2022 AMC 10A 第 25 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

RRSSTT 是坐标平面中的正方形,它们的顶点都在格点上,即两个坐标都是整数的点,并且包含各自内部。

每个正方形的底边都在 xx 轴上。RR 的左边和 SS 的右边都在 yy 轴上,且 RR 中格点数是 SS 中格点数的 94\dfrac{9}{4} 倍。TT 的顶部两个顶点在 RSR \cup S 中,且 TT 中格点数是 RSR \cup S 中格点数的 14\dfrac{1}{4}。见图,图未按比例绘制。

SS 中属于 STS \cap T 的格点所占比例,是 RR 中属于 RTR \cap T 的格点所占比例的 2727 倍。求 RRSSTT 的边长之和的最小可能值。

Let R,R, S,S, and TT be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors.

The bottom edge of each square is on the xx-axis. The left edge of RR and the right edge of SS are on the yy-axis, and RR contains 94\dfrac{9}{4} as many lattice points as does S.S. The top two vertices of TT are in RS,R \cup S, and TT contains 14\dfrac{1}{4} of the lattice points contained in RS.R \cup S. See the figure (not drawn to scale).

The fraction of lattice points in SS that are in STS \cap T is 2727 times the fraction of lattice points in RR that are in RT.R \cap T. What is the minimum possible value of the edge length of RR plus the edge length of SS plus the edge length of T?T?

336336

337337

338338

339339

340340

答案:B
知识点:格点模运算最优化
难度评级:2600
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rrRR 的每条边上的格点数; 类似地,令 ss 对应 SS,令 tt 对应 TT

注意,一个矩形中的格点数等于其宽和长方向上的格点数之积。 r2=94s2 r^2 = \dfrac{9}{4} \cdot s^2 r=32s(1)r = \dfrac{3}{2} \cdot s \tag*{(1)}

第一个条件给出 RSR \cup S 中的格点数等于两个正方形格点数之和,再减去它们在 yy 轴上的重合格点数; 这段重合边就是正方形 SS 与该轴相接之处。

因此第二个条件给出 由 (1)(1) 可知 ss22 的倍数。 把 ss 换成 2j2j,得到 为了使乘积能被 44 整除,jj 也必须被 22 整除。 再把 jj 换成 2k2k,得到 t2=14(r2+s2s) t^2 = \dfrac{1}{4}(r^2 + s^2 - s) t2=14(94s2+s2s) t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s) t2=1413s24s4 t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4} 16t2=s(13s4). 16t^2 = s(13s - 4). 16t2=2j(26j4) 16t^2 = 2j(26j - 4) 4t2=j(13j2). 4t^2 = j(13j - 2). 4t2=2k(26k2) 4t^2 = 2k(26k - 2) t2=k(13k1)(2) t^2 = k(13k - 1) \tag*{(2)}

xx 为矩形 STS \cap T 底边上的格点数,令 yy 为矩形 RTR \cap T 底边上的格点数。

这样,STS \cap T 中的格点数为 xtxt,而 RTR \cap T 中的格点数为 ytyt

第三个条件给出 xts2=27ytr2 \dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2} xs2=27y94s2 \dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} x=12y. x = 12y.

又有 t=x+y1t = x + y - 1,于是 t=13y1(3) t = 13y - 1 \tag*{(3)}

(3)(3) 可得 t1(mod13),t21(mod13). \begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13}. \end{gathered}

另一方面,由 (2)(2) 可得 因此 t2k(mod13) t^2 \equiv -k \pmod{13} k1(mod13). k \equiv -1 \pmod{13}.

同时由 (2)(2) 可知 kk 是平方数,因为它与 13k113k - 1 互质,而它们的乘积是平方数。

因此 kk 必须是满足 k1(mod13)k\equiv-1\pmod{13} 的完全平方数。 较小的正平方数 1,4,9,161,4,9,16 都不满足这一同余条件,而 k=25k=25 满足,所以 2525 是最小可能值。

由这个 k,k, 值可得 j=225=50,j = 2 \cdot 25 = 50, s=250=100,s = 2 \cdot 50 = 100,r=32100=150.r = \dfrac{3}{2} \cdot 100 = 150. 还可算得 t2=25(13251)=25324 t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324 t=518=90. t = 5 \cdot 18 = 90. 因此 r+s+t=340. r + s + t = 340. 不过题目问的是边长之和。每个正方形的边长都比每边的格点数少 11,三个正方形共要减去 3.3.

