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# A pentagon with 5 sides of equal length and 5 interior angles of equal

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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I agree with how Ian is looking at this – definitely prefer this approach over analyzing arc lengths for statement (1) and solving the quadratic equation $$d^{2} – ds – s^{2} = 0$$ with the quadratic formula for statement (2), as the official solution suggests.

Here are a few more thoughts on how to get through this beast!

Statement (1) pins down the radius of the circle, so there’s only one way to draw the pentagon, and the perimeter will be definitively greater than 26cm or not. Thankfully, we don’t have to determine whether it is or isn’t – statement (1) is sufficient.

Statement (2) demands that we determine the relationship between the side s and diagonal d in a regular pentagon. This will help us figure out what d < 8cm tells us about s. That’s driving the analysis. Of course, to figure out whether the perimeter is > 26cm, we need to figure out whether s > 5.2.

First, this definitely requires some good sketching! In the figure below, we label what we can to piece together a relationship between d and s. Here’s how you can determine the angles in the diagram.

1. There are (5 – 2) x 180˚ = 540˚ in any pentagon. In this case, since all interior angles are equal, each interior angle = 108˚.
2. Any central angle formed by connecting adjacent vertices of the pentagon to the center of the circle must be 1/5 of 360˚ = 72˚.
3. Each of the angles labeled 36˚ is an inscribed angle, and therefore equal to half of the corresponding 72˚ central angle.
4. Of course, we also need to use the fact that triangles have a total of 180˚, and opposite angles are equal.

Next, because the angles in triangle 1 match the angles in triangle 2, these two triangles are similar, and because the ratio of their sides is 1:1, these two triangles are congruent. Because the angles in triangle 3 match the angles in triangles 1 and 2, triangle 3 is similar to triangles 1 and 2. As a result, corresponding sides are in a constant ratio, and we get

$$\frac{d}{s} = \frac{s}{d – s}$$

This yields

$$s^{2} = d(d – s) = d^{2} – ds$$

Since we’re trying to figure out what d < 8 tells us about s, we can set that up as:

$$s^{2} = d^{2} – ds < 8^{2} – 8s$$

So the following must be true

$$s^{2} < 8^{2} – 8s$$

Re-arranging, the following must be true

$$s^{2} + 8s < 64$$

Since we’re trying to see whether s > 5.2cm, we can test a case near the borderline (s = 5cm) to get some insight.

If we assume s = 5, we get 25 + 40 < 64, which is a contradiction. Therefore, the assumption that s = 5 must be false.

Since s must be positive, $$s^{2} + 8s$$ strictly increases as s increases. Therefore, any s > 5 will also yield a contradiction. Therefore s is definitively NOT greater than 5.2, and statement (2) is sufficient.

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*** The following solution has been revised thanks to Ian's comment below. Thanks, Ian.

The Alternative approach to this question - since we know very little about a regular pentagon - is to try using specific numbers and more familiar shapes to assist us.
Statement (1): we can use the formula of circular area and find that the radius is 4. Now, if it were a regular HEXAGON inscribed in the circle, its side would have been equal to the radius (since it can be divided into 6 equilateral triangles), and thus its perimeter would have been 6x4=24. But since the more sides there are to a regular polygon inscribed in a circle, the larger its perimeter, then the perimeter of the pentagon must be less than 24. That's enough. Answer choices (B), (C) and (E) are eliminated.
Statement (2): As the attached figure shows, the green and the red triangles created by the diagonals are identical (that's because the pentagon's 108 angle is divided into three equal angles - opposite the same arc - of 36 degrees, so all of their angles are equal; and their largest angle is opposite the same side - the diagonal). Thus, since if the pentagon's perimeter was 26 each side would have been just a bit above 5, we'll mark all sides as 5 to see whether this is possible. Also, since the statement relates to the diagonal as less than 8, we'll try 8. If that's true, the sides of the blue triangle are 8 - 5 = 3. Now, how do we know if that's possible? The green and the blue triangles are similar triangles (having 2 identical angles of 36 degrees), so
3/5 = 5/8 We'll multiply by 5 & 8:
24 = 25
This is wrong, of course, and since the diagonal is smaller than 8 and the side is larger than 5, the left side of the equation is even smaller than 24. Thus, in order to create a correct equation, the side of the pentagon must be smaller than 5, and so its perimeter is smaller than 5x5=25 >>> smaller than 26.
So statement (2) also gives us enough information, and the correct answer is (D).
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GC_circumscribed_2.png [ 46.52 KiB | Viewed 102384 times ]

