Nine points are placed on xy-plane as shown above. What is the maximum

Question

https://gmatclub.com/forum/download/file.php?id=58982 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. Untitled.png

aamani02 · Answer

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

TestPrepUnlimited · Answer

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

hkkat · Answer

D
big triangles 4
medium triangles 4*2*2 = 16
small triangles 8*3 =24

24 + 16 + 4 = 44//

abhik1502 · Answer

will someone pls elaborate the solution ?

KarishmaB · Answer

Look at the symmetry of the nine points. To simplify our work, note that we have 3 types of points. Let's find the number of triangles we can make such that the right angle lies at each of these points.
Referring the figure above given by  TestPrepUnlimited

1. Points at the vertices (A, C, G, I) of the square
Here, for each point, we can make 4 right triangles (for A, these will be ABD, ACD, ADC and AGC)
So we get 4 * 4 = 16 triangles

2. Points on the sides of the squares (B, D, H, F)
For each point, we can make 5 triangles right angled at that point. e.g. for B, these will be BAE, BCE, ABH, CBH and DBF (because BDFH is also a square)
So we get 4 * 5 = 20 triangles

3. Point at the centre (E)
4 triangles we can see straight away such as BDE etc.
The other 4 will be AEC, CEI etc
Total we will get 8 right triangles at E.

Total = 16 + 20 + 8 = 44 triangles

Regor60 · Answer

Nobody has a problem with most of these triangles joining not just the three points at the vertices but also their intervening points ?

Posted from my mobile device

Budhaditya_Saha · Answer

This solution is extensive,
But this requires a lot of time to analyse.
How much time, do you think,, this question should take?

Nine points are placed on xy-plane as shown above. What is the maximum

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Nine points are placed on xy-plane as shown above. What is the maximum

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards