Bunuel wrote:
If the length of each of the three sides of a triangle is a positive integer, is the triangle right-angled ?
(1) The perimeter of the triangle is a prime number.
(2) The area of a circle circumscribing the triangle is \(9\pi^2\)
Official Solution:If the length of each of the three sides of a triangle is a positive integer, is the triangle right-angled ? (1) The perimeter of the triangle is a prime number.
The perimeter must equal to some
odd prime, because no sum of three positive integers is 2 (which is the only even prime).
For a right triangle we know that \(a^2+b^2=c^2\). Now, if both \(a\) and \(b\) are even or both are odd, then \(c\) will be even, and in this case \(perimeter=(a+b)+c=even+even=even\). If \(a\) and \(b\) are odd/even (or vise-versa), then \(c\) will be odd, and in this case \(perimeter=(a+b)+c=odd+odd=even\). So, IF the triangle were right-angled (with integer lengths of its sides), then the perimeter would be even, NOT odd as we are told in this statement. Therefore, the triangle is NOT right-angled. Sufficient.
(2) The area of a circle circumscribing the triangle is \(9\pi^2\)
\(The \ area \ of \ a \ circle =\pi r^2=9\pi^2\), which gives \(r=3\sqrt{\pi}\). The diameter of the circle will be twice of that, so \(6\sqrt{\pi}\). Next, we know that
a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle (the reverse is also true: if the diameter of the circle is also the triangle's side, then that triangle is a right triangle). So, IF the triangle were right-angled, then its hypotenuse would be \(6\sqrt{\pi}\), which is NOT an integer as given in the stem. Therefore, the triangle is NOT right-angled. Sufficient.
Answer: D
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