Difference between revisions of "1959 IMO Problems/Problem 5"
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== Problem == | == Problem == | ||
− | An arbitrary point <math> | + | An arbitrary point <math>M </math> is selected in the interior of the segment <math>AB </math>. The squares <math>AMCD </math> and <math>MBEF </math> are constructed on the same side of <math>AB </math>, with the segments <math>AM </math> and <math>MB </math> as their respective bases. The circles about these squares, with respective centers <math>P </math> and <math>Q </math>, intersect at <math>M </math> and also at another point <math>N </math>. Let <math>N' </math> denote the point of intersection of the straight lines <math>AF </math> and <math>BC </math>. |
− | (a) Prove that the points <math> | + | (a) Prove that the points <math>N </math> and <math>N' </math> coincide. |
− | (b) Prove that the straight lines <math> | + | (b) Prove that the straight lines <math>MN </math> pass through a fixed point <math>S </math> independent of the choice of <math>M </math>. |
− | (c) Find the locus of the midpoints of the segments <math> | + | (c) Find the locus of the midpoints of the segments <math>PQ </math> as <math>M </math> varies between <math>A </math> and <math>B </math>. |
== Solutions == | == Solutions == | ||
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=== Part A === | === Part A === | ||
− | Since the triangles <math> | + | Since the triangles <math>AFM, CBM</math> are congruent, the angles <math>AFM, CBM</math> are congruent; hence <math>AN'B </math> is a right angle. Therefore <math>N' </math> must lie on the circumcircles of both quadrilaterals; hence it is the same point as <math>N </math>. |
[[Image:1IMO5A.JPG]] | [[Image:1IMO5A.JPG]] | ||
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=== Part B === | === Part B === | ||
− | We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math> | + | We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, BCN </math> are similar. Then <math>NM </math> bisects <math>ANB </math>. |
− | We now consider the circle with diameter <math> | + | We now consider the circle with diameter <math>AB </math>. Since <math>ANB </math> is a right angle, <math>N </math> lies on the circle, and since <math>MN </math> bisects <math>ANB </math>, the arcs it intercepts are congruent, i.e., it passes through the bisector of arc <math>AB </math> (going counterclockwise), which is a constant point. |
=== Part C === | === Part C === | ||
− | Denote the midpoint of <math> | + | Denote the midpoint of <math>PQ </math> as <math>R </math>. It is clear that <math>R </math>'s distance from <math>AB </math> is the average of the distances of <math>P </math> and <math>Q </math> from <math>AB </math>, i.e., half the length of <math>AB</math>, which is a constant. Therefore the locus in question is a line segment. |
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | {{IMO box|year=1959|num-b=4|num-a=6}} |
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 19:23, 25 October 2007
Contents
[hide]Problem
An arbitrary point is selected in the interior of the segment . The squares and are constructed on the same side of , with the segments and as their respective bases. The circles about these squares, with respective centers and , intersect at and also at another point . Let denote the point of intersection of the straight lines and .
(a) Prove that the points and coincide.
(b) Prove that the straight lines pass through a fixed point independent of the choice of .
(c) Find the locus of the midpoints of the segments as varies between and .
Solutions
Part A
Since the triangles are congruent, the angles are congruent; hence is a right angle. Therefore must lie on the circumcircles of both quadrilaterals; hence it is the same point as .
Part B
We observe that since the triangles are similar. Then bisects .
We now consider the circle with diameter . Since is a right angle, lies on the circle, and since bisects , the arcs it intercepts are congruent, i.e., it passes through the bisector of arc (going counterclockwise), which is a constant point.
Part C
Denote the midpoint of as . It is clear that 's distance from is the average of the distances of and from , i.e., half the length of , which is a constant. Therefore the locus in question is a line segment.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1959 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |