Question

How do you solve the top two problems ?

Answer 1

1. It doesn't matter what the absolute sizes are; the probability is the ratio of the area of the circle to the area of the triangle.

Let's make it the unit circle, radius 1 (so apothem 1 for the triangle) so area $\pi$.

We can get the area of the triangle a couple of different ways; we'll compare.

Let's call the side of the equilateral triangle s and its height h.

Since the incenter is the centroid for an equilateral triangle, the height is three times the apothem, so h=3. The sides of the right triangle made by the altitude are s/2 and s, so

$(s/2)^2 + h^2=s^2$

$h^2 = \frac 3 4 s^2$

$s^2 = \frac 4 3 h^2$

$A =\dfrac{\sqrt{3} s^2}{4} = \dfrac{\sqrt{3} h^2}{3}$

We have h=3 so the triangle area is $3 \sqrt{3}$ and the probability we seek is

Answer: $p = \dfrac{\pi}{3 \sqrt{3}}$

That's approximately 0.605

2. P(not in circle | in triangle) = 1 - P(in circle | in triangle)

$p = 1 - \dfrac{\pi}{3 \sqrt{3}} = 1 - \pi \sqrt{3}{9}= \dfrac{9 - \pi \sqrt{3}}{9}$

Answer: $\dfrac{9 - \pi \sqrt{3}}{9}$

That's approximately 0.395

Edit: I forgot to do the area another way. I'll add that.

If we connect the tangent points of the incircle, by symmetry we get four congruent equilateral triangles. Any of the three chords makes an isosceles triangle with two radii and included angle 120 degrees. By the law of cosines, the chord length, which is the side of the small equilateral triangle, is

$s^2 = 1^2 + 1^2 - 2 (1)(1) \cos 120^\circ = 2 - (2)(-1/2) = 3$

so a total triangle area of

$4 \times \dfrac{\sqrt{3} s^2}{4}} = 3 \sqrt{3}$

which agrees with the previous calculation.