Difference between revisions of "2015 AIME II Problems/Problem 6"
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{{AIME box|year=2015|n=II|num-b=5|num-a=7}} | {{AIME box|year=2015|n=II|num-b=5|num-a=7}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
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Revision as of 21:55, 9 August 2020
Problem
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form for some positive integers
and
. Can you tell me the values of
and
?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of ." He writes down a positive integer and asks, "Can you tell me the value of
?"
Jon says, "There are still two possible values of ."
Find the sum of the two possible values of .
Solution 1 (Algebra)
We call the three roots (some may be equal to one another) ,
, and
. Using Vieta's formulas, we get
,
, and
.
Squaring our first equation we get .
We can then subtract twice our second equation to get .
Simplifying the right side:
So, we know .
We can then list out all the triples of positive integers whose squares sum to :
We get ,
, and
.
These triples give values of
,
, and
, respectively, and
values of
,
, and
, respectively.
We know that Jon still found two possible values of when Steve told him the
value, so the
value must be
. Thus, the two
values are
and
, which sum to
.
~BealsConjecture~
Solution 2 (Algebra+ Brute Force)
First things first. Vietas gives us the following:
(1)
(2)
(3)
From (2), must have odd parity, meaning
must be a multiple of 4, which implies that both sides of (2) are even. Then, from (1), we see that an odd number of
,
, and
must be odd, because we have already deduced that
is odd. In order for both sides of (2) to be even, there must only be one odd number and 2 even numbers.
Now, the theoretical maximum value of the left side of (2) is . That means that the maximum bound of
is where
(4)
which simplifies to
meaning
(5)
So now we have that from (2),
from (5), and
is odd from (2). This means that
could equal
,
, or
. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of
,
, and
, of which the last two return equal
values. Then,
AWD.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.