Packet 3: Bonus 20
Answer the following about Swiss mathematician Jean-Pierre Sydler. For 10 points each:
[10e] Sydler names a polyhedron in which only one angle between faces is not equal to this value. All four internal angles of a rectangle have this value.
ANSWER: 90 degrees [or pi/2 radians; accept right angles]
[10h] Sydler also worked on the property of “scissors congruence,” which in two dimensions is equivalent to two objects having the same value of this property. Brahmagupta’s formula can find this value for cyclic quadrilaterals.
ANSWER: area
[10m] Sydler’s work on scissors congruence contributed to the solution of the third of this German mathematician’s 23 problems. A thought experiment devised by this mathematician demonstrates that a hotel with an infinite number of rooms can always accommodate an additional guest.
ANSWER: David Hilbert [accept Hilbert’s problems; accept Hilbert’s paradox of the Grand Hotel; accept Hilbert’s hotel]
<Math — Sam Macchi> | SHOW-ME Four Packet 3
| Heard | PPB | E % | M % | H % |
|---|---|---|---|---|
| 13 | 21.54 | 100% | 46% | 69% |
Conversion
| Team | Opponent | Part 1 | Part 2 | Part 3 | Total | Parts |
|---|---|---|---|---|---|---|
| (not so) Evil Empire | washed and braindead!!! | 10 | 10 | 10 | 30 | EHM |
| ASCA (Ava, Sean, Chris, and Aiden) | Bombardino Rizzbaino and the Tung Tung Tung Trio | 10 | 10 | 10 | 30 | EHM |
| Copilot, give me a witty team name for an online quizbowl tournament | ccny potato salad | 10 | 10 | 10 | 30 | EHM |
| Fremd Larpers | son toliver | 10 | 10 | 10 | 30 | EHM |
| POA | Quaker Oats | 10 | 0 | 10 | 20 | EM |
Summary
| Tournament | Edition | Match | Heard | PPB | E % | M % | H % |
|---|---|---|---|---|---|---|---|
| Rube Goldberg XIX | 11/20/2025 | ✓ | 7 | 18.57 | 100% | 14% | 71% |
| SHOW-ME Four Nationwide HS | 11/20/2025 | ✓ | 5 | 28.00 | 100% | 100% | 80% |
| SPOOKY IX | 11/20/2025 | ✓ | 1 | 10.00 | 100% | 0% | 0% |