# Let H be a regular hexagon with an area of 360. Three distinct vertices X, Y, and Z, are picked randomly, with all possible triples of distinct vertices equally likely. Let A, B, and C be the unpicked vertices. What is the expected value (average value) of the area of the intersection of triangle ABC and triangle XYZ?

This question consists of two different parts: combinatorial and geometrical (determining areas). As Educators on eNotes are only supposed to answer one question at a time, I'll solve the combinatorial part here.

There are three different possibilities to choose three points from the geometrical point of view. All three points may be consecutive, or only two may be consecutive, or all three may be nonconsecutive. Please look at the attached picture.

For the first option, the intersection of the corresponding triangles is empty, so it makes zero contribution into the expected value.

For the second option, the intersection is a quadrilateral: denote its area as `A_4 . ` For the third option, the intersection is a hexagon: denote its area as `A_6 .`

There are `( ( 6 ) , ( 3 ) ) = ( 6 * 5 * 4 ) / ( 3 * 2 * 1 ) = 20 ` ways to select three distinct points from 6. 6 from them correspond to the option 1 (it is necessary and sufficient to choose the central vertex of a triangle), and only two to the option 3. Because of this, or independently, twelve correspond to the option 2.

This way, the expected value is `360 * 1/20 (0 + 12 A_4 + 2 A_6) = 36 (6 A_4 + A_6 ) .`