Difference between revisions of "2018 IMO Problems/Problem 1"
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The quadrilateral <math>OHQH'</math> is formed by the intersection of two pairs of equidistant lines <math>\implies</math> <math>OHQH'</math> is a parallelogram with equal heights <math>\implies</math> <math>OHQH'</math> is a rhombus. | The quadrilateral <math>OHQH'</math> is formed by the intersection of two pairs of equidistant lines <math>\implies</math> <math>OHQH'</math> is a parallelogram with equal heights <math>\implies</math> <math>OHQH'</math> is a rhombus. | ||
− | Let <math>I</math> and <math>I'</math> be the bases of the heights of the rhombus | + | Let <math>I</math> and <math>I'</math> be the bases of the heights of the rhombus from the vertex <math>O.</math> In isosceles triangles <math>\triangle ADE</math> and <math>\triangle OII'</math> the sides are parallel, hence the bases of these triangles are parallel, <math>DE||II'.</math> |
The bases of the heights <math>I</math> and <math>I'</math> divide the sides of the rhombus in the same ratio, which means that the diagonal <math>HH'</math> of the rhombus is parallel to the segments <math>HH' ||II' || DE.</math> | The bases of the heights <math>I</math> and <math>I'</math> divide the sides of the rhombus in the same ratio, which means that the diagonal <math>HH'</math> of the rhombus is parallel to the segments <math>HH' ||II' || DE.</math> | ||
The angles between the diagonal of the rhombus and its sides are the same, so <math>\angle HQO = \angle H'QO.</math> | The angles between the diagonal of the rhombus and its sides are the same, so <math>\angle HQO = \angle H'QO.</math> |
Revision as of 11:59, 15 August 2022
Let be the circumcircle of acute triangle
. Points
and
are on segments
and
respectively such that
. The perpendicular bisectors of
and
intersect minor arcs
and
of
at points
and
respectively. Prove that lines
and
are either parallel or they are the same line.
-- solution --
http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg
The diagram is certainly not to scale, but the argument is sound (I believe) and involves re-ordering the construction as specified in the original problem so that an identical state of affairs results, yet in so doing differently it is made clear that the line segments in question are parallel.
Construct a right-angled triangle ABC'. Select an arbitrary point H along the segment BC', and from point H select an arbitrary point F such that the segment HF is perpendicular to the segment AB. Mark the distance from the intersection of HF and AB to B at B" (i.e., HF is a perpendicular bisector). It follows that the triangle B"FB is isosceles. Construct an isosceles triangle FHG. Mark the distance of AB" along AC' at C". (From here, a circle can be constructed according to the sets of points A, B, F, and A, B, G. Points F and G may be repositioned to allow for these circles to coincide; also, point H may be repositioned so that point C falls on the coinciding circle, understood that HG is the other perpendicular bisector.)
Assign the angle BAC the value α. Hence, the angle FHG has the value 180°-α, and the angle HFG (also, HGF) has the value α/2. Assign the angle BFH the value β. Hence, the angle B'FB" has the value β-α/2. Consequently, the angle FB'B" has the value 180°-(β-α/2)-(90°-β) = 90°+α/2, and so too its vertical angle BB'G. As the triangle B"AC" is isosceles, and its subtended angle has the value α, the angles BB"C" and CC"B" both have the value 90°+α/2. It follows therefore that segments B"C" and FG are parallel.
(N.B. Points D and E, as given in the wording of the original problem, have been renamed B" and C" here.)
Solution 2
The essence of the proof is using a rhombus formed by the perpendicular bisectors of the segments and
and the parallelism of its diagonal and the base of the triangle formed by the perpendiculars from one vertex of the rhombus.
The perpendicular bisectors of the segments and
intersect at points
(the circumcenter
and
Let
The distance from the line
to point
is
from
to
is
Therefore, the distance between lines
and
is
Similarly, the distance between the lines
and
is
The quadrilateral is formed by the intersection of two pairs of equidistant lines
is a parallelogram with equal heights
is a rhombus.
Let and
be the bases of the heights of the rhombus from the vertex
In isosceles triangles
and
the sides are parallel, hence the bases of these triangles are parallel,
The bases of the heights
and
divide the sides of the rhombus in the same ratio, which means that the diagonal
of the rhombus is parallel to the segments
The angles between the diagonal of the rhombus and its sides are the same, so
The lines
and
are symmetrical with respect to the diameter
so
The homothety of triangles
and
centered at
implies
Therefore, these three lines are parallel to
Note that triangle
may be obtuse.