Fifteen dots are evenly spaced on the circumference of a cir

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ktrout2020: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Wed Oct 30, 2019 1:10 pm

by ktrout2020 » Sun Nov 03, 2019 10:43 am

Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Sun Nov 03, 2019 10:43 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Mon Nov 04, 2019 1:30 am

ktrout2020 wrote:Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640

Source: Magoosh

Let's first calculate the total no. of possible triangles (these include equilaterals too).

To make a triangle, we have to choose 3 dots out of 15. Thus, total no. of possible triangles = 15C3 = (15.14.13) / (1.2.3) = 455

Note that an equilateral triangle would be formed connecting 3 equally spaced dots. Let's count them.

For an equilateral triangle, out of 15 dots, there would be 15 - 3 = 12 dots that will not be connected. And 12/3 = 4 dots would be between any two connecting dots.

Taking 1st connecting dot, the second connecting dot would be 1 + 4 + 1 = 6th; similarly, the third connecting dot would be 6 + 4 + 1 = 11th

So, the following connecting dots will form equilateral triangles:

"¢ (1, 6, 11);
"¢ (2, 7, 12);
"¢ (3, 8, 13);
"¢ (4, 9, 14);
"¢ (5, 10, 15);

There are 5 equilateral triangles.

Thus, there are 455 - 5 = 450 triangles that are not equilateral triangles.

The correct answer: B

Hope this helps!

-Jay
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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 04, 2019 3:53 am

ktrout2020 wrote:Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640

Good = Total - Bad

Total:
From 15 points, the number of ways to choose 3 = 15C3 = (15*14*13)/(3*2*1) = 5*7*13

Bad:
Here, a bad combination can be used to form an equilateral triangle.
Let the 15 points on the circumference be divided into 3 equally spaced groups:
A-B-C-D-E
F-G-H-I-J
K-L-M-N-O
Equilateral triangles can be formed as follows:
A-F-K (connecting the first point in each group)
B-G-L (connecting the second point in each group)
C-H-M (connecting the third point in each group)
D-I-N (connecting the fourth point in each group)
E-J-O (connecting the last point in each group)
5 ways

Good:
Total - Bad = (5*7*13) - 5 = 5(7*13 - 1) = 5*90 = 450

The correct answer is B.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Nov 04, 2019 5:36 am

ktrout2020 wrote:Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640

Source: Magoosh

KEY CONCEPT
If we connect ANY 3 dots, we'll get a unique triangle.
Since the order in which we select the dots does not matter, we can use COMBINATIONS.
We can select 3 dots from 15 dots in 15C3 ways
15C3 = (15)(14)(12)/(3)(2)(1) = 455

Now SOME of these 455 triangles will be equilateral triangles, so we must subtract from 455 the number of those triangles that are equilateral triangles.

IMPORTANT: Since the correct answer must be less than 455, we can ELIMINATE answer choices C, D and E

At this point, we COULD determine the number of equilateral triangles that are included among the 455 triangles we've counted (the above posters have already done so).
However, we could use the answer choices to our advantage.

Answer choice A (160) suggests that there are 295 equilateral triangles among the 455 triangles we've counted (since 455 - 295 = 160)
Answer choice B (450) suggests that there are 5 equilateral triangles among the 455 triangles we've counted (since 455 - 5 = 450)

If answer A is correct, then more than half of the 455 triangles are equilateral triangles. This doesn't seem right since MOST selections of 3 points will NOT yield an equilateral triangle.
So, ELIMINATE A

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8139; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Nov 06, 2019 7:14 pm

ktrout2020 wrote:Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640

Source: Magoosh

The total number of triangles that can be formed (regardless of whether they are equilateral) is 15C3 = (15 x 14 x 13)/(3 x 2) = 5 x 7 x 13 = 455. Now let's determine the number of triangles formed that are equilateral triangles.

Let A, B, C, D, E, F, G, H, I, J, K, L, M, N, and O be the 15 points that are evenly spaced on the circumference of the circle. Every pair of consecutive points (for example, A and B, G and H, etc.) form a 360/15 = 24-degree arc. In order for 3 points to form an equilateral triangle, they must evenly spaced among themselves on the circumference of the circle also. That is, each pair of the three points have to be 360/3 = 120 degrees apart. Since 120/24 = 5, the points should be 5 spaces apart from one another. For example, if A is one of the vertices of the equilateral triangle, then the next one should be F and the last one should be K. In other words, triangle AFK is an equilateral triangle. Using the same analogy, triangles BGL, CHM, DIN, and EJO are equilateral triangles and only these 5 (including triangle AFK) are equilateral triangles.

Since 455 triangles can be formed and 5 of these are equilateral, then the number of triangles that are not equilateral is 455 - 5 = 450.

Answer: B

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