Difference between revisions of "2019 AMC 12A Problems/Problem 25"
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− | Subtracting <math>(1)</math> from <math>(3)</math>, we have <math>pq(q^2-1)=360</math>, and subtracting <math>(1)</math> from <math>(2)</math> gives <math>pq(q-1)=-360</math>. Dividing these two equations gives <math>q+1=-1</math>, so <math>q=-2</math>. Substituting back, we get <math>p=-60</math> and <math>r=60</math>. | + | Subtracting <math>(1)</math> from <math>(3)</math>, we have <math>pq(q^2-1)=360</math>, and subtracting <math>(1)</math> from <math>(2)</math> gives <math>pq(q-1)=-360</math>. Dividing these two equations gives <math>q+1=-1</math>, so <math>q=-2</math>. Substituting back, we get <math>p=-60</math> and <math>r=60</math>. |
+ | |||
+ | We will now prove that for all positive integers <math>n</math>, <math>x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60</math> via induction. Clearly the base case of <math>n=1</math> holds, so it is left to prove that <math>x_{n+1}=(-2)^{n+1}(x_0-60)+60</math> assuming our inductive hypothesis holds for <math>n</math>. Using the recurrence relation, we have <math></math>\begin{align*} | ||
+ | x_{n+1}&=180-2x_n \\ | ||
+ | &=180-2((-2)^n(x_0-60)+60) \\ | ||
+ | &=(-2)^{n+1}(x_0-60)+60 | ||
+ | \end{align*} | ||
+ | |||
+ | Our induction is complete, so for all positive integers <math>n</math>, <math>x_n=(-2)^n(x_0-60)+60</math>. Identical equalities hold for <math>y_n</math> and <math>z_n</math> | ||
The problem asks for the smallest <math>n</math> such that either <math>x_n</math>, <math>y_n</math>, or <math>z_n</math> is greater than <math>90^\circ</math>. WLOG, let <math>x_0=60^\circ</math>, <math>y_0=59.999^\circ</math>, and <math>z_0=60.001^\circ</math>. Thus, <math>x_n=60^\circ</math> for all <math>n</math>, <math>y_n=-(-2)^n(0.001)+60</math>, and <math>z_n=(-2)^n(0.001)+60</math>. Solving for the smallest possible value of <math>n</math> in each sequence, we find that <math>n=15</math> gives <math>y_n>90^\circ</math>. Therefore, the answer is <math>\boxed{\bold{(E) 15}}</math>. | The problem asks for the smallest <math>n</math> such that either <math>x_n</math>, <math>y_n</math>, or <math>z_n</math> is greater than <math>90^\circ</math>. WLOG, let <math>x_0=60^\circ</math>, <math>y_0=59.999^\circ</math>, and <math>z_0=60.001^\circ</math>. Thus, <math>x_n=60^\circ</math> for all <math>n</math>, <math>y_n=-(-2)^n(0.001)+60</math>, and <math>z_n=(-2)^n(0.001)+60</math>. Solving for the smallest possible value of <math>n</math> in each sequence, we find that <math>n=15</math> gives <math>y_n>90^\circ</math>. Therefore, the answer is <math>\boxed{\bold{(E) 15}}</math>. |
Revision as of 22:57, 9 February 2019
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
For all nonnegative integers , let
,
, and
.
Note that quadrilateral is cyclic since
; thus,
. By a similar argument,
. Thus,
. By a symmetric 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 $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*}
x_{n+1}&=180-2x_n \\
&=180-2((-2)^n(x_0-60)+60) \\
&=(-2)^{n+1}(x_0-60)+60
\end{align*}
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
.
-soyamyboya
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.