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Geometry Difficulty 5.5 AIME, harder Find the answer

3. As shown in Figure 3, let ABA B and CDC D be two perpendicular chords of a circle with center OO and radius rr, dividing the circle into four parts (each part may degenerate to a point) in clockwise order denoted as X,Y,Z,WX, Y, Z, W. Then the maximum value of SX+SZSY+SW\frac{S_{X}+S_{Z}}{S_{Y}+S_{W}} (where SUS_{U} represents the area of UU) is \qquad

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

3. π+2π2\frac{\pi+2}{\pi-2}.

Let's assume the center of the circle falls in ZZ as shown in Figure 7(a).
When the chord ABAB moves upward, the shaded area in Figure 7(b) is greater than the unshaded area to its left, so SX+SZS_{X}+S_{Z} increases, while SY+SWS_{Y}+S_{W} decreases (note that the sum of the areas of XX, YY, ZZ, and WW is a constant πr2)\left.\pi r^{2}\right). Therefore, the ratio SX+SZSY+SW\frac{S_{X}+S_{Z}}{S_{Y}+S_{W}} increases.

Thus, when point AA coincides with point CC, it is possible to achieve the maximum value.

In Figure 7(c), in the right triangle ABD\triangle ABD, the hypotenuse BDBD is the diameter, so ABD\triangle ABD has the maximum area when OAOA is the height. At this point, SZS_{Z} is maximized, and SX+SZS_{X}+S_{Z} is also maximized, with a value of 12πr2+r2\frac{1}{2} \pi r^{2}+r^{2}. Meanwhile, SY+SWS_{Y}+S_{W} is minimized, with a value of 12πr2r2\frac{1}{2} \pi r^{2}-r^{2}.
Therefore, the maximum value of SX+SZSY+SW\frac{S_{X}+S_{Z}}{S_{Y}+S_{W}} is
12πr2+r212πr2r2=π+2π2. \frac{\frac{1}{2} \pi r^{2}+r^{2}}{\frac{1}{2} \pi r^{2}-r^{2}}=\frac{\pi+2}{\pi-2} .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.