2012年IMO几何预选题第6题
设有非等腰的 △ A B C \triangle ABC △ABC, O O O 和 I I I 分别为外心和内心. 在边 A C AC AC, A B AB AB 上分别存在两点 E E E 和 F F F, 使得 C D + C E = A B CD+CE=AB CD+CE=AB, B F + B D = A C BF+BD=AC BF+BD=AC. 设 ( B D F ) (BDF) (BDF) 和 ( C D E ) (CDE) (CDE) 交于 D D D, P P P. 求证: O I = O P OI=OP OI=OP.
证明:
不妨设 C > B C>B C>B
由密克定理可知 A A A, F F F, P P P, E E E 共圆.
设射线 A I AI AI, B I BI BI, C I CI CI 除端点外分别交 ( A E F ) (AEF) (AEF), ( B F D ) (BFD) (BFD), ( C D E ) (CDE) (CDE) 于 X X X, Y Y Y, Z Z Z.
先证明 I I I, X X X, Y Y Y, Z Z Z 共于一圆, 且该圆的圆心为 O O O.
根据三弦定理有, 这等价于:
I X sin ( π 2 − A 2 ) + I Y sin ( π 2 − B 2 ) = I Z sin ( π 2 + C 2 ) IX \sin (\frac{\pi}{2}-\frac{A}{2})+IY \sin (\frac{\pi}{2}-\frac{B}{2})=IZ \sin (\frac{\pi}{2}+\frac{C}{2}) IXsin(2π−2A)+IYsin(2π−2B)=IZsin(2π+2C)
整理得:
I X cos ( A 2 ) + I Y cos ( B 2 ) + I Z cos ( C 2 ) = 0 IX \cos (\frac{A}{2})+IY \cos (\frac{B}{2})+IZ \cos (\frac{C}{2})=0 IXcos(2A)+IYcos(2B)+IZcos(2C)=0
根据三弦定理有:
( B D + B F ) sin B 2 = B Y sin B = > B Y = b 2 cos B 2 (BD+BF)\sin \frac{B}{2}=BY \sin B=>BY=\frac{b}{2\cos \frac{B}{2}} (BD+BF)sin2B=BYsinB=>BY=2cos2Bb
类似地, 可以证明 A X = a 2 cos A 2 AX=\frac{a}{2\cos \frac{A}{2}} AX=2cos2Aa, C Z = c 2 cos C 2 CZ=\frac{c}{2\cos \frac{C}{2}} CZ=2cos2Cc
所以
I Y cos ( B 2 ) = ( I B − B Y ) cos ( B 2 ) = I B cos B 2 − b 2 = ( p − b ) − b 2 IY\cos (\frac{B}{2})=(IB-BY)\cos (\frac{B}{2})=IB\cos \frac{B}{2}-\frac{b}{2}=(p-b)-\frac{b}{2} IYcos(2B)=(IB−BY)cos(2B)=IBcos2B−2b=(p−b)−2b
类似地, 可以证明
I
X
cos
(
A
2
)
=
(
p
−
a
)
−
a
2
IX\cos (\frac{A}{2})=(p-a)-\frac{a}{2}
IXcos(2A)=(p−a)−2a
I
Z
cos
(
C
2
)
=
(
p
−
c
)
−
c
2
IZ \cos(\frac{C}{2})=(p-c)-\frac{c}{2}
IZcos(2C)=(p−c)−2c
所以 I X cos ( A 2 ) + I Y cos ( B 2 ) + I Z cos ( C 2 ) = 3 p − ( a + b + c ) − a + b + c 2 = 0 IX \cos (\frac{A}{2})+IY \cos (\frac{B}{2})+IZ \cos (\frac{C}{2}) = 3p - (a+b+c) - \frac{a+b+c}{2}=0 IXcos(2A)+IYcos(2B)+IZcos(2C)=3p−(a+b+c)−2a+b+c=0
所以 I I I, X X X, Y Y Y, Z Z Z 共圆.
易知若
I
X
=
2
(
O
A
cos
∠
O
A
I
−
A
X
)
IX=2(OA \cos \angle OAI-AX)
IX=2(OAcos∠OAI−AX), 则
O
I
=
O
X
OI=OX
OI=OX
易知
∠
O
A
I
=
C
−
B
2
\angle OAI = \frac{C-B}{2}
∠OAI=2C−B,
2 O A cos ∠ O A I cos ( A 2 ) = 2 O A cos C − B 2 sin B + C 2 = O A ( sin C − sin B ) = c − b 2 OA\cos \angle OAI \cos (\frac{A}{2})=2OA \cos \frac{C-B}{2} \sin \frac{B+C}{2}=OA (\sin C - \sin B)=\frac{c-b}{2} OAcos∠OAIcos(2A)=2OAcos2C−Bsin2B+C=OA(sinC−sinB)=2c−b
2 A X cos A 2 = a AX \cos \frac{A}{2}=a AXcos2A=a
I X cos A 2 = p − 3 a 2 IX\cos \frac{A}{2}=p-\frac{3a}{2} IXcos2A=p−23a
I X cos A 2 − 2 IX\cos \frac{A}{2}-2 IXcos2A−2 [ O A cos ∠ O A I cos ( A 2 ) − A X cos A 2 OA\cos \angle OAI \cos (\frac{A}{2})-AX \cos \frac{A}{2} OAcos∠OAIcos(2A)−AXcos2A ]= p − 3 a 2 − c − b 2 − a = 0 p-\frac{3a}{2}-\frac{c-b}{2}-a=0 p−23a−2c−b−a=0
所以 I X = 2 ( O A cos ∠ O A I − A X ) IX=2(OA \cos \angle OAI-AX) IX=2(OAcos∠OAI−AX) 成立, 进而 O I = O X OI=OX OI=OX.
所以 O I = O X OI=OX OI=OX.
类似地, 还可以证明 O I = O Y OI=OY OI=OY, O I = O Z OI=OZ OI=OZ.
所以该圆的圆心为 O O O.
∠ F P X = π − A 2 \angle FPX = \pi - \frac{A}{2} ∠FPX=π−2A
∠ F P Y = π − B 2 \angle FPY= \pi - \frac{B}{2} ∠FPY=π−2B
∠ X P Y = π 2 − C 2 = ∠ X Z Y \angle XPY= \frac{\pi}{2}- \frac{C}{2} = \angle XZY ∠XPY=2π−2C=∠XZY
所以 P P P 在 ( X Y Z ) (XYZ) (XYZ) 上, 进而 O I = O P OI=OP OI=OP.
证毕.