How would I go about solving for the grayed out area?

Question

I was asked to get the grayed out area that you see in this image.

All that I'm given to solve this is the three points of the main triangle (A, B, C) and the center of the circle (X) Also the dashed lines are parallel to the sides of the main triangle and tangent to the circle. All ways I can think of, for solving this problem seem over complicated, like: rotating a copy of the main triangle, like the dashed lines imply, and then computing the intersections between the sides of both triangles to get the vertices of the hexagon shaped area and then computing that area and subtracting it from the main triangle area.

Is there a better more intuitive way of solving this?

Thanks btw.

Answer 1

The construction of the dotted line triangle appears to be the $180^\circ$ rotation of the original triangle about the incenter, or equivalently the construction of lines parallel to each side and shifted by twice the radius of the incircle.

Then the required residual triangles are all scaled-down copies of the original triangle.

The easiest general process seems to be:

• Calculate the side lengths $|AB|,|BC|,|AC|$
• Calculate the semiperimeter, $s = (|AB|+|BC|+|AC|)/2$
• Calculate the area $A_T$ either using the coordinates (bounding rectangle - three triangles) or Heron's formula $A_T = \sqrt{\strut s(s-|AB|)(s-|BC|)(s-|AC|)}$
• Calculate the incircle radius using $r_I = \dfrac{A_T}{s}$
• Calculate the three altitudes $ h_A, h_B, h_C$ from the area and side lengths e.g $h_A = \dfrac{2A_T}{|BC|}$
• Calculate the scaled-down area of each grey section, eg. $A_A = A_T\left(\dfrac{h_A-2r_I}{h_A}\right)^2$
• Sum