M22-30

Official Solution:


Is the area of a rectangle inscribed in a circle with a radius of 5 centimeters greater than 48 square centimeters?

Refer to the diagram below:



A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle comprises two right triangles, a rectangle inscribed in a circle must have its diagonal as the diameter of the circle.

Given: \(radius=5\), hence \(diameter=diagonal=10\).

Question: \(area=ab=?\) (where \(a\) and \(b\) are the sides of the rectangle).

(1) The ratio of the lengths of the rectangle's sides is 3:4.

Given that \(\frac{a}{b}=\frac{3x}{4x}\), for some positive multiplier \(x\). Therefore, \(diagonal^2=a^2+b^2=(3x)^2+(4x)^2\), and since \(diagonal=10\), then \(100=9x^2+16x^2\). Solving gives \(x=2\), hence \(a=3x=6\) and \(b=4x=8\). Therefore \(area=ab=6*8=48\). Sufficient.

Alternatively, even without calculating, one can observe that since the hypotenuse (diameter) is 10 and the legs are in the ratio of 3 to 4, then we have a 6-8-10 right triangle (Pythagorean Triple).

(2) The difference between the lengths of the rectangle's two adjacent sides is less than 3 centimeters.

Given that \(b-a \lt 3\). Square both sides: \(b^2-2ab+a^2 \lt 9\). Now, since \(diagonal=10^2=a^2+b^2\) then \(100-2ab \lt 9\), so \(ab \gt 45.5\). Therefore, \(area=ab \gt 45.5\), hence the area may or may not be more than 48. Not sufficient.


Answer: A