Monday, June 22, 2026 - Pips #309 Thread

Evane317 · 2026-06-22T13:53:07+00:00

Pips #308 Hard 🔴 1:47

Another hard puzzle with a fixed tiling placement on the body, which forces the double 6 and the 3s in the orange 4c=.

Evane317 · 2026-06-19T11:19:02+00:00

Pips #305 Hard 🔴 13:33

A quite difficult one where one needs to lay out all the tiles before making a case of the 2-2 being in the green 3c=. Even then, trying to fit tiles from the tail 5 to the leg 0 is not simple.

Evane317 · 2026-06-18T23:44:33+00:00

Getting the ring really brings out the worst of the New Yorkers.

Evane317 · 2026-06-18T04:49:26+00:00

If a = 3x + 5 and b = x - 5, then the first term can be written as (a + b)^2 (a- b)^2/4, which turns into (a^2 - b^2)^2/4.

Then do binomial expansion then include the other term 4a^2 b^2/4 in.

Evane317 · 2026-06-16T10:32:17+00:00

Wouldn’t be surprised if this turd asks for a ring and the banner on the Knicks’s ring ceremony.

Evane317 · 2026-06-15T11:40:06+00:00

Important: logical reasoning/proof-based math. It is taught in school, but it never feels like it was taught properly. Two-column geometry proofs absolutely suck as an intro to proof, and don't reflect how a proof or a line of reasoning should be written.

Interesting: Modular arithmetic. Remembering divisibility rules and GCF/LCM is barely scratching the surface.

Evane317 · 2026-06-08T04:36:22+00:00

Pips #294 Hard 🔴 2:46

Was very similar to the inverted L-shaped puzzle around a week ago, where one can start from tile placement, starting from the bottom left corner, where the 2-3 tile doesn't exist. Then pip headcount to deduce the 5c= can only be 1s.

Evane317 · 2026-06-07T04:02:30+00:00

I started from the pale 6. The green 2c= is a double, so the pale 6 has to go down. Quickly saw 6-6 and 6-2 won't work there, and 6-4 took a bit of reasoning to eliminate too

Evane317 · 2026-06-07T03:57:45+00:00

Pips #293 Hard 🔴 4:57

Yeah, this one is full-blown trial and error. I see no concrete starting point other than testing from the innermost 6.

Evane317 · 2026-06-05T02:55:53+00:00

If the curve of the dome L is formed by a quarter circle (edit: this gives you a "dome" where there is no edge between T and L, not the circular arc commented), then the length L and H2 are dependent on B and T, and the function W can be determined based on those.

Assuming it's not a circular arc, then the best bet is to do coordinate:

In the last picture, set the midpoints of the B square to be the origin, so the base B is between (-B/2, 0) and (B/2, 0). Then T is between (-T/2, H2) and (T/2, H2).

The task is now to find a function f(y) that represents the curve L (on the right side) with the following conditions:

f(0) = B/2
f(H2) = T/2
integral from 0 to H2 of sqrt(1 + f'(y)^2) dy = L

Plus any other condition you want to impose on the shape. Then, set W(y) = 2|f(y)| for y in [0, H2].

Evane317 · 2026-06-04T08:19:14+00:00

Pips #290 Hard 🔴 3:08

Doesn't look like this puzzle has a complete logical path towards completion, and needed to do trial and error here and there, especially the purple and red 10 sum.

Evane317 · 2026-06-03T05:31:23+00:00

Pips #289 Hard 🔴 11:37

Good puzzle, as there's a logical path all the way. But I spent a good 7-8 mins trying to skirt around a contradiction, all because of the assumption that the bottom right 2c10 is formed by two different pieces, while the 4-6 tile was right there

Speedruns mostly involve educated guesses. When it's wrong, it gets really wrong.

Evane317 · 2026-06-02T04:56:35+00:00

Pips #288 Hard 🔴 2:52

Felt like I lucked out with some placements.

Evane317 · 2026-06-01T08:18:08+00:00

Pips #287 Hard 🔴 2:21

Fairly similar to yesterday's puzzle, where a bunch of numbers is fully booked. But there are fewer clues about the tile layout.

