HMMT February 2025 2025 年 2 月 HMMT

February 15, 2025 2025 年 2 月 15 日

Team Round 团队赛

[20] Let $a, b$ , and $c$ be pairwise distinct positive integers such that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ is an increasing arithmetic sequence in that order. Prove that $\gcd (a, b) > 1$ .
[20] 设 $a, b$ 和 $c$ 为互不相同的正整数，且 $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ 是按此顺序的递增等差数列。证明 $\gcd (a, b) > 1$ 。
Proposed by: Srinivas Arun
提出者：Srinivas Arun
Solution 1: Observe that $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$ , so $b (a + c) = 2 a c$ , and thus $a ∣ b (a + c)$ . If we assume that $\gcd (a, b) = 1$ , then we must have $a ∣ a + c$ , so $a ∣ c$ . However, $\frac{1}{a} < \frac{1}{c}$ , so $a > c$ , contradiction. Thus, $\gcd (a, b) > 1$ , as desired.
解决方案 1：观察到 $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$ ，所以 $b (a + c) = 2 a c$ ，因此 $a ∣ b (a + c)$ 。如果我们假设 $\gcd (a, b) = 1$ ，那么我们必须有 $a ∣ a + c$ ，所以 $a ∣ c$ 。但是 $\frac{1}{a} < \frac{1}{c}$ ，所以 $a > c$ ，矛盾。因此 $\gcd (a, b) > 1$ ，如所求。

Solution 2: Observe that

\frac{2}{b} - \frac{1}{a} = \frac{1}{c}

, so

(2 a - b) c = a b

and thus

2 a - b ∣ a b

. If we assume that

\gcd (a, b) = 1

, then

\gcd (2 a - b, a) = 1

, so

2 a - b ∣ b

. Then

2 a - b ∣ (2 a - b) + b = 2 a

, so

2 a - b ∣ \gcd (2 a, b) \leq 2

. Thus

2 a - b \leq 2

. But

a > b

, contradiction. Thus,

\gcd (a, b) > 1

, as desired.
解决方案 2：观察到

\frac{2}{b} - \frac{1}{a} = \frac{1}{c}

，所以

(2 a - b) c = a b

和因此

2 a - b ∣ a b

。如果我们假设

\gcd (a, b) = 1

，那么

\gcd (2 a - b, a) = 1

，所以

2 a - b ∣ b

。然后

2 a - b ∣ (2 a - b) + b = 2 a

，所以

2 a - b ∣ \gcd (2 a, b) \leq 2

。因此

2 a - b \leq 2

。但是

a > b

，矛盾。因此

\gcd (a, b) > 1

，如所求。
2. [25] A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter 180, and area 2024.
2. [25] 多米诺是一种由一个或多个单位正方形边缘连接而成的连通图形。证明确定没有孔、周长为 180、面积为 2024 的非全等多米诺的数量。
Proposed by: Albert Wang
提出者：Albert Wang
Answer: 2 答案：2
Solution: Define the bounding box of a polyomino to be the smallest axis-aligned rectangle that contains the entire polyomino. Suppose a polyomino satisfying the given conditions has a bounding box with dimensions

w \times h

.
解法：将多格骨牌的边界框定义为包含整个多格骨牌的最小轴对齐矩形。假设一个满足给定条件的多格骨牌有一个边界框，其尺寸为

w \times h

。

Claim 1.

w + h \leq 90

. 声明 1。

w + h \leq 90

。
Proof. The polyomino has at least

2 w

horizontal edges and at least

2 h

vertical edges. Moreover, it has a perimeter of 180 . Therefore,

2 w + 2 h \leq 180

, so

w + h \leq 90

.
证明。这个多边形单元至少有

2 w

条水平边和至少

2 h

条垂直边。此外，它的周长为 180。因此，

2 w + 2 h \leq 180

，所以

w + h \leq 90

。

Claim 2. The dimensions of the bounding box are either

44 \times 46, 45 \times 45

, or

46 \times 44

.
声明 2。边界框的尺寸要么是

44 \times 46, 45 \times 45

，要么是

46 \times 44

。
Proof. Note that

h w \geq 2024

since it contains the polyomino with area 2024. Suppose for sake of contradiction that

h + w \leq 89

. Then,
证明。注意到

h w \geq 2024

，因为它包含面积为 2024 的多边形单元。假设为了矛盾，

h + w \leq 89

。那么，

(h - w)^{2} = (h + w)^{2} - 4 h w \leq 89^{2} - 4 \cdot 2024 = - 175

contradiction. Therefore,

h + w = 90

, so we can let

(h, w) = (45 + x, 45 - x)

