Difference between revisions of "1959 IMO Problems/Problem 5"
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== Solution == | == Solution == | ||
− | === Part | + | === Part A === |
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>. | 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>. | ||
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[[Image:1IMO5A.JPG]] | [[Image:1IMO5A.JPG]] | ||
− | === Part | + | === Part B === |
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 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>. | ||
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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. | 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 | + | === Part C === |
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. | 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. |
Revision as of 10:01, 30 May 2012
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 .
Solution
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.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |