Let three circles O, O_1, O_2, each externally tangent to the other two, lie on the same side of a line . Let them be tangent to at points A, A_1, A_2 respectively, where point A lies on segment A_1A_2. Denote the points of tangency of O with O_1 and O_2 as B_1 and B_2 respectively, and the point of tangency of O_1 with O_2 as C. Let line A_1C intersect line A_2B_2 at point D_1, and line A_2C intersect line A_1B_1 at point D_2. Prove that line D_1D_2 is parallel to line .

*Proof* 1. Consider the antipodal point P of A on the circle O. We notice that in the right trapezoid A_1A_2O_1O_2, we have O_1A_1 = O_1B_1 and OB_1 = OA. Therefore, align* A_1BA &= - O_1B_1A_1 - OB_1A &= - 12( - A_1O_1O) - 12( - AOO_1) &= 12( A_1O_1O + AOO_1) = 2, align* which implies that the line A_1B_1 passes through point P. Similarly, A_2B_2 also passes through point P. We have, align* D_1B_2B_1 &= PB_2B_1 = PAB_1 (P, B_2, A, B_1 concyclic) &= B_1A_1A = B_1CA_1 ( is tangent to O_1) &= B_1CD_1. align* Hence, points C, D_1, B_1, B_2 are concyclic. Similarly, D_2 also lies on the circumcircle of CB_1B_2. Therefore, D_1D_2B_1 = D_1B_2B_1 = B_1A_1A. This implies that D_1D_2 is parallel to . Proof completed. *Proof* 2. We establish a coordinate system with as the x-axis and A as the origi…

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China-TST-2023B

China 2023 geometry

Problem

Let three circles

⊙ O

⊙ O_{1}

⊙ O_{2}

, each externally tangent to the other two, lie on the same side of a line

ℓ

. Let them be tangent to

ℓ

at points

A

A_{1}

A_{2}

respectively, where point

A

lies on segment

A_{1} A_{2}

. Denote the points of tangency of

⊙ O

with

⊙ O_{1}

and

⊙ O_{2}

B_{1}

and

B_{2}

respectively, and the point of tangency of

⊙ O_{1}

with

⊙ O_{2}

C

. Let line

A_{1} C

intersect line

A_{2} B_{2}

at point

D_{1}

, and line

A_{2} C

intersect line

A_{1} B_{1}

at point

D_{2}

. Prove that line

D_{1} D_{2}

is parallel to line

ℓ

Solution

Proof 1. Consider the antipodal point

P

A

on the circle

⊙ O

. We notice that in the right trapezoid

A_{1} A_{2} O_{1} O_{2}

, we have

O_{1} A_{1} = O_{1} B_{1}

and

O B_{1} = O A

. Therefore,

∠ A_{1} B A = π - ∠ O_{1} B_{1} A_{1} - ∠ O B_{1} A = π - \frac{1}{2} (π - ∠ A_{1} O_{1} O) - \frac{1}{2} (π - ∠ A O O_{1}) = \frac{1}{2} (∠ A_{1} O_{1} O + ∠ A O O_{1}) = \frac{π}{2},

which implies that the line

A_{1} B_{1}

passes through point

P

. Similarly,

A_{2} B_{2}

also passes through point

P

. We have,

∠ D_{1} B_{2} B_{1} = ∠ P B_{2} B_{1} = ∠ P A B_{1} (P, B_{2}, A, B_{1} concyclic) = ∠ B_{1} A_{1} A = ∠ B_{1} C A_{1} (ℓ is tangent to ⊙ O_{1}) = ∠ B_{1} C D_{1} .

Hence, points

C, D_{1}, B_{1}, B_{2}

are concyclic. Similarly,

D_{2}

also lies on the circumcircle of

△ C B_{1} B_{2}

. Therefore,

∠ D_{1} D_{2} B_{1} = ∠ D_{1} B_{2} B_{1} = ∠ B_{1} A_{1} A .

This implies that

D_{1} D_{2}

is parallel to

ℓ

. Proof completed.

□

Proof 2. We establish a coordinate system with

ℓ

as the

x

-axis and

A

as the origin. Without loss of generality, let

O (0, 1)

. For

i = 1, 2

, let

r_{i}

be the radius of

⊙ O_{i}

, which corresponds to the

y

-coordinate of point

O_{i}

. From the equation

x_{O_{i}}^{2} + (r_{i} - 1)^{2} = (r_{i} + 1)^{2},

we obtain

O_{1} (- 2 r_{1}, r_{1})

and

O_{2} (2 r_{2}, r_{2})

(thus

A_{1} (- 2 r_{1}, 0)

and

A_{2} (2 r_{2}, 0)

). Furthermore, due to the tangent property of

⊙ O_{1}

and

⊙ O_{2}

, we have

[2 (r_{1} + r_{2})]^{2} + (r_{1} - r_{2})^{2} = (r_{1} + r_{2})^{2},

which simplifies to

(*) r_{1} + r_{2} = r_{1} r_{2} .

