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USA IMO 2003

United States 2003 geometry

Problem

Let

\overline{A H_{1}}

\overline{B H_{2}}

, and

\overline{C H_{3}}

be the altitudes of an acute scalene triangle

A B C

. The incircle of triangle

A B C

is tangent to

\overline{B C}

\overline{C A}

, and

\overline{A B}

T_{1}, T_{2}

, and

T_{3}

, respectively. For

k = 1, 2, 3

, let

P_{i}

be the point on line

H_{i} H_{i + 1}

(where

H_{4} = H_{1}

) such that

H_{i} T_{i} P_{i}

is an acute isosceles triangle with

H_{i} T_{i} = H_{i} P_{i}

. Prove that the circumcircles of triangles

T_{1} P_{1} T_{2}

T_{2} P_{2} T_{3}

T_{3} P_{3} T_{1}

pass through a common point.

Solution

First Solution. (By Po-Ru Loh) We begin by showing that points

O_{3}, H_{2}, T_{2}

, and

O_{3}

lie on a cyclic. We will prove this by establishing

∠ O_{3} O_{1} H_{2} = ∠ O_{3} T_{2} C = ∠ O_{3} T_{2} H_{2}

. To find

∠ O_{3} O_{1} H_{2}

, observe that triangles

H_{2} A H_{3}

and

H_{2} H_{1} C

are similar. Indeed, quadrilateral

B H_{3} H_{2} C

is cyclic so

∠ H_{2} H_{3} A = ∠ C

, and likewise

∠ C H_{1} H_{2} = ∠ A

. Now,

O_{1}

and

O_{3}

are corresponding incenters of similar triangles, so it follows that triangles

H_{2} A O_{1}

and

H_{2} H_{1} O_{3}

are also similar, and hence are related by a spiral similarity about

H_{2}

. Thus,

\frac{A H _{2}}{H _{1} H _{2}} = \frac{O _{1} H _{2}}{O _{3} H _{2}}

and

∠ A H_{2} H_{1} = ∠ A H_{2} O_{1} + ∠ O_{1} H_{2} H_{1} = ∠ O_{1} H_{2} H_{1} + ∠ H_{1} H_{2} O_{3} = ∠ O_{1} H_{2} O_{3} .

It follows that another spiral similarity about

H_{2}

takes triangle

H_{2} A H_{1}

to triangle

H_{2} O_{1} O_{3}

. Hence

∠ O_{3} O_{1} H_{2} = ∠ H_{1} A H_{2} = 9 0^{\circ} - ∠ C

. We wish to show that

∠ O_{3} T_{2} C = 9 0^{\circ} - ∠ C

as well, or in other words,

T_{2} O_{3} ⊥ B C

. To do this, drop the altitude from

O_{3}

B C

and let it intersect

B C

D

. Triangles

A B C

and

H_{1} H_{2} C

are similar as before, with corresponding incenters

I

and

O_{3}

. Furthermore,

I T_{2}

and

O_{3} D

also correspond. Hence,

C T_{2} / T_{2} A = C D / D H_{1}

, and so

T_{2} D ∥ A H_{1}

. Thus,

T_{2} D ⊥ B C

, and it follows that

T_{2} O_{3} ⊥ B C

.

Having shown that

O_{1} H_{2} T_{2} O_{3}

is cyclic, we may now write

∠ O_{1} T_{2} O_{3} = ∠ O_{1} H_{2} O_{3}

. Since triangles

H_{2} A O_{1}

and

H_{2} H_{1} O_{3}

are related by a spiral similarity about

H_{2}

, we have

∠ O_{1} H_{2} O_{3} = ∠ A H_{2} H_{1} = 18 0^{\circ} - ∠ B,

by noting that

A B H_{2} H_{1}

is cyclic. Likewise,

∠ O_{2} T_{3} O_{1} = 18 0^{\circ} - ∠ C and ∠ O_{3} T_{1} O_{2} = 18 0^{\circ} - ∠ A,

and so

∠ O_{1} T_{2} O_{3} + ∠ O_{2} T_{3} O_{1} + ∠ O_{3} T_{1} O_{2} = 36 0^{\circ}

. Therefore,

∠ T_{3} O_{1} T_{2}

∠ T_{1} O_{2} T_{3}

, and

∠ T_{2} O_{3} T_{1}

of hexagon

O_{1} T_{2} O_{3} T_{1} O_{2} T_{3}

also sum to

36 0^{\circ}

. Now let

H

be the intersection of circles

ω_{1}

and

ω_{2}

. Then

∠ T_{2} H T_{3} = 18 0^{\circ} - \frac{1}{2} ∠ T_{3} O_{1} T_{2}

and

∠ T_{3} H T_{1} = 18 0^{\circ} - \frac{1}{2} ∠ T_{1} O_{2} T_{3}

. Therefore,

∠ T_{1} H T_{2} = 36 0^{\circ} - ∠ T_{2} H T_{3} - ∠ T_{3} H T_{1} = \frac{1}{2} ∠ T_{3} O_{1} T_{2} + \frac{1}{2} ∠ T_{1} O_{2} T_{3} = 18 0^{\circ} - \frac{1}{2} ∠ T_{1} O_{3} T_{2},

