Problem 1 from the IMO 2006

Here is Problem 1 from the IMO 2006:

  Let ABC be a triangle with incenter I. A point P in the interior of the triangle satisfies ∠PBA + ∠PCA = ∠PBC + ∠PCB. Show that AP ≥ AI, and that equality holds if and only if P = I.

The solution below is by Vo Duc Dien

  problem #1 from 2006 IMO

Chase the angles:

 

∠BIC = 180 - ½ (∠ABC + ∠ACB).

 

∠BPC = 180 - (∠PBC + ∠PCB).

The problem gives us

 ∠PBA + ∠PCA= ∠PBC + ∠PCB
  = ½ (∠ABC + ∠ACB).

Therefore, ∠BPC = ∠BIC.

Draw a circle with center at A that passes through point I and intersects AC at K.

(1) ∠MPK = 360° - ∠BPC - ∠MPB - ∠KPC

But

 ∠BPC= ∠BIC,
 ∠MPB= ∠MIB - ∠IMP - ∠IBP,
 ∠KPC= ∠KIC + ∠IKP + ∠ICP,
 ∠IBP= ½ ∠ABC - ∠PBC,
 ∠ICP= ∠PCB - ½ ∠ACB.

So that (1) becomes

 ∠MPK=360° - ∠BIC - ∠MIB + ∠IMP + ∠IBP - ∠KIC - ∠IKP - ∠ICP
  =360° - ∠BIC - ∠MIB + ∠IMP + ½∠ABC - ∠PBC - ∠KIC - ∠IKP + ½∠ACB - ∠PCB.

And since ½(∠ABC + ∠ACB ) = ∠PBC + ∠PCB,

 ∠MPK= (360° - ∠BIC - ∠MIB - ∠KIC) + ∠IMP - ∠IKP
 ∠MIK= 360° - ∠BIC - ∠MIB - ∠KIC
  = 90°
 ∠MPK= 90° + ∠IMP - ∠IKP
  = 90° + ∠IMP - ∠ILP

Further, ∠ILP > ∠IMP implies ∠IMP - ∠ILP < 0, or

  ∠MPK < 90°.

Therefore, point P is outside the circle with center A implying AP > AI.

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