AMC10 2025 B

AMC10 2025 B · Q12

AMC10 2025 B · Q12. It mainly tests Triangles (properties), Area & perimeter.

The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T, R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon. Which of the following is true?

下图显示了一个等边三角形、一个内角为$60^\circ$的菱形，以及一个正六边形，每一个都包含一些相互相切的同余圆盘。分别用$T, R,$ 和$H,$表示每种情况下圆盘总面积与包围多边形面积的比率。以下哪项正确？

(A) \ T=H=R \ T=H=R

(B) \ H<R=T \ H<R=T

(D) \ H<R<T \ H<R<T

(E) \ H<T<R \qquad \ H<T<R \qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

To solve this problem, we can dissect the different shapes into equilateral triangles. We can keep figure $T$ the same, and then divide the rhombus $R$ into two equilateral triangles (as a result of one of the angles equaling $60^\circ$), with one circle in each that is the incircle to its triangle. We can do the same for the hexagon $H$, dividing it into six equilateral triangles with one circle in each, and each circle is tangent to the triangle that encloses it. We can see that $H = R$, since the proportion of the shaded circles to the unshaded portions in the rhombus and the hexagon is the same. Both are equal to an equilateral triangle with a circle inscribed in it; therefore they have an equal ratio. We can use this to estimate that the ratio of the circle to the triangle is the largest for the first figure, as we can obviously see this through intuition and observation. Furthermore, there is no answer choice for $T < H = R$, and it is evidently not $H = R = T$. Hence, the answer to the question is as follows: $\boxed{\textbf{(C)}\ H = R < T}.$

要解此题，我们可以将不同形状分解成等边三角形。我们保持图$T$不变，然后将菱形$R$分成两个等边三角形（因为其中一个角等于$60^\circ$），每个三角形内有一个圆，是其内切圆。对于六边形$H$，我们同样将其分成六个等边三角形，每个三角形内有一个圆，每个圆都与包围它的三角形相切。我们可以看到$H = R$，因为菱形和六边形中阴影圆盘与非阴影部分的比率相同。两者都等于一个内接圆的等边三角形的比率；因此它们有相同的比率。我们可以用此估计第一个图中圆与三角形的比率最大，这可以通过直观观察明显看出。此外，没有$T < H = R$的选项，显然也不是$H = R = T$。因此，答案如下： $\boxed{\textbf{(C)}\ H = R < T}$。

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