One way to pack a 100 by 100 square with 10000 circles, each of diameter 1 , is to put them in 100 rows with 100 circles in each row. If the circles are repacked so that the centres of any three tangent circles form an equilateral triangle, what is the maximum number of additional circles that can be packed?
(A) 647
(B) 1442
(C) 1343
(D) 1443
(E) 1344
One way to pack a 100 by 100 square with 10000 circles, each of diameter 1 , is to put them in 100 rows with 100 circles in each row. If the circles are repacked so that the centres of any three tangent circles form an equilateral triangle, what is the maximum number of additional circles that can be packed?
(A) 647
(B) 1442
(C) 1343
(D) 1443
(E) 1344

## Solution

Remove one circle from every second row and shift to form the given configuration. Label the diagram as shown. Since each circle has diameter 1, $\triangle P Q R$ and $\triangle P X Y$ are equilateral triangles with sides of length 1 .

In $\triangle P Q R$, altitude $P S$ bisects side $Q R$. Use the Pythagorean Theorem to find $P S$.

$$
\begin{aligned}
P S^{2} & =(1)^{2}-\left(\frac{1}{2}\right)^{2} \\
& =\frac{3}{4} \\
P S & =\frac{\sqrt{3}}{2}
\end{aligned}
$$

Also, $X Z=\frac{\sqrt{3}}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-286.jpg?height=485&width=439&top_left_y=229&top_left_x=1385)

Since all radii have length $\frac{1}{2}$, then $P U=\frac{\sqrt{3}}{2}-\frac{1}{2}$ and $T U=\frac{1}{2}+\frac{\sqrt{3}}{2}+\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=\sqrt{3}$. This tells us that two rows of circles require a height of $\sqrt{3}$ before a third row begins.

Since $\frac{100}{\sqrt{3}}=57.7$, we can pack 57 double rows, each containing 100+99=199 circles.

Can we pack one final row of 100 circles? Yes. The square has sides of length 100 and our configuration of 57 double rows requires a height of $57 \sqrt{3}$ before the next row begins. Since $100-57 \sqrt{3}>1$, and since the circles each have diameter 1 , there is room for one final row of 100 circles.

The number of circles used in this new packing is $57(199)+100=11443$

Thus, the maximum number of extra circles that can be packed into the square is $11443-10000=1443$.

ANSWER: (D)