Question

设

\triangle ABC

是一个锐角不等边三角形，外接圆为

\omega

。

\omega

在

B

与

C

处的切线相交于

T

。令

X

与

Y

分别为

T

到直线

AB

与

AC

的垂足。已知

BT = CT = 16

、

BC = 22

，且

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

。求

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.

答案

答案：717

知识点：切线圆内接四边形正弦定理坐标几何

难度评级：3370

小提示：

切线-弦定理给出

\angle ABT = 180^\circ - C

，所以

TX = 16 \sin C

；且

AXTY

是以

AT

为直径的圆内接四边形，所以

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

大提示：

将

B

和

C

放在

(\mp 11, 0)

，于是

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

。此时

TX^2 + TY^2

与

XY^2

都是

A

的

y

坐标的一次函数。

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.

解答：

由切线-弦定理，

\angle TBC = A

，所以

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

，且

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

；同理

TY = 16 \sin B

。又

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

，所以

A

、

X

、

T

、

Y

共圆，且直径为

AT

，从而

XY = AT \sin A

。由正弦定理，

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

、

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

、

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

，题设条件变为

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

取

B = (-11, 0)

、

C = (11, 0)

。因为

TB = 16

，且

T

在

BC

的垂直平分线上，得

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

。圆心为

O = (0, k)

，且

OB \perp BT

，由此

121 - k\sqrt{135} = 0

，所以

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

，并且

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

。对圆

\omega

上的

A = (x, y)

，展开

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

，得到

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

。因此

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

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

代入后，

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

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

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

，得到

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

。于是

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

，且

\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}\text{。}

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 第 15 题

答案：717

难度评级：3370

小提示：

大提示：

解答：

其他年份的第 15 题