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China Girls' Mathematical Olympiad

China geometry

Problem

As shown in Fig. 8.1,

⊙ O

is the inscribed circle touching side

B C

△ A B C

at point

M

, and points

D, E

are on the segments

A B, A C

, respectively, satisfying

D E ∥ B C

;

⊙ O_{1}

is the inscribed circle of

△ A D E

tangent to side

D E

at point

N

;

O_{1} B, D O

intersect at point

F

, and

O_{1} C, E O

intersect at point

G

. Please prove that

M N

divides segment

F G

equally. (posed by Bian Hongping)

Fig. 8.1

Solution

A B = A C

, then the graph is symmetric about the bisector of

∠ B A C

and the conclusion is obvious. So we may assume that

A B > A C

. As shown in Fig. 8.2, let

L

be the midpoint of

B C

, line

O_{1} L

intersecting with

F G

R

, and

O_{1} N

be extended to intersect with

B C

K

. Draw line

A T

that is perpendicular to

B C

T

and intersects with line

D E

S

. Connecting

A O

, obviously

O_{1}

is on the segment

A O

. By Menelaus' theorem, we have

Fig. 8.2

\frac{O _{1} F}{F B} \cdot \frac{B D}{D A} \cdot \frac{A O}{O O _{1}} = 1, \frac{O _{1} G}{GC} \cdot \frac{C E}{E A} \cdot \frac{A O}{O O _{1}} = 1. \overset{◯}{1}

Since

D E ∥ B C

, we have

\frac{B D}{D A} = \frac{C E}{E A}

. So

\frac{O _{1} F}{F B} = \frac{O _{1} G}{GC}

, which means that

F G ∥ B C

. Then

\frac{F R}{GR} = \frac{B L}{C L} = 1

. Therefore,

R

is the midpoint of

F G

. We now only need to prove that

M, R, N

are collinear. For this purpose, by the inverse of Menelaus' theorem, we only need to prove that

\frac{O _{1} R}{R L} \cdot \frac{L M}{M K} \cdot \frac{K N}{N O _{1}} = 1. \overset{◯}{2}

Since

F R ∥ B L

, we have

\frac{O _{1} R}{R L} = \frac{O _{1} F}{F B} = \frac{O O _{1}}{A O} \cdot \frac{A D}{D B}

(the second equality is justified by ①). So we only need to prove that

\frac{O O _{1}}{A O} \cdot \frac{A D}{D B} \cdot \frac{L M}{M K} \cdot \frac{K N}{N O _{1}} = 1. \overset{◯}{3}

Since

O_{1} K ⊥ D E

O M ⊥ B C

A T ⊥ B C

D E ∥ B C

, then lines

O_{1} K

O M

A T

are parallel. By the theorem of dividing the segments into proportional by parallel lines, we have

\frac{O O _{1}}{A O} = \frac{M K}{M T}

. Substituting it into ③, we then only need to prove

\frac{A D}{D B} \cdot \frac{L M}{M T} \cdot \frac{K N}{N O _{1}} = 1. \overset{◯}{4}

Since

D E ∥ B C

K N ⊥ D E

S T ⊥ B C

, quadrilateral

K N S T

is a rectangle and then

K N = S T

. Furthermore, from

D S ∥ B T

we have

\frac{A D}{D B} = \frac{A S}{S T}

. Substituting these results into ④, we now only need to prove

\frac{L M}{M T} = \frac{N O _{1}}{A S} . \overset{◯}{5}

Let

B C = a

A C = b

A B = c

. We have

B M = \frac{a + b - c}{2} (the property of an inscribed circle), B L = \frac{a}{2},

B T = c cos ∠ A B C = c \cdot \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c} = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a} .

Then

\frac{L M}{M T} = \frac{B L - B M}{B T - B M} = \frac{\frac{c - b}{2}}{\frac{c ^{2} - b ^{2} + a ( c - b )}{2 a}} = \frac{a}{a + b + c} .

On the other hand,

\frac{N O _{1}}{A S} = \frac{\frac{2 S _{△ A D E}}{A D + D E + A E}}{\frac{2 S _{△ A D E}}{D E}} = \frac{D E}{A D + D E + A E} = \frac{a}{a + b + c} .

Therefore, ⑤ holds. The proof is complete.

---

Alternative solution.

Let the radii of

⊙ O

and

⊙ O_{1}

r

and

r_{1}

, respectively. Obviously,

O_{1}

O

and

A

are collinear.

