Le Salon d'Emmanuel Chanel

正方形の面積を X 、 AH = AK = a とすると、
X = 2a^2 , AB = 6 , AC = 8
三平方の定理より、
BC^2 = AB^2 + AC^2 = 6^2 + 8^2 = 100 = 10^2 より、 BC = 10.
また、 角AKH = 角AHK = 角BHI = 45度
BH = 6 - a , CK = 8 -a
sin角B = cos角C = 4/5
(sinθ)^2 + (cosθ)^2 = 1 より、 (sin角C)^2 = 1 - 16/25 = 9/25 = (3/5)^2 などから
cos角B = sin角C = 3/5
https://hatsudy.com/jp/law-of-sines.html を見て、
正弦定理より
a / sinB = BI / sin 45度
a*√2 / (4/5) = BI / (1/√2)
a*5√2/4 = BI√2
a*5/4 = BI
BI = 5a/4
余弦定理より
HI^2 = BH^2 + BI^2 + 2BH*BI*cosB
2a^2 = (6-a)^2 + (5a/4)^2 + (6-a)(5a/4)*(3/5)
自分でも計算してみたが、 アメリカ人の数学者の友達( [url=http://mazack.org/]PhD Michael Mazack[/url])がチャットで私に見せた[url=https://www.wolframalpha.com/input?i=solve+2a%5E2+%3D+%286-a%29%5E2+%2B+%285a%2F4%29%5E2++%2B+%286-a%29%285a%2F4%29*%283%2F5%29+for+a&lang=ja]上のWolfram Alpha の計算結果[/url]は、
a = 4 (√37 - 5) ( a = -4 (5 + √37) もあったが、 a > 0 なので棄却)
X = 2*a^2 は、[url=https://www.wolframalpha.com/input?i=solve+32*%28sqrt%2837%29+-+5%29%5E2&lang=ja]Wolfram Alpha の計算結果[/url]は、
[b][color=#0000FF][size=200]X = 64 (31 - 5√37) = 1984 - 320√37[/size][/color][/b]
私が自分で手計算した答えは [b][color=red]196-20√73[/color][/b] で、計算ミスしていたよう。

It's introduced on my Facebook friend's account on Facebook. The post itself wasn't on the global share. But it would be found on the Japanese net for the entrance exams of secondary schools or of colleges.
I found a way to solve this question but I don't have confidence in my calculations on the solution.
Can you solve this question? I wonder how many foreigners can solve it. So I posted it.
[b][color=blue][size=150]What's the area of the square among HIJK on this figure?[/size][/color][/b]
[attachment=0]376280715_6726886040696862_5135486137143371148_n.jpg[/attachment]