Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 \(\frac{4}{a}\).

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.

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.

(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.

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.

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 .

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.

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.

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.

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.