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Multiple Choice Geometry Challenge Problem and Explanation

Try the following problem. And, remember, don’t give up. If you think a certain approach might work, try it. Giving up, however, will never get you the answer.

A twelve-sided polygon consists of vertices A – L. How many lines can be drawn between any two vertices, such that a line is neither repeated, nor redundant, with any sides of the polygon?

(A)  45

(B)  54

(C)  55

(D)  66

(E)  110


Find a Pattern

Start small. That’s right. Instead of a massive dodecahedron (the name of a twelve-sided polygon) ask yourself the same question, but with a square. A square will yield two such lines. And, a pentagon, a five-sided figure, will yield five such lines (you can quickly draw this out on a scratch paper). Finally, a hexagon will yield nine such lines.

At this point, you should notice that we are adding a number that was one greater than the number we added in the previous case. For the pentagon we added 3 to the total, for the hexagon, 4. For a seven-sided figure, we would therefore add another 5. For an octagon, 6.


Now, we have a sequence of numbers 2 + 3 + 4 + 5 + 6…. notice that the final number I am adding is always two less than the number of sides. Therefore, for a twelve-sided figure, I will have to continue the series all the way up to 10: 2 + 3 + 4… + 9 + 10 = 54. And there’s our answer: (B).


A Totally Different Method

This is a great, intuitive way to solve problems. Of course, it is not the most elegant way, but requires some brute force. But that’s okay – as long as you get the answer in a reasonable amount of time, that is the important part.

But, let’s think of this problem in a totally different way. I am choosing from twelve letters. Each line has two vertices; so, therefore, I am choosing two from 12. The order doesn’t matter – meaning that line AG is the same as line GA.

Now, we use the combinations approach. And, notice I say approach, and not formula, because the formula for combinations is needlessly cumbersome and potentially intimidating (watch my videos on Combinations/Permutations).

Using the combinations approach we get: (12)(11)/2 = 66. Now, we have to account for the redundancies. Meaning that there are lines which are already part of the polygon, i.e. we wouldn’t be drawing a line from A to B, because Line AB is already part of the polygon. How many lines are there in a twelve-sided figure? 12, of course.

So, 66 -12 = 54.  The answer.



Now, we can turn this into a formula, but I am not going to do so. The key to the GRE is not to cram a bunch of formulas into your head, most of which you’ll probably never see. The key is how to approach problems. In this case, we found a pattern, and, from that pattern, we were able to extrapolate an answer. Developing that skill is far more important than memorizing formulas.

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2 Responses to Multiple Choice Geometry Challenge Problem and Explanation

  1. Erin August 24, 2013 at 9:34 am #

    Hi Chris –

    How do you know which lines to count? For instance, if vertice a and d are diagonal, how do you know not to count lines that could connect by cutting across the shape? Maybe I’m totally off but that’s what threw me off on the question.

    • Karan July 31, 2014 at 9:56 am #

      Hey Erin, the reason we should not exclude intersecting lines from the count is because the question statement does not mention that constraint.

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