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

Revision as of 12:32, 20 April 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

@@ Line 9: / Line 9: @@
 [[File:2016IMOQ1Solution.jpg|600px]]
-Given  $∠ D A C = ∠ D C A = ∠ C A D$ , it follows that  $A B ∥ C D$ . Extend  $D C$  to intersect  $A B$  at  $G$ , making  $∠ G F A = ∠ G F B = ∠ C F D$ . Triangles  $△ C D F$  and  $△ A G F$  are similar. Also,  $∠ F D C = ∠ F G A = 90^{\circ}$  and  $∠ F B C = 90^{\circ}$ , which implies points  $D$ ,  $C$ ,  $B$ , and  $F$  are concyclic.
+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.
-Furthermore,  $∠ B F C = ∠ F B A + ∠ F A B = ∠ F A E = ∠ A F E$ . 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.
+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.
-It is also given that  $A D = D B$  and  $∠ D A F = ∠ D B F = ∠ F X D$ . Additionally,  $∠ M F X = ∠ F X D = ∠ F X M$  and  $F E ∥ M D$  with  $E F = F M = M D = D E$ , making  $E F M D$  a rhombus. Consequently,  $∠ F B D = ∠ M B D = ∠ M X F = ∠ D X F$  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  $△ F E M$ , and  $E M = F X = B D$ .
+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

2016 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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Difference between revisions of "2016 IMO Problems/Problem 1"

Revision as of 12:32, 20 April 2024

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