Difference between revisions of "2016 IMO Problems/Problem 1"
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Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, | 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>\implies</math> <math>BF = 2.cos(2\alpha) = FA</math>, |
− | <math>DA = \frac{AC}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}</math> and | + | <math>\implies</math> <math>DA = \frac{AC}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}</math> and |
− | <math>DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1</math>. | + | <math>\implies</math> <math>DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1</math>. |
So <math>MX = DE = 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, | So <math>MX = DE = 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, |
Revision as of 12:09, 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 |