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official gmac geometry problem

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official gmac geometry problem

by fangtray » Fri May 11, 2012 7:17 am
A rectangle is inscribed in a circle of radius r. If the rectangle is NOT a square, which of the following could be the perimeter of the rectangle?
a. 2r* root(3)
b. 2r* [root(3) + 1]
c. 4r*root(2)
d. 4r*root(3)
e. 4r*[root(3)+1]
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by Mike@Magoosh » Fri May 11, 2012 12:21 pm
Hi, there! I'm happy to help with this.

This is a tricky one. First of all, see the diagram I attached. I'm going to use the notation that a & b are the sides of the rectangle, and of course the diagonal is a diameter with length 2r. One crucial point is to recognize that the Pythagorean Theorem is satisfied by a, b, and 2r.

One way would be to go through a laborious calculate whether each possible perimeter produces an a & b pair that would satisfy a^2 + b^2 = (2r)^2. That's the long way.

Instead, notice, we have a bunch of sqrt(2) and sqrt(3) floating around, which suggests that special triangles are in play. Let's see what happens if we assume the a-b-2r triangle is a special triangle.

There are two special triangles
1) The Isosceles Right Triangle, 45-45-90, 1-1-sqrt(2)
2) The Half Equilateral Triangle, 30-60-90, 1-2-sqrt(3)

We can't have the a-b-2r triangle as a 45-45-90 triangle, because that would produce a square, which is prohibited in the question.

If the a-b-2r triangle is a 30-60-90 triangle, then the hypotenuse is 2r. The short leg (opposite the 30 degree angle) is half the length of the hypotenuse, so a = r. The longer leg (opposite the 60 degree angle) is sqrt(3) times the length of the short leg, so b = r*sqrt(3). (If you are not familiar with special triangles, that's definitely a topic to review before taking the GMAT!)

So the combination of a = r, b = r*sqrt(3), and hypotenuse = 2r automatically satisfies the Pythagorean theorem, because it's in the ratios of the 30-60-90 triangle. If half the rectangle were this triangle, then the perimeter would be

perimeter = 2a + 2b = 2r + 2sqrt(3)*r = 2r*(1 + sqrt(3))

This is choice B, the correct answer.

Basically, if there weren't a trick to solving this --- here the "trick" is using special triangles --- then the GMAT couldn't give you this questions, because ultimately to answer it in general you would need trigonometry and calculus, well beyond what you are expected to know for the GMAT. Also, I cannot underscore enough how important the special triangles are, in a wide variety of Quantitative questions.

FWIW, here's a blog on the Pythagorean theorem
https://magoosh.com/gmat/2012/the-pythag ... -the-gmat/

Does all this make sense? Please let me know if you any more questions.

Mike :)
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by fangtray » Fri May 11, 2012 3:16 pm
Mike@Magoosh wrote:Hi, there! I'm happy to help with this.

This is a tricky one. First of all, see the diagram I attached. I'm going to use the notation that a & b are the sides of the rectangle, and of course the diagonal is a diameter with length 2r. One crucial point is to recognize that the Pythagorean Theorem is satisfied by a, b, and 2r.

One way would be to go through a laborious calculate whether each possible perimeter produces an a & b pair that would satisfy a^2 + b^2 = (2r)^2. That's the long way.

Instead, notice, we have a bunch of sqrt(2) and sqrt(3) floating around, which suggests that special triangles are in play. Let's see what happens if we assume the a-b-2r triangle is a special triangle.

There are two special triangles
1) The Isosceles Right Triangle, 45-45-90, 1-1-sqrt(2)
2) The Half Equilateral Triangle, 30-60-90, 1-2-sqrt(3)

We can't have the a-b-2r triangle as a 45-45-90 triangle, because that would produce a square, which is prohibited in the question.

If the a-b-2r triangle is a 30-60-90 triangle, then the hypotenuse is 2r. The short leg (opposite the 30 degree angle) is half the length of the hypotenuse, so a = r. The longer leg (opposite the 60 degree angle) is sqrt(3) times the length of the short leg, so b = r*sqrt(3). (If you are not familiar with special triangles, that's definitely a topic to review before taking the GMAT!)

So the combination of a = r, b = r*sqrt(3), and hypotenuse = 2r automatically satisfies the Pythagorean theorem, because it's in the ratios of the 30-60-90 triangle. If half the rectangle were this triangle, then the perimeter would be

perimeter = 2a + 2b = 2r + 2sqrt(3)*r = 2r*(1 + sqrt(3))

This is choice B, the correct answer.

Basically, if there weren't a trick to solving this --- here the "trick" is using special triangles --- then the GMAT couldn't give you this questions, because ultimately to answer it in general you would need trigonometry and calculus, well beyond what you are expected to know for the GMAT. Also, I cannot underscore enough how important the special triangles are, in a wide variety of Quantitative questions.

FWIW, here's a blog on the Pythagorean theorem
https://magoosh.com/gmat/2012/the-pythag ... -the-gmat/

Does all this make sense? Please let me know if you any more questions.

Mike :)
the mike! i got the right answer when doing this problem, because i thought: what if this rectangle makes a 30/60/90 triangle and this is some rule that i did not learn, turns out it did, but it doesn't have to be a 30/60/90 triangle.. but i like the logic that you used in looking at the answer choices first and seeing what kind of answers we have.. but given that its not a special triangle, how would you do it?
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by Mike@Magoosh » Fri May 11, 2012 3:50 pm
fangtray wrote:i got the right answer when doing this problem, because i thought: what if this rectangle makes a 30/60/90 triangle and this is some rule that i did not learn, turns out it did, but it doesn't have to be a 30/60/90 triangle.. but i like the logic that you used in looking at the answer choices first and seeing what kind of answers we have.. but given that its not a special triangle, how would you do it?
Dear fangtray:
If the triangle is not a special triangle, then this problem becomes (a) considerably more time consuming, and (b) considerably more advanced, such that a calculator becomes necessary. For both those reasons, it would be something no longer in the scope of the GMAT.

The short answer is: in a circle, the rectangle with the largest perimeter is the square. The square has a diagonal of 2r, so each side of the square is r*sqrt(2). From that, we see the perimeter of the square would be 4r*sqrt(2). Therefore, any non-square rectangle in the circle would have to have a perimeter that is smaller than 4r*sqrt(2).

Knowing that fact but having no calculator poises some significant challenge. For example, from the above analysis, we know 2r*(1 + sqrt(3)) is a possible perimeter, so it follows that 2r*(1 + sqrt(3)) must be less than 4r*sqrt(2), but without a calculator, that's far from easy to verify. If the possible perimeters were expressed, for example, in decimal form, or in terms of pi, it would be essentially impossible to do without a calculator. Basically, we are in a realm in which only rare math-freak types can function without a calculator.

The special triangle trick is a quintessential GMAT trick. If we remove that, and look at the general case, we are really no longer anywhere near the GMAT realm.

Does all that make sense?

Mike :)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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