Editing Openai/6957b604-b0d8-8003-945b-9e6b2d974265


{{DISPLAYTITLE:Cone Development and Track}}

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== gpt-5-2-thinking ==
=== User: Uploaded an image ===
Uploaded an image

=== Assistant: Unroll develop the cone into a circular sector. ===
Unroll (develop) the cone into a circular sector.

1) Cone development
* Slant height L=60L=60L=60
* Base radius R=20R=20R=20
* Sector angle

θ=arc lengthL=2πR60=40π60=2π3=120∘\theta=\frac{\text{arc length}}{L}=\frac{2\pi R}{60}=\frac{40\pi}{60}=\frac{2\pi}{3}=120^\circθ=Larc length=602πR=6040π=32π=120∘
Point BBB is 10 units up the slant from AAA, so its slant distance from the apex is

60−10=5060-10=5060−10=50
2) “Around the mountain once”
Cut the cone along the generator through AAA (and BBB). To go once around, in the development you join AAA to the copy of BBB in the adjacent sector, i.e. rotated by 120∘120^\circ120∘.

Set coordinates in the development:

A=(60,0),B′=(50cos⁡120∘, 50sin⁡120∘)=(−25, 253)A=(60,0),\quad B'=\bigl(50\cos120^\circ,\ 50\sin120^\circ\bigr)=(-25,\ 25\sqrt3)A=(60,0),B′=(50cos120∘, 50sin120∘)=(−25, 253)
The shortest track is the straight segment AB′AB'AB′. It goes uphill then downhill because it passes closest to the apex somewhere in the middle.

3) Split at the highest point
The highest point corresponds to the point on segment AB′AB'AB′ closest to the origin (apex). That is the perpendicular projection of the origin onto the line through AAA and B′B'B′.

Let v=B′−A=(−85, 253)v=B'-A=(-85,\,25\sqrt3)v=B′−A=(−85,253). A point on the line is A+tvA+t vA+tv. Choose ttt so (A+tv)⋅v=0(A+t v)\cdot v=0(A+tv)⋅v=0:

t=−A⋅vv⋅v=−(60,0)⋅(−85,253)(−85)2+(253)2=51009100=5191t=-\frac{A\cdot v}{v\cdot v}
= -\frac{(60,0)\cdot(-85,25\sqrt3)}{(-85)^2+(25\sqrt3)^2}
=\frac{5100}{9100}=\frac{51}{91}t=−v⋅vA⋅v=−(−85)2+(253)2(60,0)⋅(−85,253)=91005100=9151
So the “top” point is between AAA and B′B'B′, and the downhill length is the segment from that top point to B′B'B′, i.e.

(1−t) ∣v∣=(1−5191)(−85)2+(253)2=40919100(1-t)\,|v|=\left(1-\frac{51}{91}\right)\sqrt{(-85)^2+(25\sqrt3)^2}
=\frac{40}{91}\sqrt{9100}(1−t)∣v∣=(1−9151)(−85)2+(253)2=91409100
=4091⋅1091=4009191=40091=\frac{40}{91}\cdot 10\sqrt{91}=\frac{400\sqrt{91}}{91}=\boxed{\frac{400}{\sqrt{91}}}=9140⋅1091=9140091=91400
So the downhill track length is 40091\boxed{\frac{400}{\sqrt{91}}}91400 (option 4).

== References ==
<references />


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