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International Mathematical Olympiad

geometry

Problem

Let

A B C

be a triangle, and let

ℓ_{1}

and

ℓ_{2}

be two parallel lines. For

i = 1, 2

, let

ℓ_{i}

meet the lines

B C

C A

, and

A B

X_{i}

Y_{i}

, and

Z_{i}

, respectively. Suppose that the line through

X_{i}

perpendicular to

B C

, the line through

Y_{i}

perpendicular to

C A

, and finally the line through

Z_{i}

perpendicular to

A B

, determine a non-degenerate triangle

Δ_{i}

. Show that the circumcircles of

Δ_{1}

and

Δ_{2}

are tangent to each other.

Solution

Throughout the solutions,

\Varangle (p, q)

will denote the directed angle between lines

p

and

q

, taken modulo

18 0^{\circ}

. Let the vertices of

Δ_{i}

D_{i}, E_{i}, F_{i}

, such that lines

E_{i} F_{i}

F_{i} D_{i}

and

D_{i} E_{i}

are the perpendiculars through

X, Y

and

Z

, respectively, and denote the circumcircle of

Δ_{i}

ω_{i}

. In triangles

D_{1} Y_{1} Z_{1}

and

D_{2} Y_{2} Z_{2}

we have

Y_{1} Z_{1} ∥ Y_{2} Z_{2}

because they are parts of

ℓ_{1}

and

ℓ_{2}

. Moreover,

D_{1} Y_{1} ∥ D_{2} Y_{2}

are perpendicular to

A C

and

D_{1} Z_{1} ∥ D_{2} Z_{2}

are perpendicular to

A B

, so the two triangles are homothetic and their homothetic centre is

Y_{1} Y_{2} \cap Z_{1} Z_{2} = A

. Hence, line

D_{1} D_{2}

passes through

A

. Analogously, line

E_{1} E_{2}

passes through

B

and

F_{1} F_{2}

passes through

C

. The corresponding sides of

Δ_{1}

and

Δ_{2}

are parallel, because they are perpendicular to the respective sides of triangle

A B C

. Hence,

Δ_{1}

and

Δ_{2}

are either homothetic, or they can be translated to each other. Using that

B, X_{2}, Z_{2}

and

E_{2}

are concyclic,

C, X_{2}, Y_{2}

and

F_{2}

are concyclic,

Z_{2} E_{2} ⊥ A B

and

Y_{2}, F_{2} ⊥ A C

we can calculate

\Varangle (E_{1} E_{2}, F_{1} F_{2}) = \Varangle (E_{1} E_{2}, X_{1} X_{2}) + \Varangle (X_{1} X_{2}, F_{1} F_{2}) = \Varangle (B E_{2}, B X_{2}) + \Varangle (C X_{2}, C F_{2}) = \Varangle (Z_{2} E_{2}, Z_{2} X_{2}) + \Varangle (Y_{2} X_{2}, Y_{2} F_{2}) = \Varangle (Z_{2} E_{2}, ℓ_{2}) + \Varangle (ℓ_{2}, Y_{2} F_{2}) = \Varangle (Z_{2} E_{2}, Y_{2} F_{2}) = \Varangle (A B, A C) \neq \equiv 0, (1)

and conclude that lines

E_{1} E_{2}

and

F_{1} F_{2}

are not parallel. Hence,

Δ_{1}

and

Δ_{2}

are homothetic; the lines

D_{1} D_{2}

E_{1} E_{2}

, and

F_{1} F_{2}

are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by

H

. For

i = 1, 2

, using (1), and that

A, Y_{i}, Z_{i}

and

D_{i}

are concyclic,

\Varangle (H E_{i}, H F_{i}) = \Varangle (E_{1} E_{2}, F_{1} F_{2}) = \Varangle (A B, A C) = \Varangle (A Z_{i}, A Y_{i}) = \Varangle (D_{i} Z_{i}, D_{i} Y_{i}) = \Varangle (D_{i} E_{i}, D_{i} F_{i}),

H

lies on circle

ω_{i}

. The same homothety that maps

Δ_{1}

Δ_{2}

, sends

ω_{1}

ω_{2}

as well. Point

H

, that is the centre of the homothety, is a common point of the two circles, That finishes proving that

ω_{1}

and

ω_{2}

are tangent to each other.

---

Alternative solution.

As in the first solution, let the vertices of

Δ_{i}

D_{i}, E_{i}, F_{i}

, such that

E_{i} F_{i}

F_{i} D_{i}

and

D_{i} E_{i}

are the perpendiculars through

X_{i}, Y_{i}

and

Z_{i}

, respectively. In the same way we conclude that

(A, D_{1}, D_{2}), (B, E_{1}, E_{2})

and

(C, F_{1}, F_{2})

are collinear. The corresponding sides of triangles

A B C

and

D_{i} E_{i} F_{i}

are perpendicular to each other. Hence, there is a spiral similarity with rotation

\pm 9 0^{\circ}

that maps

A B C

D_{i} E_{i} F_{i}

; let

M_{i}

be the centre of that similarity. Hence,

\Varangle (M_{i} A, M_{i} D_{i}) = \Varangle (M_{i} B, M_{i} E_{i}) = \Varangle (M_{i} C, M_{i} F_{i}) = 9 0^{\circ}

. The circle with diameter

A D_{i}

passes through

M_{i}, Y_{i}, Z_{i}

, so

M_{i}, A, Y_{i}, Z_{i}, D_{i}

are concyclic; analogously (

M_{i}, B, X_{i}, Z_{i}, E_{i}

) and (

M_{i}, C, X_{i}, Y_{i}, F_{i}

) are concyclic. By applying Desargues' theorem to triangles

A B C

and

D_{i} E_{i} F_{i}

we conclude that the lines

A D_{i}, B E_{i}

and

B F_{i}

are concurrent; let their intersection be

H

. Since

(A, D_{1}, D_{2}), (B, E_{1}, E_{2})

and

(C, F_{1}, F_{2})

are collinear, we obtain the same point

H

for

i = 1

and

i = 2

. By

\Varangle (C B, C H) = \Varangle (C X_{i}, C F_{i}) = \Varangle (Y_{i} X_{i}, Y_{i} F_{i}) = \Varangle (Y_{i} Z_{i}, Y_{i} D_{i}) = \Varangle (A Z_{i}, A D_{i}) = \Varangle (A B, A H)

, point

H

lies on circle

A B C

. Analogously, from

\Varangle (F_{i} D_{i}, F_{i} H) = \Varangle (F_{i} Y_{i}, F_{i} C) = \Varangle (X_{i} Y_{i}, X_{i} C) = \Varangle (X_{i} Z_{i}, X_{i} B) = \Varangle (E_{i} Z_{i}, E_{i} B) = \Varangle (E_{i} D_{i}, E_{i} H)

, we can see that point

H

lies on circle

D_{i} E_{i} F_{i}

as well. Therefore, circles

A B C

and

D_{i} E_{i} F_{i}

intersect at point

H

. The spiral similarity moves the circle

A B C

to circle

D_{i} E_{i} F_{i}

, so the two circles are perpendicular. Hence, both circles

D_{1} E_{1} F_{1}

and

D_{2} E_{2} F_{2}

are tangent to the radius of circle

A B C

H

Techniques

HomothetySpiral similarityDesargues theoremAngle chasingTangentsCyclic quadrilaterals