AMC10 2025 B

AMC10 2025 B · Q20

AMC10 2025 B · Q20. It mainly tests Pythagorean theorem, Coordinate geometry.

Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below. The diameter of the small circle can be written as $(\sqrt a+b)(\sqrt c+d)$, where $a$, $b$, $c$, and $d$ are integers. What is $a+b+c+d$?

四个全等的半圆内切于边长为1的正方形中，它们的直径在正方形的边上，每个直径的一端在正方形的顶点，相邻半圆相互切线。如图所示，一个以正方形中心为中心的小圆与四个半圆相切。小圆的直径可以写成$(\sqrt a+b)(\sqrt c+d)$，其中$a$、$b$、$c$、$d$是整数。求$a+b+c+d$？

(A) 3 3

(B) 5 5

(D) 9 9

(E) 11 11

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let the radius of the large semicircles be $r$, and let the diameter of the inner be $D$. Draw lines as follows: By the Pythagorean Theorem, $r^2+(1-r)^2=(2r)^2$, so $2r^2+2r-1=0$. Hence $r=\frac{\sqrt{3}-1}{2}$ (we take the positive solution). Now apply the Pythagorean Theorem on the quadrilateral’s altitude down to get $(1-2r)^2+1=(2r+D)^2$. Solving yields $2r+D=\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$. Then $D=\sqrt{6}-\sqrt{2}-\sqrt{3}+1=(\sqrt{3}-1)(\sqrt{2}-1)$. Hence the answer is $3-1+2-1=\boxed{\textbf{(A) } 3}$.

设大半圆的半径为$r$，内圆的直径为$D$。画如下线：由勾股定理，$r^2+(1-r)^2=(2r)^2$，所以$2r^2+2r-1=0$。因此$r=\frac{\sqrt{3}-1}{2}$（取正解）。现在在四边形的垂线到底边应用勾股定理得到$(1-2r)^2+1=(2r+D)^2$。求解得到$2r+D=\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$。则$D=\sqrt{6}-\sqrt{2}-\sqrt{3}+1=(\sqrt{3}-1)(\sqrt{2}-1)$。因此答案为$3-1+2-1=\boxed{\textbf{(A) } 3}$。

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