Originally posted by DavidTutorexamPAL on 26 Apr 2019, 14:42.
Last edited by DavidTutorexamPAL on 06 May 2019, 00:05, edited 3 times in total.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
DavidTutorexamPAL wrote:
Statement (2): If it were a SQUARE inscribed in the circle, the diagonal would have been (root2) times the side, so with a diagonal of 8 the side would have been 8/(root2), which is 4(root2) >>> 4x1.3 >>> 5.2. But since the more sides there are to a regular polygon inscribed in a circle, the smaller its side, then the side of the pentagon must be less than 5.2, and thus its perimeter must be less than 5x5.2,which is less than 26. That's enough. Answer choice (A) is also eliminated.

I mentioned in my post above that this solution isn't correct (or if it is, I have not correctly understood what you're trying to say), and in case it's confusing for test takers, I think I should explain. For one thing, I think there's a logical issue with your inequalities. If s is the side of the square, and p the side of the pentagon, you seem to have concluded from these two inequalities:

s > 5.2
s > p

that 5.2 > p is true. There's no way from those two inequalities to compare p and 5.2.

Nor is it even the case that we can directly compare the square and pentagon you're describing. It's true that if you inscribe a square and a pentagon in the same circle, the square will have longer sides. But we can't inscribe them in the same circle here, if the square and pentagon both have a diagonal of 8. The pentagon is too big to fit in the circle you'd use for the square.

There are other ways to prove quickly in this question that p < 4√2 (and maybe you have and I haven't followed) but that isn't enough to answer the question if we genuinely need to prove that the perimeter is less than 26, which is why I had to resort to similar triangles in my post above.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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dng992 wrote:
your sentences dont even make sense, why would i subscribe to your exam service

"But since the more sides there are to a regular polygon inscribed in a circle,
the smaller its perimeter, [ok so youre saying more sides smaller perimeter]
then the perimeter of the pentagon must be less than 24. That's enough" [pentagon has fewer sides then hexagon, less sides, larger perimeter]

triggers students looking for solutions and seeing this

This is a typo, it should say 'larger' and not smaller.
Thanks for pointing this out, fixed!

Note also that a potentially shorter way to address stmt (2) is to notice that drawing a height in the 'green triangle' in my answer above (or the triangle labeled '1' in Ian's answer) divides this triangle into two 36-54-90 triangles. This is very nearly a 30-60-90 triangle and so we have a decent approximation for the length of the diagonal. In particular, if the perimeter were exactly 26, each side of the pentagon would be 5.2, meaning that the hypotenuse of our 36-54-90 triangle is also 5.2. Then half of the diagonal, which is the side in front of the 54-degree angle, is a bit shorter than sqrt(3)*5.2/2, and doing the math gives that the diagonal must be a bit shorter than 9 (more precisely, a bit shorter than 8.84). Since (2) tells us the diagonal is shorter than 8, and since 8 is quite a ways off from our approximation of 9, then the perimeter must be shorter than 26, which is sufficient.
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First of all, I want to thank all of you guys for the detailed answer but I feel that the method used for the second statement would not be suitable during the exam.

