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Difference between revisions of "2018 IMO Problems/Problem 1"

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Let <math>\Gamma</math> be the circumcircle of acute triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> are on segments <math>AB</math> and <math>AC</math> respectively such that <math>AD = AE</math>. The perpendicular bisectors of <math>BD</math> and <math>CE</math> intersect minor arcs <math>AB</math> and <math>AC</math> of <math>\Gamma</math> at points <math>F</math> and <math>G</math> respectively. Prove that lines <math>DE</math> and <math>FG</math> are either parallel or they are the same line.
 
Let <math>\Gamma</math> be the circumcircle of acute triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> are on segments <math>AB</math> and <math>AC</math> respectively such that <math>AD = AE</math>. The perpendicular bisectors of <math>BD</math> and <math>CE</math> intersect minor arcs <math>AB</math> and <math>AC</math> of <math>\Gamma</math> at points <math>F</math> and <math>G</math> respectively. Prove that lines <math>DE</math> and <math>FG</math> are either parallel or they are the same line.
  
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== Solution ==
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http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg
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Solution!
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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 follows that the line segments in question are parallel.
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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.)
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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.
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(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 20:50, 12 July 2018

Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ 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 follows 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 BFB 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 BAC is isosceles, and its subtended angle has the value α, the angles BBC and CCB both have the value 90°+α/2. It follows therefore that segments BC 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.)

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