3. As shown in Figure 3, let and be two perpendicular chords of a circle with center and radius , dividing the circle into four parts (each part may degenerate to a point) in clockwise order denoted as . Then the maximum value of (where represents the area of ) is
Solution
3. .
Let's assume the center of the circle falls in as shown in Figure 7(a).
When the chord moves upward, the shaded area in Figure 7(b) is greater than the unshaded area to its left, so increases, while decreases (note that the sum of the areas of , , , and is a constant . Therefore, the ratio increases.
Thus, when point coincides with point , it is possible to achieve the maximum value.
In Figure 7(c), in the right triangle , the hypotenuse is the diameter, so has the maximum area when is the height. At this point, is maximized, and is also maximized, with a value of . Meanwhile, is minimized, with a value of .
Therefore, the maximum value of is
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