M09-07

Question

In the picture below, where  a > 0 , is the area of the triangle  ABC  greater than 1?
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m09-7.gif

(1)  \angle ABC \lt 90^\circ 
 
(2) Perimeter of the triangle  ABC  is greater than  \frac{4}{a}

Bunuel · Accepted Answer

Official Solution:

In the picture below, where  a > 0 , is the area of the triangle  ABC  greater than 1? 
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m09-7.gif  
 
We are given an isosceles triangle ABC with base  AC = \frac{2}{a}  and height  a^2 .
 
The area of the triangle is  \frac{1}{2}*	ext{base}*	ext{height} = \frac{1}{2}*\frac{2}{a}*a^2 = a . So, the question essentially asks whether  a > 1 .
 
(1)  \angle ABC < 90^\circ . 
 
Assume  \angle ABC = 90^\circ . Then, the hypotenuse is  AC = \frac{2}{a} . Since triangle ABC is an isosceles right triangle (with angles  45^\circ-45^\circ-90^\circ  and side ratios of  1:1:\sqrt{2} ), the legs  BC  and  AB  are equal:  \frac{\frac{2}{a}}{\sqrt{2}}=\frac{2}{a\sqrt{2}} =\frac{\sqrt{2}}{a} .
 
However,  BC = \frac{\sqrt{2}}{a}  also equals  \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 . Thus, we find that  a^6 = 1  or  a = 1  (since  a  is positive in the diagram). Now, if we increase  a , then the base  \frac{2}{a}  will decrease and the height  a^2  will increase, making angle ABC smaller than 90. If we decrease  a , the base  \frac{2}{a}  will increase and the height  a^2  will decrease, making the angle ABC greater than 90. Since 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} > \frac{4}{a} ;
 
 \sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a} ;
 
 a^6 \gt 0 ;
 
 a > 0 . However, we don't know whether  a > 1  is true. Not sufficient.

Answer: A

Boycot · Answer

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?

Bunuel · Answer

Yes, that's correct.

Hugoba · Answer

I made an "educated guess" in this one: (1) For statement 1 if 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

bcai48 · Answer

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?

wwambugu14 · Answer

How do we come to the conclusion that ABC is isosceles

Bunuel · Answer

Notice that BC and BA are mirror images of each other around the y-axis, so they must be equal.

manishcmu · Answer

\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.

Bunuel · Answer

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.).

Kinjo17 · Answer

Another solution for (1):

If ABC = 90°, then:

(a^2)^2 = (1/a)*(1/a)  
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.

pantera07 · Answer

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.

jaskiratsingh07 · Answer

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 ?

Bunuel · Answer

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.

smyarga · Answer

Let O is the origin. Triangle ABC is isosceles with base AC=\frac{2}{a} and height BO=a^2 . So, the area is S=\frac{1}{2}*a^2*\frac{2}{a}=a . The question is: whether a>1? (1) If \angle ABC<90^{\circ} , then \angle OBC<45^{\circ} , and therefore \angle BCO>45^{\circ} . If one angle in a triangle is larger than another angle in a triangle, then the side opposite the larger angle will be longer than the side opposite the smaller angle. We obtain that BO>CO , or a^2>\frac{1}{a} , or a^3>1 , or a>1 . Sufficient. (2) P=BC+AB+AC>\frac{4}{a}=2*AC . Therefore we have the triangle inequality BC+AB>AC , that is true for any triangle, and doesn't give us any new information. Insufficient. The answer is (A)

GMATCoachBen · Answer

bcai48  Yes, this is an excellent method! It can also be helpful to draw the angle as acute or obtuse to see the difference.

Kali123 · Answer

HI Bunuel,

Could you explain how you got a>0 from this

\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a} ;
 
 a^6 \gt 0 ;
 
 a \gt 0

Bunuel · Answer

\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a} ;
 
Square both side:  \frac{1+a^6}{a^2} \gt \frac{1}{a^2} ;

Reduce by 1/a^2:  1+a^6>1

a^6 \gt 0 ;
 
 a \gt 0

kaladin123 · Answer

I think this is a  high-quality  question and I agree with the explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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