and so

H

lies on the circle

ω_{3}

as well. Hence, circles

ω_{1}

ω_{2}

, and

ω_{3}

share a common point, as wanted.

Second Solution. (By Anders Kaseorg) Note that

A H_{2} = A B cos ∠ A

and

A H_{3} = A C cos ∠ A

, so triangles

A H_{2} H_{3}

and

A B C

are similar with ratio

cos ∠ A

. Thus, since

O_{1}

is the incenter of triangle

A H_{2} H_{3}

A O_{1} = A I cos ∠ A

. If

X_{1}

is the intersection of segments

A I

and

T_{2} T_{3}

,

we have

∠ I X_{1} T_{2} = ∠ A T_{2} I = 9 0^{\circ}

, and so

X_{1} I = T_{2} I cos ∠ T_{2} I A = A I cos^{2} ∠ T_{2} I A = A I sin^{2} \frac{∠ A}{2} = A I \cdot \frac{1 - cos ∠ A}{2} = \frac{A I - A O _{1}}{2} = \frac{O _{1} I}{2} .

Hence

O_{1} X_{1} = X_{1} I

, so

O_{1}

is the reflection of

I

across line

T_{2} T_{3}

, and

O_{1} T_{2} = I T_{2} = I T_{3} = O_{1} T_{3}

. Therefore,

O_{1} T_{2} I T_{3}

, and similarly

O_{2} T_{3} I T_{1}

and

O_{3} T_{1} I T_{2}

, are rhombi with the same side length

r

, implying that circles

ω_{1}, ω_{2}, ω

have the same radius

r

. We also conclude that

O_{1} T_{2} = T_{3} I = O_{2} T_{1}

and

O_{1} T_{2} ∥ T_{3} I ∥ O_{2} T_{1}

, and so

O_{1} O_{2} T_{1} T_{2}

is a parallelogram. Hence the midpoints of

O_{1} T_{1}

and

O_{2} T_{2}

(similarly

O_{3} T_{3}

) are the same point

P

, and

O_{1} O_{2} O_{3}

is the reflection of

T_{1} T_{2} T_{3}

across

P

. If

H

is the reflection of

I

across

P

, we have

O_{1} H = O_{2} H = O_{3} H = r

, that is,

H

is a common point of the three circumcircles.

Third Solution. We use directed lengths (along line

B C

, with

C

B

as the positive direction) and directed angles modulo

18 0^{\circ}

in this proof. (For segments not lying on line

B C

, we assume its direction as the direction of its projection on line

B C

.) We claim that

ω_{i}

i = 1, 2, 3

, all pass through

H

, the orthocenter of triangle

T_{1} T_{2} T_{3}

. Without loss of generality, it suffices to prove that

T_{1} P_{1} T_{2} H

is cyclic. If

A B = A C

, then

T_{1} = H_{1} = P_{1}

and the case is trivial. Let

A B = c

B C = a

C A = b

∠ B A C = α

∠ C B A = β

, and

∠ A C B = γ

.

Let

Q

be the intersection of lines

H P_{1}

and

B C

. Note that

∠ H T_{2} T_{1} = 9 0^{\circ} - ∠ T_{2} T_{1} T_{3} = 9 0^{\circ} - [18 0^{\circ} - ∠ T_{3} T_{1} B - ∠ C T_{1} T_{2}] = 9 0^{\circ} - [18 0^{\circ} - (9 0^{\circ} - \frac{β}{2}) - (9 0^{\circ} - \frac{C}{2})] = \frac{α}{2} .

(Likewise,

∠ T_{2} T_{1} H = β /2

.) Thus to prove that

T_{1} P_{1} T_{2} H

is cyclic is equivalent to prove that

∠ Q P_{1} T_{1} = α /2

. Let

Q_{H}

and

Q_{P}

be the respective feet of perpendiculars from

H

and

P_{1}

to line

B C

. Because

∠ A H_{1} B = ∠ A H_{2} B = 9 0^{\circ}

A B H_{1} H_{2}

is cyclic, and so

∠ T_{1} H_{1} P_{1} = ∠ B H_{1} P_{1} = α

. Thus triangles

A T_{3} T_{2}

and

H_{1} T_{1} P_{1}

are similar, implying that

∠ Q_{P} P_{1} T_{1} = 9 0^{\circ} - ∠ P_{1} T_{1} H_{1} = 9 0^{\circ} - (9 0^{\circ} - \frac{∠ T _{1} H _{1} P _{1}}{2}) = \frac{α}{2} .

