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D. S_D9
Posted: 2023-11-20 12:13:29 Solutions will be accepted throughFor convenience, since segment A C is divided into four equal parts, we introduce coordinates scaled so that A C = 4 . We place the origin at the center of the (semi)ellipse, which is to say the midpoint between the foci F and G . Thus the equation of the ellipse will have the…
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D. S_D8
Posted: 2023-11-19 21:09:52 Solutions will be accepted throughPMP participants Chris Bistryski of Monroe, WA, William Jones of FCI Loretto, William Keehn, and Jesse Waite submitted solutions to this problem. As Mr. Keehn pointed out, it’s smoothest to work with polar coordinates; see Figure 2 for an illustration of the coordinates ( r , θ ) of a general point M on the…
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D. S_D7
Posted: 2023-11-19 19:31:00 Solutions will be accepted throughWe received a solution from William Keehn and a partial solution from Jesse Waite of Fort Leavenworth. Both submitters realized that one straightforward way to control the numbers of sides of the cross-sections is to tip back the top-right vertex of the tetrahedron in the figure associated with the problem, straight toward its leftmost horizontal…
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D. S_D6
Posted: 2023-11-19 15:45:55 Solutions will be accepted throughWe received solutions to this Dilemma from Matthew Helmer (Pacific Lutheran University), William Jones (PMP), and Jesse Waite (PMP), and partial solutions from Christopher Havens (PMP Founder) and Sonny Kim (PMP). As Jesse pointed out, the rectangles look like dominoes (length twice the width), but since we want the exact answer, we’d better not rely…
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D. S_D4
Posted: 2022-07-02 12:46:47 Solutions will be accepted throughAs submitter Jesse Waite noted, a good first step in attacking one of these sequences is to write out (in symbols) the rule defining it, known as its recurrence relation. For example, if we denote the first series connected with addition by c 0 , c 1 , c 2 , … , its recurrence…
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D. S_D3
Posted: 2022-07-01 20:58:46 Solutions will be accepted throughThe Prisoner’s Dilemma received a solution to the first part by reader Jesse Waite from Ft. Leavenworth, Kansas. Jesse realized that there are eight vertices in a cube, so that it’s necessary to find eight sums of three of the numbers equaling zero. In addition, each number is used exactly twice in the eight sums…