Difference between revisions of "2011 AMC 10B Problems/Problem 25"
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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, 2012,</math> and <math>2013</math>. For <math>n \ge 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, 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>( T_n )</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> | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
− | + | ||
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. | By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. | ||
Line 24: | Line 24: | ||
Solving gives: | Solving gives: | ||
− | <math>x= \frac{a}{2} - 1</math> | + | <math>x= \frac{a}{2} -1</math> |
<math>y = \frac{a}{2}</math> | <math>y = \frac{a}{2}</math> | ||
Line 32: | Line 32: | ||
Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>. | Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>. | ||
− | <math>T_3</math> can easily be | + | <math>T_3</math> can easily be derived from this as the sides still differ by 1 hence the above solutions still work (now with <math>a=1006</math>). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinitely until the side lengths no longer form a triangle. |
Subbing in gives <math>T_3</math> with sides <math>502, 503, 504</math>. | Subbing in gives <math>T_3</math> with sides <math>502, 503, 504</math>. | ||
Line 51: | Line 51: | ||
<math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these lengths do not make a | <math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these lengths do not make a | ||
− | triangle as < | + | triangle as |
+ | <cmath>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</cmath> | ||
+ | |||
Likewise, you could create an equation instead of listing all the triangles to <math>T_{11}</math>. | Likewise, you could create an equation instead of listing all the triangles to <math>T_{11}</math>. | ||
− | The sides of a triangle <math>T_{k}</math> would be < | + | The sides of a triangle <math>T_{k}</math> would be |
− | We then have < | + | <cmath>\frac{503}{2^{k-3}} - 1, \frac{503}{2^{k-3}}, \frac{503}{2^{k-3}} + 1</cmath> |
+ | We then have | ||
+ | <cmath>503 - 2^{k-3} + 503 > 503 + 2^{k-3}</cmath> | ||
+ | <cmath>1006 - 2^{k-3} > 503 + 2^{k-3}</cmath> | ||
+ | <cmath>503 > 2^{k-2}</cmath> | ||
+ | <cmath>9 > k-2</cmath> | ||
+ | <cmath>k < 11</cmath> | ||
Hence, the first triangle which does not exist in this sequence is <math>T_{11}</math>. | Hence, the first triangle which does not exist in this sequence is <math>T_{11}</math>. | ||
− | Hence the perimeter is < | + | Hence the perimeter is |
+ | <cmath>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</cmath>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Proceeding similarly to the first solution, we have that sides of each triangle are of the form <math>a, a+1, a+2</math> for some number <math>a</math>. Also, note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that <math>a + a + 1 < a+2 \Rightarrow a<1</math>. Then, the perimeter would be <math>a + a + 1 + a + 2 = 3a + 3 < 6</math>. So, to have a proper triangle, we have <math>\frac{3018}{2^{k}} > 6 \Rightarrow 2^k < 503 \Rightarrow 2^{k} < 512</math>. The first triangle to not work would have perimeter <math>\frac{3018}{256} = \frac{1509}{128}</math>, thus the answer is <math>\boxed{\textbf{(D)} \frac{1509}{128}}</math>. | ||
==See Also== | ==See Also== | ||
− | + | Identical problem to the [[2011 AMC 12B Problems/Problem 22]]. | |
{{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:39, 9 November 2024
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 1
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.
Hence and and . Let and gives three equations:
(where for the first triangle.)
Solving gives:
Subbing in gives that has sides of .
can easily be derived from this as the sides still differ by 1 hence the above solutions still work (now with ). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinitely until the side lengths no longer form a triangle.
Subbing in gives with sides .
has sides .
has sides .
has sides .
has sides .
has sides .
has sides .
has sides .
would have sides but these lengths do not make a triangle as
Likewise, you could create an equation instead of listing all the triangles to .
The sides of a triangle would be
We then have
Hence, the first triangle which does not exist in this sequence is .
Hence the perimeter is .
Solution 2
Proceeding similarly to the first solution, we have that sides of each triangle are of the form for some number . Also, note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that . Then, the perimeter would be . So, to have a proper triangle, we have . The first triangle to not work would have perimeter , thus the answer is .
See Also
Identical problem to the 2011 AMC 12B Problems/Problem 22.
2011 AMC 10B (Problems • Answer Key • Resources) | ||
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