Difference between revisions of "2016 IMO Problems/Problem 1"

Revision as of 12:16, 19 May 2024

Problem

Triangle $BCF$ has a right angle at $B$ . Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$ . Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$ . Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$ . Let $M$ be the midpoint of $CF$ . Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ 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 $\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha$ . And WLOG, $MF = 1$ . Hence, $CF = 2$ ,

$\implies$ $BF = CF.cos(2\alpha) = 2.cos(2\alpha) = FA$ ,

$\implies$ $DA = \frac{AC}{2cos(\alpha)} = \frac{CF+FA}{2cos(\alpha)} = \frac{2+2cos(2\alpha)}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}$ and

$\implies$ $DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1$ .

So $MX = DE = 1$ which means $B$ , $C$ , $X$ and $F$ are concyclic. We know that $DE || MC$ and $DE = 1 = MC$ , so we conclude $MCDE$ is parallelogram. So $\angle AME = \alpha$ . That means $MDEA$ is isosceles trapezoid. Hence, $MD = EA = 1$ . By basic angle chasing,

$\angle MBF = \angle MFB = 2\alpha$ and $\angle MXD = \angle MDX = 2\alpha$ and we have seen that $MB = MF = MD = MX$ , so $BFDX$ is isosceles trapezoid. And we know that $ME$ bisects $\angle FMD$ , so $ME$ is the symmetrical axis of $BFDX$ .

$B$ and $X$ , $D$ and $E$ are symmetrical respect to $ME$ . Hence, the symmetry of $BD$ with respect to $ME$ is $FX$ . And we are done $\blacksquare$ .

~EgeSaribas

@@ Line 1: / Line 1: @@
-Pr♦❜❧❡♠ ✶✳ ❚r✐❛♥❣❧❡ BCF ❤❛s ❛ r✐❣❤t ❛♥❣❧❡ ❛t B✳ ▲❡t A ❜❡ t❤❡ ♣♦✐♥t ♦♥ ❧✐♥❡ CF s✉❝❤ t❤❛t FA = FB ❛♥❞ F ❧✐❡s ❜❡t✇❡❡♥ A ❛♥❞ C✳ P♦✐♥t D ✐s ❝❤♦s❡♥ s✉❝❤ t❤❛t DA = DC ❛♥❞ AC ✐s t❤❡ ❜✐s❡❝t♦r ♦❢ ∠DAB✳ P♦✐♥t E ✐s ❝❤♦s❡♥ s✉❝❤ t❤❛t EA = ED ❛♥❞ AD ✐s t❤❡ ❜✐s❡❝t♦r ♦❢ ∠EAC✳ ▲❡t M ❜❡ t❤❡ ♠✐❞♣♦✐♥t ♦❢ CF✳ ▲❡t X ❜❡ t❤❡ ♣♦✐♥t s✉❝❤ t❤❛t AMXE ✐s ❛ ♣❛r❛❧❧❡❧♦❣r❛♠ ✭✇❤❡r❡ AM k EX ❛♥❞ AE k MX✮✳ Pr♦✈❡ t❤❛t ❧✐♥❡s BD✱ FX✱ ❛♥❞ ME ❛r❡ ❝♦♥❝✉rr❡♥t✳
+==Problem==
+Triangle <math>BCF</math> has a right angle at <math>B</math>. Let <math>A</math> be the point on line <math>CF</math> such that <math>FA=FB</math> and <math>F</math> lies between <math>A</math> and <math>C</math>. Point <math>D</math> is chosen so that <math>DA=DC</math> and <math>AC</math> is the bisector of <math>\angle{DAB}</math>. Point <math>E</math> is chosen so that <math>EA=ED</math> and <math>AD</math> is the bisector of <math>\angle{EAC}</math>. Let <math>M</math> be the midpoint of <math>CF</math>. Let <math>X</math> be the point such that <math>AMXE</math> is a parallelogram. Prove that <math>BD,FX</math> and <math>ME</math> are concurrent.
+[[File:2016IMOQ1.jpg|800px]]
+==Solution==
+[[File:2016IMOQ1Solution.jpg|600px]]
+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 <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>,
+<math>\implies</math> <math>BF = CF.cos(2\alpha) = 2.cos(2\alpha) = FA</math>,
+<math>\implies</math> <math>DA = \frac{AC}{2cos(\alpha)} = \frac{CF+FA}{2cos(\alpha)} = \frac{2+2cos(2\alpha)}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}</math> and
+<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,
+<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>.
+<math>B</math> and <math>X</math>, <math>D</math> and <math>E</math> are symmetrical respect to <math>ME</math>. Hence, the symmetry of <math>BD</math> with respect to <math>ME</math> is <math>FX</math>. And we are done <math>\blacksquare</math>.
+~EgeSaribas
+==See Also==
+{{IMO box|year=2016|before=First Problem|num-a=2}}

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Difference between revisions of "2016 IMO Problems/Problem 1"

Revision as of 12:16, 19 May 2024

Contents

Problem

Solution

Solution 2

See Also