Difference between revisions of "2011 AMC 12B Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | Let <math>T_1</math> be a triangle with | + | Let <math>T_1</math> be a triangle with side lengths <math>2011</math>, <math>2012</math>, and <math>2013</math>. For <math>n \geq 1</math>, if <math>T_n = \triangle ABC</math> and <math>D, E</math>, and <math>F</math> are the points of tangency of the incircle of <math>\triangle ABC</math> to the sides <math>AB</math>, <math>BC</math>, and <math>AC</math>, respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE</math>, and <math>CF</math>, if it exists. What is the perimeter of the last triangle in the sequence <math>\left(T_n\right)</math>? |
<math>\textbf{(A)}\ \frac{1509}{8} \qquad \textbf{(B)}\ \frac{1509}{32} \qquad \textbf{(C)}\ \frac{1509}{64} \qquad \textbf{(D)}\ \frac{1509}{128} \qquad \textbf{(E)}\ \frac{1509}{256}</math> | <math>\textbf{(A)}\ \frac{1509}{8} \qquad \textbf{(B)}\ \frac{1509}{32} \qquad \textbf{(C)}\ \frac{1509}{64} \qquad \textbf{(D)}\ \frac{1509}{128} \qquad \textbf{(E)}\ \frac{1509}{256}</math> | ||
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For <math>n = 10</math>, perimeter is <math>\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}</math> | For <math>n = 10</math>, perimeter is <math>\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | This problem is a PoP(Power of Point) Bash exercise. What I did to solve this problem was look at recursive perimeters. The first ever perimeter is <math>P_1 = 2011 + 2012 + 2013</math>. Next, by PoP, inscribing the circle gives us three new lengths, namely <math>AD, BE, CF</math>. Denote <math>AD</math> as <math>x_1</math> and homogeneously the others. Then, <math>x_1 + x_2 = 2011, x_2 + x_3 = 2012</math>, and <math>x_1 + x_3 = 2013</math>. If we add all these equations up and divide by <math>2</math>, we get <math>x_1 + x_2 + x_3 = \frac{2011 + 2012 + 2013}{2}</math>. Writing this with <math>P_1 = 2011 + 2012 + 2013</math>, we get that our new perimeter, <math>P_2</math>, is indeed equal to <math>\frac{P_1}{2}</math>. Similarly, by the same concept, we get that <math>P_3 = \frac{P_1}{4}</math> and the pattern keeps going. In general, I found that for each new perimeter <math>P_n</math>, <math>P_n = \frac{P_1}{2^{n-1}}</math>. Now, substituting in the numerical value of <math>P_1</math>, we get that <math>P_n = \frac{2011 + 2012 + 2013}{2^{n-1}}</math>. If you keep dividing the numerator and denominator by 2, I got that: <math>P_n = \frac{1509}{2^{n-3}}</math>. This representation of a new perimeter in terms of <math>n</math> looks very similar to the option choices, so we're on the right path. Now, all we need to do is find out the last value of <math>n</math> when the sequence stops working. By the Triangular Inequality, we may be able to finish this off. Now, I won't go into depths here, but listing out terms and using the Triangular Inequality gives us: | ||
+ | First Sequence: | ||
+ | <math>x_1 < 2012</math> | ||
+ | |||
+ | <math>x_2 < 2013</math> | ||
+ | |||
+ | <math>x_3 < 2011</math>. | ||
+ | The ordered triple for this set of <math>(x_1, x_2, x_3)</math> is <math>(1005, 1006, 1007)</math> if you solve the PoP system of equations we got earlier. If you keep listing out terms, we can come up with the general form, and that is: | ||
+ | <math>x_1 < \frac{503}{2^{n-4}}</math> | ||
+ | |||
+ | <math>x_2 < \frac{503}{2^{n-4}} + 1</math> | ||
+ | |||
+ | <math>x_3 < \frac{503}{2^{n-4}} - 1</math>. | ||
+ | The ordered triple for this set of <math>(x_1, x_2, x_3)</math> is <math>(\frac{503}{2^{n-4}}, \frac{503}{2^{n-4}} - 1, \frac{503}{2^{n-4}} + 1)</math>. Now, if we plug in these values of <math>x_1, x_2</math>, and <math>x_3</math> into the inequalities, we see that the first two are always satisfied, but the last one is only satisfied when: | ||
+ | <math>2^{n-2} < 503</math>(You will get this if you simplify the last inequality). The last <math>n</math> for this to be satisfied is when <math>n = 10</math>. If we go up to our general representation of <math>P_n</math>, we see that <math>P_n = \frac{1509}{2^{n-3}}</math>. Plugging in <math>10</math> because this is the last <math>n</math>(and also the last triangle), we our final answer of <math>\frac{1509}{2^{7}} = \boxed{\frac{1509}{128}}</math> or <math>\boxed{D}</math>. | ||
+ | |||
+ | ~ilikemath247365 | ||
== See also == | == See also == | ||
+ | Identical problem to the [[2011 AMC 10B Problems/Problem 25]]. | ||
+ | |||
{{AMC12 box|year=2011|num-b=21|num-a=23|ab=B}} | {{AMC12 box|year=2011|num-b=21|num-a=23|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:37, 17 February 2025
Contents
[hide]Problem
Let be a triangle with side lengths
,
, and
. For
, if
and
, and
are the points of tangency of the incircle of
to the sides
,
, and
, respectively, then
is a triangle with side lengths
, and
, if it exists. What is the perimeter of the last triangle in the sequence
?
Solution
Answer: (D)
Let ,
, and
Then ,
and
Then ,
,
Hence:
Note that and
for
, I claim that it is true for all
, assume for induction that it is true for some
, then
Furthermore, the average for the sides is decreased by a factor of 2 each time.
So is a triangle with side length
,
,
and the perimeter of such is
Now we need to find when fails the triangle inequality. So we need to find the last
such that
For , perimeter is
Solution 2
This problem is a PoP(Power of Point) Bash exercise. What I did to solve this problem was look at recursive perimeters. The first ever perimeter is . Next, by PoP, inscribing the circle gives us three new lengths, namely
. Denote
as
and homogeneously the others. Then,
, and
. If we add all these equations up and divide by
, we get
. Writing this with
, we get that our new perimeter,
, is indeed equal to
. Similarly, by the same concept, we get that
and the pattern keeps going. In general, I found that for each new perimeter
,
. Now, substituting in the numerical value of
, we get that
. If you keep dividing the numerator and denominator by 2, I got that:
. This representation of a new perimeter in terms of
looks very similar to the option choices, so we're on the right path. Now, all we need to do is find out the last value of
when the sequence stops working. By the Triangular Inequality, we may be able to finish this off. Now, I won't go into depths here, but listing out terms and using the Triangular Inequality gives us:
First Sequence:
.
The ordered triple for this set of
is
if you solve the PoP system of equations we got earlier. If you keep listing out terms, we can come up with the general form, and that is:
.
The ordered triple for this set of
is
. Now, if we plug in these values of
, and
into the inequalities, we see that the first two are always satisfied, but the last one is only satisfied when:
(You will get this if you simplify the last inequality). The last
for this to be satisfied is when
. If we go up to our general representation of
, we see that
. Plugging in
because this is the last
(and also the last triangle), we our final answer of
or
.
~ilikemath247365
See also
Identical problem to the 2011 AMC 10B Problems/Problem 25.
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 | |
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