In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which of the

Divide the pentagon into triangles:

From triangle PQR, PQ+QR > PR,
or, PR < 5 ---(1)

From triangle PRS, PR + RS > PS
or, PR + 4 > PS ---(2)
from (1) and (2), PS < 9 ---(3)

From triangle PST, PS + ST > PT
or, PS + 5 > PT ---(4)
from (3) and (4), PT < 14

So, PT can not be 15. So, C is the answer.

I followed the same logic and eliminated 15. My concern was about 5 and 10, the official solution contains quite a long and detailed explanation but should I bother? Is there any need in a question like that and dig further - proving that 10 or 5 can be the answer?

Hi Erjan,

Please find a picture attached for easy understanding.

Divide the pentagon PQRST into three triangles- triangle PQR, triangle PRS and triangle PST.

The range for the lengths of PR, PS and PT comes from one of the properties of triangles that the length of the third side will be less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.

I hope this helps.

Aiena.

Solution:

This question can be answered by using the triangle inequality, which states that the length of the third side of a triangle is less than the sum of the lengths of the other two sides.

We can create 3 triangles in the pentagon: PQR, RST, and PRT.

For triangle PQR, two side lengths are 3 and 2, so the third side length (PR) must be less than 5.

For triangle RST, two side lengths are 4 and 5, so the third side length (RT) must be less than 9.

For triangle PRT, we know that the length of PR must be less than 5 and the length of RT must be less than 9. Thus, the length of the third side, PT, must be less than 14.

Of the possible lengths of PT, which are 5, 10, and 15, we see that only 5 and 10 meet the criteria of the triangle inequality.

Answer: C

Posted from my mobile device

Hello ScottTargetTestPrep,
I was wondering whether you could clarify how did you find the minimum values, in your solution you mention that PT must be less than 14, but how do we know whether we could accept both 5 and 10, also you have mentioned that the length of PR must be less than 5 and the length of RT must be less than 9 so basically aren't we talking about 4+8=12 so shouldn't PT be less than 12?

We can actually use the triangle inequality to obtain lower bounds for the length of a side using the other two side lengths. If a, b and c are the side lengths of a triangle, then the triangle inequality tells us that c < a + b must be true. That's the upper bound. Now, let's use the triangle inequality for sides a and b:

a < b + c

c > a - b,

and

b < a + c

c> b - a

Notice that either a - b or b - a is non-negative and the above inequalities show that c must be longer than each of those differences. Thus, not only the length of a side must be shorter than the sum of the lengths of the remaining two sides, but also the length of a side must be longer than the positive difference between the lengths of the remaining sides.

Applying the above to the side PR, we see that 1 < PR < 5. So, PR can be anything between 1 and 5. Similarly, applying the same argument to side RT, we see that 1 < RT < 9. Finally, let's look at PT. We already know PT must be less than 14, but how short can PT be? Well, notice that 4 is a possible value for both PR and RT. If both PR and RT are 4, then PT must be longer than 4 - 4 = 0; i.e. PT can be made as short as we want (by choosing PR = RT and choosing the angle between PR and RT small enough).

To answer your second question, notice that it is not stated anywhere in the question that the distance between P and R is an integer. Similarly, the distance between R and T is not necessarily an integer either. If those distances were required to be integers, then indeed the maximum value of PR would be 4 and the maximum value of RT would be 8 and PT would have to be less than 4 + 8 = 12. However, as long as PR < 5 is satisfied, PR can be chosen as close to 5 as we want (for instance, we can choose PR to be 4.999). Similarly, RT can be chosen as close to 9 as we want (for instance, we can choose RT to be 8.999 or even closer). That is why PT can be greater than 12; in fact, PT can be made as close to 14 as we want by choosing PR and RT accordingly. However, PT can never equal 14 or anything greater than 14.

The following quadrilateral is viable:


If we separate the sides of 5 and 3 just a bit, we get:


The figure above illustrates that the length of PT can be any nonzero value less than the sum of the other four sides.

I disagree. You ended up with the correct answer, but it's a good thing the answer choices didn't include things like 2 or 3 as options.

PT can be anything between 0 and 14, exclusive.

Imagine deleting the segment PT and making each of the vertices Q, R, and S into hinges. You'd have PQ-hinge-QR-hinge-RS-hinge-ST-hinge. Now, imagine stretching the whole thing out with each of the hinges open to alllllllmost 180 degrees. That would make the whole thing really close to a straight line of length 14, but not quite straight and not quite stretched to 14. Hence, the upper bound of 14 for PT, which you got right. But now imagine moving your contraption around so that P and T touch. Could you make that happen? Sure you could!!! But we need a side PT in order to have a pentagram, so move P and T apart by just the tiniest amount. 0 is our lowest bound, but 0.0000000001 would be just fine!

