7. In an equilateral triangle, 3 circles are inscribed such that each touches two sides and the inscribed circle $k$ of the triangle. Determine the ratio of the area of circle $k$ to the sum of the areas of these three inscribed circles.
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-25.jpg?height=688&width=763&top_left_y=1015&top_left_x=401)

Sketch:

$1 \text{POINT}$

Let point $S$ be the center of the inscribed circle. Thus, point $S$ is the intersection of the angle bisectors.

$|\measuredangle B A C|=60^{\circ}$, and line $AS$ is the angle bisector of $\measuredangle B A C$, so $|\measuredangle E A S|=30^{\circ}$.

1 POINT

Point $E$ is the point of tangency, $\overline{S E} \perp \overline{A B} \Rightarrow|\measuredangle A E S|=90^{\circ}$.

Point $F$ is the point of tangency, $\overline{G F} \perp \overline{A B} \Rightarrow|\measuredangle A F G|=90^{\circ}$.

1 POINT

Triangles $\triangle A F G$ and $\triangle A E S$ are right triangles with angles $30^{\circ}$ and $60^{\circ} \Rightarrow$ the length of the hypotenuse is twice the length of the shorter leg: $|A G|=2|G F|,|A S|=2|S E|$. Let $R$ be the radius of the larger circle, and $r$ be the radius of the smaller circle.

$R=|S E|=|S D|$, and since $|A S|=2|S E|$, it follows that $|A D|=R$.

1 POINT

$r=|G D|=|G F|=|G H|$, and since $|A G|=2|G F|$, it follows that $|A H|=r$.

1 POINT

$R=|A D|=|A H|+|H G|+|G D|=r+r+r=3 r$

1 POINT

Area of the larger circle: $P_{V}=R^{2} \pi=(3 r)^{2} \pi=9 r^{2} \pi$.

$1 \text{POINT}$

Area of the smaller circle: $P_{M}=r^{2} \pi$.

1 POINT

We calculate the desired ratio $\frac{P_{V}}{3 P_{M}}=\frac{9 r^{2} \pi}{3 r^{2} \pi}=\frac{3}{1}=3: 1$.

The ratio of the area of circle $k$ to the sum of the areas of the three inscribed circles is 3:1.

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-26.jpg?height=700&width=785&top_left_y=564&top_left_x=356)

Let $a$ be the length of the side of the equilateral triangle $A B C$.

The height $\overline{A G}$ belongs to the angle bisector of $\measuredangle B A C$, $R$ is the radius of the circle inscribed in triangle $\mathrm{ABC}$ and is one-third the length of the height $\overline{A G}$, so $R=\frac{a \sqrt{3}}{6}$.

The length $\overline{A S}$ is the radius of the circle circumscribed around triangle $A B C$ and its length is two-thirds the length of the height $\overline{A G}$, so $|A S|=\frac{a \sqrt{3}}{3}$.

1 POINT

Triangles AED and AFS are right triangles with congruent acute angles $\left(30^{\circ}\right.$ and $\left.60^{\circ}\right)$, so by the AA similarity criterion, they are similar.

1 POINT

Therefore, for their side lengths, $\frac{|A S|}{|A D|}=\frac{R}{r}$, i.e., $\frac{\frac{a \sqrt{3}}{3}}{x+r}=\frac{\frac{a \sqrt{3}}{6}}{r}$, which simplifies to $r=x$.

The height $\overline{A G}$ has a length of $\frac{a \sqrt{3}}{2}$ and it holds that $x+2 r+2 R=\frac{a \sqrt{3}}{2}$, and further $r=\frac{a \sqrt{3}}{18}$.

2 POINTS

The area of the three inscribed circles that touch the larger circle is $P_{3}=3 \cdot r^{2} \pi=3 \cdot\left(\frac{a \sqrt{3}}{18}\right)^{2} \pi=\frac{a^{2} \pi}{36}$, and the area of the larger circle is $P=R^{2} \pi=\left(\frac{a \sqrt{3}}{6}\right)^{2} \pi=\frac{a^{2} \pi}{12}$.