. Then,

2025 - x^{2} = h w \geq

2024 implies that

x \in {- 1, 0, 1}

, as desired.
矛盾。因此，

h + w = 90

，所以我们可以让

(h, w) = (45 + x, 45 - x)

。然后，

2025 - x^{2} = h w \geq

2024 意味着

x \in {- 1, 0, 1}

，如所希望的那样。

In the first and third cases, the bounding box has area 2024, so it must be the entire polyomino, giving us the

44 \times 46

rectangle (and its rotation) as a possible answer. In the second case, the bounding box has area 2025, so one cell must be removed to form the polyomino. Removing the corner cell yields a polyomino with perimeter 180, and removing any other cells yields a polyomino with perimeter greater than 180. Therefore, the only other possibility is a

45 \times 45

square missing a corner. Thus the answer is 2 .
在第一和第三个情况下，边界框的面积为 2024，因此它必须是整个多面体，给我们

44 \times 46

矩形（及其旋转）作为可能的答案。在第二个情况下，边界框的面积为 2025，所以必须移除一个单元格来形成多面体。移除角落单元格会产生一个周长为 180 的多面体，而移除其他任何单元格产生的多面体周长都大于 180。因此，唯一的其他可能性是缺少一个角落的

45 \times 45

正方形。所以答案是 2。
3. [30] Let

ω_{1}

and

ω_{2}

be two circles intersecting at distinct points

A

and

B

. Point

X

varies along

ω_{1}

, and point

Y

ω_{2}

is chosen such that

A B

bisects the angle

∠ X A Y

. Prove that as

X

varies along

ω_{1}

, the circumcenter of

△ A X Y

(if it exists) varies along a fixed line.
3. [30] 设

ω_{1}

和

ω_{2}

是两个在

A

和

B

处相交的不同点的圆。点

X

沿着

ω_{1}

变化，点

Y

在

ω_{2}

上被选择，使得

A B

平分角

∠ X A Y

。证明当

X

沿着

ω_{1}

变化时，

△ A X Y

的 circumcenter（如果存在）沿着一条固定直线变化。
Proposed by: Pitchayut Saengrungkongka
提议者：Pitchayut Saengrungkongka

Solution 1: 解决方案 1：

Let

O_{1}, O_{2}

, and

O

be the centers of

ω_{1}, ω_{2}

, and the circumcircle of

△ A X Y

, respectively.
设

O_{1}, O_{2}

和

O

分别是

ω_{1}, ω_{2}

的中心和

△ A X Y

的外接圆的圆心。
We claim that triangle

O O_{1} O_{2}

is isosceles with

O O_{1} = O O_{2}

, and thus in particular

O

always lies on the perpendicular bisector of

O_{1} O_{2}

.
我们断言三角形

O O_{1} O_{2}

是等腰三角形，并且

O O_{1} = O O_{2}

，因此

O

特别地总在

O_{1} O_{2}

的垂直平分线上。
To this end, observe that

O O_{1} ⊥ A X

and

O_{1} O_{2} ⊥ A B

, so

∠ O O_{1} O_{2} = ∠ X A B

. Analogously,

∠ O O_{2} O_{1} = ∠ Y A B

. So indeed

O O_{1} O_{2}

is isosceles, and we are done.
为此，观察到

O O_{1} ⊥ A X

和

O_{1} O_{2} ⊥ A B

，所以

∠ O O_{1} O_{2} = ∠ X A B

。类似地，

∠ O O_{2} O_{1} = ∠ Y A B

。因此

O O_{1} O_{2}

确实是等腰三角形，我们完成了证明。
Remark. One may also consider the antipodes of

A

ω_{1}

and

ω_{2}

for an equivalent but more natural angle-chase.
注记。也可以考虑

A

在

ω_{1}

和

ω_{2}

上的对径点，以获得一个等价但更自然的角 chase。

Solution 2: 解决方案 2：

Let

A^{'}

be the

A

-antipode in circle

(A X Y)

. It suffices to show that

A^{'}

lies on a fixed line. We will show that this line is one that is parallel to

A B

.
设

A^{'}

是圆

(A X Y)

