Difference between revisions of "2018 IMO Problems/Problem 1"
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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. | 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 | + | 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 -- but not H -- will likely need repositioning so that these circles will coincide. Also, their position should allow for point C to fall on the circle as well, 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 (as well as HGF) has the value α/2. Assign the angle BFH the value β. Hence, the angle B'FB | + | Assign the angle BAC the value α. Hence, the angle FHG has the value 180°-α, and the angle HFG (as well as 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 | + | (N.B. Points D and E, as given in the wording of the original problem, have been renamed B" and C" here.) |
Revision as of 21:01, 12 July 2018
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.
http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg
Solution!
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 -- but not H -- will likely need repositioning so that these circles will coincide. Also, their position should allow for point C to fall on the circle as well, 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 (as well as 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.)