D E // B C \Rightarrow \frac{A B}{B D} = \frac{A C}{C E} \frac{O F}{F D} = \frac{S _{△ B O O_{1}}}{S _{△ D B O_{1}}} = \frac{\frac{1}{2} ( A B s i n \frac{A}{2} ) \cdot O O _{1}}{\frac{1}{2} r _{1} \cdot B D} \frac{O G}{GE} = \frac{S _{△ O O_{1} C}}{S _{△ E O O_{1}}} = \frac{\frac{1}{2} ( A C s i n \frac{A}{2} ) \cdot O O _{1}}{\frac{1}{2} r _{1} \cdot C E} \Rightarrow F G ∥ D E ∥ B C . ⎭ ⎬ ⎫ \Rightarrow \frac{O F}{F D} = \frac{O G}{GE}

Connect

O N

and extend it to intersect with

B C

K

. If

∠ A B C = ∠ A C B

, then by symmetry, the proposition holds. So we may assume that

∠ A B C < ∠ A C B

in the following. We connect

O M, O_{1} M, O B, M D, D O_{1}

, respectively (see Fig. 8.3). Since

O_{1} N ∥ O M

, we have

Fig. 8.3

S_{△ O N M} = S_{△ M O O_{1}} = \frac{1}{2} r \cdot O O_{1} \cdot sin \frac{C - B}{2}, ①

\frac{O G}{GE} = \frac{O F}{D F} = \frac{S _{△ B O O_{1}}}{S _{△ B D O_{1}}} = \frac{\frac{1}{2} B O \cdot O O _{1} \cdot sin \frac{C}{2}}{\frac{1}{2} \cdot B D \cdot D O _{1} \cdot sin \frac{B}{2}}

= \frac{r \cdot O O _{1} \cdot sin \frac{C}{2}}{r _{1} \cdot B D \cdot cos \frac{B}{2}} ②

(r = B O \cdot cos \frac{B}{2}, r_{1} = D O_{1} \cdot sin \frac{B}{2}),

S_{△ D M N} - S_{△ M E N} = \frac{1}{2} N K \cdot D N - N E = \frac{1}{2} B D \cdot sin B \cdot (r_{1} cot \frac{B}{2} - r_{1} cot \frac{C}{2}) . ③

From ② and ③, we have

\frac{O G}{GE} S_{△ D M N} - S_{△ M E N} = \frac{1}{2} r_{1} \cdot B D \cdot sin B \cdot (cot \frac{B}{2} - cot \frac{C}{2}) \cdot \frac{r \cdot O O _{1} \cdot sin \frac{C}{2}}{r _{1} \cdot B D \cdot cos \frac{B}{2}} = r \cdot O O_{1} \cdot sin \frac{B}{2} \cdot sin \frac{C}{2} \cdot (cot \frac{B}{2} - cot \frac{C}{2}) = r \cdot O O_{1} \cdot sin \frac{C - B}{2} .

Combining it with ①, we get

\frac{O G}{GE} (S_{△ D M N} - S_{△ M E N}) = 2 S_{△ M O N} .

Since

\frac{O G}{GE} : 2 = \frac{O G}{O E} : (\frac{D F}{O D} + \frac{E G}{O E})

, then

\frac{O F}{O D} \cdot S_{△ M N D} - \frac{O G}{O E} \cdot S_{△ M E N} = (\frac{D F}{O D} + \frac{E G}{O E}) \cdot S_{△ M O N}, \frac{O F \cdot S _{△ M N D} - D F \cdot S _{△ M O N}}{O D} = \frac{O G \cdot S _{△ M E N} + E G \cdot S _{△ M O N}}{O E} . ④

On the other hand,

S_{△ N M G} = \frac{S _{△ M E N} \cdot O G + E G \cdot S _{△ M O N}}{O E} .

Therefore, S_{△ N M G} = \frac{O F \cdot S _{△ M N D} - D F \cdot S _{△ M O N}}{O D} .

In the same way,

S_{△ N M F} = \frac{O F \cdot S _{△ M N D} - D F \cdot S _{△ M O N}}{O D} .

By ④, we have

S_{△ N M G} = S_{△ N M F}

. Therefore,

M N

divides segment

F G

equally. The proof is complete.

Techniques

Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleMenelaus' theoremTangentsTriangle trigonometryDistance chasingTrigonometry