practice problems pending

Q1. (combinatorial geometry problem)

There is a 21-sided regular polygon with vertices (A1, A2 .... A21). Triangles are formed by joining these vertices.

a) How many of these triangles are acute-angled?

b) How many of these triangles are right-angled?

c) How many of these triangles are obtuse-angled?

d) How many of these triangles are equilateral?

e) How many of these triangles are isosceles?

S1.
Regular polygon can be inscribed in a circle. Each arc being of equal length.
For e.g. equilateral triangle will create 3 arcs of 120 degrees each.
So 21-gon will create 21 equal-sized arcs.
Total possible triangles = 21C3 = 21.20.19/6 = 19.70 = 1330

b) 0 right angle triangles. Why?
By Thales' theorem, diameter inscribes right angle on circumference.
And a diameter divides circle into 2 equal parts.
In an odd sided regular polygon, we cannot have equal number of arcs on both sides of the diameter.

c) By inscribed angle theorem, for an obtuse angle on circumference, the intercepted arc should be more than 180 (because inscribed angle is half of that).
So all the vertices have to be in one half of the circle.
So we start by vertex 1 and draw a diameter which will land between vertex 11 and 12.
This diameter will have 10 vertices on each side(excluding 1).
Let's say they are numbered 1,2,3... clockwise.

Now we will choose 2 vertices from 2,3...11 and join them with 1 to create an obtuse triangle.
10C2 = 45 ways.
Similarly for each vertex: 45*21 = 945 ways to create an obtuse triangle.

But why didn't we choose vertices from the other half containing 21,20...12?
That will create duplicates.
When we start with 12, that will also create 1,21,12.

How does our solution ensure duplicates don't occur?

If we look ahead only(clockwise), then
1 will join with 2,3...11
2 will join with 3,4...12 so it will never find 1.
3 will join with 4,5..13 so it will never find 1,2.
and so on..

a) acute angle triangles = total - right - obtuse = 1330 - 0- 945 = 385

d) for equilateral, equal spacing is required which is 7 here.
So triangles will be (1,8,15) (2,9,16) ... (7,14,21) and then it will repeat. So 7 is the answer.

e) For isosceles:
From 1 we can find 20/2 = 10 equidistant pairs like (2,21) (3,20) ...
Same for every other, so total: 21*10 = 210
But this will capture equilateral triangles also.
And each equilateral triangle will be counted thrice(one for each of its vertices).
So 210 - 7*3 = 189 if you want only isosceles and not equilateral, else 210-7*2 = 196.

 Q2. 

S2.
Total: 11!/6!5!
Routes via CD: 5!/2!3!*5!3!2!
Answer: [1] - [2]