Question

A triangle ABC has 2 points marked on the side BC, 5 points marked on the side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have atleast one vertex common with the triangle ABC?

• A. 256
• B. 237
• C. 207
• D. 127

Answer 1

The correct option is D

127

Total points including the vertices = 13
Out of these 13 points,
5 points on side AB ( including B and A) are collinear
7 points on side CA (including C and A) are collinear
4 points on side BC (including B and C) are collinear
thus, total triangles possible ${}^{13}$C${}_3$ - ${}^5$C${}_3$ - ${}^7$C${}_3$ - ${}^4$C${}_3$ = 237
Out of these,
we should exclude those triangles which do not share the vertices A or B or C
Such triangles = ${}^{13-3}$C${}_3$ - ${}^{5-2}$C${}_3$ - ${}^{7-2}$C${}_3$ - 0 = 109
We should also exclude the triangle ABC itself.
So, possible new triangles = 237 - 109 - 1 = 127.