To answer the question we need the length of the Pentagon's sides, therefore we do not need the exact solution but just to know that given the information provided we would be able to determine that length.
I would state that the second statement is sufficient just by applying the following rule: you can determine each of the measures of triangle 1 (referring to Ian sketch) if you know the length of one side (given by statement two) and the width of two angles (which can be derived given the fact that it is a regular pentagon).
Now we know that is possible to determine the sides of the pentagon and therefore answering the question, so statement 2 is SUFFICIENT
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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Is there a good solution as to why statement 2 is sufficient? Most of the solutions for St 2 is difficult to implement in test-environment.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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If you are good at trigonometry, you can use this approach

Let side of the pentagon is x and diagonal 8cm
then x=4/cos(36)
Now, if we know value of cos 36, we can find x very easily.

sin(108) = sin(72+36)
Sin(pi-72)=sin72 co36 + cos72 sin36 {sin(A+B)= sinA cosB + cosA sinB)
sin72= 2sin36 cos36 cos36 + (2cos^2(36) -1) sin36 [sin2A= 2sinAcosA and cos2A= 2cos^2 (A) - 1]
2sin36cos36 = sin36 {2cos^2(36)+2cos^2(36) -1}
2cos36= 4cos^2(36) -1
4cos^2(36) - 2cos36 -1 =0
Let cos36=y
4y^2 - 2y -1 = 0
we know cos36 is positive, hence we will consider only positive root of this equation.
By solving above equation, we will get
cos36= 0.8 nearly
Hence x= 4/0.8 =5
Now we know that diagonal is less than 8, hence length of any side of pentagon is less than 5.
Perimeter will be less than 25

OR

There is one more approach, though it's not accurate one, to find approximate value of cos36
We know cos 30 is 3^{1/2}/3 = 0.866
and cos 45= 1/(2)^1/2 = 0.7
Use interpolation to find the cos 36, and you will get value equals to 0.8

I know cosine curve in not linear, hence interpolation technique is not a good approach, but it gives nearly exact value in this case.
If you're running out of time in exam, you can use it. Otherwise i won't recommend it.

souvikgmat1990 wrote:
Is there a good solution as to why statement 2 is sufficient? Most of the solutions for St 2 is difficult to implement in test-environment.

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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sebastianm wrote:
First of all, I want to thank all of you guys for the detailed answer but I feel that the method used for the second statement would not be suitable during the exam.

To answer the question we need the length of the Pentagon's sides, therefore we do not need the exact solution but just to know that given the information provided we would be able to determine that length.
I would state that the second statement is sufficient just by applying the following rule: you can determine each of the measures of triangle 1 (referring to Ian sketch) if you know the length of one side (given by statement two) and the width of two angles (which can be derived given the fact that it is a regular pentagon).
Now we know that is possible to determine the sides of the pentagon and therefore answering the question, so statement 2 is SUFFICIENT

Hey @sebastinam,
Unfortunately this doesn't work, as (2) gives you a range of potential diagonal lengths and not a specific length. Then you have a range of potential pentagon side lengths, and without knowing its lower bound cannot know if the perimeter is larger than 26 or not. So you need to find the lower bound by using one of the methods of calculation offered above.

Best,
David
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Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

The real question is whether the side length of this regular pentagon is greater than 5.2.

(1) This tells us all we could ever hope to know about the circle, and therefore the regular pentagon therein inscribed. Sufficient.

(2) The ratio of a regular pentagon's diagonal to its side length is equal to the Golden ratio, which is quite close to 1.6 = 8/5. If the diagonal is less than 8, then the side length is less than 5. Sufficient.

