \, |v| \, \cos{\theta}$, where $\theta$ is the angle between $\vec{u} \text{ and } \vec{v}$

note triangles BPA and CPA are 5-12-13 right triangles ... circle radius is rather simple to determine.

thanks, I got it now

I found 104 e 50. The answers are 104 e -50 (why minus?)

$\displaystyle \vec{AB}= -\vec{BA}$?

Originally Posted by Plato

From the given, it follows that $|BP|=|PC|=5~$ or $\Delta BAC$ is isosceles.

I saw, thanks

$\vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\pi - \angle{ABC}) = 13 \cdot 10 \cdot \left(-\dfrac{5}{13}\right) = -50$

that's why

Originally Posted by skeeter

$\vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\pi - \angle{ABC}) = 13 \cdot 10 \cdot \left(-\dfrac{5}{13}\right) = -50$

that's why

why pi?


$\displaystyle \vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\angle{ABC}) = 13 \cdot 10 \cdot \left(\dfrac{5}{13}\right) = 50$

[QUOTE=provasanteriores;915999]why pi?/QUOTE]

see diagram ...