2016 IMO Problems/Problem 1
Contents
Problem
Triangle has a right angle at . Let be the point on line such that and lies between and . Point is chosen so that and is the bisector of . Point is chosen so that and is the bisector of . Let be the midpoint of . Let be the point such that is a parallelogram. Prove that and are concurrent.
Solution
The Problem shows that \(\angle DAC = \angle DCA = \angle CAD\), it follows that \(AB \parallel CD\). Extend \(DC\) to intersect \(AB\) at \(G\), we get \(\angle GFA = \angle GFB = \angle CFD\). Making triangles \(\triangle CDF\) and \(\triangle AGF\) similar. Also, \(\angle FDC = \angle FGA = 90^\circ\) and \(\angle FBC = 90^\circ\), which points \(D\), \(C\), \(B\), and \(F\) are concyclic.
And \(\angle BFC = \angle FBA + \angle FAB = \angle FAE = \angle AFE\). Triangle \(\triangle AFE\) is congruent to \(\triangle FBM\), and \(AE = EF = FM = MB\). Let \(MX = EA = MF\), then points \(B\), \(C\), \(D\), \(F\), and \(X\) are concyclic.
Finally \(AD = DB\) and \(\angle DAF = \angle DBF = \angle FXD\). \(\angle MFX = \angle FXD = \angle FXM\) and \(FE \parallel MD\) with \(EF = FM = MD = DE\), making \(EFMD\) a rhombus. And \(\angle FBD = \angle MBD = \angle MXF = \angle DXF\) and triangle \(\triangle BEM\) is congruent to \(\triangle XEM\), while \(\triangle MFX\) is congruent to \(\triangle MBD\) which is congruent to \(\triangle FEM\), so \(EM = FX = BD\).
~Athmyx
Solution 2
Let . And WLOG, . Hence, ,
,
and
.
So which means , , and are concyclic. We know that and , so we conclude is parallelogram. So . That means is isosceles trapezoid. Hence, . By basic angle chasing,
and and we have seen that , so is isosceles trapezoid. And we know that bisects , so is the symmetrical axis of .
and , and are symmetrical respect to . Hence, the symmetry of with respect to is . And we are done .
~EgeSaribas
See Also
2016 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |