## English Issue, December 2002 | ||||

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## Solutions of problems B

**B. 3463.** *The diameter of a semicircular sponge is *20 cm* long. Find the area wiped by the sponge in the corner of the floor of a room if the endpoints of the diameter slide along the two walls enclosing a right angle.*

**Solution.** The sponge is a semicircular disc of diameter *AB*. If *A* is moving along the *x*-axis and *B* is moving along the *y*-axis then the circle of diameter *AB* passes through *O*. Therefore, no point of the wiped region is farther away from the point *O* than 20 cm, no points outside the quadrant of radius 20 cm centred at *O* can be reached by the sponge.

The points of the quadrant of radius 10 cm centred at *O* are covered by the semicircles whose diameters lie along the *x* and *y* axes (*Figure 2*).

Figure 1 | Figure 2 | Figure 3 |

If *P* is a point of the ``ring'' bounded by the two arcs and *OP* intersects the larger arc at *R*, let *R*_{1} and *R*_{2} denote the projections of *R* onto the axes. Then *OR*_{1}*RR*_{2} is a rectangle with diagonals *RO*=*R*_{1}*R*_{2}=20 cm. If *C* is the centre of the rectangle, then *OC*=10 cm, *C* lies on the smaller arc, and *P* lies on the segment *CR*. The point *P* is thus contained in the right-angled triangle *R*_{1}*RR*_{2}, which is entirely contained in the semicircle of diameter *R*_{1}*R*_{2} passing through *R*. Thus the point *P* is wiped by the sponge.

Therefore, the area wiped by the sponge consists of the points of the closed quadrant of radius 20 cm centred at *O*, and hence the area wiped is

\(\displaystyle \frac{1}{4}\cdot20^2\pi=100\pi~\rm cm^2. \)