1959 IMO Problems/Problem 5
Contents
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.
Solution 2
Part a)
Notice that and .
.
Now notice that . Considering as , this gives and thus . But notice that , which means that . Therefore points are collinear. Now and . Therefore, and thus points are collinear. Therefore, AF and BC intersect at N.
Part b)
Construct the bisector of arc AB above AB. Call it X. . Now which means N lies on the circle with AB as diameter.
. Therefore since M and X are on the same side of , passes through wherever we choose on .
Part c)
Let the midpoint of be . Let be the midpoint of . Let be the midpoint of . Let be the foot of the perpendicular from onto . Therefore by Midpoint Theorem, . Therefore the distance is a constant and thus the locus is a straight line parallel to at a distance (to , of course) of .
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 |