Question

有一个四面体，它有两个等腰三角形面，边长分别为

5\sqrt{10}

、

5\sqrt{10}

、

10

，还有两个等腰三角形面，边长分别为

5\sqrt{10}

、

5\sqrt{10}

、

18

。四面体的四个顶点都在一个以

S

为球心的球面上，四个面都与一个以

R

为球心的球相切。距离

RS

可写成

\frac{m}{n}

，其中

m

和

n

是互质正整数。求

m + n

。

Consider a tetrahedron with two isosceles triangle faces with side lengths

5\sqrt{10},

5\sqrt{10},

and

10

and two isosceles triangle faces with side lengths

5\sqrt{10},

5\sqrt{10},

and

18.

The four vertices of the tetrahedron lie on a sphere with center

S,

and the four faces of the tetrahedron are tangent to a sphere with center

R.

The distance

RS

can be written as

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

答案

答案：223

知识点：立体几何球坐标几何对称性

难度评级：2990

小提示：

边长

10

与

18

的边彼此相对，另外四条边都是

5\sqrt{10}

。把这两条相对边放在互相垂直的水平轴上；两个球心都在竖直对称轴上。

The edges are

10

and

18

opposite each other plus four of

5\sqrt{10}.

Place those two edges on perpendicular horizontal axes; both centers lie on the vertical symmetry axis.

大提示：

这两条边在竖直方向相距

12

。在对称轴上分别求两个球心：

S

到各顶点等距，

R

到两个相邻面等距。

The two edges sit

12

apart vertically. Find each center on the axis: equal distances to vertices for

S,

equal distances to two adjacent faces for

R.

解答：

四个面的边长总多重集为

\{5\sqrt{10} \times 8,\ 10 \times 2,\ 18 \times 2\}

，且每条边属于两个面，所以四面体

ABCD

有相对边

AB = 10

和

CD = 18

，另外四条边都等于

5\sqrt{10}

。把各点放置为

\begin{aligned} &A = (-5, 0, 12), \\ &B = (5, 0, 12), \\ &C = (0, -9, 0), \\ &D = (0, 9, 0) \end{aligned}

这是相容的，因为

AC^2 = 25 + 81 + 144

= 250

= (5\sqrt{10})^2

。该构型关于

x \to -x

和

y \to -y

对称，所以两个球心都在

z

轴上。

对

S = (0, 0, s)

，令到

A

与到

C

的距离相等，得

25 + (12 - s)^2 = 81 + s^2

，所以

s = \frac{11}{3}

。对

R = (0, 0, t)

，面

ABC

的平面为

4y - 3z + 36 = 0

，面

ACD

的平面为

12x + 5z = 0

，等距条件给出

\frac{36 - 3t}{5} = \frac{5t}{13} \implies t = \frac{117}{16}

并且由两个镜面对称可知此点到四个面的距离都相等（距离为

\frac{45}{16}

）。

因此

RS = \frac{117}{16} - \frac{11}{3}

= \frac{351 - 176}{48} = \frac{175}{48}

，已经最简，所以

m + n = 175 + 48 = 223

。

The four faces have side multiset

\{5\sqrt{10} \times 8,\ 10 \times 2,\ 18 \times 2\},

and each edge lies on two faces, so the tetrahedron

ABCD

has

AB = 10

and

CD = 18

as opposite edges and the other four edges equal to

5\sqrt{10}.

Place

\begin{aligned} &A = (-5, 0, 12), \\ &B = (5, 0, 12), \\ &C = (0, -9, 0), \\ &D = (0, 9, 0), \end{aligned}

which is consistent since

AC^2 = 25 + 81 + 144

= 250

= (5\sqrt{10})^2.

The configuration is symmetric under

x \to -x

and under

y \to -y,

so both centers lie on the

z

-axis.

For

S = (0, 0, s),

equating distances to

A

and

C

gives

25 + (12 - s)^2 = 81 + s^2,

so

s = \frac{11}{3}.

For

R = (0, 0, t),

face

ABC

has plane

4y - 3z + 36 = 0

and face

ACD

has plane

12x + 5z = 0,

so equal distances require

\frac{36 - 3t}{5} = \frac{5t}{13} \implies t = \frac{117}{16},

and by the two mirror symmetries this point is equidistant (at distance

\frac{45}{16}

) from all four faces.

Therefore

RS = \frac{117}{16} - \frac{11}{3}

= \frac{351 - 176}{48} = \frac{175}{48},

which is in lowest terms, so

m + n = 175 + 48 = 223.

2026 AIME II 第 12 题

答案：223

难度评级：2990

小提示：

大提示：

解答：

其他年份的第 12 题