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Difference between revisions of "1963 IMO Problems/Problem 2"

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==Solution==
 
==Solution==
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Let <math>\omega_1</math> be the circle with diameter <math>AB</math>, and let <math>\omega_2</math> be the circle with diameter <math>AC</math>.  Then the locus is simply the set of points inside either <math>\omega_1</math> or <math>\omega_2</math>, but not both.
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To see this, suppose the right angle's ray that does not pass through <math>A</math> intersects segment <math>BC</math> at <math>X</math>.  Then the right angle's vertex must lie on the circle with diameter <math>AX</math>.  So, for a particular <math>X</math>, the desired locus is a circle with diameter <math>AX</math>.  Accounting for all possible <math>X</math>, the total locus is the union of the circumferences of all circles that have a diameter <math>AX</math>, where <math>X</math> is some point on <math>BC</math>. 
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As <math>X</math> moves from <math>B</math> to <math>C</math>, the motion of the circle with diameter <math>AX</math> is continuous and fluid.  Any point <math>P</math> lying within <math>\omega_1</math> but outside <math>\omega_2</math> will eventually be intersected by this moving circle, since it went from inside to outside the circle.  This applies similarly to all points inside <math>\omega_2</math> but outside <math>\omega_1</math>.  Also, the points inside both <math>\omega_1</math> and <math>\omega_2</math> are never intersected by this moving circle, as it always stays inside.
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(This proof sucks and needs some formalism)
  
 
==See Also==
 
==See Also==

Latest revision as of 04:49, 19 February 2019

Problem

Point $A$ and segment $BC$ are given. Determine the locus of points in space which are the vertices of right angles with one side passing through $A$, and the other side intersecting the segment $BC$.

Solution

Let $\omega_1$ be the circle with diameter $AB$, and let $\omega_2$ be the circle with diameter $AC$. Then the locus is simply the set of points inside either $\omega_1$ or $\omega_2$, but not both.

To see this, suppose the right angle's ray that does not pass through $A$ intersects segment $BC$ at $X$. Then the right angle's vertex must lie on the circle with diameter $AX$. So, for a particular $X$, the desired locus is a circle with diameter $AX$. Accounting for all possible $X$, the total locus is the union of the circumferences of all circles that have a diameter $AX$, where $X$ is some point on $BC$.

As $X$ moves from $B$ to $C$, the motion of the circle with diameter $AX$ is continuous and fluid. Any point $P$ lying within $\omega_1$ but outside $\omega_2$ will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside $\omega_2$ but outside $\omega_1$. Also, the points inside both $\omega_1$ and $\omega_2$ are never intersected by this moving circle, as it always stays inside.

(This proof sucks and needs some formalism)

See Also

1963 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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