Easy, next.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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shobhitkh wrote:
Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

Hi Bunuel and VeritasKarishma, Can anyone of you please provide a good explanation for this question? Thanks

Here is the thing about this question and why I took time to get back - It is an OG problem so I expected a good, quick solution. I did invest a bit of time in this but unfortunately couldn't come up with anything satisfactory. Then I checked with another expert but still no luck.
The fastest way was trigonometry though I had to approximate the value of cos 36 about which I wasn't happy at all.
The better solution is the one Ian suggested above (using similar triangles). It just seems a bit too elaborate as the only solution to an official problem.
Also, I certainly cannot rely on knowing some obscure fact about pentagon to solve an official question.
All in all, I hope this was an experimental question that was later discarded, but found its way into the OG. In case we find some more such questions, we might need to reassess our understanding of the official problems (but I will wait for @Bunuel's input before making my mind about it)
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Here is the thing about this question and why I took time to get back - It is an OG problem so I expected a good, quick solution. I did invest a bit of time in this but unfortunately couldn't come up with anything satisfactory. Then I checked with another expert but still no luck.
The fastest way was trigonometry though I had to approximate the value of cos 36 about which I wasn't happy at all.
The better solution is the one Ian suggested above (using similar triangles). It just seems a bit too elaborate as the only solution to an official problem.
Also, I certainly cannot rely on knowing some obscure fact about pentagon to solve an official question.

For that very reason, I wondered if the question contained a typo, and meant to ask about a perimeter of 28 (instead of 26). Then you could do the problem easily, just using estimates with 30-60-90 and 45-45-90 triangles. I haven't seen OG2020 yet, but I gather from a post above that the official solution uses the quadratic formula, so they don't seem to have a fast solution for this either. The fastest way, as someone pointed out above, is to use the 'golden ratio', but I didn't post that solution because there's no way test takers could be expected to know anything about 'the golden ratio' (and since this is the only GMAT question you'll ever see where it would be useful, don't bother learning about it).

It is important to understand why Statement 1 is sufficient alone - a few people have asked about it since the first solutions were posted. From Statement 1, we know we have a circle with radius 4. There is only one size of regular pentagon you could fit in that circle, with the corners of the pentagon precisely on the circumference of the circle. Since only one size of pentagon can fit in the circle, and since we know exactly how big the circle is, there must be some way to work out how big that pentagon is too. We don't care what that method is (even if that method were just "take a ruler and measure it", that would be fine), because it's a DS question, and we don't actually need to answer the question; we only need to know that the question can be answered.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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Essentially the question is to test if given statements are sufficient enough to find the side of the pentagon.

Do you have sufficient data to determine the side of the pentagon? The perimeter can be >26cm(YES) and for all other cases, NO. In each case you have a definite answer. There is no ambiguity.

Statement1: area of circle is 16pi sq.cm. So radius of the circle is 4cm. Since the pentagon is cyclic, the distance between vortex of pentagon and center of the circle is 4cm. As all sides of pentagon are of equal length, each side forms an isosceles triangle with center as the 3rd vortex (subtends 72 degrees, i.e., =360/5). Hence, length (unique not multiple) of the side of the pentagon can be determined. Doesn't matter if it is >26cm or <26cm or =26cm.
Therefore Statement1 is sufficient.

Statement2: All diagonals are of equal length, because pentagon is cyclic and each side is of equal length. Since two diagonals (8cm each) subtend 36 degrees (angle subtended by a chord in the major arc is half the angle subtended by the chord at the center), one can determine the side (unique not multiple) of the pentagon. Hence this statement too is sufficient. Doesn't matter if it is >26cm or <26cm or =26cm.

IMO no complex math or trigonometry is required to solve this question.
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My approach for Solution for Statement 2:

Given: Diagonal of pentagon is less than 8 centimetres. We can safely say that the diagonal of the circle will be greater than the diagonal of pentagon. Let's take the smallest such possible value for diagonal of circle, ie. 8 cm. Hence the perimeter of circle is 8x3.14 = 25 ish. Since pentagon is inscribed inside the circle, its perimeter will be lesser than that of the circle. Hence the perimeter of the pentagon will be lesser than 26.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
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Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

