2016 AIME II Problems/Problem 1
Problem
Initially Alex, Betty, and Charlie had a total of peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats
of his peanuts, Betty eats
of her peanuts, and Charlie eats
of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
Solution 1
Let be the common ratio, where
. We then have
. We now have, letting, subtracting the 2 equations,
, so we have
or
, which is how much Betty had. Now we have
, or
, or
, which solving for
gives
, since
, so Alex had
peanuts.
Solution by Shaddoll
Solution 2 (Quadratic Formula)
Let be Alex's peanuts and
the common ratio. Then we have
. Adding
to both sides and factoring,
For the common difference, . Simplifying,
. Factoring,
, so
. Then,
. Substitute
in
to get
. Simplifying, expanding, and applying the quadratic formula,
\[a=150\pm\frac{\sqrt{300^2-4(144^2)}{2}\] (Error compiling LaTeX. Unknown error_msg)
Taking out from under the radical leaves
Since Alex's peanut number was the lowest of the trio, and
, Alex initially had
peanutes.
(Solution by BJHHar)
Solution 3
Let the initial numbers of peanuts Alex, Betty and Charlie had be ,
, and
respectively.
Let the final numbers of peanuts, after eating, be
,
, and
.
We are given that . Since a total of
peanuts are eaten, we must have
.
Since
,
, and
form an arithmetic progression, we have that
and
for some integer
.
Substituting yields
and so
. Since Betty ate
peanuts, it follows that
.
Since ,
, and
form a geometric progression, we have that
and
.
Multiplying yields
.
Since
, it follows that
and
for some integer
.
Substituting yields
, which expands and rearranges to
.
Since
, we must have
, and so
.
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |