Difference between revisions of "1959 IMO Problems/Problem 5"
(IMO box) |
Sblnuclear17 (talk | contribs) m (→See Also) |
||
(6 intermediate revisions by 4 users not shown) | |||
Line 9: | Line 9: | ||
(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>. | (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>. | ||
− | == | + | == Solution == |
=== Part A === | === Part A === | ||
Line 19: | Line 19: | ||
=== Part B === | === Part B === | ||
− | We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, | + | We observe that <math> \frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, BCM</math> are similar. Then <math>NM </math> bisects <math>ANB </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. | 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. | ||
Line 30: | Line 30: | ||
{{alternate solutions}} | {{alternate solutions}} | ||
+ | == See Also == | ||
+ | |||
+ | |||
+ | Quadrados e Circulos circunscritos / IMO 1959-#5 | ||
+ | Link do vídeo: https://youtu.be/UNcHD5JI6wU | ||
{{IMO box|year=1959|num-b=4|num-a=6}} | {{IMO box|year=1959|num-b=4|num-a=6}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:Geometric Construction Problems]] |
Latest revision as of 01:40, 2 January 2023
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
Quadrados e Circulos circunscritos / IMO 1959-#5 Link do vídeo: https://youtu.be/UNcHD5JI6wU
1959 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |