Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 2)
Line 18: Line 18:
  
 
== Solution 2 ==
 
== 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>.
+
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>.
  
 
~EgeSaribas
 
~EgeSaribas

Revision as of 10:46, 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.

2016IMOQ1.jpg

Solution

2016IMOQ1Solution.jpg

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, $CF = 2$, $BF = 2.cos(2\alpha) = FA$, $DA = {AC}{2cos(\alpha)} = {1+cos(2\alpha}{cos(\alpha)}$ and $DE = AE = {AD}{2cos(\alpha)} = {1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1$. So $MX = 1$ 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 $BFDX$. İt is clear that the symmetry of $BD$ with respect to $ME$ is $FX$. And we are done $\blacksquare$.

~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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_IMO_Problems/Problem_1&oldid=219503"