Please find the diagram attached herewith.
Lets see why option B is also sufficient.
Triangle ABC and Triangle BCF and Trinagle FDE are 36 , 36 , 108 degree triangle.
( I explained the reason in the diagram attached )
So I am keeping it brief here.
Angle BAC = 108 degree ( interior angle of a pentagon)
Since AB = AC (sides of a regular pentagon are equal ) , Angle ABC = angle ACB = (180 - 108) / 2 = 72 / 2 = 36 degree
ANgle BDE = 108 degree ( interior angle of a pentagon)
Angle DBE = Angle DFB = (180 degree - angle BDE ) / 2 = (180 - 108 )/ 2 = 72 / 2 = 36 degree
Angle CBE = 108 - ( angle ABC + angle DBE) = 108 - 72 = 36 degree
Similarly , Angle BCD = 36 degree
ANgle BFC = 180 - ( CBE + BCD) = 180 - ( 36 + 36 ) = 108 degree

Say , The length of each diagonal = c
So BE = c

Now Triangle ABC and traingle BCF are congruent .( because they have two angles equal and one side common So ASA condition satisifes)
ANgle ABC = ANgle CBE = 36 degree
Angle ACB = angle BCF = 36 degree
SIde BC common
So Triangle ABC and traingle BCF are congruent.
SO the corresponding sides AB = BF
and AC = CF

Say AB = a
So , BF = a
Now , FE = BE - BF = c - a

SImilarly FD = c - a
DE = a

Now Triangles ABC and triangle FDE are similar. ( All the angles are same for them...36 , 36 , 108 )
So corresponding sides are of same proportion...
So , c/a = a / c-a
so c^2 - a*c = a^2
For the maximum length of the diagonal i.e 8 cm,

64 - 8*a = a^2
Now Say , side of the pentagon = 5 cm
it does not satisfy the equation above as (64 - 8 * 5 ) is smaller than 5^2 = 25 cm

So 'a' has to a bit smaller than 5 cm.
Now if the diagonal is less than 8 cm , then it will be even less than 64 . And side i.e the value of 'a' will be even smaller.
Even if the side was 5 cm , the perimeter of the pentagon would have been 5 *5 = 25 cm , which is smaller than 26 cm.
If the sides are even smaller , the perimeter will be even smaller and hence less than 26 cm.

Hence , Option B is also sufficient.

Please give me KUDO s if you liked my explanation.
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1. R= 4 . Circumference of circle is 25.12. so obviously Pentagon perimeter will be less than that as it's inscribed in circle.

2. Diagonal length less than 8 cms . What is intended here is : let's take the largest diagonal whose length is approx. 8 cm we can some how still find the lengths of sides using sine rule --> a/sinA = b/sinB=c/sinC.. as it forms a iscosceles triangle with 108-36-36 --> a=b=a and c=7.999~=8 (say).

So as it's a data sufficiency question without solving overly we can quickly tick off this one too..

So Choose D .

But need to get this thought under 2 minutes is quite challenging task but somehow I would mark D if I am on this question on exam day without wasting time as I know I can't think so much in 2 mins. I would be thinking why specifically 8 and imagine that (here all diagonals of Pentagon are chords of circle) chord length will always be less than diameter length and assume the diameter to be 8 and radius to be 4 i.e., the same as 1st option.

I DON'T MIND TO GET A KUDOS.

Posted from my mobile device
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal [#permalink]
Hi experts Bunuel IanStewart VeritasKarishma DavidTutorexamPAL SimplyBrilliant chetan2u ,

A quick thought about statement 2.

Can't we just solve it using the triangle property that 3rd side of a triangle has to be greater than the difference and smaller than the sum of other two sides ?

I mean that if we take a diagonal here which has 2 sides of pentagon as the other two of its sides to be the two sides of a triangle (3rd being the diagonal itself), we know that diagonal (lets say 7.999) has to be greater than 0 and less than 2 times the side of pentagon which means the maximum the side of a pentagon can be is ~4 (half of 8 or 7.99999) which means that the perimeter is going to be less than 20 and hence sufficient.
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