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17 Nov 2020, 10:07
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D
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Difficulty:
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19% (02:11) correct
81%
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Nine points are placed on xy-plane as shown above. What is the maximum number of right angled triangles which can be formed by joining any three of the points?
A. 18
B. 36
C. 40
D. 44
E. 48
Are You Up For the Challenge: 700 Level Questions
Nine dots are placed on a piece of paper. Find the maximum number of right angled triangles which can be formed by joining any three dots.
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Untitled.png [ 7.24 KiB | Viewed 12310 times ]
Question is how many right angle triangles can be formed joining 'any three points" it doesn't have to be consecutive 3 points.
All points are equally spaced,
Let's see how many Square/ rectangles it can form
4, 2by2 squares ==> 1*4*4=16 right angled Triangles
1, 4by4 square ==> 1×(4+4) =8 (4@corner, 4@ center)
2, 4by2 rectangles==>2×(4+1) = 10 ( 4 at corners, one at each long sides of rectangle - triangles formed at the center point that we hv already consideredin 4by4 square)
2, 2by4 rectangles same as above rectangle ==> 10
So total right angled triangles are 44
IMO answer is D
Posted from my mobile device
All points are equally spaced,
Let's see how many Square/ rectangles it can form
4, 2by2 squares ==> 1*4*4=16 right angled Triangles
1, 4by4 square ==> 1×(4+4) =8 (4@corner, 4@ center)
2, 4by2 rectangles==>2×(4+1) = 10 ( 4 at corners, one at each long sides of rectangle - triangles formed at the center point that we hv already consideredin 4by4 square)
2, 2by4 rectangles same as above rectangle ==> 10
So total right angled triangles are 44
IMO answer is D
Posted from my mobile device
17 Nov 2020, 13:33
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We can start with \(9C3 = \frac{9*8*7 }{ 3!} = 3*4*7 = 84\) random options and eliminate the selections that do not form a right triangle. (See labeled graph below).
If we let AB be the base of a triangle, then selecting points F or I as the third point will not form a right triangle. We can do this for any of the small line segments, there are 6 horizontal and 6 vertical resulting in 12 of these bases. Thus 12*2 = 24 cases have been eliminated.
There are 6 lines (e.g. ABC) and two diagonals we cannot choose so 8 options are also gone.
The other scenario we haven't considered is the triangle ACH (can be rotated so that's 4 scenarios), and triangle DBI (can be rotated so again 4 scenarios).
Finally 84 - 24 - 8 - 4 - 4 = 44 total.
Ans: D
If we let AB be the base of a triangle, then selecting points F or I as the third point will not form a right triangle. We can do this for any of the small line segments, there are 6 horizontal and 6 vertical resulting in 12 of these bases. Thus 12*2 = 24 cases have been eliminated.
There are 6 lines (e.g. ABC) and two diagonals we cannot choose so 8 options are also gone.
The other scenario we haven't considered is the triangle ACH (can be rotated so that's 4 scenarios), and triangle DBI (can be rotated so again 4 scenarios).
Finally 84 - 24 - 8 - 4 - 4 = 44 total.
Ans: D
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LabeledGraph.png [ 12.1 KiB | Viewed 11605 times ]