0<PT<14

Answer choice C.

This is a great example of a question that many test-takers spend more time on than they need to and run the risk on missing because they focus on the geometry, the math. Forget the math. Eyeball geometry questions. Use your reason. Visualize things and see if you can picture manipulating them. It's a form of ballparking.

ThatDudeKnowsBallparking

Bunuel MartyTargetTestPrep pls tell the concept of sum of any side of polygon must be less than sum of all other- pls share a linkn where it is explained as to where a,ll it is appplicable and how

If you look at the paragraph in my post above that begins with “Imagine,” that explains why.

To really make the concept sink in, try drawing a polygon with one side that is longer than the sum of the others. I promise you’ll quickly understand the concept without needing any formal mathematical proof!!

Posted from my mobile device

This problem, especially the graph is very misleading. if 10 can be the side, than the graph should change a lot. we can clearly see that angles Q, R and S are clearly less then 180 and P is acute. Sides Q and R seem somewhat parallel. Q+QR=3+2=5. using acute angles and these measures it looks like side PT might be only 5. if you want to fit 10 in this, than you have to change the graph by a lot.

The question might be made less misleading by just not drawing the graph and just saying that 4 sides of normal (all angles less then 180 degrees) pentagon are these, and what can the fifth side be.

I believe there should be a difference between "not drawn to scale" and "completely irrelevant to the problem".

Bunuel
If there was a choice that is 10 only, we would have to check the minimum as well, right?

Tanchat

Here's a bonus question for you!! (Hint: it has been answered a few times in the thread already, but see if you can figure it out before you go look above.)

What would be the minimum?

ScottTargetTestPrep

Thank you for your helpful explanation. To clarify, I realize that the "bonus" question of finding the minimum is mentioned on this thread. But to confirm, the minimum can be any value greater than 0? For triangles, the minimum has to be greater than the positive difference of the two other sides, so I was curious if there was a similar rule for pentagons. Thank you

There is no rule specifically. But that is why we default to using the triangle inequality theorem.

GIVEN:

• A pentagon PQRST
• PQ = 3, QR = 2, RS = 4, and ST = 5

TO ANSWER:

• Which of the lengths 5, 10, and 15 could be the value of PT?

APPROACH:
We will construct small triangles within the pentagon by drawing diagonals - by joining one pair of vertices at a time. On each triangle thus constructed, we will apply the triangle inequality theorem to find the range of length of the diagonal.
As we go on this way, we will eventually be able to find the missing side we set out to find.


WORKING OUT:
Triangle PQR: Join P and R to get triangle PQR.

• Using Triangle Inequality Theorem,
  
  • |PQ – QR| < PR < PQ + QR
  • 1 < PR < 5 -----------(1)

Triangle PRS: Join P and S to get triangle PRS.

• Using Triangle Inequality Theorem,
  
  • |PS – RS| < PS < PR + RS
    
    • The lower limit of PS, |PR – RS|, will be minimum when the length of PR is closest to the length of RS.
      
      • Now, RS = 4 and 1 < PR < 5 (from (1), so, lengths of PR and RS will be closest when PR is 4. (4 is a possibility since 1 < PR < 5)
        
        • Thus, PS > |PR – RS| = 0 ---------(2)
    • The upper limit of PS, PR + RS = PR + 4, will be maximum when the length of PR is as large as possible.
      
      • Since PR < 5, PR + 4 < 5 + 4 = 9 ------------(3)
    • So, from (2) and (3), 0 = PS < 9. ------------- (4)

Triangle PST: (This is formed automatically by the constructions for the previous triangles.)

Using Triangle Inequality Theorem,

• |PS – ST| < PT < PS + ST
  
  • The lower limit of PT, |PS – ST|, will be minimum when the length of PS is closest to the length of ST.
    
    • Now, ST = 5 and 0 = PS < 9 (from (4), so, lengths of PS and ST will be closest when PS is 5. (5 is a possibility since 0 = PS < 9)
    • Thus, PT > |PS – ST| = 0 ---------(5)
• The upper limit of PT, PS + ST = PS + 5, will be maximum when the length of PS is as large as possible.
  
  • Since PS < 9, PS + 5 < 9 + 5 = 14 ------------(6)
• So, from (5) and (6), 0 = PT < 14. ------------- (7)

Final Answer:
Of the lengths 5, 10, and 15, the value of PT can only be 5 or 10.


Correct Answer: Choice C


TAKEAWAYS:

1. Triangle inequality theorem: If A and B are two sides of a triangle, then third side C satisfies |A – B| < C < A + B.
2. Repeatedly applying the triangle inequality theorem on a polygon can help us calculate any missing side when other sides are given.

Hope this helps!


Shweta Koshija
Quant Product Creator, e-GMAT