Twelve points are spaced evenly around a circle, lettered from A to L.

Question

Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices.

What is the value of N?

Attachment:

Screenshot 2025-02-18 at 3.33.35 PM.png [ 46.39 KiB | Viewed 2348 times ]

Since if we chose a particular vertice e.g. A as apex of the triangle, we can create 5 isosceles triangles and there are 12 such vertices.
N <= 5*12 = 60

But equilateral triangles 12/3 = 4 are each counted thrice.
Therefore, we have to reduce 4*2 = 8 triangles from our calculation

N = 60 - 8 = 52

IMO B

mikemcgarry · Accepted Answer

Dear Uma, I'm happy to respond. :-)

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fskilnik · Answer

My solution is presented in the image attached.

chetan2u · Answer

Hi...

I will just tell a logical method ...

There are 12 points at EQUAL interval..
So the isosceles triangle can have EQUAL sides of length.
Let's work on the central vertex.

A as the vertex between EQUAL sides..
The EQUAL sides can be 1 unit, two unit, 3 unit, 4 unit or 5 unit...
6 unit will be the diameter so not possible.
So A as central vertex can make 5 isosceles triangles.

Thus 12 vertices from A to L can make 12*5=60 such isosceles triangles.
 
However 4 units results in an equilateral triangle.
So we are counting the same triangle THREE times. 
So actual different triangles with 4 unit as length = 12/3=4..
So we have 12-4=8 extra triangles counted..
 
Total = 60-8=52...

B

umabharatigudipalli · Answer

Hi, 60 Isosceles triangles make sense, but would u help me out why did u subtract 8?

Thanks in advance, Uma

fskilnik · Answer

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It´s not (just a matter of) business (although I really "earn my living" teaching quant GMAT since 2000): I believe in the bold statement above. (Yes, bold in the two senses, LoL.)

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Fabio.

theoryguy · Answer

I solved this problem a little differently.

I remembered from another problem that the total number of equilateral triangles using evenly spaced points in a circle = (# of points on the circle)/3, which equals 4.

I then subtracted 4 from every answer and choose B because the remaining value, 48, was the only answer divisible by 3.

Kinshook · Answer

Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices.

What is the value of N?

Attachment:

Screenshot 2025-02-18 at 3.33.35 PM.png [ 46.39 KiB | Viewed 2348 times ]

Since if we chose a particular vertice e.g. A as apex of the triangle, we can create 5 isosceles triangles and there are 12 such vertices.
N <= 5*12 = 60

But equilateral triangles 12/3 = 4 are each counted thrice.
Therefore, we have to reduce 4*2 = 8 triangles from our calculation

N = 60 - 8 = 52

IMO B

Twelve points are spaced evenly around a circle, lettered from A to L.

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Twelve points are spaced evenly around a circle, lettered from A to L.

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