## Task 6B - 281246B

Determine the maximum number of squares with side length 1 that can fit into a given square with side length 1.99, without extending beyond its edges and without overlapping each other.
The sought maximum number is 1.

Let $ABCD$ be the given square with side length 1.99. Clearly, one can place a square of side length 1 within $ABCD$. However, it is not possible to place two such squares, because any two such squares within $ABCD$ would overlap.

To prove this, we show that any square of side length 1 that lies within $ABCD$ must contain the center $M$ of $ABCD$ in its interior.

Assume, for the sake of contradiction, that this is not the case for some square $PQRS$.

We denote the lines through $P, Q$ and $Q, R$ and $R, S$ and $S, P$ by $p, q, r$, and $s$, respectively. Then $M$ lies on the boundary or outside of at least one of the two strips between $p$ and $r$ or between $q$ and $s$. Thus, $M$ has a distance of at least 1 from at least one of the lines $p, q, r$, or $s$, without loss of generality, from $p$. Therefore, $p$ lies entirely outside the incircle $k$ of $ABCD$, since this circle has a radius of $\frac{1}{2} \cdot 1.99 = 0.995 = AE = AH$.

Thus, without loss of generality, let $X$ be on $AH$ and $Y$ be on $AE$.

Among all lines parallel to $p$ and passing through a point on the segment $XH$, there must be exactly one that touches $k$, since $p$ itself lies outside $k$, and the parallel through $H$ to $p$ intersects the circle $k$ at two points, as it does not stand perpendicular to the radius $MH$. This $k$-touching line $u$ intersects the segment $XH$ at an interior point $U$ and thus the segment $YE$ at an interior point $V$, and its point of tangency with $k$ is $W$. Since $AX < AU$, by the intercept theorem, it follows that $XY < UV$. By triangle inequalities and the theorem of the equality of tangent segments, we finally get

$$
1 = PQ \leq XY < UV < \frac{1}{2}(AU + AV + UW + VW) = \frac{1}{2}(AU + AV + UH + VE) = AE = \frac{1}{2} \cdot 1.99
$$

This contradiction shows that our assumption was false, thus completing the proof.

Taken from [1]

### 9.31 XXIX. Olympiad 1989

### 9.31.1 First Round 1989, Class 12