This one tripped me up more than I expected.
Set up the problem with coordinates, identify the two quarter-circles as regions defined by inequalities, and find the area of their intersection using geometric decomposition. Use symmetry to simplify the calculation and express the final answer exactly in terms of π and √3.
Pro tip: After deriving the exact answer, verify it by checking that the overlap area is less than the area of one quarter-circle and greater than zero, and consider mentioning how you would generalize the approach to other shapes or sizes.
Place the square with corners at (0,0), (1,0), (1,1), (0,1). Write the equations of the two quarter-circles: one centered at (0,0) with radius 1 (x^2 + y^2 ≤ 1, x≥0, y≥0) and one centered at (1,1) with radius 1 ((x-1)^2 + (y-1)^2 ≤ 1, x≤1, y≤1).
The overlap is the set of points satisfying both inequalities. By symmetry, the region is symmetric about the line y = x, and its boundary consists of arcs of the two circles and possibly the square's edges.
Solve the system x^2 + y^2 = 1 and (x-1)^2 + (y-1)^2 = 1. Subtract to get x + y = 1. Substitute to find the intersection points: (1/2, 1/2) is not on the circles; actually solving yields points ((1±√3)/2, (1∓√3)/2) but only one lies in the square: ((1-√3)/2, (1+√3)/2) is outside? Check: The valid intersection inside the square is at ( (1-√3)/2, (1+√3)/2 )? Wait, compute: x+y=1, x^2+y^2=1 => 2x^2 -2x +1 =1 => 2x(x-1)=0 => x=0 or x=1. So intersections are (0,1) and (1,0). But those are corners of the square. However, the quarter-circles also intersect at (1/2, 1/2)? No, (1/2,1/2) is not on either circle. Actually, the two quarter-circles overlap in a lens shape whose boundary includes arcs from both circles and the line segment from (0,1) to (1,0)? Let's re-evaluate: The first quarter-circle is the set of points in the square within distance 1 from (0,0). The second is within distance 1 from (1,1). Their intersection is the set of points in the square that are within distance 1 from both corners. The boundary of this intersection consists of the arc of the first circle from (0,1) to (1,0) (the part inside the square) and the arc of the second circle from (0,1) to (1,0). So the intersection is the region bounded by these two arcs. The arcs meet at (0,1) and (1,0). So the intersection points are (0,1) and (1,0).
The overlap region can be seen as the union of two circular segments. Alternatively, use the formula for the area of intersection of two circles of equal radius r with distance d between centers. Here r=1, d=√2. The area is 2r^2 cos^{-1}(d/(2r)) - (d/2)√(4r^2 - d^2). Plug in: 2*1*cos^{-1}(√2/2) - (√2/2)√(4-2) = 2*(π/4) - (√2/2)*√2 = π/2 - 1. So area = π/2 - 1.
Check that the area is positive and less than π/4 (area of one quarter-circle). π/2 - 1 ≈ 0.5708, which is less than π/4 ≈ 0.7854. Present the answer as π/2 - 1.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the definition: a flush is any five cards of the same suit, and a straight flush is a flush that also forms a straight. Then compute the total number of flushes and subtract the number of straight flushes, and finally divide by the total number of 5-card hands. Present the counting formula explicitly as (C(13,5) - 10) * 4 / C(52,5).
Pro tip: Mention that the 10 straight flushes per suit include the ace-high straight (10-J-Q-K-A) and the wheel (A-2-3-4-5), and that these are distinct. Also, note that the total number of 5-card hands is C(52,5) = 2,598,960, which helps verify the final probability.
Clarify that a flush is five cards of the same suit, and a straight flush is a flush that also forms a straight. The desired event is a flush that is not a straight flush.
For each suit, there are C(13,5) ways to choose 5 cards. Multiply by 4 suits to get the total number of flushes: 4 * C(13,5).
In each suit, there are 10 possible straights (A-2-3-4-5 through 10-J-Q-K-A). So there are 4 * 10 = 40 straight flushes.
The number of flushes that are not straight flushes is 4 * C(13,5) - 40 = 4 * (C(13,5) - 10).
Divide the desired count by the total number of 5-card hands, C(52,5). Write the final formula: P = [4 * (C(13,5) - 10)] / C(52,5).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic puzzle and I'd seen it before, which helped.
Recognize this as the classic two-egg problem and explain that the optimal strategy balances linear and binary search by dropping the first marble at decreasing intervals. Derive the minimum worst-case drops by solving the triangular number equation n(n+1)/2 >= 100, yielding n=14.
Pro tip: Mention that the optimal first drop is at floor 14, and if it breaks, use the second marble to linearly search the 13 floors below; if it doesn't, jump 13 floors to 27, then 12, etc. This shows you understand the adaptive nature of the strategy.
Restate the constraints: two marbles, 100 floors, find the highest safe floor, minimize worst-case drops. Confirm that marbles can be reused if they don't break.
Explain that with one marble you must go linearly, and with unlimited marbles you could binary search. With two marbles, you need a hybrid strategy that balances the number of drops for the first marble and the linear search for the second.
Let the first drop be at floor n. If it breaks, you need at most n-1 more drops with the second marble. If it doesn't, the next drop should be at floor n + (n-1), then n + (n-1) + (n-2), etc. The total floors covered must be at least 100, so solve n(n+1)/2 >= 100.
Compute n=14 because 14*15/2 = 105 >= 100, while 13*14/2 = 91 < 100. So the worst-case number of drops is 14.
Drop the first marble at floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (intervals decreasing by 1). When it breaks, use the second marble to linearly search the floors between the previous safe floor and the breaking floor.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.