Difference between revisions of "2016 IMO Problems/Problem 1"
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| + | == Solution 2 == | ||
| + | Let <math>\angle(FBA) = \angle(FAB) = \angle(FAD) = \angle(FCD) = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, <math>BF = 2.cos(2\alpha) = FA</math>, <math>DA = AC/2cos(\alpha) = (1+cos(2\alpha)/cos(\alpha)</math> and <math>DE = AE = AD/2cos(\alpha) = (1+cos(2\alpha)/2.(cos(\alpha))^2 = 1</math>. So <math>MX = 1</math> which means <math>B</math>, <math>C</math>, <math>X</math> and <math>F</math> are concyclic. We know that <math>DE // MC</math> and <math>DE = 1 = MC</math>, so we conclude <math>MCDE</math> is parallelogram. So <math>\angle(AME) = \alpha</math>. That means <math>MDEA</math> is isosceles trapezoid. Hence, <math>MD = EA = 1</math>. By basic angle chasing, <math>\angle(MBF) = \angle(MFB) = 2\alpha</math> and <math>\angle(MXD) = \angle(MDX) = 2\alpha</math> and we have seen that <math>MB = MF = MD = MX</math>, so <math>BFDX</math> is isosceles trapezoid. And we know that <math>ME</math> bisects <math>\angle(FMD)</math> so <math>ME</math> is the symmetrical axis of <math>BFDX</math>. İt is clear that the symmetry of <math>BD</math> with respect to <math>ME</math> is <math>FX</math>. And we are done <math>\blacksquare</math>. | ||
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| + | ~EgeSaribas | ||
==See Also== | ==See Also== | ||
{{IMO box|year=2016|before=First Problem|num-a=2}} | {{IMO box|year=2016|before=First Problem|num-a=2}} | ||
Revision as of 10:37, 19 May 2024
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 . İt is clear that 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 | ||