On the other hand, we have

A B_{1} = \frac{1 \cdot A O _{1} + r _{1} \cdot A O}{r _{1} + 1},

which yields

B_{1} (\frac{- 2 r _{1}}{r _{1} + 1}, \frac{2 r _{1}}{r _{1} + 1})

. Similarly, we have

B_{2} (\frac{2 r _{2}}{r _{2} + 1}, \frac{2 r _{2}}{r _{2} + 1})

and

C (\frac{2 ( r _{1} r _{2} - r _{2} r _{1} )}{r _{1} + r _{2}}, \frac{2 r _{1} r _{2}}{r _{1} + r _{2}})

y = \frac{\frac{2 r _{1}}{r _{1} + 1}}{\frac{- 2 r _{1}}{r _{1} + 1} + 2 r _{1}} (x + 2 r_{1}) = \frac{1}{r _{1}} x + 2,

and the equation of line

C A_{2}

is given by

y = \frac{\frac{2 r _{1} r _{2}}{r _{1} + r _{2}}}{\frac{2 ( r _{1} r _{2} - r _{2} r _{1} )}{r _{1} + r _{2}} - 2 r _{2}} (x - 2 r_{2}),

which can be combined to obtain

y = \frac{2 r _{1} r _{2}}{2 r _{1} r _{2} ( r _{1} - r _{2} ) - 2 r _{2} ( r _{1} + r _{2} )} [r_{1} (y - 2) - 2 r_{2}] .

Rearranging the terms, we obtain a linear equation in terms of

y

, with the constant term

4 r_{1} r_{2} (r_{1} + r_{2}),

which is a symmetric expression in terms of

r_{1}

and

r_{2}

. The coefficient of

y

in this case is (using () to convert the expression into a homogeneous form) $2 {r_{1} r_{2} r_{1} - r_{1} r_{2} (r_{1} - r_{2}) + r_{2} (r_{1} + r_{2})} = 2 {(r_{1} + r_{2})^{2} r_{1} - r_{1} r_{2} (r_{1} - r_{2}) + r_{2} (r_{1} + r_{2})} = 2 {r_{1} r_{1} + 2 r_{1} r_{2} + r_{2} r_{1} - r_{1} r_{2} + r_{2} r_{1} + r_{1} r_{2} + r_{2} r_{2}} = 2 {r_{1} r_{1} + 2 r_{1} r_{2} + 2 r_{2} r_{1} + r_{2} r_{2}},$ which is also a symmetric expression in terms of $r_{1}$ and $r_{2}$ . Therefore, we know that the $y$ -coordinates of points $D_{1}$ and $D_{2}$ have the same expression, which proves the proposition. $□$

Proof 3.* (In fact, it is not necessary for circle

O

to be tangent to line

ℓ

.) It is easy to see that

O, B_{1}, O_{1}

are collinear,

O, B_{2}, O_{2}

are collinear,

O_{1}, C, O_{2}

are collinear,

O B_{1} = O B_{2}

O_{1} A_{1} = O_{1} B_{1}

O_{2} A_{2} = O_{2} B_{2}

, and

O_{1} A_{1}, O_{2} A_{2}

are both perpendicular to

ℓ

. Now, we have:

∠ C D_{1} B_{2} = ∠ C A_{1} A_{2} + ∠ A_{1} A_{2} B_{2} = ∠ C A_{1} B_{1} + ∠ B_{1} A_{1} A_{2} + ∠ A_{1} A_{2} B_{2} = \frac{1}{2} ∠ C O_{1} B_{1} + \frac{1}{2} ∠ A_{1} O_{1} B_{1} + \frac{1}{2} ∠ A_{2} O_{2} B_{2} = (9 0^{\circ} - ∠ O_{1} B_{1} C) + \frac{1}{2} ∠ O_{1} O O_{2} = (∠ O B_{1} C - 9 0^{\circ}) + 9 0^{\circ} - ∠ O B_{1} B_{2} = ∠ C B_{1} B_{2} .

Therefore,

C, D_{1}, B_{1}, B_{2}

are concyclic. Similarly,

C, D_{2}, B_{2}, B_{1}

are concyclic. Hence,

C, D_{1}, B_{1}, B_{2}, D_{2}

are concyclic. Furthermore,

∠ C D_{1} D_{2} = ∠ C B_{1} D_{2} = 18 0^{\circ} - ∠ A_{1} B_{1} C = ∠ C A_{1} A_{2}

. Thus, we have

D_{1} D_{2} // ℓ

□

Techniques

TangentsCartesian coordinatesHomothetyAngle chasing