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

Revision as of 10:04, 27 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 $∠ D A C = ∠ D C A = ∠ C A D$ , it follows that $A B ∥ C D$ . Extend $D C$ to intersect $A B$ at $G$ , we get $∠ G F A = ∠ G F B = ∠ C F D$ . Making triangles $△ C D F$ and $△ A G F$ similar. Also, $∠ F D C = ∠ F G A = 90^{\circ}$ and $∠ F B C = 90^{\circ}$ , which points $D$ , $C$ , $B$ , and $F$ are concyclic.

And $∠ B F C = ∠ F B A + ∠ F A B = ∠ F A E = ∠ A F E$ . Triangle $△ A F E$ is congruent to $△ F B M$ , and $A E = E F = F M = M B$ . Let $M X = E A = M F$ , then points $B$ , $C$ , $D$ , $F$ , and $X$ are concyclic.

Finally $A D = D B$ and $∠ D A F = ∠ D B F = ∠ F X D$ . $∠ M F X = ∠ F X D = ∠ F X M$ and $F E ∥ M D$ with $E F = F M = M D = D E$ , making $E F M D$ a rhombus. And $∠ F B D = ∠ M B D = ∠ M X F = ∠ D X F$ and triangle $△ B E M$ is congruent to $△ X E M$ , while $△ M F X$ is congruent to $△ M B D$ which is congruent to $△ F E M$ , so $E M = F X = B D$ .

~Athmyx

Solution 2

Let $\angle FBA = \angle FAB = \angle FAD = \angle FCD = \angle DAE = \angle ADE = \alpha$ . 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 18: / Line 18: @@
 == 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>,
+Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \angle DAE = \angle ADE = \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>,

2016 IMO (Problems) • Resources
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Difference between revisions of "2016 IMO Problems/Problem 1"

Revision as of 10:04, 27 May 2024

Contents

Problem

Solution

Solution 2

See Also