Difference between revisions of "1959 IMO Problems/Problem 5"
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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>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. |
Revision as of 12:17, 20 July 2018
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 |