Posts

• 2025-03-20 23:05:11 +0000

  Dr Stone S4 11

  

• 2025-03-10 21:37:26 +0000

  Salaryman ga Isekai ni Ittara Shitennou ni Natta Hanashi 11

  

• 2025-03-09 05:47:32 +0000

  Kisaki Kyouiku kara Nigetai Watashi 08

  

• 2025-03-08 20:54:24 +0000

  Fuguushoku Kanteishi ga Jitsu wa Saikyou Datta 09

  

• 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!!

  

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