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27 Aug 2021, 08:00
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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 \(9\pi^2\)
M36-09
(1) The perimeter of the triangle is a prime number.
(2) The area of a circle circumscribing the triangle is \(9\pi^2\)
M36-09
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23 Sep 2021, 03:44
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Bunuel
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
General Discussion
27 Aug 2021, 09:01
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(1) If the perimeter of the triangle is a prime number, the number must be odd (there is no triangle with the three sides being a positive integer whose perimeter is 2) so the three sides must be odd or two even and the other odd.
As they must complies the formula:
\(a^2=b^2+c^2\)
If the three are odd the formula never complies (odd+odd=even).
If two even and the other odd the formula never complies (even+even=even, even-even=even).
SUFFICIENT
(2) If the triangle were right angle the hypotenuse would be equal to 2r.
\(A=9\pi^2\)
\(r=3\sqrt{\pi}\)
\(2r=6\sqrt{\pi}\)
As 2r is not a positive integer, 2r can not be the hypotenuse, the triangle is not right angled.
SUFFICIENT
IMO D
As they must complies the formula:
\(a^2=b^2+c^2\)
If the three are odd the formula never complies (odd+odd=even).
If two even and the other odd the formula never complies (even+even=even, even-even=even).
SUFFICIENT
(2) If the triangle were right angle the hypotenuse would be equal to 2r.
\(A=9\pi^2\)
\(r=3\sqrt{\pi}\)
\(2r=6\sqrt{\pi}\)
As 2r is not a positive integer, 2r can not be the hypotenuse, the triangle is not right angled.
SUFFICIENT
IMO D
