From F.S 1, we get a valid triangle for AC=5 and BC=4, and the smallest side is AB.
Again, for BC=2.1, we get another valid triangle and this time, the smallest side is BC. As we get 2 different answers, Insufficient.

From F.S 2, we know that AC+BC = 13-3 = 10. Now, we know that the difference of 2 sides must be less than the third side of a valid triangle.

Thus, as have |AC-BC|<3. Replacing the value of AC, we get \(|10-BC-BC|<3 \to -3<10-2BC<3 \to 3.5<BC<6.5\). Similarly, we will get the same range for AC. Thus, as both of them are greater than 3.5, the least value WILL BE for the side AB=3. Sufficient.

B.
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Dear mau5, could you please elaborate on why |AC-BC|<3?

Thanks!
Cheers
J

This is due to the triangle inequality and comes directly from mau5's previous statement:



I'll explain the sufficiency of (2) a bit differently to see if it helps.

Suppose AB isn't the shortest side, and there is a shorter side BC. So let BC = 3. Then AC must be 7 (since perimeter is 13). A 3-3-7 isn't a valid triangle due to the triangle inequality (try to draw a triangle with sides 3, 3, and 7!).
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Stat.(1): you can only plug in BC, which, according to the Third Side Rule must be smaller than 8 (the sum of AB and AC) and greater than 2 (their difference). Plug in BC=2.5 and BC becomes the smallest side; plug in BC=7 and AB is the smallest side. Hence, Stat.(1)->IS->BCE.

Stat.(2) tells you BC+AC=10. Plug in BC=5 AC=5 and AB becomes the smallest side. But can you plug in less than 3 (less than AB)? If you plug in BC=3, AC must be 7 and the Third Side Rule is broken (AB+BC<AC). Any BC smaller than 3 would also fail. So AB must be the smallest side. Hence, Stat.(2)->S->B.

AB = 3 and BC+CA=10...considering the general rules that sum of 2 sides is greater than the 3rd side and difference between 2 sides is less than the third side...what I get are the 2 combinations for BC and CA ... those are 5+5 and 4+6...so we can have triangles 5,5,3 and 3,4,6...

hence, if I am not missing something, answer to this question would be C (I need both the statements to answer it).

The question asks what is the shortest side of triangle ABC? In both your examples the shortest side is 3. Isn't it? Also, notice that we are NOT told that the lenghts of the sides are integers, so there could be many more combinations than you wrote.

The point is that from (2) it follows that no side can be less than or equal AB=3, because if any of the sides is, then the third side must be more than or equal to 7, which would be more than the sum of the other two sides (3 and less than or equal 3).

Does this make sense?
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Gosh...what was I thinking You are absolutely correct. Thanks!!

(1) AC=5
BC could be equal to 3, thus there will be no shortest side and ABC will be isosceles. Insuff
(2) P=13
above mentioned example is not valid since BC=3 and AC must be 7, that is not possible as AC must be less then BC+AB
So after testing all cases it is becoming clear that AB is the shortest side. Suff.

When you say valid triangle, what do you mean?



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