Difference between revisions of "2005 Canadian MO Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | {{ | + | (a) Let <math>H</math> be the convex hull of <math>S</math>, and choose any three vertices <math>V_1, V_2, V_3</math> of <math>H</math>. |
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| + | Now, draw a line <math>l_1</math> through <math>V_1</math> such that all points of <math>S</math> lie on one side of <math>l_1</math>. (This is possible because <math>H</math> is convex and contains <math>S</math>.) Define <math>l_2</math> and <math>l_3</math> similarly. | ||
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| + | For <math>i = 1, 2, 3</math>, let <math>m_i</math> be the line perpendicular to <math>l_i</math> passing through <math>V_i</math>, and let <math>m_i</math> hit the circle at point <math>P_i</math> on the side of <math>l_i</math> opposite <math>S</math>. | ||
| + | Now, each <math>P_i</math> is closer to <math>V_i</math> than any other point in <math>S</math>. | ||
| + | Since vertices <math>V_1, V_2, V_3 \in H</math> are also in <math>S</math>, we are done. | ||
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| + | (b) Describe the circle as the curve <math>r = 1</math> in polar coordinates. We will construct a set <math>S</math> with an arbitrary number of elements satisfying the desired property. Let <math>V_1 = \left(\frac{1}{2}, 0\right), V_2 = \left(\frac{1}{2}, \frac{\pi}{3}\right), V_3 = \left(\frac{1}{2}, \frac{2\pi}{3}\right) \in S</math>. It is easy to see that for any point <math>P</math> on the circle, <math>P</math> is at most <math>\frac{\sqrt{3}}{2}</math> away from one of the <math>V_i</math>. Therefore, if we also include points <math>Q_i</math> on any of the line segments <math>OV_i</math> (<math>O</math> is the center), where <math>OQ_i < 1 - \frac{\sqrt{3}}{2},</math> then no point can be chosen on the circle which is closer to one of the <math>Q_i</math> then the <math>P_i</math>. | ||
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| + | Thus <math>S</math> is as desired. | ||
==See also== | ==See also== | ||
Latest revision as of 12:35, 9 June 2011
Problem
Let 
be a set of 
points in the interior of a circle.
- Show that there are three distinct points
and three distinct points 
on the circle such that 
is (strictly) closer to 
than any other point in 
, 
is closer to 
than any other point in and 
is closer to 
than any other point in . - Show that for no value of
can four such points in (and corresponding points on the circle) be guaranteed.
and three distinct points 
on the circle such that 
is (strictly) closer to 
than any other point in 
is closer to 
than any other point in 
is closer to 
than any other point in 
can four such points in Solution
(a) Let 
be the convex hull of , and choose any three vertices 
of .
Now, draw a line 
through 
such that all points of lie on one side of . (This is possible because is convex and contains .) Define 
and 
similarly.
For 
, let 
be the line perpendicular to 
passing through 
, and let hit the circle at point 
on the side of opposite .
Now, each is closer to than any other point in .
Since vertices 
are also in , we are done.
(b) Describe the circle as the curve 
in polar coordinates. We will construct a set with an arbitrary number of elements satisfying the desired property. Let 
. It is easy to see that for any point 
on the circle, is at most 
away from one of the . Therefore, if we also include points 
on any of the line segments 
(
is the center), where 
then no point can be chosen on the circle which is closer to one of the then the .
Thus is as desired.
See also
| 2005 Canadian MO (Problems) | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 4 |
