I need a help for this problem

Question

Given an isosceles triangle. Find the locus of the points inside the triangle such that the distance from that point to the base equals to the geometric mean of the distances to the sides. Any ideas please?

Answer 1

In ⊿ABC, AB = AC, M is the midpoint of BC, and D is the in-center.

The special case result --- B and C (limiting cases), and D are part of the said locus.

We suspect that the required locus is the circle passing through B, D, C.

To verify, we need to

(1) Let F be a point on that circle such that FX, FY, FZ are the perpendiculars from F to AB, AC and BC respectively.

(2) We need to show $FX.FY = FZ^2$.

Hope you can take it from here. (I have included in FB, FC, ZX and ZY in the figure. They are needed in the verification process.)

Answer 2

we can deduct the result without guess.

$x=p\sin{\alpha}=q\sin{\beta},y=psin{\alpha_1},z=q\sin{\beta_1},\alpha+\alpha_1=\beta+\beta_1,x^2=yz \implies p\sin{\alpha}*q\sin{\beta}=psin{\alpha_1}*q\sin{\beta_1} \implies \sin{\alpha}*\sin{\beta}=sin{\alpha_1}*\sin{\beta_1} \implies \cos{(\alpha-\beta)}-\cos{(\alpha+\beta)}=\cos{(\alpha_1-\beta_1)}-\cos{(\alpha_1+\beta_1)}$

note :$\alpha-\beta=-(\alpha_1-\beta_1) \implies \cos{(\alpha-\beta)}=\cos{(\alpha_1-\beta_1)} \implies \cos{(\alpha+\beta)}=\cos{(\alpha_1+\beta_1)} \implies (\alpha+\beta=\alpha_1+\beta_1) \cap (\alpha+\alpha_1=\beta+\beta_1 )\implies \alpha_1=\beta,\alpha=\beta_1$

now we know the locus is a circle which tangent both sides at the two ends,and must pass through the $I$(which is trivial when the point is move to middle). and luckily we do know that the circle that pass through these 3 points is tangent both sides. so the circle is the one.