How solve with a short way a problem on the sequences for students of a high school? The equilateral triangle has side length $a=1$; suppose that the number of circles in the last row
(i.e. the circles tangent to the base of the triangle) is $n$ (the figure shows the special case where $n=10$), that of the penultimate row $n-1$ and so on, up to the first row, which contains only one circle.
a. Express as a function of $n$ the area an of the region of the plane occupied altogether by the circles.
b. When $n\to +\infty$, do the circles fill the triangle completely?
The picture is:

Solution: Neighbouring circles are externally tangent to each other, so if $r$ is their common radius and we consider $k$ of them consecutive, from left to right just to fix the ideas, the distance between the left edge of the first circle and the right edge of the last one is $2rk$. If we consider all the circles on the base (lower side), let us suppose them to be $n$, and let us consider the first one on the left, tangent to both sides of the vertex on the lower left, and the last one on the right, tangent to both sides of the vertex on the lower right, there remain on the base between them $n-2$, which must fill exactly the space left between the edge on the right of the first circle and the one on the left of the last circle. How is this space calculated? The edge on the right of the first circle is projected onto the base at a point at distance $x+r$ from the vertex on the left, and the same is done for the edge on the right of the last circle, at distance $x+r$ from the vertex on the right of the base; the space sought is therefore $a-2(x+r)$, and must be filled exactly by $n-2$ circles of radius $r$, from which $(n-2)(2r)=a-2(x+r)$. Of course, not for every radius $r$ we have the desired set of circles, only for the $r$ for which the relation $(n-2)(2r)=a-2(x+r)$ is verified by a natural $n$.
In an angle of $\pi/3$ radians consider a circle of radius $r$ tangent to both sides (therefore with its centre on the bisector); let $x$ be the distance of the point of tangency from the vertex of the angle; then we have $r/x=\tan\pi/6=1/\sqrt3$, therefore $x=r\sqrt3$. If we consider the two circles at the extremity of the base of the triangle, there must be $n-2$ circles of radius $r$ between them, from which the relation $(n-2)(2r)=a-2(x+r)$, where $a$ is the side of the equilateral triangle (in our case $a=1$, but let $a$ be left).
Therefore we have
$$(n-2)2r=a-2r\,(\sqrt3+1)\iff r=\frac a{2(n+\sqrt3-1)}.$$
Each circle has area $\pi\,r^2=\pi a^2/(4(n+\sqrt3-1)^2)$ and there are $1+2+\dots+n=n(n+1)/2$ circles, so the total area of the circles is
$$a_n=\frac{\pi\,n(n+1)}{8(n+\sqrt3-1)^2}\,a^2$$
as given in the problem. The area of the equilateral triangle is $(\sqrt3/4)\,a^2$, the ratio between the areas is therefore:
$$\frac{a_n}{(\sqrt3/4)\,a^2}=\dfrac\pi{2\,\sqrt3}\,\dfrac{n^2+n}{(n+\sqrt3-1)^2}$$
which tends, increasing, to $\pi/(2\sqrt3)\approx 0,907$: even in the limit for $r\to 0^+$, equivalently for $n\to\infty$, the circles cover a little more than $90\%$ of the area of the triangle.
My question: If I were to explain the problem to my high school students ($17-18$ years old) step by step, it would take at least an hour of class time to do it, excluding explanations. Is there a faster and more immediate solution?
 A: An idea for $b)$: The density of the hexagonal lattice circle packing (in the plane) can be calculated quite easily:

If the circles have radius $R$, then the hexagons pictured have sidelength $2R$, thus area* $6 \cdot \frac{\sqrt{3}}{4} 4R^2 = 6 \sqrt{3} R^2$, whereas each hexagon is covered by an area equivalent to three circles**, so $3 \pi R^2$. Thus the circles cover $ \frac{3 \pi R^2}{6 \sqrt{3} R^2} = \frac{\pi}{2\sqrt{3}} \approx 0.906$ of the plane.
This should suffice as an argument that the circles cannot cover 100% of the area of the triangle (for example, when the radius of the circles is small enough, your students can imagine that away from the edges of the triangle, the packing looks like a plane packing so it can't cover the whole triangle.)
In fact, formalizing this argument (neglecting the boundary contributions when the radius becomes smaller and smaller) you get that the circles should cover exactly $\frac{\pi}{2\sqrt{3}}$ of the area of the triangle (which is what you also showed).

*) The area of the hexagon is 6 times the area of an equilateral triangle of side $2R$.
**) to see this, note that the hexagon contains 1 full circle and 6 identical "pieces" of a circle. It should be pretty easy to show that the hexagon is regular, so each angle is $120^\circ$; Thus each circle "piece" is a third of a circle, so the areas of the pieces give $6 \cdot \frac{1}{3} = 2$ additional circles.
