Question

给定两点A(1,2),B(3,4),若p在x轴上移动,使角APB达到最大的p点的横坐标

Answer 1

p(x,0),
X=1,tanAP1B=(3-1)/4=1/2
X=3,tanAP2B=(3-1)/2=1
所以：∠AP1B<∠AP2B
X≠1,X≠3
斜率KPA=2/(1-X),KPB=4/(3-X)
t=tanBPA=|KPA-PKB|/(1+KPA*KPB)
=2|x+1|/(x^2-4x+11) >0
x<1
tx^2-(4t+2)x+11t-2=0
(4t+2)^2-4t(11t-2)>=0
0<=|t|<=1
x>1
tx^2-(4t-2)x+11t+2=0
(4t-2)^2-4t(11t+2)>=0
0<=|t|<=1
所以：t=tanAPB=1,x=3
∠APBmax=45