Openai/6919ca73-5614-8009-bb7f-6ff22c9d2193

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gpt-5-t-mini

User: solve this

solve this

Assistant: Answer: x80∘\displaystyle x80^\circx80∘.

Answer: x=80∘\displaystyle x=80^\circx=80∘.

Sketch of a short solution (clean numeric / trig setup):

Place the square with A=(0,1),B=(1,1),C=(1,0),D=(0,0)A=(0,1),B=(1,1),C=(1,0),D=(0,0)A=(0,1),B=(1,1),C=(1,0),D=(0,0). Let E=(e,0)E=(e,0)E=(e,0) and F=(1,1−t)F=(1,1-t)F=(1,1−t) (so t∈(0,1)t\in(0,1)t∈(0,1)). Angle at AAA between AE→=(e,−1)\overrightarrow{AE}=(e,-1)AE=(e,−1) and AF→=(1,−t)\overrightarrow{AF}=(1,-t)AF=(1,−t) is 45∘45^\circ45∘, so

cos⁡45∘=(e)(1)+(−1)(−t)e2+11+t2=e+te2+11+t2.\cos45^\circ=\frac{(e)(1)+(-1)(-t)}{\sqrt{e^2+1}\sqrt{1+t^2}}=\frac{e+t}{\sqrt{e^2+1}\sqrt{1+t^2}}.cos45∘=e2+1​1+t2​(e)(1)+(−1)(−t)​=e2+1​1+t2​e+t​. Angle at FFF between the vertical BFBFBF and FE→=(e−1,−(1−t))\overrightarrow{FE}=(e-1,-(1-t))FE=(e−1,−(1−t)) gives

cos⁡70∘=1−t(1−e)2+(1−t)2.\cos70^\circ=\frac{1-t}{\sqrt{(1-e)^2+(1-t)^2}}.cos70∘=(1−e)2+(1−t)2​1−t​. Solving these two equations (one can do it numerically or with algebraic manipulation) yields

e≈0.17632698,t≈0.70020754.e\approx0.17632698,\qquad t\approx0.70020754.e≈0.17632698,t≈0.70020754. Finally the desired angle at EEE (between EA→=(−e,1)\overrightarrow{EA}=(-e,1)EA=(−e,1) and EF→=(1−e, 1−t)\overrightarrow{EF}=(1-e,\,1-t)EF=(1−e,1−t)) evaluates to

∠AEF≈80∘.\angle AEF \approx 80^\circ.∠AEF≈80∘. So x=80∘x=80^\circx=80∘.