上的

A

-对跖点。只需证明

A^{'}

位于一条固定直线上。我们将证明这条直线平行于

A B

。
Let

M

be the second intersection of line

A B

with circle

(A X Y)

, and let

N

be the antipode of

M

on this circle. Since

A M A^{'} N

is a rectangle with

N

lying on the line through

A

perpendicular to

A B

, it suffices to show that

N

is fixed (independent of

X

and

Y

).
设

M

是直线

A B

与圆

(A X Y)

的第二个交点，设

N

是该圆上

M

的对跖点。由于

A M A^{'} N

是一个矩形，且

N

位于通过

A

且垂直于

A B

的直线上，只需证明

N

是固定的（与

X

和

Y

无关）。
To this end, take an inversion at

A

with arbitrary radius, denoting images with

∙ \mapsto ∙

.
为此，在

A

处进行任意半径的 inversion，用

∙ \mapsto ∙

表示像。
Observe that

X^{*}

and

Y^{*}

lie on the fixed lines

ℓ_{1} = ω_{1}^{*}

and

ℓ_{2} = ω_{2}^{*}

. Let

ℓ

be the line through

A

perpendicular to

A B^{*}

, and suppose that

ℓ_{1}

and

ℓ_{2}

intersect

ℓ

P

and

Q

, respectively.
观察到

X^{*}

和

Y^{*}

位于固定直线

ℓ_{1} = ω_{1}^{*}

和

ℓ_{2} = ω_{2}^{*}

上。设

ℓ

是通过

A

且垂直于

A B^{*}

的直线，并且假设

ℓ_{1}

和

ℓ_{2}

分别与

ℓ

相交于

P

和

Q

。
Since

∠ X A B = ∠ B A Y

, we have

∠ X^{*} A B^{*} = ∠ B^{*} A Y^{*}

. Circles

ω_{1}, ω_{2}

, and

(A X Y)

are mapped to lines

B^{*} X^{*}, B^{*} Y^{*}

, and

X^{*} Y^{*}

. As

A N ⊥ A B

, it follows that

N^{*}

is the intersection of

X^{*} Y^{*}

and

ℓ

.
由于

∠ X A B = ∠ B A Y

，我们有

∠ X^{*} A B^{*} = ∠ B^{*} A Y^{*}

。圆

ω_{1}, ω_{2}

和

(A X Y)

被映射到直线

B^{*} X^{*}, B^{*} Y^{*}

和

X^{*} Y^{*}

。作为

A N ⊥ A B

，可知

N^{*}

是

X^{*} Y^{*}

和

ℓ

的交点。

Finally, observe that

(N^{*}, A; P, Q) \overset{B^{*}}{=} (N^{*}, M^{*}; X, Y)

is a harmonic bundle, as

A M^{*}

bisects

∠ X^{*} A Y^{*}

and

∠ M^{*} A N^{*} = 90^{\circ}

. Since

A, P

, and

Q

are fixed, so is

N^{*}

. Thus

N

is fixed, and

O

lies on the perpendicular bisector of

A N

, which is also fixed.
最后，观察到

(N^{*}, A; P, Q) \overset{B^{*}}{=} (N^{*}, M^{*}; X, Y)

是一个调和束，因为

A M^{*}

平分

∠ X^{*} A Y^{*}

和

∠ M^{*} A N^{*} = 90^{\circ}

。由于

A, P

和

Q

是固定的，所以

N^{*}

也是固定的。因此

N

是固定的，而

O

位于

A N

的垂直平分线上，这也是固定的。
Remark. An alternative approach to the last paragraph is to recall Blanchet’s theorem, which states that

P Y^{*}, Q X^{*}

, and

A B^{*}

are concurrent. By Ceva and Menelaus, we get that

(N^{*}, A; P, Q) = - 1

.
注记。对于最后一段话的另一种方法是回忆 Blanchet 定理，该定理指出

P Y^{*}, Q X^{*}

和

A B^{*}

共点。根据 Ceva 和 Menelaus 定理，我们得到

(N^{*}, A; P, Q) = - 1

。
Remark. If one projects the kite/harmonic quadrilateral

N X M Y

from

A^{'}

onto the line through the antipodes defined in the previous remark, we obtain that

A^{'} N

(a line parallel to

A B

) passes through the midpoint of the two antipodes, directly finishing.
备注。如果将风筝/调和四边形

