留学考试中数学的答题格式(用英语答题!!!)

发布网友发布时间：2023-08-20 02:51

共1个回答

热心网友时间：2024-11-23 11:29

Here is an example:
"Call the two rays R (contains P) and S (contains A,B) and let the angle between them be α.
I tried this in Cartesian form, but geometry is probably better.

WLOG, choose origin O and R as the +x-axis.
The points are A(a,0) B(b,0)
Let's scale by 1/a such that A' is mapped to (1,0) and B' is mapped to (b/a,0). Let's call that (k,0). k is some constant.)

Let P be distance r along R (strictly, that distance has been scaled by 1/a too).
We are trying to find r (r is going to depend on α and k)

Coords of APB are thus A(1,0) P(r cos α, r sin α) B(k,0)
Let angle APB = θ, and apply the Cosine Rule.
Sidelengths are
|PA| = √((r cos α -k)² + (r sin α)²) = √(r² +k² -2kr cos α)
|PB| = √((r cos α -1)² + (r sin α)²) = √(r² +1 -2r cos α)
|AB| = (k-1)

You can get a (nasty) expression for cos θ in terms of parameter r and constants k, α.
Then find the r which maximizes θ, by minimizing cos θ (regardless whether 0 < θ < π/2 or π/2 < θ < π (cosine goes -ve)).

This will get the right answer, it'll just be awkward.
A change of notation might make it clearer.

Anyway, for what it's worth:
cos θ = |PA|²+|PB|²-|AB|² / 2|PA||PB|
= ((r² +k² -2kr cos α) +(r² +1 -2r cos α) -(k-1)²) / (r² +k² -2kr cos α)*(r² +1 -2r cos α)
and so on"