USAMO 1972 #2, #3 July 21, 2009Posted by lumixedia in Problem-solving.
Tags: combinatorics, contest math, geometry, olympiad math, USAMO, USAMO 1972
I think I might as well just start going through the USAMOs in chronological/numerical order.
USAMO 1972 #2. A given tetrahedron is isosceles, that is , , . Show that the faces of the tetrahedron are acute-angled triangles.
My (also the official) solution. By SSS congruency, we have . Then . But we must have and similarly for the other two pairs of angles at , so none of , , can be or greater. So all angles in are acute and therefore all angles in all the triangles are acute.
The official solution just asserts that the angles at satisfy the triangle inequality. This seems intuitive, but I am having an unexpected amount of trouble trying to prove it. My initial attempt to transform it into the regular triangle inequality failed; Damien suggested using vectors, which is probably a good idea, but I haven’t worked it out yet and I need to get back to my project soon. Anyway.
USAMO 1972 #3. A random number selector can only select one of the nine integers , and it makes these selections with equal probability. Determine the probability that after selections (), the product of the numbers selected will be divisible by .
My (also the official) solution. After selections, the total number of possible distinct sequences is , the number of sequences not containing an even number is , the number of sequences not containing a is , and the number of sequences containing neither an even number nor a is . So by the principle of inclusion-exclusion, the desired probability is
That would’ve been about an AMC-12 #15 today…