Difference between revisions of "2019 AMC 12A Problems/Problem 25"
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~Dhillonr25 | ~Dhillonr25 | ||
− | + | == Video Solution by Richard Rusczyk == | |
https://artofproblemsolving.com/videos/amc/2019amc12a/497 | https://artofproblemsolving.com/videos/amc/2019amc12a/497 |
Latest revision as of 23:22, 3 November 2024
Contents
[hide]Problem
Let be a triangle whose angle measures are exactly
,
, and
. For each positive integer
, define
to be the foot of the altitude from
to line
. Likewise, define
to be the foot of the altitude from
to line
, and
to be the foot of the altitude from
to line
. What is the least positive integer
for which
is obtuse?
Solution 1
For all nonnegative integers , let
,
, and
.
Note that quadrilateral is cyclic since
; thus,
. By a similar argument,
. Thus,
. By a similar argument,
and
.
Therefore, for any positive integer , we have
(identical recurrence relations can be derived for
and
). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to
(and the coefficient of
is
). Hence, we let
. We will solve for
,
, and
by iterating the recurrence to obtain
,
, and
. Letting
respectively, we have
Subtracting from
, we have
, and subtracting
from
gives
. Dividing these two equations gives
, so
. Substituting back, we get
and
.
We will now prove that for all positive integers ,
via induction. Clearly the base case of
holds, so it is left to prove that
assuming our inductive hypothesis holds for
. Using the recurrence relation, we have
Our induction is complete, so for all positive integers ,
. Identical equalities hold for
and
.
The problem asks for the smallest such that either
,
, or
is greater than
. WLOG, let
,
, and
. Thus,
for all
,
, and
. Solving for the smallest possible value of
in each sequence, we find that
gives
. Therefore, the answer is
.
Solution 2
We start from Solution 1 until we reach the recurrence relation Iterate this again, to get
Subtract the two, getting
This recurrence has characteristic equation
Now, write
We obtain similar recursions for
that can be easily solved by getting
from the original recursive formula and then using those two values to solve for
and
Then proceed with the last paragraph of Solution 1.
Solution 3 (If you're short on time)
We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of and
(we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometric sequence with an integer common ratio. From here, we get that
and we need
. The first power of two greater than
is
thus our answer is
.
~Dhillonr25
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2019amc12a/497
~ dolphin7
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.