Difference between revisions of "2012 USAJMO Problems/Problem 4"
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Let <math>\alpha</math> be an irrational number with <math>0 < \alpha < 1</math>, and draw a circle in the plane whose circumference has length 1. Given any integer <math>n \ge 3</math>, define a sequence of points <math>P_1</math>, <math>P_2</math>, <math>\dots</math>, <math>P_n</math> as follows. First select any point <math>P_1</math> on the circle, and for <math>2 \le k \le n</math> define <math>P_k</math> as the point on the circle for which the length of arc <math>P_{k - 1} P_k</math> is <math>\alpha</math>, when travelling counterclockwise around the circle from <math>P_{k - 1}</math> to <math>P_k</math>. Supose that <math>P_a</math> and <math>P_b</math> are the nearest adjacent points on either side of <math>P_n</math>. Prove that <math>a + b \le n</math>. | Let <math>\alpha</math> be an irrational number with <math>0 < \alpha < 1</math>, and draw a circle in the plane whose circumference has length 1. Given any integer <math>n \ge 3</math>, define a sequence of points <math>P_1</math>, <math>P_2</math>, <math>\dots</math>, <math>P_n</math> as follows. First select any point <math>P_1</math> on the circle, and for <math>2 \le k \le n</math> define <math>P_k</math> as the point on the circle for which the length of arc <math>P_{k - 1} P_k</math> is <math>\alpha</math>, when travelling counterclockwise around the circle from <math>P_{k - 1}</math> to <math>P_k</math>. Supose that <math>P_a</math> and <math>P_b</math> are the nearest adjacent points on either side of <math>P_n</math>. Prove that <math>a + b \le n</math>. | ||
− | ==Solution== | + | === Solution=== |
+ | Use mathematical induction. For <math>n=3</math> it is true because one point can't be closest to <math>P_3</math> in both ways, and that <math>1+2\le 3</math>. Suppose that for some <math>n</math>, the nearest adjacent points <math>P_a</math> and <math>P_b</math> on either side of <math>P_n</math> satisfy <math>a+b \le n</math>. Then consider the nearest adjacent points <math>P_c</math> and <math>P_d</math> on either side of <math>P_{n+1}</math>. It is by the assumption of the nearness we can see that either <math>c=a+1</math>, <math>d=b+1</math>, or one of <math>c</math> or <math>d</math> equals two <math>1</math>. Let's consider the following two cases. | ||
+ | |||
+ | (i) Suppose <math>a+b=n</math>. | ||
+ | |||
+ | Since the length of the arc <math>P_nP_a</math> is <math>\{(a-n)\alpha\}</math> (where <math>\{x\}</math> equals to <math>x</math> subtracted by the greatest integer not exceeding <math>x</math>) and length of the arc <math>P_bP_n</math> is <math>\{(n-b)\alpha\} = \{a\alpha\}</math>, we now consider a point <math>P_0</math> which is defined by <math>P_1</math> traveling clockwise on the circle such that the length of arc <math>P_0P_1</math> is <math>\alpha</math>. We claim that either <math>P_0</math> is on the interior of the arc <math>P_nP_a</math> or on the interior of the arc <math>P_bP_n</math>. Algebraically, it is equivalent to either <math> \{0-n\alpha\} < \{(a-n)\alpha\}</math> or <math>\{n\alpha -0 \} < \{a\alpha\}</math>. Suppose the former fails, i.e. <math> \{0-n\alpha\} \ge \{(a-n)\alpha\}</math>. Then suppose <math>-n\alpha = m_1 + r_1</math> and <math>(a-n)\alpha = m_2 + r_2</math>, where <math>m_1</math>, <math>m_2</math> are integers and <math>1> r_1 \ge r_2 \ge 0</math>. We now have | ||
+ | <cmath>\{n\alpha\} = \{-m_1-1 + (1 -r_1)\}=1-r_1</cmath> and <cmath>a\alpha = \{m_2-m_1-1 + (1-r_2+r_1)\} = 1-r_2+r_1>1-r_1</cmath> | ||
+ | Therefore <math>P_0</math> is either closer to <math>P_n</math> than <math>P_a</math> on the <math>P_a</math> side, or closer to <math>P_n</math> than <math>P_b</math> on the <math>P_b</math> side. In other words, <math>P_1</math> is the closest adjacent point of <math>P_{n+1}</math> on the <math>P_{a+1}</math> side, or the closest adjacent point of <math>P_{n+1}</math> on the <math>P_{b+1}</math> side. Hence <math>P_c</math> or <math>P_d</math> is <math>P_1</math>, therefore <math>c+d \le n+1</math>. | ||
+ | |||
+ | (ii) Suppose <math>a+b\le n-1</math> | ||
+ | Then either <math>c+d = (a+1)+(b+1) \le n+1</math> when <math>c=a+1</math> and <math>d=b+1</math>, or <math>c+d \le 1+n</math> when one of <math>P_c</math> or <math>P_d</math> is <math>P_1</math>. | ||
+ | |||
+ | In either case, <math>c+d\le n+1</math> is true. | ||
+ | |||
+ | --[[User:Lightest|Lightest]] 20:27, 6 May 2012 (EDT) | ||
==See Also== | ==See Also== |
Revision as of 20:27, 6 May 2012
Problem
Let be an irrational number with , and draw a circle in the plane whose circumference has length 1. Given any integer , define a sequence of points , , , as follows. First select any point on the circle, and for define as the point on the circle for which the length of arc is , when travelling counterclockwise around the circle from to . Supose that and are the nearest adjacent points on either side of . Prove that .
Solution
Use mathematical induction. For it is true because one point can't be closest to in both ways, and that . Suppose that for some , the nearest adjacent points and on either side of satisfy . Then consider the nearest adjacent points and on either side of . It is by the assumption of the nearness we can see that either , , or one of or equals two . Let's consider the following two cases.
(i) Suppose .
Since the length of the arc is (where equals to subtracted by the greatest integer not exceeding ) and length of the arc is , we now consider a point which is defined by traveling clockwise on the circle such that the length of arc is . We claim that either is on the interior of the arc or on the interior of the arc . Algebraically, it is equivalent to either or . Suppose the former fails, i.e. . Then suppose and , where , are integers and . We now have and Therefore is either closer to than on the side, or closer to than on the side. In other words, is the closest adjacent point of on the side, or the closest adjacent point of on the side. Hence or is , therefore .
(ii) Suppose Then either when and , or when one of or is .
In either case, is true.
--Lightest 20:27, 6 May 2012 (EDT)
See Also
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |