Maths Olympiad Prep

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Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

In the diagram, line segment PSPS has length 4. Points QQ and RR are on line segment PSPS. Four semi-circles are drawn on the same side of PSPS. The diameters of these semi-circles are PSPS, PQPQ, QRQR, and RSRS. The region inside the largest semi-circle and outside the three smaller semi-circles is shaded.
Figure 0
What is the area of a square whose perimeter equals the perimeter of the shaded region?

Pick one

Solution

The perimeter of the shaded region consists of four pieces: a semi-circle with diameter PSPS, a semi-circle with diameter PQPQ, a semi-circle with diameter QRQR, and a semi-circle with diameter RSRS. We note that since a circle with diameter dd has circumference equal to πd\pi d, then, not including the diameter itself, the length of a semi-circle with diameter dd is 12πd\frac{1}{2}\pi d. Suppose that the diameter of semi-circle PQPQ is xx, the diameter of semi-circle QRQR is yy, and the diameter of semi-circle RSRS is zz. [[IMAGE0]] We are given that the diameter of the semi-circle PSPS is 4. Since PQ+QR+RS=PSPQ+QR+RS=PS, then x+y+z=4x+y+z=4. Thus, the perimeter of the shaded region is 12π(PQ)+12π(QR)+12π(RS)+12π(PS)=12πx+12πy+12πz+12π(4)=12π(x+y+z+4)=12π(4+4)=4π\tfrac{1}{2}\pi (PQ) + \tfrac{1}{2}\pi (QR) + \tfrac{1}{2}\pi (RS) + \tfrac{1}{2}\pi (PS) = \tfrac{1}{2}\pi x + \tfrac{1}{2}\pi y + \tfrac{1}{2}\pi z + \tfrac{1}{2}\pi (4) = \tfrac{1}{2}\pi(x+y+z+4) = \tfrac{1}{2}\pi(4+4) = 4\pi We want to determine the area of the square whose perimeter equals this perimeter (that is, whose perimeter is 4π4\pi). If a square has perimeter 4π4\pi, then its side length is 14(4π)=π\frac{1}{4}(4\pi) = \pi, and so its area is π2\pi^2. Because this problem is multiple choice, then we should get the same answer regardless of the actual lengths of PQPQ, QRQR and RSRS, since we are not told what these lengths are. Therefore, we can assign to these lengths arbitrary values that satisfy the condition PQ+QR+RS=4PQ+QR+RS=4. For example, if PQ=QR=1PQ=QR=1 and RS=2RS=2, then we can calculate the perimeter of the shaded region to be 12π(PQ)+12π(QR)+12π(RS)+12π(PS)=12π(1)+12π(1)+12π(2)+12π(4)=12π(1+1+2+4)=4π\tfrac{1}{2}\pi (PQ) + \tfrac{1}{2}\pi (QR) + \tfrac{1}{2}\pi (RS) + \tfrac{1}{2}\pi (PS) = \tfrac{1}{2}\pi (1) + \tfrac{1}{2}\pi (1) + \tfrac{1}{2}\pi (2) + \tfrac{1}{2}\pi (4) = \tfrac{1}{2}\pi(1+1+2+4) = 4\pi and obtain the same answer as above.

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