Posts

• 2024-12-15 20:58:07 +0000

  Tensei Kizoku, Kantei Skill de Nariagaru - 23

  haha what’s this

  

• 2024-11-23 20:30:57 +0000

  Mahoutsukai ni Narenakatta Onnanoko no Hanashi 08

  

  I love how they deliberately chose a temperature such that it has the same value in both Celsius and Fahrenheit.

• 2024-11-05 03:18:40 +0000

  Chio-chan no Tsuugakuro 12

  So I was rewatching Chio-chan no Tsuugakuro, and I got completely nerd-sniped by this geometry problem.

  

  I don’t think my solution is the most elegant, but here it is.

  I have made the assumption that the line labelled “3” is parallel to the line labelled “A”. I will also make the notational change to use lowercase “a” instead of uppercase “A” to represent the required length.

  Label the points and angles as follows.

  

  By congruence, \( \alpha = \alpha’ \).

  Considering \(\triangle CBD\), \(\alpha + \beta + \gamma + \delta = \pi \); considering straight line HBG, \(\alpha’ + \beta + \gamma’ + \delta = \pi \). Therefore \(\gamma = \gamma’\).

  Lengths \(CA\) and \(AD\) are found by Pythagoras’ Theorem, and lengths \(BF\) and \(FG\) by congruence.

  Note that \(\triangle HBA \sim \triangle HGJ\), so

  \[\frac{b}{b+\sqrt{7}+\sqrt{40}} = \frac{3}{a}\]

  which gives

  \[a = \frac{3}{b} \left(b+\sqrt{40}+\sqrt{7}\right). \tag{1}\]

  Consider \(\triangle HBA\), we have

  \[\cos(\beta+\gamma) = \frac{3}{b},\]

  so

  \[\cos\beta\cos\gamma - \sin\beta\sin\gamma = \frac3b.\]

  The sines and cosines of \(\beta\) and \(\gamma\) can be obtained by considering \(\triangle ABC\) and \(\triangle ABD\):

  \[\frac{3}{4} \cdot \frac{\sqrt{40}}{7} - \frac{\sqrt{7}}{4} \cdot \frac{3}{7} = \frac{3}{b}\]

  which rearranges to give

  \[\frac{3 \left( \sqrt{40}-\sqrt{7} \right)}{28} = \frac3b \tag{2}\]

  and

  \[b = \frac{28\left(\sqrt{40}+\sqrt{7}\right)}{33}. \tag{3}\]

  Substitute (2) and (3) into (1), we get

  \[a = \frac{3 \left( \sqrt{40}-\sqrt{7} \right)}{28} \cdot \frac{61\left(\sqrt{40}+\sqrt{7}\right)}{33} = \frac{183}{28}.\]
• 2024-10-27 14:36:56 +0000

  Shokugeki no Souma S4 03

  

  smh…

• 2024-10-19 01:59:44 +0100

  Mahoutsukai ni Narenakatta Onnanoko no Hanashi - 03

  

  Can’t see much detail but where are white’s pawns?

• 2024-09-26 23:33:56 +0100

  Koi wa Futago de Warikirenai - 12

  WTF you got it right last episode!!

  

• 2024-09-22 20:43:28 +0100

  Koi wa Futago de Warikirenai - 11

  

  They fixed the starting position!

  

  We have a Ruy Lopez. Nice, looks like they actually looked it up.

  

  Berlin Defence. Having high hopes now…

  Uhh… Girl, that’s not how you castle…

• 2024-09-16 21:41:43 +0100

  Koi wa Futago de Warikirenai - 10

  Talk about having a chess-themed episode, then completing fucking up every opening position smh

  

  Can’t they just open lichess.org or chess.com or Wikipedia or Google Images and just look at it while drawing?

  

  Or read this fucking book that clearly has the queen on the d file and king on the e file?

  

  OK this is actually plausible, if white could bait black to take the knight, then a mad queen could have been possible at some point.

• 2024-08-11 01:57:11 +0100

  Senpai wa Otokonoko 03

  USODA!!!

• 2024-07-14 23:30:30 +0100

  Koi wa Futago de Warikirenai 01

  

  The classic “king on d file” again…

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