SOLUTION


In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which of the lengths 5, 10, and 15 could be the value of PT ?

(A) 5 only
(B) 15 only
(C) 5 and 10 only
(D) 10 and 15 only
(E) 5, 10, and 15

The length of any side of a triangle must smaller than the sum of the other two sides.

The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

PQ + QR + RS + ST = 3 + 2 + 4 + 5 = 14, so the length of the fifths side can not be more than 14.

Answer: C (5 and 10 only).
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The pentagon can be split in to 3 triangles: PQR, PRS and PST
Consider Triangle PQR:
Since PQ= 3, QR = 2; we can say that 1<PR<5 - Based on the property of triangles: The length of any side of a triangle must be smaller than the sum of the other 2 sides and greater than the difference of the other 2 sides.

Consider Triangle PRS:
We can say that 1<PS<9;

Consider Triangle PST:
We can say that PT is definitely less than 15. So, eliminate B, D and E.
4<PT<14;

From the answer choices, 5 and 10 hold true.

Ans is (C).

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.

Since PQ + QR + RS + ST = 14, PT has to be < 14. Hence C

Bunuel, thanks for your explanation! I understand the solution and I agree, but I have one concern.

On the figure drawn we can see that
1) the direction of lines PQ, QR and RS is to the right from the point P. The sum of of these lines is only 9.
2) the direction of line ST is opposite (or to the left / back to point P).

So, if taking into account this fact, it appears that the line TP is <=9.

Is this reasoning incorrect only because it is written that the figure is drawn to scale?

Divide pentagon into three triangles: Triangle RST, Triangle QRT, Triangle PQT
In RST, RS = 4, ST =5; RT has to be less than their sum, and greater than their difference; therefore, RT could be 8-2; just start with 8, and if it does not work, go back and try 7-2
In QRT, QR = 2, RT = 8, and QT is unknown. QT < 10; QT >6; therefore, QT = 9-7
In PQT, PQ = 3, QT = 9, PT unknown. PT<12; PT>6; therefore, 15 cannot be PT, thus eliminating answer choices B, D, and E
Suppose QT is 8; PT<11; PT>3, so it could still be 5, or 10
Suppose QT is 7; PT<10; PT>4. 10 is eliminated here, but since if QT equals 8, 10 is still possible, as with 5. Supposing RT is 2, then QT<4; if QT is 3, then PT<6; therefore, 5 has to be the answer since we used the smallest possible outcomes. Ten could be eliminated, however, under other conditions, it is permissible.


The only answer choice that was eliminated across every possible condition is 15. The conditions showed us that it can be 5 or 10.
Therefore, the answer is (C) 5 and 10 only


A helpful general rule: questions that involve ranges, work with extremes such as biggest or smallest.

Hope this helps

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?

Bunuel I would much appreciate your comment

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.

PQ + QR + RS + ST = 14, PT has to be < 14. SoC

Would you not say this is a 700 question? I find it quite tedious

You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty: 600-700 Level.
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Bunuel there is typo in your sentence

The length of any side of a triangle must be smaller than the sum of the other two sides. "BE" is missing

Hi Bunuel

Can we derive a general rule: length of any side of a polygon (including regular polygons) must be < sum (n-1) sides.
Also, can we derive a rule for the minimum? Say > difference of two smallest lengths?

Thanks.

does this rule work for quadrilateral or hexagon or any other polygon ??????????????
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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
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This is how i solved this one:
* Split the Pentagon into 3 different Triangles
* PQT, QRT and RST

Start with RST-> RT < RS +ST which leads us to RT < 9
Now move on to Triangle QRT: QT < QR+RT which leads us to QT < 11
Now move on to the last Triangle PQT: PT < PQ + QT which leads us to PT < 14
5 & 10 are ok

Hi Bunuel, does this property hold true for all polygons? Regardless of them being regular or irregular?

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?