定角定弦轨迹问题

问题1

如果 $\angle MON=60^\circ$ ，点 $A、B$ 为射线 $OM、ON$ 上的动点( $A、B$ 不与点 $O$ 重合)，且 $AB=4\sqrt{3}$ ，在 $\angle MON$ 内部， $\triangle AOB$ 的外部有一点 $P$ ，且 $AP=BP，\angle APB=120^\circ$ ，求证明点 $P$ 在 $\angle MON$ 的平分线上

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问题2

如图，边长为 $2$ 的正方形 $ABCD$ 内接于 $\odot O$ ， $EF$ 是 $\odot O$ 的一条直径，点 $P$ 在 $AD$ 上， $\angle EPF=120^\circ$ ，则 $PE+PF$ 的最大值为

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

问题3

在 $RT△ABC$ 中， $∠BAC=90^{\circ}$ , ${AB}={AC}, \; {BC}=3 \sqrt{2}$ ，点 $D$ 是 $AC$ 边上一动点，连接 $BD$ ，以 $AD$ 为直径的圆与 $BD$ 相交的另一点为 $E$ ，求 $CE$ 的最小值

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

问题4

(2016武昌区月考)圆 $O$ 的半径为 $3$ ，在 $RT△ABC$ 中， $B$ 在圆 $O$ 上， $\angle A=30^\circ$ ，点 $C$ 在圆 $O$ 内，当点 $A$ 在圆上运动时，求 $OC$ 最小值

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问题5

(18槐阴一模)在 $RT\triangle ABC$ 中， $\angle ABC=90^\circ$ ， $\triangle ACB=30^\circ$ ， $BC=2\sqrt{3}$ ， $\triangle ADC$ 与 $\triangle ABC$ 关于 $AC$ 对称，点 $E$ 、 $F$ 分别是边 $DC$ 、 $BC$ 上的任意一点，且 $DE=CF$ ， $BE、DE$ 相交于点 $P$ ，求 $CP$ 的最小值

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问题6

$\triangle ABC$ 中， $AC=\sqrt{3}$ ， $\angle B=\dfrac{\pi}{3}$ ， $BD=1$ ， $CD=AD$ ，求 $AD$

典型的定弦定角轨迹问题
当 $D$ 与 $O$ 重合时， $AD=1$
当 $D$ 在 $O$ 下方时，

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问题7

向量 $\boldsymbol a$ 与 $\boldsymbol b$ 的夹角为 $60^\circ$ , $|2\boldsymbol b+\boldsymbol a|=1$ , 求 $|\boldsymbol b+2\boldsymbol a|$ 的范围

图形自己动手, 但要注意变式题

已知 $|\boldsymbol c|=1$ , 向量 $2\boldsymbol c+\boldsymbol d$ 与向量 $\boldsymbol c+2\boldsymbol d$ 的夹角为 $60^\circ$ , $|2\boldsymbol b+\boldsymbol a|=1$ , 求 $|\boldsymbol d|$ 的范围

问题1 GeoGebra 绘图代码

O=(0,sqrt(3))
h:Ray(O,(-1,0))
i:Ray(O,(1,0))
A=point(segment(O,(-3.3,(-3.3+1)*sqrt(3))))
d:Circle(A,4*sqrt(3))
B=Intersect(d,i,1)
P=Intersect(x=0,PerpendicularBisector(A,B))
c:Circle(A,P,B)
Ray(O,P)

问题2 GeoGebra 部分代码

A=(-1,1)
B=(-1,-1)
C=(1,-1)
D=(1,1)
polyline({A,B,C,D,A})
c:x^2+y^2=2
d:x^2+y^2=2/3
O=(0,0)

问题3 GeoGebra 绘图代码

B=(-1.5*sqrt(2),0)
C=-B
A=(0,1.5*sqrt(2))
t = slider(0,1,0.02)
D=A+t*(C-A)
cir:circle(0.5*(A+D),D)
E=intersect(cir, line(B,D),2)
locus(E, t)
{segment(B,D),segment(C,E)}
polyLine({A,B,C,A})
ans = length(C-E)

问题5 GeoGebra 绘图代码

B=(0,0)
C=(sqrt(3),0)
A=(0,1)
D=Reflect(B,Line(A,C))
t=slider(0,1,0.01)
t=0.2
E = D + t*(C-D)
F = C + t*(B-C)
h=polyline(A,B,C,A,D,C)
f=segment(B,E)
g=segment(D,F)
P=intersect(f,g)
c:CircularArc(A,B,D)

问题6 GeoGebra 绘图代码

A= (sqrt(3)/2,-1/2)
eq1:x^(2)+y^(2)=1
C= (-sqrt(3)/2,-1/2)
B=Point(eq1)
O=(0,0)
f=Segment(B,C)
g=Segment(C,A)
h=Segment(B,A)
s=Segment((0,-1),(0,1))
D=Intersect(s,h)