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Re M0907
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15 Sep 2014, 23:39
Official Solution: Given isosceles triangle ABC with \(\text{base}=AC=\frac{2}{a}\) and \(\text{height}=a^2\). \(\text{area}=\frac{1}{2}*\text{base}*\text{height}=\frac{1}{2}*\frac{2}{a}*a^2=a\). So, the question basically ask whether \(a \gt 1\)? (1) \(\angle ABC \lt 90^\circ\). Assume \(\angle ABC=90^\circ\) then hypotenuse is \(AC=\frac{2}{a}\), as ABC becomes isosceles right triangle (\(454590=11\sqrt{2}\)) then the \(\text{leg}=BC=AB=\frac{2}{a \sqrt{2}}=\frac{\sqrt{2}}{a}\). But \(BC=\frac{\sqrt{2}}{a}\) also equals to \(\sqrt{(\frac{1}{a})^2+(a^2)^2}\), so we have \(\frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2}\), which leads to \(2=1+a^6\). So, finally we have that \(a^6=1\) or \(a=1\) (as \(a\) per diagram is positive). Now, if we increase \(a\) then the base \(\frac{2}{a}\) will decrease and the height \(a^2\) will increase thus making angle ABC smaller than 90 and if we decrease \(a\) then the base \(\frac{2}{a}\) will increase and the height \(a^2\) will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, \(a\) must be more than 1. Sufficient. (2) Perimeter of the triangle \(ABC\) is greater than \(\frac{4}{a}\). \(P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a} \gt \frac{4}{a}\); \(\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}\); \(a^6 \gt 0\); \(a \gt 0\). Hence we don't know whether \(a \gt 1\) is true. Not sufficient. Answer: A
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Re: M0907
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20 Sep 2014, 01:05
There is another approach for (1).
If ABC is a isosceles right triangle with a median to hypotenuse than we have two small right isosceles triangles. Therefore a^2=1/a => a^3=1 => a=1. The subsequent steps are the same.
Is it correct?



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20 Sep 2014, 10:57



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Re: M0907
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11 Oct 2015, 09:08
I made an "educated guess" in this one:
(1) For statement 1 if <ABC=90 then the triangle is well defined (there is only one point in yaxis that can make <ABC=90). So you can probably find the area in this situation and decide whether a>1 or not.
(2) For statement 2 you can slide the points trough the axis to make different triangles. Thus you can probably slide the points to get a>1 or a<1.
letter A



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Re: M0907
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02 Mar 2016, 06:49
Thinking any simpler way to solve this!



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Re: M0907
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02 Mar 2016, 07:36
rezaulnsu wrote: Thinking any simpler way to solve this! Not really. Ease of solution is dependent on how comfortable are you with your concepts on triangles. Refer to Bunuel 's solution above ( m09183815.html#p1414284) You can clearly see that if \(\angle {ABC} = 90\), then a =1 . Now start pulling the points A and C away from (0,0). You will see that \(\angle {ABC}\) starts becoming obtuse. Do the opposite and move the points A and C towards (0,0) and you will clearly see that \(\angle {ABC}\) becomes acute (<90). As 90 was the critical point (i.e. gave you a=1), any movement away or towards the origin will be sufficient to provide you a yes or no (unambiguously for the question asked!). Thus Statement 1 is sufficient. Hope this helps.



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Re: M0907
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17 Mar 2016, 21:40
Looks like a simpler way for those comfortable with trigonometric ratios.
1) ABC < 90, so angle OBC < 45. In triangle OBC, for angle OBC, tan OBC = OC/OB (both given directly). Therefore tan OBC = 1/a^3 . As tan continuously increases between 0 and 90 degrees, we can say that if angle OBC < 45 then Tan OBC < 1 (tan 45=1). Therefore we get 1/a^3 < 1 => a^3>1 => a>1 .
2) Same explanation as above.



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Re: M0907
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17 Apr 2017, 15:22
Perhaps faster way for determining that (1) is sufficient.
If ABC is 90, then BCA and BAC are 45 each. Since the Yaxis is a bisector, then the triangle is split into two right, isosceles triangles. Therefore a^2 = 1/a >> a = 1.
Is this correct?



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Re: M0907
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17 Apr 2017, 21:47
Boycot wrote: There is another approach for (1).
If ABC is a isosceles right triangle with a median to hypotenuse than we have two small right isosceles triangles. Therefore a^2=1/a => a^3=1 => a=1. The subsequent steps are the same.
Is it correct? How is a^2 = 1/a in this solution?



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Re: M0907
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19 Apr 2017, 10:57
It is given that the height of the triangle is a^2 and the width of the triangle is 2/a.
If we assume that the height also splits the triangle into two isosceles right triangles, then a^2 must equal 1/a.



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Re: M0907
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31 May 2017, 10:44
How do we come to the conclusion that ABC is isosceles



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31 May 2017, 11:25



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Re M0907
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02 Aug 2017, 00:29
I think this is a highquality question and the explanation isn't clear enough, please elaborate.



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Re: M0907
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26 Aug 2017, 21:41
\(\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}\); \(a^6 \gt 0\);
Can you square both sides and maintain inequality. Lets say LHS is 1 and RHS is 2 so that 1 > 2 . When you square this , it becomes 1 > 4 which is incorrect.



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27 Aug 2017, 02:51



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Another solution for (1):
If ABC = 90°, then:
(a^2)^2 = (1/a)*(1/a) [Euclides] therefore, a = 1, which means 1/a = 1.
As ABC < 90, then 1/a < 1 (because a smaller angle at B creates a shorter base length), so: a > 1.



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From question stem: We want to find if a>1?
From Statement 1: If we assume ABC = 90. From properties of triangles (Median through a right angle bisecting the hypotenuse), BO=AO=OC So a^2=1/a, So a^3=1, and a=1. Now we know that ABC<90 degree, so in reality either a>1 or a<1 (whatever the value is does not matter, the statement is sufficient to answer the question if a>1, as the value will be constant). Statement 1 is sufficient.
Statement 2: Insufficient.



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Re: M0907
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12 Jun 2018, 22:34
a^6>0 ;
a>0. Hence we don't know whether a>1a>1 is true. Not sufficient.
How did we get to this step ?



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