The question is about geometry.

Question

Let λ be a positive real number. Consider a semicircle of center O and diameter AB. Choose points C and D (C is between D and B) on the semicircle and let ∠ AOD=2α and ∠ BOC=2β.

The point X on the line CD is such that XD/XC=λ. Prove that when α and β satisfy tan(α)=tan(β)+(√3)/2, all lines through X perpendicular to CD pass through a fixed point.

Answer 1

Let's approach this problem step-by-step:

1) First, let's visualize the problem. We have a semicircle with center O and diameter AB. Points C and D are on the semicircle, with ∠AOD = 2α and ∠BOC = 2β.

2) Point X is on line CD such that XD/XC = λ.

3) We need to prove that when tan(α) = tan(β) + (√3)/2, all lines through X perpendicular to CD pass through a fixed point.

4) Let's consider the perpendicular bisector of CD. This line will pass through O and intersect CD at a point, let's call it Y.

5) Now, let's focus on triangles OYD and OYC:

- OY is common to both triangles

- OD = OC (radii of the semicircle)

- YD = YC (Y is the midpoint of CD)

Therefore, these triangles are congruent.

6) This means that ∠OYD = ∠OYC = 90°

7) Now, in triangle OYD:

tan(α) = YD / OY

8) Similarly, in triangle OYC:

tan(β) = YC / OY

9) Given condition: tan(α) = tan(β) + (√3)/2

Substituting: YD/OY = YC/OY + (√3)/2

10) This implies: YD - YC = (√3/2)OY

11) But YD = YC (Y is midpoint), so:

0 = (√3/2)OY

This is only possible if OY = 0, meaning O and Y coincide.

12) This means that O is on CD.

13) Now, consider any perpendicular to CD through X. This line will form a right-angled triangle with OX as hypotenuse.

14) In this right-angled triangle:

sin(∠XOD) = XD/OX = λ/(λ+1)

(because XD/XC = λ, so XD/(XD+XC) = λ/(λ+1))

15) This angle is constant for any perpendicular through X, which means all these perpendiculars pass through a fixed point on the circle.

Therefore, when tan(α) = tan(β) + (√3)/2, O lies on CD, and all lines through X perpendicular to CD pass through a fixed point on the circle.