N X M Y

从

A^{'}

投影到前一条备注中定义的通过对跖点的直线上，我们得到

A^{'} N

（一条平行于

A B

的直线）通过两个对跖点的中点，直接完成。
4. [35] Jerry places at most one rook in each cell of a

2025 \times 2025

grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them.
4. [35] Jerry 在

2025 \times 2025

个单元格的网格的每个单元格中最多放置一个车。如果两辆车在同一行或同一列并且它们之间没有其他车，则一辆车会攻击另一辆车。
Determine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.
确定，并证明 Jerry 可以在网格上放置的最大车数，使得没有车攻击 4 辆其他车。
Proposed by: Arul Kolla
提出者：Arul Kolla
Answer: 8096 答案：8096
Solution 1: The answer is

2024 \times 4 = 8096

. More generally, for an

n \times n

grid, the answer is

4 n - 4

. Call a rook that attacks at most 3 other rooks good.
解法 1：答案是

2024 \times 4 = 8096

。更一般地，对于一个

n \times n

的网格，答案是

4 n - 4

。攻击最多 3 个其他车的车被称为好车。
We use the following observation in both parts of the solution: a rook on the border of the grid must be good.
我们在两部分的解法中都使用了以下观察：网格边上的车必须是好车。
Lower Bound: Place rooks on all

4 n - 4

border cells of the grid. By the above observation, every rook is good.
下界：在网格的所有

4 n - 4

边界单元格上放置车。根据上述观察，每辆车都是好车。
Upper Bound: Consider any valid placement of rooks, and assume there exists a rook that is not on the border. We can move this rook to the border via a usual rook move, since this rook is good and thus the path to one of the four border cells in its row or column must be empty.
上界：考虑任何有效的车放置，假设存在一个不在边框上的车。我们可以通过常规的车移动将这个车移动到边框上，因为这个车是好的，因此在它的行或列到四个边框细胞之一的路径必须是空的。
After this move, we claim the placement of rooks is still valid. Indeed:
在这个移动之后，我们声称车的放置仍然是有效的。确实：

the moved rook is now on the border, so by the observation above, it must be good;
移动的车现在在边框上，所以根据上述观察，它必须是好的；
any rook that used to attack this rook cannot attack more rooks after the move, so such rooks must still be good;
任何以前攻击这个车的车在移动后不能攻击更多的车，所以这样的车仍然必须是好的；
any rook attacked in the final position must be either:
任何在终局位置被攻击的车必须是：
opposite the direction moved, in which case it attacked the moved rook both before and after the move (so is still good), or
与移动方向相反，在这种情况下，它在移动前后都攻击了被移动的车（所以仍然有效），或者
perpendicular to the direction moved, in which case it is a border rook and must always be good.
与移动方向垂直，在这种情况下，它是一边车，并且必须总是有效。

By repeating the above process, we can always move from any good position to one where all rooks are on the border. This implies that the number of rooks in any good position is at most

4 n - 4

. When

n = 2024

, the answer is

4 n - 4 = 8096

.
通过重复上述过程，我们可以从任何有效位置移动到所有车都在一边的位置。这意味着任何有效位置中的车数量最多为

4 n - 4

。当

n = 2024

时，答案是

4 n - 4 = 8096

。

Solution 2: Consider the set of all rooks which are either the leftmost or rightmost in their row, or the topmost or bottommost in their column. Note that this set must include every rook, as any rook not in this set attacks a rook in all 4 directions.
解法 2：考虑所有在其行中是最左或最右，或在其列中是最上或最下的车。注意这个集合必须包括每辆车，因为任何不在该集合中的车会向四个方向攻击车。
Each column contributes at most 2 rooks to this set, and each row contributes at most 2 rooks. We can safely ignore the top and bottom rows in this count, as any rook in the top or bottom row is already the topmost or bottommost rook in its column. Thus the number of rooks in the set is at most

2 \cdot (2025 + 2025 - 2) = 8096

, which can be constructed as seen before.
每一列最多贡献 2 个车到这个集合中，每一行最多贡献 2 个车。我们可以安全地忽略这个计数中的顶部和底部行，因为任何位于顶部或底部行的车已经是其所在列的最顶部或最底部车。因此，这个集合中的车数最多为

2 \cdot (2025 + 2025 - 2) = 8096

，这可以像之前看到的那样构造。