Difference between revisions of "1963 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | + | 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 03:49, 19 February 2019
Problem
Point and segment are given. Determine the locus of points in space which are the vertices of right angles with one side passing through , and the other side intersecting the segment .
Solution
Let be the circle with diameter , and let be the circle with diameter . Then the locus is simply the set of points inside either or , but not both.
To see this, suppose the right angle's ray that does not pass through intersects segment at . Then the right angle's vertex must lie on the circle with diameter . So, for a particular , the desired locus is a circle with diameter . Accounting for all possible , the total locus is the union of the circumferences of all circles that have a diameter , where is some point on .
As moves from to , the motion of the circle with diameter is continuous and fluid. Any point lying within but outside will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside but outside . Also, the points inside both and 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 |