Question

Sea

\triangle ABC

un triángulo acutángulo escaleno con circunferencia circunscrita

\omega.

Las tangentes a

\omega

en

B

y

C

se cortan en

T.

Sean

X

y

Y

las proyecciones de

T

sobre las rectas

AB

y

AC,

respectivamente. Suponga que

BT = CT = 16,

BC = 22,

y

TX^2 + TY^2 + XY^2 = 1143.

Halle

XY^2.

Let

\triangle ABC

be an acute scalene triangle with circumcircle

\omega.

The tangents to

\omega

at

B

and

C

intersect at

T.

Let

X

and

Y

be the projections of

T

onto lines

AB

and

AC,

respectively. Suppose

BT = CT = 16,

BC = 22,

and

TX^2 + TY^2 + XY^2 = 1143.

Find

XY^2.

Respuesta

Respuesta: 717

Conceptos:recta tangente cuadrilátero cíclico ley de los senos geometría analítica

Nivel de dificultad: 3370

Pista pequeña:

El ángulo tangente-cuerda da

\angle ABT = 180^\circ - C,

así que

TX = 16 \sin C;

y

AXTY

es cíclico con diámetro

AT,

así que

XY = AT \sin A

The tangent-chord angle gives

\angle ABT = 180^\circ - C,

so

TX = 16 \sin C;

and

AXTY

is cyclic with diameter

AT,

so

XY = AT \sin A

Pista grande:

Coloque

B

y

C

en

(\mp 11, 0),

de modo que

T = (0, -\sqrt{135}).

Entonces

TX^2 + TY^2

y

XY^2

son ambos lineales en la coordenada

y

de

A.

Place

B

and

C

at

(\mp 11, 0),

so

T = (0, -\sqrt{135}).

Then

TX^2 + TY^2

and

XY^2

are both linear in the

y

-coordinate of

A.

Solución:

Por el ángulo tangente-cuerda,

\angle TBC = A,

así que

\angle ABT = B + A = 180^\circ - C

y

TX = TB \sin\angle ABT = 16 \sin C;

de manera similar

TY = 16 \sin B.

Además

\angle AXT = \angle AYT = 90^\circ,

así que

A,

X,

T,

y

Y

están sobre una circunferencia de diámetro

AT,

de donde

XY = AT \sin A.

Por la ley de senos,

\sin A = \frac{11}{R},

\sin B = \frac{AC}{2R},

y

\sin C = \frac{AB}{2R},

de modo que la condición dada se convierte en

\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}

Coloque

B = (-11, 0)

y

C = (11, 0).

Como

TB = 16

y

T

está en la mediatriz de

BC,

obtenemos

T = (0, -\sqrt{135}).

El circuncentro es

O = (0, k)

con

OB \perp BT,

lo que da

121 - k\sqrt{135} = 0,

así que

k = \frac{121}{\sqrt{135}}

y

R^2 = 121 + k^2 = \frac{30976}{135}.

Para

A = (x, y)

sobre

\omega,

desarrollando

x^2 + (y - k)^2 = R^2

se obtiene

x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.

Por lo tanto

\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}

\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}

Sustituyendo,

64(AB^2 + AC^2) + 121\,AT^2

= \frac{92928}{\sqrt{135}}\,y + 61952

= 1143 \cdot \frac{30976}{135}

produce

y = \frac{291}{\sqrt{135}}.

Entonces

AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},

y

\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}

By the tangent-chord angle,

\angle TBC = A,

so

\angle ABT = B + A = 180^\circ - C

and

TX = TB \sin\angle ABT = 16 \sin C;

similarly

TY = 16 \sin B.

Also

\angle AXT = \angle AYT = 90^\circ,

so

A,

X,

T,

and

Y

lie on a circle with diameter

AT,

whence

XY = AT \sin A.

Using the law of sines,

\sin A = \frac{11}{R},

\sin B = \frac{AC}{2R},

and

\sin C = \frac{AB}{2R},

so the given condition becomes

\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}

Place

B = (-11, 0)

and

C = (11, 0).

Since

TB = 16

and

T

lies on the perpendicular bisector of

BC,

we get

T = (0, -\sqrt{135}).

The circumcenter is

O = (0, k)

with

OB \perp BT,

which gives

121 - k\sqrt{135} = 0,

so

k = \frac{121}{\sqrt{135}}

and

R^2 = 121 + k^2 = \frac{30976}{135}.

For

A = (x, y)

on

\omega,

expanding

x^2 + (y - k)^2 = R^2

gives

x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.

Therefore

\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}

\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}

Substituting,

64(AB^2 + AC^2) + 121\,AT^2

= \frac{92928}{\sqrt{135}}\,y + 61952

= 1143 \cdot \frac{30976}{135}

yields

y = \frac{291}{\sqrt{135}}.

Then

AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},

and

\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}

2020 AIME II Problema 15

Respuesta: 717

Nivel de dificultad: 3370

Pista pequeña:

Pista grande:

Solución:

El Problema 15 en otros años