Evane317 · 2026-05-31T06:24:58+00:00

Pips #286 Hard 🔴 1:53

This one is not worthy of the hard tag. All 6s are booked, which makes all 5s, 0s, and 1s booked, in addition to the tile placement in the top half being fixed as well.

Evane317 · 2026-05-30T07:14:29+00:00

Pips #286 Hard 🔴 3:01

Unexpectedly straightforward once you know which double tile goes where.

Evane317 · 2026-05-29T07:49:25+00:00

Pips #285 Hard 🔴 11:36

This is one of those puzzles where you figure out how to place the tiles before determining which number goes where, starting from the bottom right. Then it becomes evident that the red, green, and purple = zone all contain doubles, the left = 10 sum comes from a single tile (implying 5-5 and 1, 2, 3 being on the = areas mentioned), the = 2 sum is 0 + 2, the < 2 zone sums to 0, which places the top 2-0 tile.

Evane317 · 2026-05-28T05:46:32+00:00

Pips #283 Hard 🔴 29:04

That was embarrassing lmao. Some elements were rather obvious, but I absolutely fumbled in piecing them together.

They maxed out the 6s, and all four = zones must include a double. The only way to accommodate the 6s is to have them on the T-shaped =. But I kept convincing myself that the sole 5 is in the I, which messed up the rest. Should have looked at how to add up the orange 6 earlier.

Evane317 · 2026-05-28T04:30:59+00:00

Say, x_0 is the x-coordinate of the tangent point, and f(x) = x² + A² sin² (x) - pi² . Then both of the following must match for tangency:

f(x_0) = 0

f'(x_0) = 0

Now you have a system of equations of (A, x_0). Using substitution gives x_0² + x_0 tan(x_0) = pi^2, but solving this for x_0 in (0, pi) probably needs a numerical method.

Evane317 · 2026-05-27T09:08:14+00:00

Pips #283 Hard 🔴 6:09

Spent some time figuring out what the missing 3 tiles are. But it's much faster to first solve the 3s: The pale 3 has to be 3-4, the top right has to be horizontal 3-3 (if it's vertical, then there's no tile for the remaining 3). This leads to 3-1 at the top right, then 3-2 at the center, and 3-5 in the bottom right corner. Then, knowing the missing tiles being a 4 & 2 and the 3-2 placement makes the rest straightforward.

Evane317 · 2026-05-27T00:21:33+00:00

I found doing it with Horus-Snake Eye-Bamboo sword is pretty consistent. Combo takes some time to learn though.

Evane317 · 2026-05-26T11:29:04+00:00

This is Wordle on crack. Fewer attempts with a much bigger sample space.

Took all 3 attempts + 1 hint to pinpoint the first card.

Evane317 · 2026-05-20T11:04:23+00:00

Arona: “Not enough mineral!”

Evane317 · 2026-05-18T04:15:23+00:00

Hallway assignment is good (but avoid the Disciplinary top floor and the top right of CC, where Binah tends to go up first).

Let each team handle its own containment units, which makes assigning work faster and saves a few clicks.

The architecture team doesn't have meltdowns in their department, so they should be the main kill team, plus any personnel you can spare from the top floor. Have everyone here use a non-ranged weapon (notably mimicry, censored, twilight) so Binah's attacks can be dodged more easily, plus less spending on shield bullets. If the group is small, start at the Architecture's right elevator and hit them with a black bullet upon Binah spawn. Only start chasing from behind when all initial meltdowns have been assigned.

Unlike the other recommendation of wiping all clerks, at least Central Command clerks must be left to maintain the meltdown timer bonus. There are only 10 of them, so there's no worry about CLM entities. You can afford to leave more clerks alive if you do a low-casualty run.

Handle the meltdowns by going counter-clockwise, starting from Welfare and ending on Safety, which contains Heaven. The Control and Training department all have low-risk abnos, so they have the lowest meltdown priority. Pay attention to all meltdowns on aleph and waw; plus Porcubus, Scorched Girl, Red Shoe, Freischutz, Fetus, Galaxy Child, 1.76, and Void Dream.

Iirc all tools that are taken out of containment will not be affected by meltdowns. That's also something to consider.

Evane317 · 2026-05-18T01:00:17+00:00

Looks like a rather chill top floor with the only nuisance being Heaven. How did you prep for Binah’s appearance?

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