Therefore, to prove that

∠ Q_{P} P_{1} T_{1} = α /2

, we have now reduced to proving that

Q_{P} = Q_{H}

, or

\frac{T _{1} Q _{P}}{T _{1} H _{1}} = \frac{T _{1} Q _{H}}{T _{1} H _{1}} . (1)

Note that

T_{1} H_{1} = P_{1} H_{1} and \frac{T _{1} Q _{P}}{T _{1} H _{1}} = 1 - \frac{Q _{P} H _{1}}{T _{1} H _{1}},

that is,

\frac{T _{1} Q _{P}}{T _{1} H _{1}} = 1 - \frac{Q _{P} H _{1}}{P _{1} H _{1}} = 1 - cos ∠ T_{1} H_{1} P_{1} = 1 - cos α . (2)

On the other hand, applying the Law of Cosines to triangle

A B C

gives

T_{1} H_{1} = T_{1} C - H_{1} C = \frac{a + b - c}{2} - b cos γ = \frac{a + b - c}{2} - \frac{a ^{2} + b ^{2} - c ^{2}}{2 a} = \frac{a ( b - c ) - ( b ^{2} - c ^{2} )}{2 a},

T_{1} H_{1} = \frac{( b - c ) ( a - b - c )}{2 a} = \frac{( c - b ) ( b + c - a )}{2 a} . (3)

Now we calculate

T_{1} Q_{H}

. Because

H

is the orthocenter of triangle

T_{1} T_{2} T_{3}

∠ T_{1} H T_{2} = 18 0^{\circ} - ∠ H T_{2} T_{1} - ∠ T_{2} T_{1} H = (9 0^{\circ} - ∠ H T_{2} T_{1}) + (9 0^{\circ} - ∠ T_{2} T_{1} H) = ∠ T_{2} T_{1} T_{3} + ∠ T_{3} T_{2} T_{1} = 18 0^{\circ} - ∠ T_{1} T_{3} T_{2} .

Applying the Law of Sines to triangle

T_{1} T_{2} H

and applying the Extended Law of Sines to triangle

T_{1} T_{2} T_{3}

gives

\frac{T _{1} H}{sin ∠ H T _{2} T _{1}} = \frac{T _{1} T _{2}}{sin ∠ T _{1} H T _{2}} = \frac{T _{1} T _{2}}{sin ∠ T _{1} T _{3} T _{2}} = 2 r,

and consequently,

T_{1} H = 2 r sin \frac{α}{2} .

Because

∠ Q_{H} T_{1} H = ∠ C T_{1} T_{2} + ∠ T_{2} T_{1} H = (9 0^{\circ} - \frac{γ}{2}) + \frac{β}{2} = 9 0^{\circ} + \frac{β - γ}{2},

we obtain

T_{1} Q_{H} = T_{1} H cos ∠ H T_{1} Q_{H} = 2 r sin \frac{α}{2} sin \frac{γ - β}{2} . (4)

Combining equations (1), (2), (3), and (4), we conclude that it suffices to prove that

1 - cos α = \frac{4 a r sin \frac{α}{2} sin \frac{γ - β}{2}}{( c - b ) ( b + c - a )} . (5)

Applying the fact

\frac{sin \frac{α}{2}}{cos \frac{α}{2}} = tan \frac{α}{2} = \frac{r}{A T _{2}} = \frac{2 r}{b + c - a},

and applying the Law of Sines to triangle

A B C

, (5) becomes

1 - cos α = \frac{2 sin α sin ^{2} \frac{α}{2} sin \frac{γ - β}{2}}{cos \frac{α}{2} ( sin γ - sin β )} . (6)

By the Double-angle formulas,

1 - cos α = 2 sin^{2} \frac{α}{2}

and

sin α = 2 sin \frac{α}{2} cos \frac{α}{2}

and so (6) reads

sin γ - sin β = 2 sin \frac{α}{2} sin \frac{γ - β}{2} .

By the Difference-to-product formulas, the last equation reduces to

2 cos \frac{β + γ}{2} sin \frac{γ - β}{2} = 2 sin \frac{α}{2} sin \frac{γ - β}{2},

which is evident.

Techniques

Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleSpiral similarityCyclic quadrilateralsTangentsTrigonometryAngle chasing