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I thought it was C, because the only way for the triangle to be isosceles given both conditions is for BC = AC , and that's not possible because side AC + BC (assuming they are the same length) would equal AB. And the third side of the triangle must be greater than the sum of the other two sides or less than the difference of the other two sides. So given both conditions, should we be able to say for sure that the triangle is not isosceles?

I thought it was C, because the only way for the triangle to be isosceles given both conditions is for BC = AC , and that's not possible because side AC + BC (assuming they are the same length) would equal AB. And the third side of the triangle must be greater than the sum of the other two sides or less than the difference of the other two sides. So given both conditions, should we be able to say for sure that the triangle is not isosceles?

The OG says it's E though. Please help.

Your reasoning is correct answer should be C, not E.

Is triangle ABC an isosceles?

Attachment:

Screen Shot 2012-05-08 at 1.05.37 PM.png [ 7.39 KiB | Viewed 6429 times ]

(1) AB/BC = 2 --> \(AB\neq{BC}\). Not sufficient on its own. (2) x≠y --> \(AB\neq{AC}\). Not sufficient on its own.

(1)+(2) Since \(AB\neq{BC}\) and \(AB\neq{AC}\), then the only way ABC to be isosceles is when \(AC=BC\). But in this case as given that AB=2BC then AB=BC+BC=BC+AC which is not possible because the length of any side of a triangle must be smaller than the sum of the other two sides. So, \(AC\neq{BC}\), which means that ABC is not isosceles. Sufficient.

I thought it was C, because the only way for the triangle to be isosceles given both conditions is for BC = AC , and that's not possible because side AC + BC (assuming they are the same length) would equal AB. And the third side of the triangle must be greater than the sum of the other two sides or less than the difference of the other two sides. So given both conditions, should we be able to say for sure that the triangle is not isosceles?

The OG says it's E though. Please help.

Answer should be E even if we cosider both 1) & 2) consider the below examples:

Case 1--> x= 30, y=100 & third angle= 50 Case 2--> x=45, y=90 & third angle =45

here both 1) & 2) are satisfied but there is contradiction in result, i.e. in one case traingle is issoceles, in other it is not, for the same conditions. Hence E
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I thought it was C, because the only way for the triangle to be isosceles given both conditions is for BC = AC , and that's not possible because side AC + BC (assuming they are the same length) would equal AB. And the third side of the triangle must be greater than the sum of the other two sides or less than the difference of the other two sides. So given both conditions, should we be able to say for sure that the triangle is not isosceles?

The OG says it's E though. Please help.

Answer should be E even if we cosider both 1) & 2) consider the below examples:

Case 1--> x= 30, y=100 & third angle= 50 Case 2--> x=45, y=90 & third angle =45

here both 1) & 2) are satisfied but there is contradiction in result, i.e. in one case traingle is issoceles, in other it is not, for the same conditions. Hence E

When considering the statements together the red scenario is not possible. If x=z then it would mean that AC=BC. But in this case as given that AB=2BC then AB=BC+BC=BC+AC which is not possible because the length of any side of a triangle must be smaller than the sum of the other two sides.
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