这个值可以达到:由方程 (3)(3)y=7y=7,从而 x=84.x=84. 因为 xs=100x\le s=100yr=150,y\le r=150,每边有 9090 个格点的正方形 TT 可以跨过 yy 轴形成所需的重叠,而且它的上方两个顶点位于 RS.R\cup S. 中。

因此所求答案为 3403=337.340 - 3 = 337.

所以正确答案是 B

Let rr be the number of lattice points on the side length of R.R. Similarly define ss for SS and tt for T.T. Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length.

The first condition gives us that r2=94s2 r^2 = \dfrac{9}{4} \cdot s^2 r=32s(1)r = \dfrac{3}{2} \cdot s \tag*{(1)}

The number of lattice points in RSR \cup S is the sum of the lattice points in each of the regions, but there is overlap along the yy-axis where SS touches it.

The second condition, therefore, yields t2=14(r2+s2s) t^2 = \dfrac{1}{4}(r^2 + s^2 - s) t2=14(94s2+s2s) t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s) t2=1413s24s4 t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4} 16t2=s(13s4). 16t^2 = s(13s - 4). From (1),(1), we get that ss is a multiple of 2.2. We can substitute ss with 2j2j to get 16t2=2j(26j4) 16t^2 = 2j(26j - 4) 4t2=j(13j2). 4t^2 = j(13j - 2). For the product to be divisible by 4,4, jj must be divisible by 2.2. We can again substitute jj with 2k2k to get 4t2=2k(26k2) 4t^2 = 2k(26k - 2) t2=k(13k1)(2) t^2 = k(13k - 1) \tag*{(2)}

Let xx be the number of lattice points along the bottom of the rectangle formed by STS \cap T and yy be the number of lattice points along the bottom of the rectangle formed by RT.R \cap T.

Using these variables, we get that the number of lattice points in STS \cap T is xtxt and in RTR \cap T is yt.yt.

The third condition gives us that xts2=27ytr2 \dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2} xs2=27y94s2 \dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} x=12y. x = 12y.

We also know that t=x+y1t = x + y - 1 (accounting for overlap), and this yields t=13y1(3) t = 13y - 1 \tag*{(3)}

(3)(3) gives us that t1(mod13),t21(mod13). \begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13}. \end{gathered}

However, by (2),(2), we get that t2k(mod13) t^2 \equiv -k \pmod{13} k1(mod13). k \equiv -1 \pmod{13}.

By (2),(2), we also get that kk is a perfect square since it is relatively prime to 13k1,13k - 1, and they must multiply to a perfect square.

Thus kk must be a perfect square satisfying k1(mod13).k\equiv-1\pmod{13}. The smaller positive squares 1,4,9,161,4,9,16 do not satisfy this congruence, while k=25k=25 does, so 2525 is the least possible value.

From this value of k,k, we get that j=225=50,j = 2 \cdot 25 = 50, s=250=100,s = 2 \cdot 50 = 100, and r=32100=150.r = \dfrac{3}{2} \cdot 100 = 150. We can also find that t2=25(13251)=25324 t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324 t=518=90. t = 5 \cdot 18 = 90. Therefore, r+s+t=340. r + s + t = 340. The question, however, asked for the sum of the side lengths. The side lengths of the squares are 11 less than the number of lattice points on the side, so we have to subtract 3.3.

This value is attainable: equation (3)(3) gives y=7y=7 and hence x=84.x=84. Since xs=100x\le s=100 and yr=150,y\le r=150, a square TT with 9090 lattice points per side can straddle the yy-axis with the required overlaps, and its top vertices lie in RS.R\cup S.

Therefore, the desired answer is 3403=337.340 - 3 = 337.

Thus, B is the correct answer.

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