A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #42 #. What is the area of the triangle's incircle?

Question

A triangle has vertices A, B, and C.  Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/4  #, and the triangle's area is #42 #.  What is the area of the triangle's incircle?

Violet Aldrich · Accepted Answer

#text{incircle_area} = {2 pi mathcal{A} tan^2(A/2) tan^2(C/2)}/{ tan A(tan(A/2) + tan(C/2))^2} approx 19.6399#

The incircle is tangent to all the sides. The incenter, its center, is the meet of the angle bisectors.

When I have trouble getting started I pin my triangle to the Cartesian plane. #C(0,0), A(a,0), B(b,c)#.

Let's express our facts;

#A=\pi/8#

#B=pi/4#

#C = \pi-pi/8-pi/4={5pi}/8#

# tan A = c/a #

# tan C = c/b #

Let's call the area #mathcal{A=42}#.

# mathcal{A} = 1/2 a c #

That's three equations in three unknowns. We mostly care about #a#.

# 2 mathcal{A} = ac = a (a tan A)=a^2 tan A #

# a = \sqrt{ {2 mathcal{A} }/ tan A} #

Tangents are slopes. The perpendicular bisector of C has equation

#y = x tan (C/2)#

The perpendicular bisector of A has equation

#y = -(x-a) tan (A/2)#

Those meet at the incenter, when

#x tan(C/2) = -x tan(A/2) + a tan (A/2)#

#x = {a tan(A/2)}/{tan(A/2) + tan(C/2)}#

#y = {a tan(A/2) tan(C/2)}/{tan(A/2) + tan(C/2)}#

The incenter is #(x,y)# and the incircle is tangent to AC, the x axis, so the radius is #y# and the incircle's area is #pi y^2#.

Let's put everything together.

#text{incircle_area} = {2 pi mathcal{A} tan^2(A/2) tan^2(C/2)}/{ tan A(tan(A/2) + tan(C/2))^2}#

There's probably some simplification to be done here, but let's plug in the numbers to see if I've gone off the deep end. I'd guess from my rough sketch an area a little less than half, around 20.

Feeding it to Alpha gets an approxmate incircle area of #19.6399#, closer to my guess than I would have guessed. Hard to believe I didn't make a mistake getting this far, but let's proceed.

That's motivation to work out the exact answer but I'm getting length warnings, so let's stop here.

Avery Allen · Answer

undefined To find the area of the triangle's incircle, you can use the formula:

$$ \text{Area} = \pi \cdot r^2 $$

Where $ r $ is the radius of the incircle. The radius can be found using the formula:

$$ r = \frac{\text{Area of triangle}}{\text{Semiperimeter of triangle}} $$

$$ \text{Semiperimeter} = \frac{a + b + c}{2} $$

Given that the area of the triangle is 42 and the angles at vertices A and B are $ \frac{\pi}{8} $ and $ \frac{\pi}{4} $ respectively, you can use trigonometric relationships to find the side lengths of the triangle.

Once you have the side lengths, you can find the semiperimeter and then the radius of the incircle using the formulas mentioned above. Finally, use the formula for the area of the incircle to calculate its area.