M09-07

Official Solution:


In the picture below, where \(a > 0\), is the area of the triangle \(ABC\) greater than 1?



We are given an isosceles triangle ABC with base \(AC = \frac{2}{a}\) and height \(a^2\).

The area of the triangle is \(\frac{1}{2}*\text{base}*\text{height} = \frac{1}{2}*\frac{2}{a}*a^2 = a\). 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