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M09-07

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New post 16 Sep 2014, 00:39
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On the picture below, is the area of the triangle \(ABC\) greater than 1?

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(1) \(\angle ABC \lt 90^\circ\)

(2) Perimeter of the triangle \(ABC\) is greater than \(\frac{4}{a}\)

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Re M09-07  [#permalink]

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New post 16 Sep 2014, 00:39
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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 (\(45-45-90=1-1-\sqrt{2}\)) 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: M09-07  [#permalink]

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New post 20 Sep 2014, 02:05
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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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New post 20 Sep 2014, 11:57
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?


Yes, that's correct.
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Re: M09-07  [#permalink]

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New post 11 Oct 2015, 10:08
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 y-axis 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.

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Re: M09-07  [#permalink]

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New post 02 Mar 2016, 07:49
Thinking any simpler way to solve this!
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Re: M09-07  [#permalink]

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New post 02 Mar 2016, 08:36
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 (m09-183815.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: M09-07  [#permalink]

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New post 17 Mar 2016, 22:40
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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New post 17 Apr 2017, 16:22
Perhaps faster way for determining that (1) is sufficient.

If ABC is 90, then BCA and BAC are 45 each. Since the Y-axis 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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New post 17 Apr 2017, 22:47
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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New post 19 Apr 2017, 11:57
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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New post 31 May 2017, 11:44
How do we come to the conclusion that ABC is isosceles
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New post 31 May 2017, 12:25
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New post 02 Aug 2017, 01:29
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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New post 26 Aug 2017, 22:41
\(\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}\);

\(a^6 \gt 0\);


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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New post 27 Aug 2017, 03:51
manishcmu wrote:
\(\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}\);

\(a^6 \gt 0\);


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.


We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Here the left hand side is the square root of a number, which cannot be negative. The right hand side is 1/a. Since 1/a is to the right of 0, it must be positive.

Manipulating Inequalities (adding, subtracting, squaring etc.).
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New post 27 Nov 2017, 17:40
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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New post 22 Feb 2018, 16:57
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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New post 12 Jun 2018, 23:34
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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New post 12 Jun 2018, 23:44
jaskiratsingh07 wrote:
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 ?


From the picture we can see that a must be a positive number (1/a is to the right of zero hence positive, so a is positive too). Now, the even power of any number is positive or 0: x^(even) >= 0. Thus, a^6 just means that a is not 0. Therefore, a^6 > 0 does not add anything to what we already knew: a > 0.
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Re: M09-07 &nbs [#permalink] 12 Jun 2018, 23:44
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