Inside an ellipse with semi-axes a=100 cm and b=10 cm a…
Inside an ellipse with semi-axes $a=100$ cm and $b=10$ cm a rectangle is placed symmetrically with respect to the centre of the ellipse, with its longer side parallel to the axis $a$; its sides are $10$ cm and $3$ cm. In addition, four circles of radius $4.3$ cm are placed inside the ellipse, not intersecting the ellipse, the rectangle or each other. Determine the probability that:
$1^{\circ}$ a random point, whose position is equally likely anywhere inside the ellipse, lies in one of the circles;
$2^{\circ}$ a circle of radius $5$ cm drawn around that point as centre intersects at least one side of the rectangle.
Explanation of Method
The point is uniform in the ellipse, whose area is $\pi ab$. In $1^{\circ}$ the favourable region is the union of the four circles (they do not overlap). In $2^{\circ}$ the circle of radius $5$ meets the boundary of the rectangle exactly when its centre is at distance at most $5$ from the rectangle; this "5 cm neighbourhood" of a rectangle is the rectangle plus four strips plus four quarter-discs. We check that this region lies inside the ellipse and divide its area by $\pi ab$.
Step 1: The sample space
\[ \text{area of the ellipse} = \pi ab = \pi\cdot100\cdot10 = 1000\pi\approx 3141.6\ \text{cm}^2. \]
Step 2: $1^{\circ}$ — the point lies in one of the circles
The four circles do not overlap, so their total area is
\[ 4\pi(4.3)^2 = 73.96\,\pi\ \text{cm}^2, \qquad \boxed{P_1 = \frac{73.96\,\pi}{1000\,\pi} = 0.07396\approx0.074.} \]
Step 3: $2^{\circ}$ — when does the circle meet a side of the rectangle?
- If the point is inside the rectangle, its distance to the nearest side is at most $1.5$ cm $\lt5$ cm, so the circle crosses a side.
- If the point is outside, the circle meets the rectangle's boundary exactly when the distance from the point to the rectangle is at most $5$ cm.
- The circle can never contain the whole rectangle (the rectangle's half-diagonal is $\sqrt{5^2+1.5^2}\approx5.22\gt5$), so "meets the rectangle" really means "crosses a side".
So the favourable region is the set of points within $5$ cm of the rectangle.
Step 4: Area of the 5 cm neighbourhood
| Part | Area (cm$^2$) |
|---|---|
| the rectangle $10\times3$ | $30$ |
| two strips $10\times5$ along the long sides | $100$ |
| two strips $3\times5$ along the short sides | $30$ |
| four quarter-discs of radius $5$ at the corners | $25\pi$ |
| total | $160+25\pi\approx238.54$ |
It lies inside the ellipse: it reaches $x=\pm10$ and $y=\pm6.5$, while the ellipse at $x=10$ still has half-height $10\sqrt{1-0.01}\approx9.95$.
Step 5: The probability
\[ \boxed{P_2 = \frac{160+25\pi}{1000\pi}\approx 0.0759.} \]
Step 6: Remark
The four circles and the rectangle do not change the sample space: the point may fall anywhere in the ellipse, including inside the rectangle or a circle. The circles matter only in part $1^{\circ}$.
Result
| Part | Probability |
|---|---|
| $1^{\circ}$ in one of the circles | $4\pi\cdot4.3^2/(1000\pi)=0.07396$ |
| $2^{\circ}$ 5 cm circle meets a side | $(160+25\pi)/(1000\pi)\approx0.0759$ |